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\begin{document}
%\frontmatter
\title{Ordinal Analiz}
\subtitle{Aksiyomatik K\"umeler Kuram\i\ Dersi}
\author{David Pierce}
\date{21 \c Subat 2018 (February 21)}
%\date{8 \c Subat 2016}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\.Istanbul\\
\url{dpierce@msgsu.edu.tr}\\
\url{http://mat.msgsu.edu.tr/~dpierce/}}

\maketitle

\frontmatter

\selectlanguage{english}

\chapter*{Preface}

In the present text,
I attempt to develop set theory on the model of calculus,
so that any student who can learn the latter can learn the former.
I shall explain later what this means in practical terms.

\section*{Mathematics as such}

Meanwhile, in more theoretical terms,
the text attempts to bridge the gap between mathematics as natural science
and mathematics as logic.
Under the former conception,
mathematics should agree with the world
(or with other mathematics courses);
under the latter conception,
mathematics should be logically derivable from axioms that may be plausible,
but are in any case accepted for the nonce.

Euclid recognized the gap, and he has given us in the \emph{Elements} \cite{MR17:814b}
the earliest known attempt to bridge the gap.
A carpenter or surveyor may recognize that many of Euclid's propositions \emph{are} true;
Euclid shows why they \emph{must} be true.

Euclid has the spirit of Socrates,
who at his trial recalls the pronouncement of the oracle at Delphi
that he, Socrates, was the wisest of men \cite{Plato-Apology}.
Incredulous, Socrates investigated the men who are reputed to be wise:
\begin{quote}
After the political experts I went on to the poets\lips
on the basis that it was here that I'd catch myself red-handed,
as actually more ignorant than [they].
So, picking out those of their poetic compositions
they seemed to me to have spent most effort on,
I would ask them what they were trying to say,
with a view to learning a thing or two from them as well.
Well, Athenians, I blush to tell you the truth,
but it has to be told:
practically speaking, almost everyone present
would have better things to say
than they did about their own compositions\lips
But, men of Athens, the good craftsmen too
seemed to me to suffer from the same failing as the poets:
because they were accomplished in practising their skill,
each one of them claimed to be wisest about other things too,
the most important ones at that---%
and this error of theirs seemed to me to obscure the wisdom that they did possess.
\end{quote}
Most undergraduate mathematics has an obvious meaning in the physical world,
or at least is of use to experts who work in the world.
Even number theory has its applications to cryptography and thus to warfare and internet commerce.
But mathematics proper must be able to explain not only how it is true, but why.
\emph{How} Newton's infinitesimal calculus is true
is that it can derive the motions of the planets from an inverse-square law of gravitation
\cite{MR1965137}.
But this success did not make calculus into mathematics.
\emph{Infinitesimal} calculus became mathematics as such only after three centuries,
when Robinson founded it in logic \cite{MR1373196}.
Euclid had shown the way, or established the ideal, some two millenia earlier.

Now set theory may stand as the purest mathematics.
But it was created to explain the power of calculus,
before Robinson was born.
Alexandre Borovik notes the paradox
that though there may be nothing infinite in the world,
infinity is still useful as an approximation to the very large.
Infinity here may not be a set,
but a point to the right of all of the points on the real number line.
Since those points are called numbers, infinity may be considered a new number.
The linguist may say that we have the capacity to form an infinite number of grammatical sentences,
as in symbolic logic we may form infinitely many conjunctions $P_1\land P_2\land\dots\land P_n$.
Here the infinite number is an infinite set,
but a \emph{countable} set.
Even the \emph{uncountably} infinite is useful,
in the form of the real numbers themselves,
which compose an uncountably infinite set.
Taking inspiration from the line of real numbers,
I try to develop in the present text an analogous conception of the class of all ordinal numbers.
I even draw graphs of continuous and non-continuous ordinal-valued functions
as if they were real-valued functions.

Most students come to us without knowing what mathematics is.
For them, it is just like physics or any other course they have to take.
Their job is to satisfy their teachers.
\emph{Our} job is to induce the students to insist on satisfying themselves.
What we ask them to learn should be justified to \emph{their} standards.
Therefore they must develop their own standards.
This is true in any course of study;
but in mathematics, the standards of truth are universal.
We expect universal agreement
on whether a given theorem follows from given axioms and definitions.
In practice there may not be agreement,
but in this case we have a clear procedure for resolving the dispute.
The person who says that a theorem is true must be prepared to supply a proof,
and to explain it as needed by anybody who is seriously interested.
Teachers may demand this of students;
students must also feel free to demand this of teachers.

I once had a student in a set-theory course who thought a certain equation was true
because he had seen it in an authoritative source.
He had not understood that, in the source,
the equation was true by a definition that my own course had not adopted.
The equation was
\begin{equation}\label{eqn:0=0}
	\bigcap\emptyset=\emptyset.
\end{equation}
I had taught instead
\begin{equation}\label{eqn:0=V}
	\bigcap\emptyset=\universe;
\end{equation}
but this was a \emph{theorem} following from the natural definition
\begin{equation*}
\bigcap\bm C=\{x\colon\Forall Y(Y\in\bm C\lto x\in Y)\},
\end{equation*}
where $\bm C$ is any class.
This is what I have always taught.
With this definition in mind,
one can write a ``De Morgan law'' in the form
\begin{equation}\label{eqn:DM}
\left(\bigcap\bm C\right)\comp=\bigcup\{X\comp\colon X\in\bm C\},
\end{equation}
which is a generalization of the more familiar
\begin{equation}\label{eqn:DMn}
(A_1\cap\dots\cap A_n)\comp=A_1{}\comp\cup\dots\cup A_n{}\comp.
\end{equation}
One has to explain the notation on the right of \eqref{eqn:DM},
since the complement of a set is a proper class,
so that $\{X\comp\colon X\in\bm C\}$ is not a class unless $\bm C=\emptyset$.
One can still define
\begin{equation*}
\bigcup\{X\comp\colon X\in\bm C\}=\{x\colon\Exists Y(Y\in\bm C\land x\notin Y)\}.
\end{equation*}
When $\bm C$ is empty, one gets
\begin{equation}\label{eqn:cup0=0}
\bigcup\emptyset=\emptyset,
\end{equation}
and so \eqref{eqn:0=V} follows.
This, or rather $\left(\bigcap\emptyset\right)\comp=\emptyset$,
is also the natural interpretation of \eqref{eqn:DMn} when $n=0$.

One could say that \eqref{eqn:DMn} made no sense when $n=0$;
but this would be contrary to the spirit of generalization in mathematics:
a spirit which may lead to vacuity, but also to insight and simplicity.
In \emph{Mathematics: A Very Short Introduction} \cite[ch.\ 3, pp.\ 35--48]{MR2147526},
Gowers notes that the straight lines joining any two of $n$ points on a circle
divide the circle into $2^{n-1}$ regions when $1\leq n\leq5$.
This is not a proof that the same is true for all $n$:
\begin{quote}
In fact, with a little further reflection one can see that the number of regions
\emph{could not possibly} double every time.
For a start, 
it is worrying that the number of regions defined 
when there are $0$ points round the boundary 
is $1$ rather than $1/2$,
which is what it would have to be if it doubled when the first point was put in.
Though anomalies of this kind sometimes happen with zero,
most mathematicians would find this particular one troubling.
\end{quote}
That a certain formula does not work when $n=0$
is a sign (albeit not a conclusive one) 
that the formula will not work for other $n$ either.

It is clear what the sum $\sum_{i=1}^na_i$ means when $n$ is a counting number;
and when $n=0$, the sum should be zero.
But then the product $\prod_{i=1}^na_n$ should be $1$ when $n=0$,
since $1$ is neutral with respect to multiplication.
As a special case, we define $0!=1$.
Likewise, since the universal class is neutral with respect to intersection,
we should define \eqref{eqn:0=V} to be true.

Nonetheless, the elegance of \eqref{eqn:0=V}
may be lost on students.
When one of them lost points on an exam for writing \eqref{eqn:0=0}
instead of \eqref{eqn:0=V},
he showed me an issue of \emph{Matematik D\"unyas\i}
in which Ali Nesin said that \eqref{eqn:0=0} was correct.
Ali worked only with sets, not with proper classes.
In particular,
he defined intersections only of sets,
and these intersections should always be sets.
By fiat then,
\eqref{eqn:0=0} held,
there being no better alternative.
Seeing this equation,
apparently my student took it to be as true as an equation from physics like $\mathbf F=m\mathbf a$.
He had not learned that Arnol$'$d \cite{Arnold} was not quite right to say,
\begin{quote}
Mathematics is a part of physics. Physics is an experimental science,
a part of natural science. Mathematics is the part of physics where
experiments are cheap.
\end{quote}
That very ``cheapness'' of its experiments makes mathematics different.
In every class, using axioms and definitions, we can create a new world,
which need not be the same as the world seen in a previous class,
or in a text that we are not using.

As the student of linear algebra must learn to work with more than three dimensions,
overcoming his or her preconception that the additional dimensions have no physical meaning,
so the student of set theory must learn to work with infinite sets that are strictly larger than other infinite sets,
and even with classes that are too large to be sets at all.
The universal class $\universe$ 
might be considered as analogous to the point $\infty$ at infinity already seen in calculus.

\section*{The writing of textbooks}

I have now taught set theory three times at \textsc{metu},
and three times at Mimar Sinan.
I have always produced my own text for the course.
This text contained more than could be covered in the course,
because I wrote the text more for the teacher 
(namely myself) than for the students.
Most textbooks may be written 
so that they can replace a teacher's lectures.
I myself did aim to put in my own texts everything that lectures would cover;
but the text might be terser than the lectures.
The text would also cover more:
things that \emph{I,} at least, wanted to know,
or that that were needed to satisfy some notion of formal completeness,
though they could be skipped in class.

I write texts for many of my courses.
For this habit, 
I blame the man who taught me 
precalculus and calculus in the last two years of high school.
Donald Brown had us buy two textbooks:
Spivak \cite{Spivak-Calc} for theory,
and Salas and Hille \cite{Salas-Hille} for practice.
These could have been used for two courses called, respectively,
analysis and computations in analysis at Mimar Sinan.
In any case, the real text for Mr Brown's course
was the one that we copied down from what Mr Brown wrote on the blackboards.
I learned in this way 
that different choices could be made about how to do mathematics.
Mr Brown was making his own choices.
If I became a teacher, I could make my own choices.

In a number-theory course that I taught at \textsc{metu},
a student complained that the text was difficult to understand.
He was embarrassed when I pointed out that the text was by me.
I told him that the real course was worked out in the classroom.
If a student could follow the text by itself, that was fine;
but I was not interested in producing a text 
that would obviate a student's need to come to class.

I may not be able to produce such a text.
Perhaps nobody can,
but if they try, they end up with a bloated textbook with something for everybody,
but too much for anybody.
At least one student did praise one edition of my own set-theory text;
but she had been by far the best student in the class.

The present text is somewhat different.
It is set theory stripped down to precisely 
what I think reasonable to ask \emph{all} students to learn.
I have also changed my mind about what needs to be in the course.

I began teaching set theory at \textsc{metu} 
in order to work out some of my own concerns about mathematical rigor.
For example, 
I had noticed that the logical distinction 
between induction and recursion in the natural numbers 
was not commonly recognized.
Even otherwise-rigorous textbooks treated induction, 
strong induction, and well-ordering
as equivalent principles, 
from any one of which, 
all of the other properties of the natural numbers could be derived.
Mr Brown did this, as did Spivak; but they were in error.
I ultimately wrote an article \cite{Pierce-IR} about this.
Meanwhile, I wanted to clarify matters in my own set-theory course.

I learned that my concerns were lost on most students.
Since my students had entered university 
on the basis of their ability to solve odd computational problems,
I decided that in set theory 
they should at least be able to perform computations 
with what Cantor \cite[\S19, pp.\ 183--195]{Cantor}
calls the normal forms of ordinal numbers.

\section*{This text}

I used to want the students to learn something of symbolic logic.
However, I have now all but dropped this aspect of the course.
Students should still learn that sets can be collected into classes,
which are defined by formulas;
but the formal definition of formulas
is now relegated to \textbf{Appendix \ref{mantik}.}

I used to try to motivate set theory by the paradox
that our earliest mathematical activity is based on a proposition that everybody accepts without proof,
but is not properly an axiom.
The proposition is that no matter how we count a set,
we always get the same number.
The proposition fails when the set is infinite.
This is a sign that the proposition for finite sets is a real theorem,
not an axiom.

One might alternatively conclude that there are no infinite sets,
or that there is not really any way to count them.
I had a friend who did not believe in infinite sets;
he could have been a mathematician,
but became a lawyer instead.

In earlier set-theory classes, 
I did postpone the Axiom of Infinity as long as possible.
I did not do this in my most recent class,
because I had decided to try to motivate set theory differently this time.

One does real analysis on the basis of the axioms of a complete ordered field.
One can in fact construct a complete ordered field from the natural numbers,
as given by the Peano Axioms.
The way was shown by Dedekind \cite{MR0159773},
and Landau \cite{MR12:397m} works out the details;
%in the \emph{Foundations of Analysis};
but one need not construct the real numbers, in order to do analysis.

I have used the same idea in the present text.
The ordinals are analogous to the real numbers
in the sense that every nonempty set of them with an upper bound 
has a least upper bound.
In order to do ``ordinal analysis'' as soon as possible,
I present axioms for ordinals in \textbf{Chapter \ref{ch:ord},}
and I prove from these axioms the tools needed for ordinal arithmetic: 
ordinal induction and ordinal recursion.
Alternatively, 
one might just accept ordinal induction and recursion as axioms themselves,
or as grand theorems whose proofs are deferred,
like the Intermediate Value Theorem in some calculus books.
In Salas and Hille, the proof of the IVT is in an appendix.
In my own set-theory classes,
I may prove the theorem of ordinary recursion (recursion on $\N$ or $\upomega$),
to give a taste of what is involved,
while waving my hands over ordinal recursion.

In just a few pages,
Suppes \cite[pp.\ 195--205]{MR0349389}
gives three versions of transfinite induction,
five versions of transfinite recursion.
I give only one version of each,
a version obtained from ordinary induction or recursion by adding a limit step.

In \textbf{Chapter \ref{ch:ax}} are presented those set axioms
with which the existence of a model of the ordinal axioms can be established.
The chapter has two parts.
As presented in the previous chapter,
ordinal recursion gives us only functions from $\on$ into itself.
With the Empty-Set, Adjunction, and Replacement axioms,
we can recursively define an isomorphism
from any structure satisfying the ordinal axioms
to $\on$ as defined by von Neumann \cite{von-Neumann}.
We may henceforth assume that $\on$ has this definition.
This gives us the convenient results
\begin{align*}
	\alpha&=\{\xi\colon\xi<\alpha\},&
	\sup A&=\bigcup A.
\end{align*}
This is the result of the first section.
In the next section,
we observe that the new $\on$ is transitive and well-ordered by $\in$,
and so are all of its elements.
This will let us define it without using the ordinal axioms.

Bruno Poizat might be critical here.
In his \emph{Course in Model Theory} \cite[p.\ 163]{MR1757487},
after showing that every well-ordered set is isomorphic to a unique von Neumann ordinal,
he writes,
\begin{quote}
We meet some students who are allergic to ordinals as ``well-ordering types''
and who find the notion of von Neumann ordinals easier to digest;
that is a singular consequence of dogmatic teaching,
which confuses formalism with rigor,
and which favors technical craft to the detriment of the fundamental idea:
It takes a strangely warped mind to find the notion of a transitive set natural!
\end{quote}
I do not know whether it is really a job for a mathematician
to judge what is strangely warped!
Different people think differently, even within mathematics.
Our students may indeed be victims of dogmatic teaching;
but it is they with whom we have to work.
In real analysis, each real number can be understood as the set of all rational numbers that are less than itself;
but one need not have this understanding,
in order to \emph{do} real analysis;
and the understanding may even be a distraction.
In ordinal analysis,
we need only think of ordinals as points on a certain line.
However, 
it \emph{is} useful for our purposes if the ordering of that line is precisely set-theoretic membership,
so that every ordinal is precisely the set of ordinals that are less than itself.
For one thing, this means the ordinal itself has a cardinality.
In an early draft of the present text,
I had written a theorem in the form,
\begin{equation*}
\alpha\cdot\beta\approx\{\xi\colon\xi<\alpha\}\times\{\xi\colon\xi<\beta\}.
\end{equation*}
At some stage, this may express the theorem better than what I have now written, namely
\begin{equation*}
\alpha\cdot\beta\approx\alpha\times\beta.
\end{equation*}
But it seemed to me that maintaining a formal distinction 
between $\alpha$ and $\{\xi\colon\xi<\alpha\}$ would be perverse.

Adjunction is not one of Zermelo's original axioms \cite{Zermelo-invest},
but it follows from his axioms of Union and ``Elementary Sets''
(classes of at most two elements are sets).
I prefer to give Adjunction before having to introduce Union.
Once the Union Axiom \emph{is} introduced, along with Separation and Infinity,
we can show that von Neumann's ordinals do exist 
so as to satisfy the ordinal axioms given earlier.
However, 
this material is independent from the rest of the text,
and perhaps it can be ignored,
however paradoxical that may be for a course of axiomatic set theory.

Most of the Zermelo--Fraenkel axioms can be understood
to be that certain classes are sets.
I have never seen the Axiom of Infinity presented this way,
even in a thorough discussion of the axioms by Fraenkel and others \cite{MR0345816}.
However, since we have already introduced the ordinals,
we can let Infinity be that the class of all ordinals 
that neither are limits nor contain limits is a set.

The Power-Set Axiom appears only in \textbf{Chapter \ref{ch:card},} on cardinals,
to establish that there are uncountable sets.
Then the Axiom of Choice gives us that every set is equipollent with some ordinal.
Defining the Beths as well as the Alephs makes some nontrivial computational problems possible.
It may be a perversity of mathematics that we look for ways to give students problems;
but this is what I have done.

I never mention the Foundation Axiom.
One may raise the question of whether a set can be a member of itself;
but I see no point in declaring that it cannot
unless one is going to give the proof that such a declaration is consistent with the other axioms.

It might be said that my use of $\universe$ for the class of all sets
implies my acceptance of the Foundation Axiom.
I use this notation in the text only to point out that $\pow{\universe}=\universe$,
so that Cantor's Theorem $A\prec\pow{A}$ must somehow use that $A$ is not a class.
I do not point out that $\bigcap\emptyset=\universe$.

The possibility of computational problems with ordinals
is developed in \textbf{Chapters \ref{ch:add}, \ref{ch:mul},} and \textbf{\ref{ch:exp},}
which concern ordinal addition, multiplication, and exponentiation respectively.
The chapters are laid out in parallel, as far as possible.
Thus in \textbf{Chapter \ref{ch:add}} we establish
that each element $\upomega^2$ is of the form $\upomega\cdot k+n$,
and $n+\upomega=\upomega$.
In particular then, $\upomega^2$ is closed under addition.
In \textbf{Chapter \ref{ch:mul},}
we see that $\upomega^{\upomega}$ is closed under addition and multiplication.
Within this set, Cantor normal forms 
can be defined in close analogy with 
the the usual positional notation for counting numbers;
for any element of $\upomega^{\upomega}$ can be written as
\begin{equation*}
  \upomega^{k}\cdot m_0+\upomega^{k-1}\cdot m_1+\dots+\upomega\cdot m_{k-1}+m_k
\end{equation*}
for some $k$ in $\upomega$,
where the coefficients $m_i$ are allowed to be $0$.
The rules for addition and multiplication in $\upomega^{\upomega}$
are as challenging as those for arbitrary Cantor normal forms,
and they can be introduced before arbitrary exponentiation is worked out in \textbf{Chapter \ref{ch:exp}.}
The tools are all there;
but I think there is no need to give a rule for raising an arbitrary Cantor normal form
to the power of an arbitrary Cantor normal form.

The chapters on ordinal arithmetic establish that ordinal sums, products, and powers
are respectively equipollent with disjoint unions, cartesian products,
and sets of finitely supported functions.
It is established within each chapter 
that the corresponding operation yields only countable sets when applied to countable sets.

In \textbf{Chapter \ref{ch:card},}
the Power-Set Axiom gives us uncountable sets.
That cardinal addition and multiplication are trivial
is established by use of Cantor normal forms.

For completeness, I added the topic of cofinality to an earlier edition of the text.
It allows precise computation of infinite cardinal powers,
provided one grants the Generalized Continuum Hypothesis.
I have never had time to talk about this in class;
but the material is in \textbf{Appendix \ref{ch:cof}.}

\textbf{Chapter \ref{ch:real}} is an attempt to introduce foundational thinking in a familiar context:
the real numbers.
Given the real numbers as constituting an ordered field,
I define the natural numbers as certain real numbers
(which is what Spivak does).
Ultimately I prove the Peano Axioms as a theorem about these natural numbers.
Many details are left as exercises;
in the class itself, I had students present at the board their solutions
(or the solutions that they looked up).

One could however skip \textbf{Chapter \ref{ch:real}.}
As it is, the text introduces \emph{four} lists of axioms:
\begin{inparaenum}[(1)]
	\item 
	the axioms of a complete ordered field,
	\item
	the Peano axioms,
	\item
	axioms for the class of ordinals, and
	\item
	the Zermelo--Fraenkel axioms.
\end{inparaenum}
One could skip the first two lists
and treat the third explicitly as a theorem
whose proof is deferred.
Or one could skip all but the third,
using ``naive'' set theory to justify what is done with it.
This would take the course as close as possible to physics,
in the sense of being about something---the class of ordinals---%
that is as real as the line composed of real numbers.

\textbf{Appendix \ref{ch:let}}
lays out the different kinds of letters used as symbols in the text.
It might be desirable to compose mathematics in such a way
that it could be written out with a standard old-fashioned typewriter.
However, the present text takes advantage of the distinctions between:
\begin{compactitem}
\item
the Latin and Greek alphabets;
\item
the upper and lower cases;
\item
roman and italic ``shapes'';
\item
plain and bold ``weights,'' along with ``blackboard bold'' and curly fonts;
\item
different intervals of an alphabet.
\end{compactitem}
Letters from the beginning of an alphabet are usually constants;
from the end, variables.
This follows the convention going back to Descartes 
\cite{Descartes-Geometry} whereby, in an equation like
\begin{equation*}
ax^2+bx+c=0,
\end{equation*}
the $x$ that is the variable or ``unknown.''
The $a$, $b$, and $c$ are constant for the sake of solving the equation;
but they are still variable in the sense of having no fixed values 
outside the solution of the equation.
If they did have values that were fixed throughout the text,
they could be printed upright.
In this way, $\upomega$ is the set of natural numbers,
as opposed to $\omega$, which could be used for an arbitrary ordinal,
though to avoid confusion it is never used in this text.
A sort of exception to this rule is that $\mathscr P$ always denotes the operation of producing the power set,
while letters like $\mathscr A$ from the same font are not fixed.
Moreover, while $f$ always denotes a function,
I may use it as a variable or constant,
simply because I have no sense that there is any standard letter for a variable function.
I have the sense that ordinary language uses no variables at all;
only formalized language does.
Thus we may say $\phi(a)$, meaning implicitly that for all $a$, $\phi(a)$ is true;
or we may say $\Forall x\phi(x)$ with the same meaning.
I do find it satisfying to use $\alpha$ and $\xi$ in ordinal analysis
the way we use $a$ and $x$ in real analysis.
But making a formal distinction between constants and variables
is admittedly not all that important.
I do not want students to have to wonder about 
whether they should write $\alpha$ or $\xi$ in a particular context.

Students should however learn to distinguish between sets and classes.
A set may be called $a$, $A$, or $\mathscr A$,
depending on what features are being emphasized.
If it is important that the set has elements, it may be called $A$;
if it is important that those elements themselves have elements,
it may be called $\mathscr A$.
A class that is not known to be a set will be $\bm A$,
written with a wavy underline on the blackboard
(and not with ``blackboard bold'').

In the past,
I used only lower-case letters for sets,
so that a capital letter would always be a class.
Now I have decided it is better to follow the practice of ordinary mathematics when possible,
using lower-case letters to the left of the $\in$ sign, upper-case to the right.

\section*{The course}

If they learn nothing else from a course of axiomatic set theory,
students should learn the Russell Paradox \cite{Russell-letter}---that,
or else the Burali-Forti Paradox \cite{Burali-Forti},
which might be taken as even more integral to the course,
given that this is presented as ``ordinal analysis.''
The paradoxes are one bit of the ``mathematics as logic''
that I mentioned at the beginning.
Each can be most succinctly expressed as the theorem that a certain class is not a set.
Without classes, you have to say things like,
``Not everything that you might expect to be a set can be a set.''
You can say, ``Not every property defines a set'';
but what is a property?

Still, introducing the class as a definite concept poses difficulty.
Not every writer dare be like Levy \cite{MR1924429},
who introduces classes near the beginning of his text.
Levy is perverse in another way too, 
by formally stating the ``Axiom of Comprehension''---%
that every formula defines a set---
and then immediately proving the theorem
(called ``Russell's antinomy'') that the Axiom of Comprehension is inconsistent.

I conceive of sets as already existing.
Sets are \emph{collections,}
though not every collection can be expected to be a set.
Here I use the word ``collection'' for the most general kind of whole that has individual elements;
but the Russell Paradox keeps us from defining a ``most general'' such thing in an absolute sense:
there is no collection of all collections that do not contain themselves.
In axiomatic set theory,
we want to figure out \emph{which} collections are sets,
or ought to be sets.
Purely for our convenience, we require every member of a set to be a set itself.
One may prefer not to consider the so-called empty set as a set;
but then one might have to say things like,
``For all $a$, where $a$ is a set or $\emptyset$.''
A similar problem arises in Euclid's number theory,
where unity is not properly a number,
but sometimes is implicitly treated as one.
In any case, since we also consider sets to be in some sense ``given,''
we have no reason to think that any new collection of sets that we may form
is already one of the given sets.
This resolves the Russell Paradox.

In my 2013--4 class, I demonstrated Tarski's Undefinability Theorem \cite{Tarski-truth}
in the form, 
\emph{Not every collection of sets is even a class.}
Indeed, set theory seems to be the best context to introduce the idea of the theorem,
which originates with G\"odel \cite{Goedel-incompl}.
G\"odel showed how to treat every formula about (counting) numbers as a number itself,
so that, given a number theory,
one could write down a true statement about numbers that was not provable in the theory.
It is easier to do the same thing for set theory.
We assign, to every symbol in the language of set theory, a different set.
Considering it as a sequence of such symbols,
we assign, to every formula $\phi$ about sets,
a set $\gn{\phi}$.
Then the collection of sets $\gn{\sigma}$
such that $\sigma$ is a true sentence is not a class:
For if it were the class defined by $\phi$,
then some formula $\psi$ would define the class of $\gn{\chi}$ such that $\chi(\gn{\chi})$ is false.
In this case $\psi(\gn{\psi})$ would be true if and only if it were false.
All of this can be shown to interested students; but it is not in the present text.

Our students at Mimar Sinan have read and presented the propositions of the first book of Euclid's \emph{Elements.}
Reading this book with them caused me to recognize what they sometimes did not:
that equality is not identity.
Euclid proves that parallelograms of the same height on the \emph{same} base are equal,
before proving that parallelograms of the same height on \emph{equal} bases are equal.
Equality here is congruence: simple congruence of line segments,
and congruence of parts in the case of parallelograms.

With this example in mind,
I prefer not to take equality of sets for granted,
but to \emph{define} it as having the same elements.
In a word, equality of sets is sameness of extension.
With this approach,
one needs the axiom that equal sets are members of the same sets.
One can then prove as a theorem that equal sets are members of the same classes.
This theorem is usually taken as a logical axiom,
because equality is treated as identity.
In this case,
that sameness of extension implies equality must be taken as a set-theoretic axiom.
I have swept all of this under the rug in the present text.

In an email exchange, 
Ali Nesin allowed that he should have been clearer in his \emph{MD} article 
that the equation \eqref{eqn:0=0} was his personal convention.
Still, as I understand it, his approach to set theory may not be uncommon:
here, everything should be a set,
and its existence should be given by an explicit construction
(or at least, as in the case of a power set or a choice function, 
a construction that is explicitly justified by an axiom).
%In \emph{Set Theory: An Introduction to Independence Proofs,}
Kunen says,
\begin{quote}
If $\mathscr F=0$, then $\bigcup\mathscr F=0$ and $\bigcap\mathscr F$ ``should be'' the set of all sets,
which does not exist.
\end{quote}
This is at \cite[page 13]{MR85e:03003}; Kunen does not define classes until page 23.

I myself am not interested in giving classes the formal existence that they have in
so-called von Neumann--Bernays--G\"odel set theory,
such as is presented by Lemmon \cite{Lemmon}.
More formalism means more need to check that it agrees with our informal understanding.

For students who just want to collect enough credits to graduate,
all of these foundational concerns can be de-emphasized.
I used to think it a reasonable exam problem
to give a verbal description of a class
and ask for a formula that defines it.
But I think now that enough problems can be asked without this.
I propose that students should be able to do the following.
\begin{compactenum}
	\item 
	Add and multiply ordinals in their Cantor normal forms.
	(Exponentiation is optional.)
	\item
	Recognize equations of ordinals that are identities,
	and supply proofs.
	\item
	Supply counterexamples to ordinal equations that are not identities,
	and identify the false steps in proposed proofs that the equations \emph{are} identities.
	(I recognized late---only in 2015--6---that the students could have difficulty in finding the false steps;
	but if they cannot do it, they can hardly be said to have learned any mathematics at all.)
	\item
	Perform cardinal computations of Alephs and Beths with addition, multiplication, exponentiation, and suprema.
\end{compactenum}
Questions like, ``Prove or disprove: every set is a class''
are also standard on my exams.

\AfterBibliographyPreamble{\relscale{0.9}}

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%\bibliography{../../references}
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\def\cprime{$'$} \def\cprime{$'$} \def\cprime{$'$}
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  \def\cprime{$'$} \def\cprime{$'$} \def\cprime{$'$}
\begin{thebibliography}{10}

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V.~I. Arnol{\cprime}d.
\newblock On the teaching of mathematics.
\newblock {\em Russian Mathematical Surveys}, 53(1):229--234, 1998.

\bibitem{Burali-Forti}
Cesare Burali-Forti.
\newblock A question on transfinite numbers.
\newblock In van Heijenoort \cite{MR1890980}, pages 104--12.
\newblock First published 1897.

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\newblock Cosimo Classics, New York, 2007.
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\bibitem{MR0345816}
Abraham~A. Fraenkel, Yehoshua Bar-Hillel, and Azriel Levy.
\newblock {\em Foundations of set theory}.
\newblock North-Holland Publishing Co., Amsterdam, revised edition, 1973.
\newblock With the collaboration of Dirk van Dalen, Studies in Logic and the
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\bibitem{Goedel-incompl}
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\newblock First published 1931.

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\newblock North-Holland Publishing Co., Amsterdam, 1983.
\newblock An introduction to independence proofs, Reprint of the 1980 original.

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\newblock Chelsea Publishing Company, New York, N.Y., third edition, 1966.
\newblock Translated by F. Steinhardt; first edition 1951; first German
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\bibitem{Lemmon}
E.~J. Lemmon.
\newblock {\em Introduction to Axiomatic Set Theory}.
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\bibitem{MR1924429}
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\newblock Dover Publications Inc., Mineola, NY, 2002.
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\newblock University of California Press, Berkeley, CA, 1999.
\newblock A new translation by I. Bernard Cohen and Anne Whitman, assisted by
  Julia Budenz. Preceded by ``A guide to Newton's \emph{Principia}'' by Cohen.

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\newblock Induction and recursion.
\newblock {\em The De Morgan Journal}, 2(1):99--125, 2012.
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\newblock {\em Socrates' Defence}.
\newblock Penguin, UK, 2015.
\newblock Translated by Christopher Rowe.

\bibitem{MR1757487}
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\newblock {\em A Course in Model Theory}.
\newblock Universitext. Springer-Verlag, New York, 2000.
\newblock An introduction to contemporary mathematical logic, Translated from
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\newblock Letter to {F}rege.
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John von Neumann.
\newblock On the introduction of transfinite numbers.
\newblock In van Heijenoort \cite{MR1890980}, pages 346--354.
\newblock First published 1923.

\bibitem{Zermelo-invest}
Ernst Zermelo.
\newblock Investigations in the foundations of set theory {I}.
\newblock In van Heijenoort \cite{MR1890980}, pages 199--215.
\newblock First published 1908.

\end{thebibliography}



\mainmatter

\selectlanguage{turkish}

\tableofcontents

\listoffigures

%\mainmatter

\chapter{Ger\c cel Analiz}\label{ch:real}

\section{Tam s\i ral\i\ cisim aksiyomlar\i}

Ger\c cel say\i lar,
$\R$\label{R} \textbf{tam s\i ral\i\ cismini} olu\c sturur.
Demek ki
\begin{compactenum}[1)]
\item
$\R$, $<$ ba\u g\i nt\i s\i\ taraf\i ndan \textbf{s\i ralanm\i\c st\i r,} yani
\begin{compactenum}
\item
$<$ ba\u g\i nt\i s\i\ \textbf{yans\i mas\i zd\i r,}
\begin{equation*}
a\not< a;
\end{equation*}
\item
$<$ ba\u g\i nt\i s\i\ \textbf{ge\c ci\c slidir,}
\begin{equation*}
a<b\land b<c\lto a<c;
\end{equation*}
\end{compactenum}
\item
$<$ s\i ralamas\i\ \textbf{do\u grusald\i r,}
\begin{equation*}
a<b\lor a=b\lor a>b;
\end{equation*}
\item
$<$ do\u grusal s\i ralamas\i\ \textbf{tamd\i r,}
yani $\R$'nin bo\c s olmayan, \"usts\i n\i r\i\ olan her altk\"umesinin
en k\"u\c c\"uk \"usts\i n\i r\i\
veya \textbf{supremumu} vard\i r:
\begin{multline*}\label{A}
\Exists xx\in A\land\Exists x\Forall y(y\in A\lto y\leq x)\lto\\
\Exists x\Bigl(\Forall y(y\in A\lto y\leq x)\land{}\\
\Forall z\bigl(\Forall y(y\in A\lto y\leq z)\lto x\leq z\bigr)\Bigr);	
\end{multline*}
\item
$\R$, iki-konumlu \textbf{toplama} ve \textbf{\c carpma} i\c slemleri alt\i nda kapal\i d\i r,
ve bu i\c slemler ile $\R$ bir \textbf{cisimdir,} yani
\begin{gather*}
\begin{gathered}
	a+b=b+a,\\
	a+0=a,\\
	-a+a=0,	
\end{gathered}\qquad
\begin{gathered}
	ab=ba,\\
	a\cdot 1=a,\\
	a\neq0\lto\Exists xax=1,
	\end{gathered}\\
	a\cdot(b+c)=ab+ac;
\end{gather*}
\item
$\R$'nin s\i ralamas\i\ ve cisim yap\i s\i\ birbirine sayg\i\ g\"osterir:
\begin{gather*}
	a\leq0\liff-a\geq0,\\
	a>0\land b>0\lto a+b>0\land ab>0.
\end{gather*}
\end{compactenum}
Her iki tam s\i ral\i\ cismin birbirine izomorf oldu\u gunu,
teorem olarak
kan\i tlayabiliriz.

\section{Ger\c cel say\i lar\i n in\c sas\i}

Ger\c cel analizdeki gibi,
bu b\"ol\"umde $\R$'nin 
%tam s\i ral\i\ cisminin 
var oldu\u gunu,
aksiyom olarak kabul ediyoruz.
Fakat \emph{k\"ume aksiyomlar\i n\i} kullanarak
ger\c cel say\i lar\i\ in\c sa edebiliriz.
K\i saca
\begin{compactenum}[1)]
\item\label{Q} 
$\R$,
$\Q$ kesirli say\i lar s\i ral\i\ cisminden elde edilir,
\item\label{Z}
$\Q$, $\Z$ tamsay\i lar s\i ral\i\ halkas\i ndan elde edilir,
\item\label{N}
$\Z$, $\N$ sayma say\i lar\i\ yap\i s\i ndan elde edilir.
\end{compactenum}
Yukar\i daki in\c salar,
a\c sa\u g\i daki \c sekilde yap\i l\i r.
\begin{compactenum}
\item
Her ger\c cel say\i, \"oyle bir $A$ k\"umesi olur ki
\begin{compactenum}
\item
$\emptyset\pincluded A\pincluded\Q$,
yani $A$ bo\c s de\u gildir,
$A$'n\i n elemanlar\i\ kesirli say\i d\i rlar,
ve her kesirli say\i\ $A$'n\i n eleman\i\ de\u gildir;
\item
$\Forall x\Forall y(x\in A\land y\in\Q\land y<x\lto y\in A)$,
yani $A$'n\i n bir eleman\i ndan k\"u\c c\"uk olan her kesirli say\i\
$A$'n\i n eleman\i d\i r; ve
\item
$\Forall x\Exists y(x\in A\lto y\in A\land x<y)$,
yani $A$'n\i n en b\"uy\"uk eleman\i\ yoktur.
\end{compactenum}
\item
Her $a/b$ kesirli say\i s\i,
$\{(x,y)\in\Z\times(\Z\setminus\{0\})\colon ay=bx\}$
k\"umesi olarak tan\i mlan\i r.
\item
Benzer \c sekilde her tamsay\i,
baz\i\ $a$ ve $b$ sayma say\i lar\i\ i\c cin
$a-b$ bi\c ciminde yaz\i labilir,
ve tamsay\i n\i n kendisi
$\{(x,y)\in\N\times\N\colon a+y=b+x\}$
k\"umesi olarak tan\i mlan\i r.
\end{compactenum}
Burada $\N$ yap\i s\i n\i n \"ozelliklerini,
\emph{Peano Aksiyomlar\i ndan} t\"uretebiliriz,
ve ondan sonra $\R$'nin tam s\i ral\i\ bir cisim oldu\u gunu,
teorem olarak kan\i tlayabiliriz.

Peano Aksiyomlar\i n\i\ kullanmadan
$\N$, 
s\i f\i r olmayan sonlu olan \emph{ordinal say\i lar} taraf\i ndan olu\c sturulabilir.
B\"ol\"um \numarada{ch:ord} ordinallerin \"ozelliklerini,
aksiyom olarak verece\u giz.
B\"ol\"um \numarada{ch:ax}
her ordinali bir k\"ume olarak tan\i mlayaca\u g\i z,
ve ordinallerin
\emph{ordinal aksiyomlar\i} sa\u glad\i\u g\i n\i\,
teorem olarak kan\i tlayaca\u g\i z.
Bu \c sekilde ger\c cel analizi,
k\"umeler kuram\i nda temelle\c stirebiliriz.
Ayr\i ca ger\c el analizin
ve ordinal analizin baz\i\ ortak \"ozellikleri
olacakt\i r.

\section{Sayma say\i lar\i}

\c Simdilik, tam tersine,
ger\c cel say\i lar\i n yukar\i daki aksiyomlar\i n\i\ varsayarak
$\N$ yap\i s\i n\i\ elde edece\u giz.

$\R$'nin her $A$ altk\"umesi i\c cin,
\begin{compactenum}[1)]
\item
$1\in A$ ve
\item
$A$'n\i n her $b$ eleman\i\ i\c cin $b+1\in A$
\end{compactenum}
durumunda $A$'ya \textbf{t\"umevar\i ml\i} densin.
O zaman tan\i ma g\"ore
\begin{equation*}
\N=\bigcap\{X\included\R\colon X\text{ t\"umevar\i ml\i d\i r}\}
\end{equation*}
olsun.
Bu \c sekilde \textbf{sayma say\i s\i} olmak i\c cin gerek ve yeter ko\c sul,
$\R$'nin her t\"umevar\i ml\i\ altk\"umesinin eleman\i\ olmakt\i r.
Genelde elemanlar\i\ k\"ume olan her $\mathscr B$ k\"umesi i\c cin
\begin{equation*}
\bigcap\mathscr B=\{x\colon\Forall Y(Y\in\mathscr B\lto x\in Y)\}.
\end{equation*}
Bu yeni k\"ume, $B$'nin \textbf{kesi\c simidir.}
\"Ozel olarak
\begin{equation*}
C\cap D=\bigcap\{C,D\}.
\end{equation*}
Ayr\i ca
\begin{equation*}
\bigcap_{i\in\N}A_i=\bigcap\{A_i\colon i\in\N\}.
\end{equation*}

\begin{theorem}[T\"umevar\i m]\label{thm:ind}
$\N$ t\"umevar\i ml\i d\i r.
Ayr\i ca $\N$'nin tek t\"umevar\i ml\i\ altk\"umesi, kendisidir.
\end{theorem}

\ktk

Bu teoreme g\"ore t\"umevar\i ml\i\ kan\i tlar yap\i labilir.
Yani $\N$'nin herhangi $A$ altk\"umesi i\c cin
\begin{compactenum}[1)]
\item
$1\in A$ ve
\item
$b\in A\lto b+1\in A$
%$\Forall x(x\in A\lto x+1\in A)$
\end{compactenum}
ise, o zaman t\"umevar\i mdan $A=\N$.
Bu kan\i tta $b\in A$ varsay\i m\i,
kan\i t\i n \textbf{t\"umevar\i m hipotezidir.}

\begin{lemma}\label{lem:-}
Her sayma say\i s\i, ya $1$'dir,
ya da bir $k$ sayma say\i s\i\ i\c cin $k+1$'dir.
\end{lemma}

\klk[ (T\"umevar\i m kullan\i n.)]

Her $a$ ger\c cel say\i s\i, $(a-1)+1$ bi\c ciminde yaz\i labilir,
ama $a$'n\i n sayma say\i s\i\ oldu\u gunda bile $a-1$, sayma say\i s\i\ olmayabilir.

\begin{lemma}\label{lem:N<}
	$\N$ do\u grusal s\i ral\i d\i r, ve her $k$ eleman\i\ i\c cin
	\begin{equation*}
	k<k+1.
	\end{equation*}
	Ayr\i ca
	\begin{equation*}
	k<\ell\lto k+1<\ell+1.
	\end{equation*}
\end{lemma}

\klk[ ($\N\included\R$ oldu\u gundan $\N$,
$\R$'den baz\i\ \"ozellikleri al\i r.)]

\begin{lemma}\label{lem:1leq}
En k\"u\c c\"uk sayma say\i s\i\ vard\i r,
ve bu say\i\ $1$'dir.
\end{lemma}

\klk[ (T\"umevar\i m ve  Lemma \numarayi{lem:N<}
kullan\i n.)]

\begin{lemma}\label{lem:k<m}
Herhangi $k$ ve $m$ sayma say\i lar\i\ i\c cin
\begin{equation}\label{eqn:km}
k\leq m\liff k<m+1,
\end{equation}
yani
$\{x\in\N\colon x<m\}\cup\{m\}=\{x\in\N\colon x<m+1\}$.
\end{lemma}

\klk[ (Lemma \numarada{lem:N<}n
\eqref{eqn:km} denkli\u gi
$m<k\liff m+1\leq k$ bi\c ciminde yaz\i labilir.
Bunun bir y\"on\"u apa\c c\i kt\i r.
Di\u ger y\"on $k$ \"uzerinde t\"umevar\i m,
Lemmalar \ref{lem:-}, \ref{lem:N<}, ve \ref{lem:1leq} ile
kan\i tlanabilir.)]

\begin{theorem}[G\"u\c clu t\"umevar\i m]
$A\included\N$ olsun,
ve t\"um $k$ sayma say\i lar\i\ i\c cin
\begin{equation*}
\{x\in\N\colon x<k\}\included A\lto k\in A
\end{equation*}
olsun.  O zaman $A=\N$.
\end{theorem}

\ktk[ ($\bigl\{x\in\N\colon\{y\in\N\colon y<x\}\included A\bigr\}$ k\"umesi $B$ olsun.
Lemmalar \ref{lem:1leq} ve \ref{lem:k<m}
ve t\"umevar\i m ile $B=\N$ oldu\u gunu kan\i tlay\i n.)]

\"Orne\u gin 
g\"u\c cl\"u t\"umevar\i mdan
her sayma say\i s\i\ ya $1$'dir
ya da asal bir say\i\ taraf\i ndan b\"ol\"un\"ur.
Zira bu \"ozelli\u gi olan sayma say\i lar\i\
bir $A$ k\"umesini olu\c stursun.
Bir $m$ i\c cin $k<m$ ise $k\in A$ olsun.
E\u ger $m=1$ ise $m\in A$.
E\u ger $m$ asal ise $m\in A$.
Kalan durumda bir $k$ i\c cin $1<k<m$ ve $k\divides m$.
(Burada
O zaman $k\in A$, ama $k\neq1$,
dolay\i s\i yla bir $p$ asal\i\ i\c cin $p\divides k$,\label{p}
ve sonu\c c olarak $p\divides m$ ve $m\in A$.
G\"u\c cl\"u t\"umevar\i mdan $A=\N$.

\begin{theorem}[\.Iyis\i ralama]
$\N$ \textbf{iyis\i ral\i d\i r,} yani $\N$'nin bo\c s olmayan her altk\"umesinin en k\"u\c c\"uk eleman\i\ vard\i r.
\end{theorem}

\ktk[ ($A\included\N$ olsun,
ama $A$'n\i n en k\"u\c c\"uk eleman\i\ olmas\i n.
G\"u\c cl\"u t\"umevar\i m ile $\N\setminus A=\N$ kan\i tlay\i n.)]

\section{G\"ondermeler}

E\u ger $f$, tan\i m k\"umesi $A$ olan
ve de\u ger k\"umesi $B$ olan bir g\"ondermeyse,
bu durum
\begin{equation*}
  f\colon A\to B
\end{equation*}
c\"umlesiyle ifade edilebilir.
Ayr\i ca $f$'nin kendisinin yerine
\begin{equation*}
x\mapsto f(x)
\end{equation*}
isimi kullan\i labilir;
\Sekle{fig:rec}, Al\i\c st\i rma \numaraya{ex:xx'} ve \Teoreme{thm:a+x} bak\i n.

E\u ger $f\colon A\to B$, $C\included A$, $g\colon C\to B$, ve $C$'nin her $d$ eleman\i\ $g(d)=f(d)$ ise,
o zaman
\begin{equation*}
g=f\restriction C;
\end{equation*}
\Teoremin{thm:ord-rec} kan\i t\i na bak\i n.
E\u ger $A$'dan $B$'ye giden birebir ve \"orten g\"onderme varsa,
bu g\"onderme bir \textbf{e\c slemedir,}
ve verilen k\"umeler \textbf{e\c sleniktir;}
bu durum
\begin{equation*}
  A\approx B
\end{equation*}
c\"umlesiye ifade edilir.
\Teoreme{thm:a+b} bak\i n.

Herhangi $A$ ve $B$ k\"umelerinin \textbf{kartezyan \c carp\i m\i} vard\i r.
Tan\i ma g\"ore
\begin{equation*}
A\times B=\{(x,y)\colon x\in A\land y\in B\}.
\end{equation*}
Burada
\begin{equation*}
(a,b)=(c,d)\liff a=c\land b=d.
\end{equation*}
E\u ger $f\colon A\to B$ ise,
o zaman $f$, $A\times B$ \c carp\i m\i n\i n
\begin{equation*}
\bigl\{\bigl(x,f(x)\bigr)\colon x\in A\bigr\}
\end{equation*}
altk\"umesini belirtir.

Bir $A$ k\"umesinde tek-konumlu bir i\c slem,
$A$'dan kendisine giden bir g\"ondermedir;
iki-konumlu bir i\c slem,
$A\times A$ \c carp\i m\i ndan $A$'ya giden bir g\"ondermedir.

Ger\c cel analiz ve say\i lar kuram\i nda
tan\i m k\"umesi $\N$ olan g\"ondermeler tan\i mlan\i p kullan\i l\i r.
\"Orne\u gin $x\mapsto x!$ g\"ondermesi i\c cin
\begin{align*}
	1!&=1,&(k+1)!&=(k+1)\cdot k!
\end{align*}
\textbf{\"ozyineli} tan\i m\i\ verilir.
Bu tan\i m neden ge\c cerli midir?

Tan\i m\i n ge\c cerlili\u gi i\c cin
$\N$ t\"umevar\i ml\i\ olmal\i d\i r,
ama bunu \Teoremde{thm:ind}n biliyoruz.
Ayr\i ca $\N$, 
ger\c cel say\i lar\i n \c carpmas\i\ alt\i nda kapal\i\ olmal\i d\i r.

\begin{theorem}
T\"um $a$ ve $b$ ger\c cel say\i lar\i\ i\c cin
\begin{equation*}
a\in\N\land b\in\N\lto a+b\in\N\land a\cdot b\in\N.
\end{equation*}
\end{theorem}

\ktk

\c Simdi $1\in\N$,
ve ayr\i ca $\N$'nin herhangi $k$ eleman\i\ i\c cin
e\u ger $k!\in\N$ ise,
o zaman $(k+1)\cdot k!\in\N$.
Bu \c sekilde $x\mapsto x!$ g\"ondermesi
tan\i mlanabilir mi?
\begin{compactitem}
\item
T\"umevar\i m veya g\"u\c cl\"u t\"umevar\i m ile 
bir k\"umenin $\N$ oldu\u gu kan\i tlanabilir;
ama $\{x\in\N\colon x!\text{ tan\i mlan\i r}\}$,
iyi tan\i mlanm\i\c s bir k\"ume de\u gildir.
\item
\.Iyis\i ralama ile $\N$'nin bo\c s olmayan bir alt\"umesinin eleman\i\ bulunabilir;
ama $x\mapsto x!$ g\"ondermesi, $\N$'nin bir eleman\i\ de\u gildir.
\end{compactitem}
Ba\c ska bir teoreme ihtiyac\i m\i z vard\i r.


\begin{theorem}[\"Ozyineleme]\label{thm:rec}
Bir $A$ k\"umesi i\c cin
\begin{compactenum}[1)]
\item
$b\in A$,
\item
$f\colon A\to A$ 
\end{compactenum}
olsun.
O zaman $\N$'den $A$'ya giden bir ve tek bir $g$ g\"ondermesi i\c cin
\begin{compactenum}[1)]
\item
$g(1)=b$,
\item
her $k$ sayma say\i s\i\ i\c cin $g(k+1)=f(g(k))$.
\end{compactenum}
\Sekle{fig:rec} bak\i n.
\begin{figure}
\relscale{1.1}
\begin{equation*}
\xymatrix@=2cm{\{1\}\ar[d]_g\ar[r]^{\included}&\N\ar[d]_g\ar[r]^{x\mapsto x+1}&\N\ar[d]^g\\
\{b\}\ar[r]_{\included}&A\ar[r]_f&A}
\end{equation*}
\caption{\"Ozyineleme}\label{fig:rec}
\end{figure}
\end{theorem}

\"Orne\u gin $A=\N\times\N$,
$b=(1,1)$, ve
\begin{equation*}
f(x,y)=\bigl(x+1,(x+1)\cdot y\bigr)
\end{equation*}
olsun.
O zaman bir ve tek bir $g$ g\"ondermesi i\c cin
$g$'nin tan\i m k\"umesi $\N$, $g(1)=(1,1)$, ve $g(k+1)=f(g(k))$.
\c Simdi $g(k)$, $(g_1(k),g_2(k))$ olarak yaz\i ls\i n.
T\"umevar\i mdan $g_1(k)=k$.  Bundan dolay\i
\begin{equation*}
g_2(k+1)=(k+1)\cdot g_2(k).
\end{equation*}
Ayr\i ca $g_2(1)=1$.
B\"oylece $k!$, $g_2(k)$ olarak tan\i mlanabilir.

\begin{proof}[\"Ozyineleme Teoremi kan\i t\i.]
Bir ve tek bir $h$ g\"ondermesi i\c cin
\begin{compactenum}[1)]
\item
g\"ondermenin tan\i m k\"umesi, $\N$'nin tek-elemanl\i\ $\{1\}$ altk\"umesidir, ve
\item
$h(1)=b$.
\end{compactenum}
Bu g\"onderme $h_1$ olsun.

T\"umevar\i m hipotezi olarak
bir $m$ sayma say\i s\i\ i\c cin,
bir ve tek bir $h$ g\"ondermesi i\c cin
\begin{compactenum}[1)]
\item
g\"ondermenin tan\i m k\"umesi, $\N$'nin $m$ elemanl\i\ $\{1,\dots,m\}$ altk\"umesi olsun,
\item
$h(1)=b$ olsun, ve
\item
$k<m$ ise $h(k+1)=f(h(k))$ olsun.
\end{compactenum}
Bu g\"onderme $h_m$ olsun.
O zaman
bir ve tek bir $h$ g\"ondermesi i\c cin
\begin{compactenum}[1)]
\item
g\"ondermenin tan\i m k\"umesi, $\{1,\dots,m+1\}$ altk\"umesidir,
\item
$h(1)=b$, ve
\item
$k<m+1$ ise $h(k+1)=f(h(k))$.
\end{compactenum}
Zira b\"oyle bir $h$ varsa,
o zaman $h\restriction\{1,\dots,m\}$ ve $h_m$ g\"ondermelerinin
\"ozellikleri ayn\i d\i r,
dolay\i s\i yla $h_m$ g\"ondermesinin biricikli\u ginden $h\restriction\{1,\dots,m\}=h_m$.
Bu \c sekilde $h$'nin tan\i m\i
\begin{equation}\label{eqn:h}
h(x)=\begin{cases}
	h_m(x),&\text{ $x\leq m$ durumunda,}\\
	f(h_m(m)),&\text{ $x=m+1$ durumunda}
\end{cases}
\end{equation}
olabilir.
Ayr\i ca $h\restriction\{1,\dots,m\}=h_m$ olmal\i d\i r,
dolay\i s\i yla $h$'nin kendisi,
\eqref{eqn:h} e\c sitli\u gini sa\u glamal\i d\i r.
Bu $h$ g\"ondermesi $h_{m+1}$ olsun.

T\"umevar\i mdan,
her $n$ sayma say\i s\i\ i\c cin,
$\{1,\dots,n\}$ k\"umesinden giden bir be tek bir $h_n$ g\"ondermesi i\c cin
$h_n(1)=b$ ve $k<n$ ise $h_n(k+1)=f(h_n(k))$.
Ayr\i ca
\begin{equation*}
h_{m+1}(m+1)=f(h_m(m)).
\end{equation*}
\c Simdi $g(x)=h_x(x)$ olsun.  O zaman
\begin{gather*}
g(1)=h_1(1)=b,\\
g(k+1)=h_{k+1}(k+1)=f(h_k(k))=f(g(k)).
\end{gather*}
Bu \c sekilde $g$, istedi\u gimiz gibidir.
Bir $h$ g\"ondermesinin istedi\u gimiz \"ozelli\u gi varsa
\begin{equation*}
h(1)=b=g(1),
\end{equation*}
ve $h(m)=g(m)$ ise
\begin{equation*}
	h(m+1)=f(h(m))=f(g(m))=g(m+1).
\end{equation*}
Bu \c sekilde her $k$ sayma say\i s\i\ i\c cin $h(k)=g(k)$,
dolay\i s\i yla $h=g$.	
\end{proof}

\begin{sloppypar}
  Baz\i\ yap\i larda t\"umevar\i m kullan\i labilir,
  ama \"ozyineleme kullan\i lamaz.
  \"Orne\u gin $p$ asal ise,\label{p2}
  Fermat Teoremine g\"ore herhangi $a$ tamsay\i s\i\ i\c cin
  \begin{equation}\label{eqn:Fermat}
    a^p\equiv a\pmod p.
  \end{equation}
  T\"umevar\i m ile bu teoremi kan\i tlayabiliriz,
  zira $1^p\equiv1$, 
  ve ayr\i ca $b^p\equiv b$ ise, o zaman
  \begin{align*}
    (b+1)^p
    &=b^p+pb^{p-1}+\binom p2b^{p-2}+\dots+\binom p{p-2}b^2+pb+1\\
    &\equiv b^p+1\equiv b+1\pmod p,
  \end{align*}
  \c c\"unk\"u $0<k<p$ ise $k\divides\binom pk$.
  Neden bu kan\i t ge\c cerlidir?
  Say\i lar kuram\i ndan
  \begin{multline}\label{eqn:+.mod}
    a\equiv a_1\land b\equiv b_1\\
    \lto a+b\equiv a_1+b_1\land ab\equiv a_1b_1\pmod p.
  \end{multline}
  $\Zmod p$, tamsay\i lar\i n $p$'ye g\"ore kalanda\c sl\i k s\i n\i flar\i\ k\"umesi olsun.
  Bu \c sekilde
  \begin{align*}
    \Zmod p&=\{[x]\colon x\in\Z\},&
	  [k]&=\{x\in\Z\colon x\equiv k\pmod p\}.
  \end{align*}
  O zaman $\Zmod p=\{[1],\dots,[p]\}$.
  Ayr\i ca \eqref{eqn:+.mod} gerektirmesine g\"ore
  \begin{align*}
    [a]+[b]&=[a+b],&
    [a][b]&=[ab]
  \end{align*}
  tan\i mlar\i\ ge\c cerlidir, \c c\"unk\"u
  \begin{equation*}
    [a]=[a_1]\land[b]=[b_1]\lto[a+b]=[a_1+b_1]\land[ab]=[a_1b_1].
  \end{equation*}
  \c Simdi $A\included\Zmod p$ ve $[1]\in A$ olsun,
  ve $[k]\in A$ ise $[k+1]\in A$ olsun.
  O zaman t\"umevar\i mdan $A=\Zmod p$.
  Zira $B=\{x\in\N\colon[x]\in A\}$ olsun.
  O zaman $1\in B$, \c c\"unk\"u $[1]\in A$.
  Ayr\i ca $k\in B$ ise,
  o zaman $[k]\in A$, dolay\i s\i yla $[k+1]\in A$ ve $k+1\in B$.
  T\"umevar\i mdan $B=\N$.
  \"Ozel olarak $\{1,\dots,p\}\included B$,
  dolay\i s\i yla $\Zmod p=A$.
\end{sloppypar}
Yukar\i daki g\"osterdi\u gimize g\"ore $[1]^p=[1]$,
ve $[b]^p=[b]$ ise $[b+1]^p=[b+1]$.
O zaman t\"umevar\i mdan her $a$ tamsay\i s\i\ i\c cin $[a]^p=[a]$,
yani \eqref{eqn:Fermat} kalanda\c sl\i\u g\i\ do\u grudur.

B\"oylece $\Zmod p$ yap\i s\i nda t\"umevar\i m y\"ontemi ge\c cerlidir;
ama \"ozyineleme y\"ontemi ge\c cerli de\u gildir.
\"Orne\u gin $\Zmod3$ yap\i s\i nda hi\c cbir tek-konumlu $g$ i\c slemi i\c cin
\begin{align*}
	g([1])&=[2],&g([k+1])&=[k][2]
\end{align*}
olmaz, \c c\"unk\"u olursa
\begin{align*}
	g([2])&=[4]=[1],&
	g([3])&=[2],&
	g([4])&=[1],
\end{align*}
ama $[4]=[1]$ oldu\u gundan $g([4])=g([1])=[2]$,
ve $[2]\neq[1]$.

\begin{xca}\sloppy
\"Ozyineleme Teoreminin yukar\i daki kan\i t\i,
$\N$'nin hangi \"ozelliklerini kullan\i r?
\end{xca}

\section{Peano Aksiyomlar\i}

\"Ozyineleme Teoreminin ba\c ska bir kan\i t\i\ vard\i r.

\begin{proof}[\"Ozyineleme Teoremi ikinci kan\i t\i.]
Birinci kan\i ttaki gibi,
istedi\u gimiz \"ozellikleri olan bir g\"onderme varsa,
tek bir \"ornek vard\i r.

\c Simdi elemanlar\i\ g\"onderme olan
bir $\mathscr C$ k\"umesini tan\i mlayaca\u g\i z.\label{curly-C}
$\mathscr C$'nin her $h$ eleman\i\ i\c cin,
\begin{compactenum}[1)]
\item
$h$'nin tan\i m k\"umesi 
$\N$'nin bir altk\"umesidir, ve
\item
herhangi $\ell$ sayma say\i s\i\ i\c cin,
$h(\ell)$ tan\i mlan\i rsa, o zaman
\begin{compactenum}
\item 
ya $\ell=1$ ve $h(\ell)=b$,
\item
ya da bir $k$ sayma say\i s\i\ i\c cin $\ell=k+1$,
$h(k)$ tan\i mlan\i r, ve
\begin{equation*}
  h(\ell)=f(h(k)).
\end{equation*}
\end{compactenum}
\end{compactenum}
Lemma \ref{lem:-} sayesinde istedi\u gimiz gibi $g$ g\"ondermesi varsa
$\mathscr C$'nin eleman\i d\i r.
Her $k$ sayma say\i s\i\ i\c cin,
$A$'n\i n bir ve tek bir $d$ eleman\i\ i\c cin,
$\mathscr C$'nin bir $h$ eleman\i\ i\c cin $h(k)=d$ g\"osterece\u giz.
Bu \c sekilde $g(k)=d$ tan\i mlanabilir.

Yukar\i daki \"ozelli\u gi olan $k$ sayma say\i lar\i,
$E$ k\"umesini olu\c stursun.
Tan\i m k\"umesi $\{1\}$ olan
bir $h$ g\"ondermesi i\c cin $h(1)=b$.
O zaman $h\in\mathscr C$.
Ayr\i ca $\mathscr C$'nin herhangi $h$ eleman\i\ i\c cin $h(1)$ tan\i mlan\i rsa,
o zaman $h(1)=b$ olmal\i d\i r, 
\c c\"unk\"u hi\c c $k$ sayma say\i s\i\ i\c cin $k+1=1$ de\u gildir.
Bu \c sekilde $1\in E$.

\c Simdi $k\in E$ olsun.
O zaman $A$'n\i n bir ve tek bir $d$ eleman\i\ i\c cin,
$\mathscr C$'nin bir $h$ eleman\i\ i\c cin $h(k)=d$.
\begin{compactenum}
\item
E\u ger $h(k+1)$ tan\i mlan\i rsa,
o zaman $\mathscr C$'nin tan\i m\i na g\"ore
$h(k+1)=f(d)$, 
\c c\"unk\"u $k+1\neq1$, ve ayr\i ca herhangi $\ell$ sayma say\i s\i\ i\c cin
e\u ger $\ell+1=k+1$ ise, o zaman $\ell=k$.
\item
E\u ger $h(k+1)$ tan\i mlanmazsa, 
o zaman yeni bir $h^*$ g\"ondermesi i\c cin
\begin{equation*}
h^*(x)=\begin{cases}
	h(x),&\text{ e\u ger $h(x)$ tan\i mlan\i rsa,}\\
	f(d),&\text{ e\u ger $x=k+1$.}
\end{cases}
\end{equation*}
O zaman $h^*\in\mathscr C$ ve $h^*(k+1)=f(d)$.
\end{compactenum}
Bu \c sekilde, her durumda,
%$A$'n\i n $f(d)$ eleman\i\ i\c cin,
$\mathscr C$'nin bir $h$ eleman\i\ i\c cin $h(k+1)=f(d)$.

M\"umk\"umse $d^*\in A$, $d^*\neq f(d)$ olsun, 
ama $\mathscr C$'nin bir $h$ eleman\i\ i\c cin $h(k+1)=d^*$ olsun.
O zaman $k+1\neq1$ oldu\u gundan bir $\ell$ sayma say\i s\i\ i\c cin
$\ell+1=k+1$, $h(\ell)$ tan\i mlan\i r, ve $d^*=f(h(\ell))$.
Ama bu durumda $\ell=k$, dolay\i s\i yla $h(\ell)=d$ ve $d^*=f(d)$.

Sonu\c c olarak $k+1\in E$.  T\"umevar\i m ile $E=\N$.
\end{proof}

\begin{sloppypar}
  Yukar\i daki kan\i t, sadece $\N$'nin a\c sa\u g\i daki \"ozelliklerini kullan\i r:
  \begin{compactenum}
  \item
    $1\in\N$.
  \item
    $k\in\N$ ise $k+1\in\N$.
  \item
    T\"umevar\i m y\"ontemi ge\c cerlidir.
  \item
    Her $k$ sayma say\i s\i\ i\c cin
    $1\neq k+1$.
  \item
    T\"um $k$ ve $\ell$ sayma say\i lar\i\ i\c cin
    $k+1=\ell+1$ ise $k=\ell$.
  \end{compactenum}
  Bu \"ozelliklere \textbf{Peano Aksiyomlar\i} denir.
  Peano Aksiyomlar\i,
  $\N$'de iki-konumlu toplama i\c sleminin tan\i mland\i\u g\i n\i\ varsaymaz;
  sadece tek-konumlu $x\mapsto x+1$ i\c slemi vard\i r.
  Ama \"ozyineleme y\"ontemiyle 
  $\N$'de toplama ve \c carpma i\c slemlerini tan\i mlayabiliriz:
  \begin{align*}
    a+(b+1)&=(a+b)+1,&a\cdot 1&=a,&a\cdot(b+1)&=ab+a.
  \end{align*}
  T\"umevar\i m ve kalan Peano Aksiyomlar\i\ ile
  toplaman\i n ve \c carpman\i n \"ozelliklerini kan\i tlayabiliriz;
  ayr\i ca $\N$'nin s\i ralamas\i n\i\ tan\i mlay\i p
  \"ozelliklerini kan\i tlayabiliriz.
  Ondan sonra
  yukar\i daki gibi $\Z$, $\Q$ ve $\R$ yap\i lar\i n\i\ elde edebiliriz.
\end{sloppypar}
Tam tersine tam s\i ral\i\ cisim aksiyomlar\i n\i\ kullanarak
$\N$ yap\i s\i n\i\ in\c sa ettik
ve onun
Peano Aksiyomlar\i n\i\ sa\u glad\i\u g\i n\i\ teorem olarak kan\i tlad\i k.
(\Teorem{thm:ind} ve Lemmalar \ref{lem:N<} ve \numaraya{lem:1leq} bak\i n.)

\begin{sloppypar}
  Sayma say\i lar\i na s\i f\i r\i\ ekleyerek do\u gal say\i lar\i\ elde ederiz.
  Do\u gal say\i lar, \emph{sonlu ordinallerdir.}
  Sonsuz ordinaller de vard\i r.
  Ordinallerin aksiyomlar\i n\i\ kullanarak
  toplama ve \c carpma i\c slemlerini tan\i mlayay\i p
  \"ozelliklerini kan\i tlayaca\u g\i z.
  Ondan sonra
  k\"ume aksiyomlar\i n\i\ kullanarak
  ordinalleri in\c sa edece\u giz.
  Bu \c sekilde bildi\u gimiz t\"um matematik,
  k\"ume aksiyomlar\i\ taraf\i ndan gerektirilir.
\end{sloppypar}

\chapter{Ordinal say\i lar}\label{ch:ord}

K\"umeler kuram\i m\i zda her k\"umenin her eleman\i,
bir k\"ume olacakt\i r.
(Bu \c sekilde
``elemanlar\i\ k\"ume olan k\"ume'' ifadesi
gereksiz k\i l\i nacakt\i r.)
Bir k\"ume bo\c s olacakt\i r,
ve bu k\"ume,
\begin{equation*}
\emptyset
\end{equation*}
olarak yaz\i l\i r.
Ayr\i ca
\begin{equation}\label{eqn:0=emptyset}
0=\emptyset
\end{equation}
tan\i mlayaca\u g\i z.
Herhangi $a$ k\"umesi i\c cin
\begin{equation}\label{eqn:a'=}
a'=a\cup\{a\}
\end{equation}
tan\i mlayaca\u g\i z.
Ayr\i ca
\begin{align}\label{eqn:1=0'}
	1&=0',&2&=1',&3&=2',&4&=3',&&\dots
\end{align}
olacakt\i r.
O zaman $0$, $1$, $2$, $3$, \lips, 
\textbf{do\u gal say\i} olacaklard\i r.
Do\u gal say\i lar, sonlu \textbf{ordinal} olacaklard\i r,
ama sonsuz ordinaller de var olacakt\i r.
\"Orne\u gin
\begin{equation}\label{eqn:omega=}
\upomega=\{0,1,2,\dots\},
\end{equation}
ve $\upomega$, en k\"u\c c\"uk sonsuz ordinal olacakt\i r.

\c Simdilik \eqref{eqn:0=emptyset}, \eqref{eqn:a'=}, ve \eqref{eqn:omega=}
tan\i mlar\i n\i\ kullanmayaca\u g\i z.
B\"ol\"um \numarada{ch:ax},
k\"ume aksiyomlar\i n\i\ kullanarak,
ordinalleri tan\i mlay\i p
\"ozelliklerini
teorem olarak kan\i tlayaca\u g\i z;
ama
\c simdilik ordinaller\i n a\c sa\u g\i da verilen \"ozelliklerini
aksiyom olarak kabul edece\u giz.

\section{K\"umeler ve s\i n\i flar}\label{sect:sets-classes}

Ordinaller bir
\begin{equation*}
\on
\end{equation*}
\textbf{s\i n\i f\i n\i} olu\c sturacakt\i r.
Biz zaten $\Zmod p$ yap\i s\i n\i\ tan\i mlamak i\c cin \emph{denklik s\i n\i flar\i} kulland\i k.
Normalde bir denklik s\i n\i f\i\ bir k\"umedir.
Asl\i nda her k\"ume bir s\i n\i ft\i r,
ama her s\i n\i f bir k\"ume de\u gildir.

Her s\i n\i f
\emph{tek serbest de\u gi\c skeni olan bir form\"ul} 
taraf\i ndan tan\i mlan\i r.
\"Orne\u gin birazdan kullanaca\u g\i m\i z
\begin{gather*}
	x\in a,\\
	x\in\bm A\liff x\in\bm B,\\
	x\in c\liff x\in d,\\
	x\notin x
\end{gather*}
ifadeleri,
serbest de\u gi\c skeni $x$ olan form\"uld\"urler.
(Form\"ullerin resmi tan\i m\i\ i\c cin,
\ref{mantik} Eki'ne bak\i n.)
E\u ger $\phi$,\label{phi} 
tek serbest de\u gi\c skeni olan bir form\"ul ise,
o zaman $\phi$'nin tan\i mlad\i\u g\i\ s\i n\i f\i n elemanlar\i,
$\phi$'yi sa\u glayan k\"umelerdir,
ve s\i n\i f\i n kendisi
\begin{equation*}
\{x\colon\phi(x)\}
\end{equation*}
olarak yaz\i labilir.
S\i n\i flar b\"uy\"uk siyah harfler de ile g\"osterece\u giz.
K\"u\c c\"uk harfler
% ve siyah olmayan b\"uy\"uk harfler,
her zaman k\"ume olacakt\i r.
\"Ornegin
\begin{equation*}
\bm A=\{x\colon\phi(x)\}
\end{equation*}
ise, o zaman her $b$ k\"umesi i\c cin
\begin{equation*}
b\in\bm A\liff\phi(b).
\end{equation*}
\emph{Her k\"ume bir s\i n\i fa e\c sittir.}
\"Ozel olarak her $a$ k\"umesi i\c cin
\begin{equation*}
a=\{x\colon x\in a\}.
\end{equation*}
Genelde elemanlar\i\ ayn\i\ olan s\i n\i flar ve k\"umeler e\c sittir:
\begin{gather*}
\bm A=\bm B\liff\Forall x(x\in\bm A\liff x\in\bm B),\\
c=d\liff\Forall x(x\in c\liff x\in d).	
\end{gather*}

\begin{theorem}[Russell Paradoksu]
Her s\i n\i f bir k\"umeye e\c sit de\u gildir.
\"Orne\u gin $\{x\colon x\notin x\}$ s\i n\i f\i\ bir k\"umeye e\c sit de\u gildir.
\end{theorem}

\begin{proof}
$x\notin x$ form\"ul\"u $\phi(x)$ olarak yaz\i ls\i n.  E\u ger $\{x\colon\phi(x)\}=a$ ise,
o zaman her $b$ k\"umesi i\c cin
\begin{equation*}
b\in a\liff\phi(b).
\end{equation*}
\"Ozel olarak
$a\in a\liff\phi(a)$,
yani
\begin{equation*}
a\in a\liff a\notin a;
\end{equation*}
ama bu bir \c celi\c skidir.
Bu \c sekilde $\{x\colon x\notin x\}$ s\i n\i f\i,
bir $a$ k\"umesine e\c sit olamaz.
\end{proof}

\"Oklid'de e\c sitlik, ayn\i l\i k de\u gildir.
\.Ikizkenar bir \"u\c cgenin iki e\c sit kenar\i\ vard\i r.
Bu kenarlar iki oldu\u gundan birbiriyle ayn\i\ de\u gildir.
Ama e\c sit s\i n\i flar ayn\i\ olarak d\"u\c s\"un\"ulebilir.

\section{Ordinallerin \"ozellikleri}

K\"u\c c\"uk Yunan harfleri her zaman ordinal g\"osterecektir.
\"Ozel olarak $\alpha$, $\beta$, $\gamma$, $\delta$, ve $\theta$,\label{minus-gr}
sabit ordinaldirler,
ama $\xi$, $\eta$, ve $\zeta$, ordinal de\u gi\c skendirler.
\"Ornegin
\begin{equation*}
\{\xi\colon\phi(\xi)\}
=\{\eta\colon\phi(\eta)\}
=\{\zeta\colon\phi(\zeta)\}
=\{x\colon x\in\on\land\phi(x)\}.
\end{equation*}
\c Simdilik aksiyom olarak kabul edece\u gimiz
$\on$'nin \"ozellikleri a\c sa\u g\i dad\i r.
\begin{enumerate}
\item\label{on-props}
En az bir ordinal vard\i r.
\item
$\on$ iyis\i ral\i d\i r.
\item
Her ordinal i\c cin, daha b\"uy\"uk ordinal vard\i r.
\item
$\on$'nin herhangi altk\"umesinin $\on$'de olan \"usts\i n\i r\i\ vard\i r.
\item
Herhangi $\alpha$ i\c cin $\{\xi\colon\xi<\alpha\}$ s\i n\i f\i\ bir k\"umedir.
\item
Herhangi $\bm F$ tek-konumlu ordinal i\c slemi i\c cin,
herhangi $\alpha$ i\c cin $\{\bm F(\xi)\colon\xi<\alpha\}$ s\i n\i f\i\ bir k\"umedir.
\item
Bir $\alpha$ i\c cin $\{\xi\colon\xi<\alpha\}$ k\"umesi sonsuzdur.
\end{enumerate}
Asl\i nda her $\alpha$ ordinali,
$\{\xi\colon\xi<\alpha\}$ k\"umesinin kendisi olarak tan\i mlanabilecektir;
ama \c simdilik bu tan\i m\i\ kullanmay\i p
sadece yukar\i daki yedi \"ozelli\u gi kullanaca\u g\i z.

\begin{theorem}[Burali-Forti Paradoksu]\label{thm:BFP}
$\on$ k\"ume de\u gildir.
\end{theorem}

\begin{proof}
Her ordinalin daha b\"uy\"u\u g\"u oldu\u gundan
$\on$'nin en b\"uy\"uk eleman\i\ yoktur,
dolay\i s\i yla
$\on$'nin $\on$'de olan \"usts\i n\i r\i\ yoktur.
$\on$'nin her altk\"umesinin \"usts\i n\i r\i\ oldu\u gundan
$\on$'nin kendisi k\"ume olamaz.
\end{proof}

\c Simdi $0$,
en k\"u\c c\"uk ordinal olarak tan\i mlans\i n,
ve herhangi $\alpha$ ordinali i\c cin
\begin{equation}\label{eqn:a'=min}
\alpha'=\min\{\xi\colon\alpha<\xi\}
\end{equation}
tan\i mlans\i n.
Bu \c sekilde $\alpha'$, 
$\alpha$'n\i n \textbf{ard\i l\i,} yani
$\alpha$'dan b\"uy\"uk olan ordinallerin en k\"u\c c\"u\u g\"ud\"ur.
\c Simdi yukar\i daki \eqref{eqn:1=0'} tan\i mlar\i n\i\ kullanabiliriz.
Ne s\i f\i r ne bir ard\i l olan ordinal, bir \textbf{limittir.}

\begin{theorem}\label{thm:'}
S\i f\i r olmayan bir $\alpha$ ordinalinin limit olmas\i\ i\c cin 
gerek ve yeter ko\c sul,
\begin{equation*}
\beta<\alpha\lto\beta'<\alpha.
\end{equation*}
\end{theorem}

\ktk

E\u ger $\{\xi\colon\xi<\alpha\}$ k\"umesi sonlu ise,
$\alpha$'ya da \textbf{sonlu} densin;
di\u ger durumda, \textbf{sonsuz.}
O zaman en k\"u\c c\"uk sonsuz ordinal bir limittir,
ve her limit ordinali sonsuzdur.
En k\"u\c c\"uk limit
\begin{equation*}
\upomega
\end{equation*}
olsun.
O zaman $\{\xi\colon\xi<\upomega\}$, do\u gal say\i lar k\"umesidir.

\begin{theorem}[Ordinal T\"umevar\i m]
$\bm A\included\on$ olsun.
E\u ger
\begin{compactenum}[1)]
\item
$0\in\bm A$,
\item
Her $\beta$ i\c cin
\begin{equation*}
\beta\in\bm A\lto\beta'\in\bm A,
\end{equation*}
\item
her $\gamma$ limiti i\c cin
\begin{equation*}
\{\xi\colon\xi<\gamma\}\included\bm A\lto\gamma\in\bm A
\end{equation*}
\end{compactenum}
ise, o zaman $\bm A=\on$.
\end{theorem}

\begin{proof}
Verilen ko\c sullar alt\i nda
$\on\setminus\bm A$ fark\i n\i n en k\"u\c c\"uk eleman\i\ olamaz.
Zira m\"umk\"umse $\alpha=\min(\on\setminus\bm A)$ olsun.
\begin{asparaenum}
\item
$\alpha=0$ ise $\alpha\in\bm A$.
\item
$\alpha=\beta'$ ise $\beta<\alpha$ oldu\u gundan $\beta\in\bm A$,
ama bu durumda $\beta'\in\bm A$, yani $\alpha\in\bm A$.
\item
Varsay\i m\i m\i za g\"ore
$\beta<\alpha$ ise $\beta\in\bm A$.
Bu \c sekilde
\begin{equation*}
\{\xi\colon\xi<\alpha\}\included\bm A.
\end{equation*}
E\u ger $\alpha$ bir limit ise,
o zaman $\alpha$ da $\bm A$'n\i n eleman\i\ olmal\i d\i r.
\end{asparaenum}

Bu \c sekilde her ordinal ya $0$, ya bir ard\i l, ya da bir limit oldu\u gundan
$\alpha\in\bm A$,
ama $\alpha=\min(\on\setminus\bm A)$ varsay\i m\i na g\"ore $\alpha\notin\bm A$.
\"Oyleyse varsay\i m imk\^ans\i zd\i r.
$\on$'nin her bo\c s olmayan altk\"umesinin en k\"u\c c\"uk eleman\i\ var oldu\u gundan
$\on\setminus\bm A=\emptyset$,
dolay\i s\i yla $\bm A=\on$.
\end{proof}

Ordinal t\"umevar\i m ile
\Teoremi{thm:ord-rec}, 
\Teoremi{thm:norm-cond},
\Teoremi{thm:0+},
ve daha sonraki teoremler 
kan\i tlayaca\u g\i z.
Ordinal t\"umevar\i m kullan\i lan bir kan\i t\i n \"u\c c ad\i m\i\ vard\i r:
\begin{compactenum}[1)]
\item
s\i f\i r ad\i m\i,
\item
ard\i l ad\i m\i, ve
\item
limit ad\i m\i.
\end{compactenum}
Ayr\i ca kan\i tta iki t\"umevar\i m hipotezi vard\i r.
Ordinal T\"umevar\i m Teoremini yazarken kulland\i\u g\i m\i z harflerde,
\begin{compactitem}
\item
ard\i l ad\i m\i n\i n hipotezi, $\beta\in\bm A$;
\item
limit ad\i m\i n\i n hipotezi, $\{\xi\colon\xi<\gamma\}\included\bm A$,
yani
\begin{equation*}
\Forall{\xi}(\xi<\gamma\lto\xi\in\bm A).
\end{equation*}
\end{compactitem}


\begin{theorem}[Ordinal \"Ozyineleme]\label{thm:ord-rec}
%$\theta$ bir ordinal, $\bm F\colon\on\to\on$, ve $\bm G\colon\pow{\on}\to\on$ olsun.
Varsay\i mlar\i m\i z,
\begin{compactenum}[1)]
\item
$\theta\in\on$, 
\item
$\bm F\colon\on\to\on$.
%\item
%$\bm G\colon\pow{\on}\to\on$.
\end{compactenum}
O zaman bir ve tek bir $\bm H$ ordinal i\c slemi i\c cin
\begin{compactenum}[1)]
\item
$\bm H(0)=\theta$,
\item
her $\beta$ ordinali i\c cin $\bm H(\beta')=\bm F(\bm H(\beta))$,
\item
her $\gamma$ limiti i\c cin
$\bm H(\gamma)=\sup\{\bm H(\xi)\colon\xi<\gamma\}$.
\end{compactenum}
\end{theorem}

\begin{proof}
Her $\alpha$ i\c cin, 
tan\i m k\"umesi $\{\xi\colon\xi\leq\alpha\}$ olan
bir ve tek bir $h_{\alpha}$ g\"ondermesi i\c cin,
\begin{compactenum}[1)]
\item
$h_{\alpha}(0)=\theta$,
\item
$\beta<\alpha$ ise $h_{\alpha}(\beta')=\bm F(h_{\alpha}(\beta))$,
\item
$\gamma\leq\alpha$ ve limit ise $h_{\alpha}(\gamma)=\sup\{h_{\alpha}(\xi)\colon\xi<\gamma\}$.
\end{compactenum}
Bunu kan\i tlamak i\c cin, ordinal t\"umevar\i m kullanaca\u g\i z.
\begin{asparaenum}
\item
$h_0$, $h_0(0)=\theta$ ile tan\i mlanabilir ve tan\i mlanmal\i d\i r.
Yani $\alpha=0$ durumunda iddia do\u grudur.
\item
E\u ger $\alpha=\delta$ durumunda iddia do\u gru ise
$h_{\delta'}$,
\begin{equation*}
h_{\delta'}(\xi)=\begin{cases}
	h_{\delta}(\xi),&\text{ $\xi\leq\delta$ durumunda,}\\
	\bm F(h_{\delta}(\delta)),&\text{ $\xi=\delta'$ durumunda}
\end{cases}
\end{equation*}
kural\i\ taraf\i ndan tan\i mlanabilir.
Ayr\i ca $h_{\delta'}$ bu \c sekilde tan\i mlanmal\i d\i r,
\c c\"unk\"u hipoteze g\"ore
\begin{equation*}
h_{\delta'}\restriction\{\xi\colon\xi\leq\delta\}=h_{\delta}
\end{equation*}
olmal\i d\i r.
Bundan dolay\i\ $\alpha=\delta'$ durumunda iddia do\u grudur.
\item
Benzer \c sekilde bir $\delta$ i\c cin $\alpha<\delta$ durumlar\i nda iddia do\u gru ise,
o zaman $\alpha<\beta<\delta$ durumlar\i nda $h_{\alpha}(\alpha)=h_{\beta}(\alpha)$.
E\u ger ayr\i ca $\delta$ bir limit ise,
o zaman $h_{\delta}$,
\begin{equation*}
h_{\delta}(\xi)=\begin{cases}
	h_{\xi}(\xi),&\text{ $\xi<\delta$ durumunda,}\\
	\sup\{h_{\xi}(\xi)\colon\xi<\delta\},&\text{ $\xi=\delta$ durumunda}
\end{cases}
\end{equation*}
kural\i\ taraf\i ndan tan\i mlanabilir ve tan\i mlanmal\i d\i r,
ve bu \c sekilde $\alpha=\delta$ durumunda iddia do\u grudur.
\end{asparaenum}

Ordinal t\"umevar\i m\i m\i z bitti.
\c Simdi $\bm H(\xi)=h_{\xi}(\xi)$ tan\i mlanabilir ve tan\i mlanmal\i d\i r.
\end{proof}

B\"ol\"umler \ref{ch:add}, \ref{ch:mul}, ve \numarada{ch:exp}
ordinal \"ozyinelemeyle ordinal toplama, 
\c carpma, ve kuvvet alma i\c slemlerini tan\i mlayaca\u g\i z.

\section{Normal i\c slemler}

\c Simdi $\bm F$,
herhangi tek-konumlu ordinal i\c slem olsun.  
Ordinal aksiyomar\i na g\"ore $\{\bm F(\xi)\colon\xi<\alpha\}$ s\i n\i f\i\
her zaman bir k\"umedir,
ve bu k\"umenin \"usts\i n\i r\i\ vard\i r.
Ayr\i ca ordinaller iyis\i ralanm\i\c s oldu\u gundan
$\{\bm F(\xi)\colon\xi<\alpha\}$
k\"umesinin \"usts\i n\i rlar\i n\i n en k\"u\c c\"u\u g\"u vard\i r,
yani k\"umenin \textbf{supremumu} vard\i r.
Bu supremum, 
\begin{align*}
	&\sup\{\bm F(\xi)\colon\xi<\alpha\},&
	\sup_{\xi<\alpha}\bm F(\xi)
\end{align*}
\c sekillerinde yaz\i labilir.

\begin{theorem}\label{thm:sup-a}
  Her $\alpha$ ordinali i\c cin
  \begin{equation*}
    \sup\{\xi\colon\xi<\alpha\}=
    \begin{cases}
      0,&\text{ $\alpha=0$ durumunda},\\
\beta,&\text{ $\alpha=\beta'$ durumunda},\\
\alpha,&\text{ $\alpha$'n\i n limit oldu\u gu durumda.}
    \end{cases}
  \end{equation*}
\end{theorem}

\ktk

\begin{xca}
  $\{\xi'\colon\xi<\alpha\}$ k\"umesinin supremumunu hesaplay\i n.
\end{xca}

E\u ger
\begin{equation*}
\alpha\leq\beta\lto\bm F(\alpha)\leq\bm F(\beta)
\end{equation*}
ise, o zaman $\bm F$ \textbf{artand\i r.}
E\u ger
\begin{equation}\label{eqn:a+x}
\alpha<\beta\lto\bm F(\alpha)<\bm F(\beta)
\end{equation}
ise, o zaman $\bm F$ \textbf{kesin artand\i r.}
E\u ger
\begin{compactenum}[1)]
\item
$\bm F$ kesin artan ve 
\item
her $\alpha$ limiti i\c cin
$\bm F(\alpha)=\sup\{\bm F(\xi)\colon\xi<\alpha\}$
\end{compactenum}
ise, 
o zaman $\bm F$'ye \textbf{normal} densin.

\begin{xca}\label{ex:xx'}
$\xi\mapsto\xi'$ i\c sleminin kesin artan olup
normal olmad\i\u g\i n\i\ g\"osterin.
\end{xca}

\begin{xca}
Normal olan bir i\c slem \"orne\u gi verin.
\end{xca}

Sonraki teoremin ilk kullan\i m\i,
\Teoremin{thm:a+x} kan\i t\i nda olacakt\i r.

\begin{theorem}\label{thm:norm-cond}
$\bm F\colon\on\to\on$ olsun.
E\u ger
\begin{compactenum}[1)]
\item
her $\alpha$ i\c cin
$\bm F(\alpha)<\bm F(\alpha')$ ve 
\item
her $\alpha$ limiti i\c cin
$\bm F(\alpha)=\sup\{\bm F(\xi)\colon\xi<\alpha\}$
\end{compactenum}
ise, 
o zaman $\bm F$ normaldir.
\end{theorem}

\begin{proof}
$\bm F$'nin kesin artan oldu\u gunu g\"ostermek yeter.
\eqref{eqn:a+x} gerektirmesini
$\beta$ \"uzerinden t\"umevar\i m kullanarak
kan\i tlayaca\u g\i z.
  \begin{asparaenum}
    \item
    $\beta=0$ ise, \eqref{eqn:a+x} iddias\i\ do\u grudur, 
\c c\"unk\"u hi\c cbir zaman $\alpha<0$ de\u gildir.  
\item
$\beta=\gamma$ durumda \eqref{eqn:a+x} 
iddia do\u gru olsun.
E\u ger $\alpha<\gamma'$ ise, o zaman $\alpha\leq\gamma$, dolay\i s\i yla
\begin{align*}
      \bm F(\alpha)
&\leq\bm F(\gamma)&&\text{[t\"umevar\i m hipotezi]}\\
&<\bm F(\gamma').&&\text{[varsay\i m]}\\
\end{align*}
\item
$\gamma$ limit
ve $\alpha<\gamma$ ise, o zaman $\alpha<\alpha'<\gamma$, dolay\i s\i yla
\begin{align*}
  \bm F(\alpha)
&<\bm F(\alpha')&&\text{[varsay\i m]}\\
&\leq\sup\{\bm F(\xi)\colon\xi<\gamma\}&&\text{[supremumun tan\i m\i]}\\
&=\bm F(\gamma).&&\text{[varsay\i m]}
\end{align*}
(Bu ad\i mda bir t\"umevar\i m hipotezi kullan\i lm\i yor.)\qedhere
  \end{asparaenum}
\end{proof}

Sonraki teoremin ilk kullan\i m\i,
\Teoremin{thm:+assoc} kan\i t\i nda olacakt\i r.

\begin{theorem}\sloppy
$\bm F\colon\on\to\on$ ve normal olsun.
O zaman $\on$'nin bo\c s olmayan her $A$ altk\"umesi i\c cin
\begin{equation}\label{eqn:F(sup)}
\bm F(\sup(A))=\sup_{\xi\in A}\bm F(\xi).
\end{equation}
\end{theorem}

\begin{proof}
$A$ k\"umesinin supremumu $\alpha$ olsun.
$\bm F$ kesin artan oldu\u gundan
$\beta\in A$ ise $\bm F(\beta)\leq\bm F(\alpha)$.
Bundan dolay\i,
e\u ger $\alpha\in A$ ise, o zaman
\begin{equation*}
  \sup_{\xi\in A}\bm F(\xi)=F(\alpha),
\end{equation*}
yani \eqref{eqn:F(sup)} do\u grudur.
\c Simdi $\alpha\notin A$ olsun.
O zaman $\alpha$ ard\i l olamaz.
$A$ bo\c s olmad\i\u g\i ndan $\alpha=0$ olamaz,
dolay\i s\i yla $\alpha$ limittir.
Bu durumda $\bm F$ normal oldu\u gundan
\begin{equation}\label{eqn:F1}
  \bm F(\alpha)=\sup_{\xi<\alpha}\bm F(\xi).
\end{equation}
Ayr\i ca
\begin{equation}\label{eqn:F2}
\sup_{\xi\in A}\bm F(\xi)\leq\sup_{\xi<\alpha}\bm F(\xi),  
\end{equation}
\c c\"unk\"u $A\included\{\xi\colon\xi<\alpha\}$.
E\u ger $\beta<\alpha$ ise, 
$A$'n\i n bir $\gamma$ eleman\i\ i\c cin $\beta\leq\gamma<\alpha$,
dolay\i s\i yla $\bm F(\beta)\leq\bm F(\gamma)\leq\sup_{\xi\in A}\bm F(\xi)$.
Bu \c sekilde
\begin{equation}\label{eqn:F3}
  \sup_{\xi<\alpha}\bm F(\xi)\leq\sup_{\xi\in A}\bm F(\xi).
\end{equation}
Sonu\c c olarak
\eqref{eqn:F1}, \eqref{eqn:F2}, ve \eqref{eqn:F3} beraber
 \eqref{eqn:F(sup)} e\c sitli\u gini tekrar gerektirir.
\end{proof}

\section{S\"ureklilik}

Normallik kavram\i n\i n yerine
ger\c cel analizden gelen s\"ureklilik kavram\i n\i\ kullanabiliriz.
Ordinallerde,
kesin artan bir i\c slemin normal olmas\i\ i\c cin
gerek ve yeter bir ko\c sul,
i\c slemin s\"urekli olmas\i d\i r.
Bu sonu\c cu kurmak, bu altb\"ol\"um\"un i\c sidir.

Tekrar $\bm F\colon\on\to\on$ olsun.  Varsa,
$\bm F$'nin bir noktadaki s\"ureklili\u gi
ger\c cel analizdeki gibi tan\i mlan\i r.
Asl\i nda e\u ger
\begin{equation*}
\beta<\bm F(\alpha)<\gamma
\end{equation*}
ko\c sulunu sa\u glayan herhangi $\beta$ ve $\gamma$ i\c cin,
baz\i\ $\delta$ ve $\theta$ i\c cin,
\begin{equation*}
\delta<\alpha<\theta
\land\Forall{\xi}(\delta<\xi<\theta\lto\beta<\bm F(\xi)<\gamma)
\end{equation*}
ise, o zaman $\bm F$, $\alpha$'da \textbf{s\"ureklidir.}
E\u ger $\bm F(\alpha)=0$ veya $\alpha=0$ ise, 
o zaman $\beta=-1$ veya $\delta=-1$ olabilir.
Benzer \c sekilde \textbf{soldan} ve \textbf{sa\u gdan} olan s\"ureklilik
tan\i mlanabilir.

\begin{lemma}
$\on$'de her tek-konumlu i\c slem,
limit olmayan her noktada s\"ureklidir
ve her noktada sa\u gdan s\"ureklidir.
\end{lemma}

\klk

Ger\c cel analizdeki gibi
$\bm F\colon\on\to\on$ ise
\begin{equation*}
\limsup_{\xi\to\alpha^-}\bm F(\xi)=\min\left\{\sup_{\eta<\xi<\alpha}\bm F(\xi)\colon\eta<\alpha\right\}
\end{equation*}
tan\i m\i n\i\ yapar\i z.

\begin{lemma}
$\bm F\colon\on\to\on$ ve artan ise	
\begin{equation*}
\limsup_{\xi\to\alpha^-}\bm F(\xi)=\sup_{\xi<\alpha}\bm F(\xi).
\end{equation*}
\end{lemma}

\klk

\begin{lemma}
$\bm F\colon\on\to\on$ olsun.
$\bm F$ bir $\alpha$ limitinde s\"ureklidir ancak ve ancak
\begin{equation*}
\limsup_{\xi\to\alpha^-}\bm F(\xi)=\bm F(\alpha).
\end{equation*}
\end{lemma}

\klk

\begin{theorem}
$\bm F\colon\on\to\on$ ve kesin artan olsun.
O zaman $\bm F$ normaldir ancak ve ancak 
her noktada s\"ureklidir.
\end{theorem}

\ktk

\begin{xca}
  S\"urekli olup normal olmayan bir i\c slem \"orne\u gi verin.
\end{xca}

\chapter{K\"ume aksiyomlar\i}\label{ch:ax}

Bu b\"ol\"umde $\on$'nin \sayfada{on-props}ki \"ozelliklerini
ve
\begin{equation}\label{eqn:pred}
\alpha=\{\xi\colon\xi<\alpha\}
\end{equation}
tan\i mlama imk\^an\i n\i,
k\"ume aksiyomlar\i ndan elde edece\u giz.
B\"ol\"umler \ref{ch:add}, \ref{ch:mul}, ve \numarada{ch:exp}
sadece \eqref{eqn:pred} e\c sitli\u gini kullanaca\u g\i z.

\section[Ordinaller varsa]{Ordinaller varsa}

Tan\i ma g\"ore
\begin{gather*}
	\emptyset=\{x\colon x\in x\land x\notin x\},\\
	a\cup\{b\}=\{x\colon x\in a\lor x=b\}.
\end{gather*}

\begin{axiom}[Bo\c s K\"ume]
$\emptyset$
bir k\"umedir.
\end{axiom}

\begin{axiom}[Biti\c stirme]
T\"um $a$ ve $b$ k\"umeleri i\c cin $a\cup\{b\}$ bir k\"umedir.
\end{axiom}

\begin{axiom}[Yerle\c stirme]
Her $\bm F$ g\"ondermesi i\c cin,
e\u ger bir $A$ k\"umesinin her $b$ eleman\i\ i\c cin
$\bm F(b)$ tan\i mlan\i rsa, o zaman
\begin{equation*}
\{\bm F(x)\colon x\in a\}
\end{equation*}
s\i n\i f\i\ bir k\"umedir.
\end{axiom}

\begin{theorem}
  $\on$'de tan\i mlanan bir $\bm F$ g\"ondermesi i\c cin,
her $\alpha$ i\c cin
\begin{equation*}
  \bm F(\alpha)=\{\bm F(\xi)\colon\xi<\alpha\},
\end{equation*}
ve sonu\c c olarak, t\"um $\alpha$ ve $\beta$ i\c cin
\begin{equation}\label{eqn:a-f(a)}
\left.\qquad
  \begin{gathered}
  \alpha<\beta\liff\bm F(\alpha)\in\bm F(\beta),\\
  \alpha\leq\beta\liff\bm F(\alpha)\included\bm F(\beta).
\end{gathered}
\qquad\right\}
\end{equation}
\end{theorem}

\begin{proof}
Yerle\c stirme Aksiyomu sayesinde,
Ordinal \"Ozyineleme Teoreminin kan\i t\i n\i n y\"ontemini kullanarak
\begin{compactenum}[1)]
\item
$\bm F(0)=\emptyset$,
\item
$\bm F(\alpha')=\bm F(\alpha)\cup\{\bm F(\alpha)\}$,
\item
$\alpha$ limit ise
\begin{equation}\label{eqn:F(a)}
\bm F(\alpha)=\{\bm F(\xi)\colon\xi<\alpha\}
\end{equation}
\end{compactenum}
ko\c sullar\i n\i\ sa\u glayan bir $\bm F$ g\"ondermesinin var oldu\u gunu kan\i tlayabiliriz.
O zaman ordinal t\"umevar\i mdan her $\alpha$ i\c cin \eqref{eqn:F(a)} do\u grudur.
O halde
\begin{gather*}
	\alpha<\beta\lto\bm F(\alpha)\in\bm F(\beta),\\
	\alpha\leq\beta\lto\bm F(\alpha)\included\bm F(\beta).	
\end{gather*}
Tersleri de do\u grudur.  
Zira $\bm F(\alpha)\in\bm F(\beta)$ ama $\beta\leq\alpha$ ise,
o zaman
\begin{equation*}
\bm F(\alpha)\in\bm F(\alpha).
\end{equation*}
Bu durumda bir $\gamma$ i\c cin $\gamma<\alpha$ ve $\bm F(\alpha)=\bm F(\gamma)$,
dolay\i s\i yla $\bm F(\gamma)\in\bm F(\gamma)$.
O halde bo\c s olmayan $\{\xi\colon\bm F(\xi)\in\bm F(\xi)\}$ s\i n\i f\i n\i n en k\"u\c c\"uk eleman\i\ yoktur,
ki bu \c celi\c skidir.
Sonu\c c olarak
\eqref{eqn:a-f(a)} do\u grudur
\end{proof}

Her $\bm A$ s\i n\i f\i\ i\c cin
\begin{equation*}
\bigcup\bm A=\{x\colon\Exists y(y\in A\land x\in y)\}
\end{equation*}
olsun.  Bu yeni s\i n\i f, $\bm A$'n\i n \textbf{bile\c simidir.}
\"Ozel olarak
\begin{equation*}
B\cup C=\bigcup\{B,C\}.
\end{equation*}

\begin{corollary}
Teoremde $\bm F$ birebirdir,
dolay\i s\i yla $\bm F(\xi)=\xi$ varsay\i labilir.
Bu durumda
\begin{gather*}
  \alpha=\{\xi\colon\xi<\alpha\},\\
\alpha<\beta\liff\alpha\in\beta\liff\alpha\pincluded\beta.
\end{gather*}
Ayr\i ca $B\included\on$ ise
\begin{equation*}
  \sup B=\bigcup B.
\end{equation*}
\end{corollary}

\ksk

\c Simdi $\on$'nin sonu\c ctaki gibi oldu\u gunu varsayal\i m.
O zaman $\on$,
$\in$ ba\u g\i nt\i s\i\ taraf\i ndan iyis\i ralanm\i\c st\i r,
ve s\i n\i f\i n her eleman\i\ s\i n\i f\i n bir altk\"umesidir.
Ayr\i ca, s\i n\i f\i n her eleman\i n\i n ayn\i\ \"ozellikleri vard\i r
(yani eleman $\in$ ba\u g\i nt\i s\i\ taraf\i ndan iyis\i ralanm\i\c st\i r,
ve eleman\i n her eleman\i\ eleman\i n bir altk\"umesidir).

Herhangi $\bm A$ s\i n\i f\i\ i\c cin
\begin{equation*}
\Forall x(x\in\bm A\lto x\included\bm A)
\end{equation*}
ise, o zaman $\bm A$ \textbf{ge\c ci\c slidir.}
Ge\c ci\c sli k\"umeler
\begin{equation*}
\Forall y\bigl(y\in x\lto\Forall z(z\in y\lto z\in x)\bigr)
\end{equation*}
form\"ul\"u taraf\i ndan tan\i mlanm\i\c s s\i n\i f\i\ olu\c sturur.
Ayr\i ca $\in$ taraf\i ndan do\u grusal s\i ralanm\i\c s k\"umeler
\begin{multline*}
\Forall y(y\in x\lto y\notin y)\land{}\\
\Forall y\Forall z\Forall u\bigl(y\in x\land z\in x\land u\in x\lto{}\\
(y\in z\land z\in u\lto y\in u)\bigr)
\land{}\\
\Forall y\Forall z(y\in x\land z\in x\lto y\in z\lor y=z\lor y\in z)
\end{multline*}
form\"ul\"u taraf\i ndan tan\i mlanm\i\c s s\i n\i f\i\ olu\c sturur.
Bu form\"ul $\phi(x)$ ise
$\in$ taraf\i ndan iyis\i ralanm\i\c s k\"umeler
\begin{multline*}
	\phi(x)\land\Forall y\Bigl(\Forall z(z\in y\lto z\in x)\land\Exists zz\in y\lto{}\\
	\Exists z\bigl(z\in y\land\Forall u(u\in y\lto z\in u\lor z=u)\bigr)\Bigr)
\end{multline*}
form\"ul\"u taraf\i ndan tan\i mlanm\i\c s s\i n\i f\i\ olu\c sturur.

\section{Ordinaller vard\i r}

\c Simdi \sayfada{on-props}ki \"ozelliklerini unutunca,
yeniden her \textbf{ordinali,}\label{on-def}
\begin{compactenum}[1)]
\item
ge\c ci\c sli ve
\item
$\in$ taraf\i ndan iyis\i ralanm\i\c s
\end{compactenum}
bir k\"ume olarak tan\i mlar\i z.
O zaman ordinaller bir s\i n\i f olu\c stururlar.
\"Onceki gibi bu s\i n\i f $\on$ olsun,
ve $\on$'nin elemanlar\i,
k\"u\c c\"uk Yunan harfleriyle yaz\i ls\i n.
Sadece k\"ume aksiyomlar\i\ kullanarak,
\sayfada{on-props}ki \"ozellikleri
teorem olarak elde edece\u giz.

\begin{theorem}\label{thm:on-trans}
$\on$ ge\c ci\c slidir,
yani her ordinalin her eleman\i\ bir ordinaldir.
\end{theorem}

\begin{proof}
$\alpha\in\on$ be $b\in\alpha$ olsun.
O zaman $b\included\alpha$, dolay\i s\i yla
$\alpha$ gibi $b$,
$\in$ taraf\i ndan iyis\i ralanm\i\c st\i r.

\c Simdi $c\in b$ olsun.
O zaman $c\in\alpha$, dolay\i s\i yla $c\included\alpha$.
\"Ozel olarak $d\in c$ ise $d\in\alpha$.
Bu durumda $d$, $c$, ve $b$, $\alpha$'n\i n eleman\i d\i rlar,
ve $d\in c$ ve $c\in b$,
dolay\i s\i yla $d\in b$ \c c\"unk\"u $\alpha$'da $\in$ ge\c ci\c slidir.
Sonu\c c olarak $c\included b$.
O halde $b$ ge\c ci\c slidir.
\end{proof}

\begin{lemma}\label{lem:on-o}
$\on$, $\in$ taraf\i ndan s\i ralanm\i\c st\i r,
yani $\on$'de $\in$, yans\i mas\i z ve ge\c ci\c slidir.
\end{lemma}

\begin{proof}
$\alpha\in\on$ olsun.  $\alpha$'da $\in$ ba\u g\i nt\i s\i\ yans\i mas\i z oldu\u gundan
$\alpha\notin\alpha$,
\c c\"unk\"u $\alpha\in\alpha$ ise $\alpha$'n\i n bir $\beta$ eleman\i\ i\c cin $\beta\in\beta$.

E\u ger $\beta\in\alpha$ ve $\gamma\in\beta$ ise, 
$\alpha$ ge\c ci\c sli oldu\u gundan $\gamma\in\alpha$.
\end{proof}

Her ordinalin bo\c s olmayan her $a$ altk\"umesinin,
$\in$ ba\u g\i nt\i s\i na g\"ore en k\"u\c c\"uk eleman\i d\i r.
Buradaki $a$'n\i n yerine bir \emph{s\i n\i f} kullanmak isteriz.
Sonraki aksiyomu kullanaca\u g\i z.

\begin{axiom}[Ay\i rma]
Her k\"umenin her alts\i n\i f\i\ bir k\"umedir.
\end{axiom}

Bu \c sekilde her $a$ k\"umesi ve $\{x\colon\phi(x)\}$ s\i n\i f\i\ i\c cin
\begin{equation*}
\{x\colon x\in a\land\phi(x)\}
\end{equation*}
s\i n\i f\i\ bir k\"umedir.
Bu k\"ume
\begin{equation*}
\{x\in a\colon\phi(x)\}
\end{equation*}
olarak yaz\i labilir.

\begin{lemma}
$\on$'de $\in$ ve $\pincluded$ s\i ralamalar\i\ ayn\i d\i r.
\end{lemma}

\begin{proof}
Kan\i t\i n iki par\c cas\i\ vard\i r.

\fbox{\mathversion{bold}$\alpha\in\beta\lto\alpha\pincluded\beta$:}
$\alpha\in\beta$ olsun.
$\beta$ ge\c ci\c sli oldu\u gundan $\alpha\included\beta$.
$\beta$'da $\in$ yans\i mas\i z oldu\u gundan $\alpha\neq\beta$.
Bu \c sekilde $\alpha\pincluded\beta$.

\fbox{\mathversion{bold}$\alpha\pincluded\beta\lto\alpha\in\beta$:}
$\alpha\pincluded\beta$ olsun.
O zaman $\beta\setminus\alpha$ k\"umesi bo\c s de\u gildir.
$\gamma=\min(\beta\setminus\alpha)$ olsun.
O zaman $\gamma\in\beta$.
Biz \fbox{\mathversion{bold}$\gamma=\alpha$} kan\i tlayaca\u g\i z.
Bu kan\i t\i n iki par\c cas\i\ vard\i r.

\fbox{\mathversion{bold}$\gamma\included\alpha$:}
$\delta\in\gamma$ olsun.
O zaman $\beta$ ge\c ci\c sli oldu\u gundan $\delta\in\beta$.
Ayr\i ca $\delta\notin\beta\setminus\alpha$, \c c\"unk\"u $\delta\in\min(\beta\setminus\alpha)$.
O halde $\delta\in\alpha$.
B\"oylece $\gamma\included\alpha$.

\fbox{\mathversion{bold}$\alpha\included\gamma$:}
$\delta\in\alpha$ olsun.
O zaman $\delta\in\beta$, \c c\"unk\"u $\alpha\pincluded\beta$,
dolay\i s\i yla $\delta\notin\beta\setminus\alpha$.
Ama $\delta\in\gamma$, $\delta=\gamma$, veya $\gamma\in\delta$;
ve son iki imk\^an olmaz.
Zira $\gamma\in\beta\setminus\alpha$ oldu\u gundan $\delta\neq\gamma$;
ve $\gamma\notin\alpha$ oldu\u gundan $\gamma\notin\delta$, \c c\"unk\"u $\delta\in\alpha$.
Bu \c sekilde $\alpha\included\gamma$.
\end{proof}

\c Simdi $\on$'nin $\in$ veya $\pincluded$ s\i ralamas\i n\i\ $<$ olarak yazabiliriz.

\begin{lemma}
$\on$'nin $<$ s\i ralamas\i\ do\u grusald\i r.
\end{lemma}

\begin{proof}
$\alpha\not\leq\beta$ olsun.
\fbox{\mathversion{bold}$\beta<\alpha$} g\"osterece\u giz.

Varsay\i mdan $\alpha\not\included\beta$,
dolay\i s\i yla $\alpha\setminus\beta\neq\emptyset$.
$\gamma=\min(\alpha\setminus\beta)$ olsun.
O zaman $\gamma\in\alpha$, yani $\gamma<\alpha$.
\fbox{\mathversion{bold}$\gamma=\beta$} g\"osterece\u giz.

\fbox{\mathversion{bold}$\gamma\included\beta$:}
$\delta\in\gamma$ olsun.
O zaman $\delta<\min(\alpha\setminus\beta)$,
ama $\delta\in\alpha$,
dolay\i s\i yla $\delta\in\beta$.

\fbox{\mathversion{bold}$\gamma\not\pincluded\beta$:}
$\gamma\in\alpha\setminus\beta$ oldu\u gundan $\gamma\notin\beta$,
yani $\gamma\not\pincluded\beta$.
\end{proof}

\begin{theorem}\label{thm:on-wo}
$\on$'nin $<$ do\u grusal s\i ralamas\i\ bir iyis\i ralamad\i r.
Asl\i nda $\on$'nin bo\c s olmayan her \emph{alts\i n\i f\i n\i n} en k\"u\c c\"uk eleman\i\ vard\i r.
\end{theorem}

\begin{proof}
$\bm A\included\on$ ve $\alpha\in\bm A$ olsun.
\begin{compactitem}
\item
$\alpha\cap\bm A=\emptyset$ ise
$\alpha=\min(\bm A)$.
\item
$\alpha\cap\bm A\neq\emptyset$ ise
$\min(\alpha\cap\bm A)=\min(\bm A)$.\qedhere
\end{compactitem}
\end{proof}

\c Simdi Teorem \ref{thm:on-trans} ve \numarada{thm:on-wo}n
$\on$ hem ge\c c\i\c sli hem $\in$ taraf\i ndan iyis\i ralanm\i\c st\i r.
Tan\i ma g\"ore $\on$'nin elemanlar\i n\i n ayn\i\ \"ozellikleri vard\i r.
Ama $\on$ $\in$ taraf\i ndan s\i ralanm\i\c s oldu\u gundan kendinin eleman\i\ olamaz.
Bu \c sekilde $\on$ k\"ume olamaz.
Bu sonu\c c, Teorem \ref{thm:BFP} olarak g\"ord\"u\u g\"um\"uz
\textbf{Burali-Forti Paradoksudur.}

\begin{theorem}
$\emptyset\in\on$ ve $\alpha\in\on$ ise $\alpha\cup\{\alpha\}\in\on$,
ve ayr\i ca her $\beta$ ordinali i\c cin
\begin{equation*}
\beta\leq\alpha\lor\alpha\cup\{\alpha\}\included\beta.
\end{equation*}
\end{theorem}

\ktk

\c Simdi
\begin{align*}
	0&=\emptyset,&
	\alpha'&=\alpha\cup\{\alpha\}
\end{align*}
tan\i mlayabiliriz.

\begin{axiom}[Sonsuzluk]
Ne limit olan ne limit i\c ceren ordinallerin olu\c sturdu\u gu s\i n\i f
bir k\"umedir.
\end{axiom}

Sonsuzluk Aksiyomu taraf\i ndan verilen k\"ume, $\upomega$'d\i r.

\begin{axiom}
Her k\"umenin bile\c simi bir k\"umedir.
\end{axiom}

\begin{theorem}\sloppy
\Sayfada{on-def} tan\i mlanm\i\c s $\on$ s\i n\i f\i,
\Sayfada{on-props}ki \"ozellikleri sa\u glar.
\end{theorem}

\ktk

\chapter{Ordinal Toplama}\label{ch:add}

\section{Tan\i m ve \"ozellikler}

\"Ozyineli tan\i ma g\"ore her $\alpha$ ordinali i\c cin
\begin{gather}\label{eqn:+0}
  \alpha+0=\alpha,\\\label{eqn:+'}
\alpha+\beta'=(\alpha+\beta)',\\\label{eqn:+lim}
\gamma\text{ limit ise }\alpha+\gamma=\sup\{\alpha+\xi\colon\xi<\gamma\}.
\end{gather}
Ordinal toplaman\i n \"ozelliklerinin \c co\u gu,
t\"umevar\i m ile kan\i tlan\i r;
ama ilk teoremimiz, t\"umevar\i mdan de\u gildir.

\begin{theorem}
  $\alpha+1=\alpha'$.
\end{theorem}

\begin{proof}
\hfill
$\begin{aligned}[t]
  \alpha+1
&=\alpha+0'&&\text{[\eqref{eqn:1=0'} tan\i m\i ndan]}\\
&=(\alpha+0)'&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}\\
&=\alpha'.&&\text{[\eqref{eqn:+0} tan\i m\i ndan]}
\end{aligned}$\hfill\vspace{-0.8\baselineskip}

\mbox{}
\end{proof}

\begin{theorem}\label{thm:0+}
Her $\alpha$ i\c cin $0+\alpha=\alpha$.
\end{theorem}

\begin{proof}
Ordinal t\"umevar\i m kullanaca\u g\i z.
  \begin{asparaenum}
    \item
E\u ger $\alpha=0$ ise
\begin{align*}
0+\alpha
&=0+0&&\text{[varsay\i mdan]}\\
&=0&&\text{[\eqref{eqn:+0} tan\i m\i ndan]}\\
&=\alpha.&&\text{[varsay\i mdan]}
\end{align*}
\item
E\u ger
\begin{equation}\label{eqn:0+b=b}
0+\beta=\beta
\end{equation}
ise, o zaman
  \begin{align*}
    0+\beta'
&=(0+\beta)'&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}\\
&=\beta'.&&\text{[\eqref{eqn:0+b=b} hipotezinden]}
  \end{align*}
\item
Bir $\alpha$ limiti i\c cin
\begin{equation}\label{eqn:0+b=b-lim}
  \Forall{\xi}(\xi<\alpha\lto0+\xi=\xi)
\end{equation}
ise, o zaman
\begin{align*}
  0+\alpha
&=\sup\{0+\xi\colon\xi<\alpha\}&&\text{[\eqref{eqn:+lim} tan\i m\i ndan]}\\
&=\sup\{\xi\colon\xi<\alpha\}&&\text{[\eqref{eqn:0+b=b-lim} hipotezinden]}\\
&=\alpha.&&\text{[\Teoremde{thm:sup-a}n]}\qedhere
\end{align*}
  \end{asparaenum}
\end{proof}

\begin{xca}
\mathversion{false}
A\c sa\u g\i daki kan\i t
nerede yanl\i\c st\i r?
\begin{falseproof}
Her $\alpha$ i\c cin $1+\alpha=\alpha'$ kan\i tlayaca\u g\i z.
\begin{asparaenum}
\item
$1+0=1=0'$.
\item
$1+\beta=\beta'$ ise, o zaman
\begin{equation*}
1+\beta'=(1+\beta)'=(\beta')'.
\end{equation*}
\item
$\gamma$ limit ve $\Forall{\xi}(\xi<\gamma\lto1+\xi=\xi')$ ise
\begin{equation*}
1+\gamma=\sup_{\xi<\gamma}(1+\xi)=\sup_{\xi<\gamma}(\xi')=\gamma'.
\end{equation*}
\end{asparaenum}

B\"oylece her $\alpha$ i\c cin $1+\alpha=\alpha'$.
\end{falseproof}
\end{xca}

\begin{theorem}\label{thm:a+x}
Her $\alpha$ ordinali i\c cin $\xi\mapsto\alpha+\xi$ normaldir.
\end{theorem}

\begin{proof}
\Teoremde{thm:norm-cond}n $\alpha+\beta<\alpha+\beta'$ g\"ostermek yeter.
Ayr\i ca
\begin{align*}
      \alpha+\beta
&<(\alpha+\beta)'&&\text{[\eqref{eqn:a'=min} tan\i m\i ndan]}\\
&=\alpha+\beta'.&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}\qedhere
\end{align*}
\end{proof}

\"Orne\u gin
\Sekle{fig:w+x} bak\i n.
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  \caption{$\eta=\upomega+\xi$ denkleminin grafi\u gi}\label{fig:w+x}  
\end{figure}
\c Sekilde
\begin{align*}
	\upomega\cdot2&=\upomega+\upomega,&
\upomega\cdot3&=\upomega\cdot2+\upomega,&
\upomega\cdot4&=\upomega\cdot3+\upomega,
\end{align*}
ve genelde,
\Teoremi{thm:rec} kullanan resmi \"ozyineli tan\i ma g\"ore,
\begin{align*}
\alpha\cdot0&=0,&
	\alpha\cdot1&=\alpha,&
	\alpha\cdot(k+1)&=\alpha\cdot k+\alpha.
\end{align*}
Bu \c sekilde $\alpha\cdot n$, ``$\alpha$'d\i r $n$ kere''
veya ``$\alpha$'n\i n $n$ kat\i d\i r.''
Ayr\i ca
\begin{equation*}
\upomega^2=\upomega\cdot\upomega=\sup_{x<\upomega}(\upomega\cdot x).
\end{equation*}

\begin{xca}
$\xi\mapsto\xi\cdot2$ g\"ondermesi kesin artan m\i d\i r?
S\"urekli midir?
\end{xca}

\begin{xca}
\mathversion{false}
A\c sa\u g\i daki kan\i t
nerede yanl\i\c st\i r?
\begin{falseproof}
Her $\alpha$ i\c cin, her $\beta$ i\c cin,
$\alpha+\beta=\beta+\alpha$ kan\i tlayaca\u g\i z.
\begin{asparaenum}
\item
$\alpha+0=\alpha=0+\alpha$.
\item
E\u ger $\alpha+\beta=\beta+\alpha$ ise, o zaman
\begin{equation*}
\alpha+\beta'=(\alpha+\beta)'=(\beta+\alpha)'=\beta'+\alpha.
\end{equation*}
\item
E\u ger $\gamma$ limit ve $\Forall{\xi}(\xi<\gamma\lto\alpha+\xi=\xi+\alpha)$ ise,
o zaman
\begin{equation*}
\alpha+\gamma
=\sup_{\xi<\gamma}(\alpha+\xi)
=\sup_{\xi<\gamma}(\xi+\alpha)
=\gamma+\alpha.
\end{equation*}
\end{asparaenum}
Bu \c sekilde her durumda $\alpha+\beta=\beta+\alpha$.
\end{falseproof}
\end{xca}

\begin{theorem}\label{thm:+assoc}
Ordinal toplama birle\c smelidir.
\end{theorem}

\begin{proof}
Her $\gamma$ i\c cin, 
t\"umevar\i m kullanarak
t\"um $\alpha$ ve $\beta$ i\c cin
  \begin{equation*}
    \alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma
  \end{equation*}
	g\"osterece\u giz.
  \begin{asparaenum}
    \item
$		\begin{aligned}[t]
\alpha+(\beta+0)
&=\alpha+\beta&&\text{[\eqref{eqn:+0} tan\i m\i ndan]}\\
&=(\alpha+\beta)+0.&&\text{[\eqref{eqn:+0} tan\i m\i ndan]}
					\end{aligned}$
\item
E\u ger
\begin{equation}\label{eqn:+assoc'}
\alpha+(\beta+\delta)=(\alpha+\beta)+\delta
\end{equation}
ise, o zaman
  \begin{align*}
    \alpha+(\beta+\delta')
		&=\alpha+(\beta+\delta)'&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}\\
&=(\alpha+(\beta+\delta))'&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}\\
&=((\alpha+\beta)+\delta)'&&\text{[\eqref{eqn:+assoc'} hipotezinden]}\\
&=(\alpha+\beta)+\delta'.&&\text{[\eqref{eqn:+'} tan\i m\i ndan]}
  \end{align*}
\item
$\delta$ limit olsun, ve
\begin{equation}\label{eqn:+assoc-lim}
\Forall{\xi}\bigl(\xi<\delta\lto
\alpha+(\beta+\xi)=(\alpha+\beta)+\xi\bigr)
\end{equation}
olsun.  
O zaman
\begin{align*}
&\phantom{{}={}}(\alpha+\beta)+\delta\\
&=\sup\{(\alpha+\beta)+\xi\colon\xi<\delta\}&&\text{[\eqref{eqn:+lim} tan\i m\i]}\\
&=\sup\{\alpha+(\beta+\xi)\colon\xi<\delta\}&&\text{[\eqref{eqn:+assoc-lim} hipotezi]}\\
&=\alpha+\sup\{\beta+\xi\colon\xi<\delta\}&&\text{[$\xi\mapsto\alpha+\xi$ normaldir]}\\
&=\alpha+(\beta+\delta).&&\text{[\eqref{eqn:+lim} tan\i m\i]}\qedhere 
\end{align*}
  \end{asparaenum}
\end{proof}

\c Simdi herhangi $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
\alpha\cdot n=\underbrace{\alpha+\dots+\alpha}_n
\end{equation*}
anla\c s\i labilir.

\begin{theorem}
$k<\upomega$ ve $\ell<\upomega$ ise
$\alpha\cdot(k+\ell)=\alpha\cdot k+\alpha\cdot\ell$.
\end{theorem}

\ktk[ (T\"umevar\i m kullan\i n.)]

\begin{theorem}\label{thm:x+a}
Her $\xi\mapsto\xi+\alpha$ g\"ondermesi artand\i r.
\end{theorem}

\begin{proof}
  $\beta\leq\gamma$ olsun.
$\alpha$ \"uzerinden t\"umevar\i m kullanarak  
\begin{equation*}
\beta+\alpha\leq\gamma+\alpha
\end{equation*}
kan\i tlayaca\u g\i z.
\begin{asparaenum}
  \item
$\beta+0=\beta\leq\gamma=\gamma+0$.
\item
$\beta+\alpha=\gamma+\alpha$ ise tabii ki
  \begin{equation*}
    \beta+\alpha'=(\beta+\alpha)'=(\gamma+\alpha)'=\gamma+\alpha'.
  \end{equation*}
$\beta+\alpha<\gamma+\alpha$ ise, 
\Teoreme{thm:'} g\"ore
\begin{equation*}
  \beta+\alpha'
  =(\beta+\alpha)'\leq\gamma+\alpha<(\gamma+\alpha)'=\gamma+\alpha'.  
\end{equation*}
\item
E\u ger $\delta$ limit ise
\begin{equation*}
\Forall{\xi}(\xi<\delta\lto\beta+\xi<\gamma+\xi)
\end{equation*}
olsun.  O zaman
  \begin{equation*}
    \beta+\delta=\sup_{\xi<\delta}(\beta+\xi)
\leq\sup_{\xi<\delta}(\gamma+\xi)=\gamma+\delta.\qedhere
  \end{equation*}
  \end{asparaenum}
\end{proof}

\section{Hesaplamalar}

Bu altb\"ol\"um\"un teoremleri
t\"umevar\i m kullanmaz.

\begin{theorem}%\label{thm:k+w=w}
  $k<\upomega$ ise $k+\upomega=\upomega$.
(\Sekle{fig:x+w} bak\i n.)
\begin{figure}
  \centering
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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(! 5.12 1.28 add                                                       7.68)
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(! 5.12 2.56 add                                                       7.68)
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  \caption{$\eta=\xi+\upomega$ denkleminin grafi\u gi}\label{fig:x+w}  
\end{figure}
\end{theorem}

\begin{proof}
  $k+\upomega=\sup\{k+x\colon x<\upomega\}=\upomega$.
\end{proof}

\begin{corollary}
$k<\upomega$ ve $1\leq n<\upomega$ ise
\begin{equation*}
k+\upomega\cdot n=\upomega\cdot n.
\end{equation*}
\end{corollary}

\ksk

\begin{comment}

\begin{corollary}
  Ordinal toplama de\u gi\c smeli de\u gildir.
\end{corollary}

\ksk
\end{comment}

\begin{theorem}[\c C\i karma]\label{thm:subtraction}
  $\alpha\leq\beta$ ise
  \begin{equation}\label{eqn:a+x=b}
    \alpha+\xi=\beta
  \end{equation}
denkleminin bir ve tek bir \c c\"oz\"um\"u vard\i r.  
\end{theorem}

\begin{proof}
Denklemin \c c\"oz\"um\"u varsa,
\Teoreme{thm:a+x} g\"ore tek \c c\"oz\"um vard\i r.

Teoremler \ref{thm:0+} ve \numarada{thm:x+a}n $\alpha+\beta\geq\beta$,
dolay\i s\i yla $\{\xi\colon\alpha+\xi\leq\beta\}$ 
s\i n\i f\i n\i n $\beta'$ \"usts\i n\i r\i\ vard\i r.
\c Simdi $\gamma$, s\i n\i f\i n\i n supremumu olsun.
O zaman
\begin{gather*}
\begin{aligned}
	\alpha+\gamma
	&=\alpha+\sup\{\xi\colon\alpha+\xi\leq\beta\}\\
	&=\sup\{\alpha+\xi\colon\alpha+\xi\leq\beta\}
	\leq\beta,
\end{aligned}\\
(\alpha+\gamma)'
		=\alpha+\gamma'
		>\beta,
\end{gather*}
dolay\i s\i yla $\alpha+\gamma=\beta$.
\end{proof}

\begin{xca}
$\alpha\leq\beta$ varsay\i nca,
$\{\xi\colon a+\xi\geq\beta\}$ s\i n\i f\i n\i n bo\c s olmay\i p
s\i n\i f\i n en k\"u\c c\"uk eleman\i n\i n \eqref{eqn:a+x=b} denkleminin
\c c\"oz\"um\"u oldu\u gunu g\"osterin.
\end{xca}

\begin{theorem}
$\upomega+\alpha=\alpha$ ancak ve ancak $\upomega^2\leq\alpha$.
\end{theorem}

\begin{proof}
$\begin{aligned}[t]
	\upomega+\upomega^2
	&=\upomega+\sup_{x<\upomega}(\upomega\cdot x)\\
	&=\sup_{x<\upomega}(\upomega+\upomega\cdot x)\\
	&=\sup_{x<\upomega}\bigl(\upomega\cdot(1+x)\bigr)\\
	&=\upomega^2.
\end{aligned}$

E\u ger $\alpha\geq\upomega^2$ ise, o zaman bir $\beta$ i\c cin $\upomega^2+\beta=\alpha$,
dolay\i s\i yla
\begin{equation*}
\upomega+\alpha
=\upomega+\upomega^2+\beta
=\upomega^2+\beta
=\alpha.
\end{equation*}
\c Simdi $\alpha<\upomega^2$ olsun.
O zaman bir $k$ do\u gal say\i s\i\ i\c cin
\begin{gather*}%\label{eqn:wk-a-w(k+1)}
\upomega\cdot k\leq\alpha<\upomega\cdot(k+1),\\%\notag
\upomega\cdot(k+1)\leq\upomega+\alpha,	
\end{gather*}
dolay\i s\i yla $\alpha<\upomega+\alpha$.
\end{proof}

Teorem sayesinde $\upomega\leq\alpha<\upomega^2$ ise,
o zaman bir $\alpha_1$ i\c cin
\begin{align*}
	\upomega+\alpha_1&=\alpha,&
	\alpha_1&<\alpha.
\end{align*}
E\u ger $\upomega\leq\alpha_1$ ise,
o zaman bir $\alpha_2$ i\c cin
\begin{align*}
	\upomega+\alpha_2&=\alpha_1,&
	\alpha_2&<\alpha_1,
\end{align*}
ve saire.
O zaman bir $k$ i\c cin
\begin{equation*}
\alpha
={\underbrace{\upomega+\dots+\upomega}_k}+\alpha_k
=\upomega\cdot k+\alpha_k.
\end{equation*}
$\on$ iyis\i ral\i\ oldu\u gundan bir $k$ i\c cin
$\alpha_k<\upomega$.
Bu \c sekilde
\begin{equation*}
\{\xi\colon\xi<\upomega^2\}
\end{equation*}
k\"umesinin her eleman\i,
$\upomega\cdot k+\ell$ bi\c ciminde yaz\i labilir.
Verilen k\"ume, toplama alt\i nda kapal\i d\i r,
ve toplama kural\i,
\begin{equation*}
(\upomega\cdot k+\ell)+(\upomega\cdot m+n)
=\upomega\cdot(k+m)+n.
\end{equation*}

\begin{xca}
$\alpha=\upomega\cdot17+6$ ve $\beta=\upomega\cdot1000+5$ ise
$\alpha+\beta$ toplam\i n\i\ hesaplay\i n.
\end{xca}

\section{Kardinaller}

\c Simdi herhangi $A$ ve $B$ k\"umeleri i\c cin
\begin{equation*}
  A\sqcup B=(A\times\{0\})\cup(B\times\{1\})
\end{equation*}
olsun; bu bile\c sim, 
$A$ ve $B$'nin \textbf{ayr\i k bile\c simidir.}
B\"ol\"um \numarada{ch:ax}n
\begin{equation*}
\alpha=\{\xi\colon\xi<\alpha\}
\end{equation*}
anla\c smas\i n\i\ kullanaca\u g\i z.

\begin{theorem}\label{thm:a+b}
  $\alpha+\beta\approx\alpha\sqcup\beta$.
\end{theorem}

\begin{proof}
  \Teoremde{thm:subtraction}n
	\begin{equation*}
	\left\{
	\begin{aligned}
(\xi,0)&\mapsto\xi,\\
(\eta,1)&\mapsto\alpha+\eta
	\end{aligned}
	\right.
	\end{equation*}
kural\i,
$\alpha\sqcup\beta$
ayr\i k bile\c siminden
$\alpha+\beta$ k\"umesine
giden bir e\c sleme tan\i mlar.
\end{proof}

Bir $A$ k\"umesi bir ordinalle e\c slenik olsun.
Tan\i ma g\"ore
\begin{equation*}
\card A=\min\bigl\{\xi\colon\xi\approx A\bigr\};
\end{equation*}
bu ordinal, $A$'n\i n \textbf{kardinalidir.}
Kardinaller, $\kappa$, $\lambda$, $\mu$, ve $\nu$\label{card}
harfleri ile g\"osterilecektir.

E\u ger $f\colon A\to B$ ve $C\included A$ ise
\begin{equation*}
f[C]=\{f(x)\colon x\in C\}
\end{equation*}
olsun.
E\u ger $f$ birebir ise, o zaman
$A$'n\i n $B$'ye bir \textbf{g\"ommesidir,}
ve
\begin{equation*}
A\approx f[A]\included B.
\end{equation*}
Bu durumda
\begin{equation*}
f\colon A\xrightarrow{\preccurlyeq}B
\end{equation*}
yazal\i m, ve \"oyle bir $f$ g\"ommesi varsa
\begin{equation*}
A\preccurlyeq B
\end{equation*}
yazal\i m.

\begin{theorem}[Schr\"oder--Bernstein]
$A\preccurlyeq B$ ve $B\preccurlyeq A$ ise
\begin{equation*}
A\approx B.
\end{equation*}
\end{theorem}

\begin{proof}
$f\colon A\xrightarrow{\preccurlyeq}B$ ve
$g\colon B\xrightarrow{\preccurlyeq}A$ olsun.
\"Ozyinelemeyle
\begin{align*}
A_0&=A,&A_{n+1}&=g[B_n],\\
B_0&=B,&B_{n+1}&=f[A_n]
\end{align*}
olsun.
O zaman
\begin{gather*}
	f[A_0\setminus A_1]=B_1\setminus B_2,\\
	g[B_0\setminus B_1]=A_1\setminus A_2,
\end{gather*}
dolay\i s\i yla $A_0\setminus A_2\approx B_0\setminus B_2$.
Benzer \c sekilde
\begin{equation*}
A_n\setminus A_{n+2}\approx B_n\setminus B_{n+2},
\end{equation*}
dolay\i s\i yla
\begin{equation*}
A\setminus\bigcap_{i<\upomega}A_i\approx B\setminus\bigcap_{i<\upomega}B_i.
\end{equation*}
Ayr\i ca
\begin{equation*}
f\left[\bigcap_{i<\upomega}A_i\right]=\bigcap_{i<\upomega}f[A_i]=\bigcap_{0<i<\upomega}B_i=\bigcap_{i<\upomega}B_i,
\end{equation*}
dolay\i s\i yla 
$\bigcap_{i<\upomega}A_i\approx\bigcap_{i<\upomega}B_i$,
ve sonu\c c olarak $A\approx B$.
\end{proof}

\begin{theorem}
$\xi\mapsto\card{\xi}$ artand\i r.
\end{theorem}

\begin{proof}
E\u ger $\alpha\leq\beta$ ama $\card{\beta}\leq\card{\alpha}$ ise,
o zaman
\begin{equation*}
\alpha\preccurlyeq\beta\approx\card{\beta}\preccurlyeq\card{\alpha}\approx\alpha,
\end{equation*}
dolay\i s\i yla $\alpha\approx\beta$.
\end{proof}

\begin{theorem}
$k<\upomega$ ise $\card k=k$.
\end{theorem}

\ktk

\begin{theorem}
$\card{\upomega+\upomega}=\upomega$.
\end{theorem}

\ktk

\begin{corollary}
$\{\xi\colon\upomega\leq\xi<\upomega^2\}$ k\"umesinin her eleman\i n\i n kardinali
$\upomega$'d\i r.
\end{corollary}

\ksk

\chapter{Ordinal \c carpma}\label{ch:mul}

\section{Tan\i m ve \"ozellikler}

\"Ozyineli tan\i ma g\"ore her $\alpha$ i\c cin
\begin{gather*}
	\alpha\cdot0=0,\\
	\alpha\cdot\beta'=\alpha\cdot\beta+\alpha,\\
	\gamma\text{ limit ise }\alpha\cdot\gamma=\sup\{\alpha\cdot\xi\colon\xi<\gamma\}.
\end{gather*}

Ordinal \c carpma hakk\i nda ilk teoremimizin bir \c s\i kk\i\ t\"umevar\i m kullanmaz;
kalanlar t\"umevar\i m kullan\i yor.

\begin{theorem}\mbox{}
\begin{compactenum}
\item
$\alpha\cdot1=\alpha$.
\item
$1\cdot\alpha=\alpha$.
\item
$0\cdot\alpha=0$.
\end{compactenum}
\end{theorem}

\ktk

\begin{theorem}
$\alpha\geq1$ ise $\xi\mapsto\alpha\cdot\xi$ i\c slemi normaldir.
\end{theorem}

\ktk
\"Orne\u gin
\Sekle{fig:wx} bak\i n.
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  \caption{$\eta=\upomega\cdot\xi$ denkleminin grafi\u gi}\label{fig:wx}  
\end{figure}
\c Sekilde
\begin{align*}
	\upomega^2&=\upomega\cdot\upomega,&
\upomega^3&=\upomega^2\cdot\upomega,&
\upomega^4&=\upomega^3\cdot\upomega,
\end{align*}
ve genelde,
\Teoremi{thm:rec} kullanan resmi \"ozyineli tan\i ma g\"ore,
\begin{align*}
\alpha^0&=1,&
	\alpha^1&=\alpha,&
	\alpha^{k+1}&=\alpha^k\cdot\alpha.
\end{align*}
Ayr\i ca
\begin{equation*}
\upomega^{\upomega}=\sup_{x<\upomega}(\upomega^x).
\end{equation*}

\begin{xca}
$\xi\mapsto\xi^2$ g\"ondermesi kesin artan m\i d\i r?
S\"urekli midir?
\end{xca}

\begin{theorem}
Ordinal \c carpma, toplama \"uzerine soldan da\u g\i l\i r.
\end{theorem}

\begin{proof}
Ordinal t\"umevar\i m ile
\begin{equation}\label{eqn:dist}
 \alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma
\end{equation}
kan\i tlayaca\u g\i z.
\begin{asparaenum}
\item
$\alpha\cdot(\beta+0)
=\alpha\cdot\beta
=\alpha\cdot\beta+0
=\alpha\cdot\beta+\alpha\cdot0$.
\item
E\u ger \eqref{eqn:dist} do\u gru ise, o zaman
\begin{align*}
	\alpha\cdot(\beta+\gamma')
	&=\alpha\cdot(\beta+\gamma)'\\
	&=\alpha\cdot(\beta+\gamma)+\alpha\\
	&=(\alpha\cdot\beta+\alpha\cdot\gamma)+\alpha\\
	&=\alpha\cdot\beta+(\alpha\cdot\gamma+\alpha)\\
	&=\alpha\cdot\beta+\alpha\cdot\gamma'.
\end{align*}
\item\c Simdi $\gamma$ limit ve
  \begin{equation*}
\Forall{\xi}(\xi<\gamma\lto\alpha\cdot(\beta+\xi)
=\alpha\cdot\beta+\alpha\cdot\xi)
  \end{equation*}
olsun.
E\u ger $\alpha=0$ ise, iddia apa\c c\i kt\i r,
dolay\i s\i yla $\alpha>0$ varsayaca\u g\i z.  
\begin{align*}
	\alpha\cdot(\beta+\gamma)
	&=\alpha\cdot\sup_{\xi<\gamma}(\beta+\xi)&&\text{[tan\i m]}\\
	&=\sup_{\xi<\gamma}\bigl(\alpha\cdot(\beta+\xi)\bigr)&&\text{[$\eta\mapsto\alpha\cdot\eta$ normaldir]}\\
	&=\sup_{\xi<\gamma}(\alpha\cdot\beta+\alpha\cdot\xi)&&\text{[t\"umevar\i m hipotezi]}\\
	&=\alpha\cdot\beta+\sup_{\xi<\gamma}(\alpha\cdot\xi)&&\text{[$\eta\mapsto\alpha\cdot\beta+\eta$ normaldir]}\\
	&=\alpha\cdot\beta+\alpha\cdot\gamma.&&\text{[tan\i m]}\qedhere
\end{align*}
\end{asparaenum}
\end{proof}

\begin{xca}
A\c sa\u g\i daki kan\i t nerede yanl\i\c st\i r?
\begin{falseproof}\noindent
\begin{compactenum}
\item
$0\cdot(\beta+\gamma)=0=0+0=0\cdot\beta+0\cdot\gamma$.
\item
E\u ger \eqref{eqn:dist} do\u gru ise, o zaman
\begin{align*}
\alpha'\cdot(\beta+\gamma)
&=\alpha\cdot(\beta+\gamma)+(\beta+\gamma)\\
&=(\alpha\cdot\beta+\alpha\cdot\gamma)+(\beta+\gamma)\\
&=(\alpha\cdot\beta+\beta)+(\alpha\cdot\gamma+\gamma)\\
&=\alpha'\cdot\beta+\alpha'\cdot\gamma.
\end{align*}
\item
E\u ger $\alpha$ limit ve
$\Forall{\xi}\bigl(\xi<\alpha\lto\xi\cdot(\beta+\gamma)=\xi\cdot\beta+\xi\cdot\gamma\bigr)$ ise
\begin{align*}
	\alpha\cdot(\beta+\gamma)
	&=\sup_{\xi<\alpha}\bigl(\xi\cdot(\beta+\gamma)\bigr)\\
	&=\sup_{\xi<\alpha}\bigl(\xi\cdot\beta+\xi\cdot\gamma)\\
	&=\sup_{\xi<\alpha}(\xi\cdot\beta)+\sup_{\xi<\alpha}(\xi\cdot\gamma)\\
	&=\alpha\cdot\beta+\alpha\cdot\gamma.
\end{align*}
\end{compactenum}
\end{falseproof}
\end{xca}

\begin{xca}
A\c sa\u g\i daki kan\i t nerede yanl\i\c st\i r?
\begin{falseproof}\noindent
\begin{compactenum}
\item
$(\alpha+\beta)\cdot0=0=0+0=\alpha\cdot0+\beta\cdot0$.
\item
E\u ger $(\alpha+\beta)\cdot\gamma=\alpha\cdot\gamma+\beta\cdot\gamma$ ise, o zaman
\begin{align*}
(\alpha+\beta)\cdot\gamma'
&=(\alpha+\beta)\cdot\gamma+(\alpha+\beta)\\
&=(\alpha\cdot\gamma+\beta\cdot\gamma)+(\alpha+\beta)\\
&=(\alpha\cdot\gamma+\alpha)+(\beta\cdot\gamma+\beta)\\
&=\alpha\cdot\gamma'+\beta\cdot\gamma'.
\end{align*}
\item
E\u ger $\gamma$ limit ve
$\Forall{\xi}\bigl(\xi<\gamma\lto(\alpha+\beta)\cdot\xi=\alpha\cdot\xi+\beta\cdot\xi\bigr)$ ise
\begin{align*}
(\alpha+\beta)\cdot\gamma
&=\sup_{\xi<\gamma}\bigl((\alpha+\beta)\cdot\xi\bigr)\\
&=\sup_{\xi<\gamma}(\alpha\cdot\xi+\beta\cdot\xi)\\
&=\sup_{\xi<\gamma}(\alpha\cdot\xi)+\sup_{\xi<\gamma}(\beta\cdot\xi)\\
&=\alpha\cdot\gamma+\beta\cdot\gamma.
\end{align*}
\end{compactenum}
\end{falseproof}
\end{xca}

\begin{theorem}\label{thm:.assoc}
Ordinal \c carpma birle\c smelidir.
\end{theorem}

\ktk

\c Simdi herhangi $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
\alpha^n=\underbrace{\alpha\cdots\alpha}_n
\end{equation*}
anla\c s\i labilir.

\begin{theorem}
$k<\upomega$ ve $\ell<\upomega$ ise
$\alpha^{k+\ell}=\alpha^k\cdot\alpha^{\ell}$.
\end{theorem}

\ktk[ (T\"umevar\i m kullan\i n.)]

\begin{theorem}
Her $\xi\mapsto\xi\cdot\alpha$ i\c slemi artand\i r.
\end{theorem}

\ktk

\section{Hesaplamalar}

\begin{lemma}
	$0<\ell<\upomega$ ise $1+\upomega^{\ell}=\upomega^{\ell}$.
\end{lemma}

\ktk

\begin{theorem}\label{thm:w^k+w^m=w^m}
$k<m<\upomega$ ise $\upomega^k+\upomega^m=\upomega^m$.
\end{theorem}

\begin{proof}
Bir $\ell$ i\c cin, $k+\ell=m$ ve $0<\ell<\upomega$,
dolay\i s\i yla
\begin{align*}
\upomega^k+\upomega^m
&=\upomega^k+\upomega^{k+\ell}\\
&=\upomega^k+\upomega^k\cdot\upomega^{\ell}\\
&=\upomega^k\cdot(1+\upomega^{\ell})\\
&=\upomega^k\cdot\upomega^{\ell}\\
&=\upomega^{k+\ell}\\
&=\upomega^m.\qedhere
\end{align*}
\end{proof}

\begin{theorem}
  $1\leq k<\upomega$ ise $k\cdot\upomega=\upomega$.
(\Sekle{fig:xw} bak\i n.)
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  \caption{$\eta=\xi\cdot\upomega$ denkleminin grafi\u gi}\label{fig:xw}  
\end{figure}

\end{theorem}

\ktk

\begin{theorem}[B\"olme]\label{thm:division}
$1\leq\alpha$ ise $(\xi,\eta)$ i\c cin
\begin{equation}\label{eqn:ax+y=b}
\alpha\cdot\eta+\xi=\beta
\land
\xi<\alpha
\end{equation}
sisteminin bir ve tek bir \c c\"oz\"um\"u vard\i r.
\end{theorem}

\begin{xca}
  Teoremi kan\i tlay\i n.
\"Orne\u gin, a\c sa\u g\i daki iddialar\i\ g\"osterin.
\begin{asparaenum}
  \item
$\alpha>0$ ise $\alpha\cdot\beta\geq\beta$.
\item
$\{\eta\colon\alpha\cdot\eta\leq\beta\}$ k\"umesinin \"usts\i n\i r\i\ vard\i r.
\item
$\sup\{\eta\colon\alpha\cdot\eta\leq\beta\}=\delta$ olsun.
O zaman $\alpha\cdot\gamma+\xi=\beta$ denkleminin 
$\gamma$ \c c\"oz\"um\"u vard\i r, ve $\delta<\alpha$.
Ayr\i ca $(\gamma,\delta)$,
\eqref{eqn:ax+y=b} sisteminin tek \c c\"oz\"um\"u vard\i r.
\end{asparaenum}
\end{xca}

\begin{theorem}
$\upomega^{\upomega}$,
\begin{equation*}
\upomega\cdot\xi=\xi
\end{equation*}
denkleminin en k\"u\c c\"uk \c c\"oz\"um\"ud\"ur.
\end{theorem}

\begin{proof}
$\begin{aligned}[t]
\upomega\cdot\upomega^{\upomega}
&=\upomega\cdot\sup_{x<\upomega}(\upomega^x)\\
&=\sup_{x<\upomega}(\upomega\cdot\upomega^x)\\
&=\sup_{x<\upomega}(\upomega^{1+x})\\
&=\upomega^{\upomega}.\end{aligned}$

\c Simdi $\alpha<\upomega^{\upomega}$ olsun.
O zaman bir $k$ do\u gal say\i s\i\ i\c cin
\begin{gather*}
	\upomega^k\leq\alpha<\upomega^{k+1},\\
	\upomega^{k+1}\leq\upomega\cdot\alpha,
\end{gather*}
dolay\i s\i yla $\alpha<\upomega\cdot\alpha$.
\end{proof}

Teorem sayesinde $\alpha<\upomega^{\upomega}$ ise,
o zaman baz\i\ $\alpha_1$ ve $a_0$ i\c cin
\begin{align*}
	\upomega\cdot\alpha_1+a_0&=\alpha,&
	\alpha_1&<\alpha,&
	a_0&<\upomega.
\end{align*}
E\u ger $\alpha_1>0$ ise,
o zaman baz\i\ $\alpha_2$ ve $a_1$ i\c cin
\begin{align*}
	\upomega\cdot\alpha_2+a_1&=\alpha_1,&
	\alpha_2&<\alpha_1,&
	a_1&<\upomega,
\end{align*}
ve saire.
O zaman bir $k$ i\c cin
\begin{align*}
	\alpha_{k+1}&=0,\\
	\alpha_k&=a_k,\\
	\alpha_{k-1}&=\upomega\cdot a_k+a_{k-1},\\
	\alpha_{k-2}&=\upomega^2\cdot a_k+\upomega\cdot a_{k-1}+a_{k-2},\\
	&\dots\\
	\alpha_1&=\upomega^{k-1}\cdot a_k+\upomega^{k-2}\cdot a_{k-1}+\dots+\upomega\cdot a_2+a_1,\\
	\alpha&=\upomega^k\cdot a_k+\upomega^{k-1}\cdot a_{k-1}+\dots+\upomega^2\cdot a_2+\upomega\cdot a_1+a_0.
\end{align*}
Burada baz\i\ $a_i$ s\i f\i r olabilir.
S\i f\i r terimler silinirse,
o zaman bir $n$ i\c cin, 
\begin{equation*}
\upomega>b_0>b_1>\dots>b_{n-1}
\end{equation*}
ko\c sulunu sa\u glayan
baz\i\ $b_i$ i\c cin, ve baz\i\ $c_i$ sayma say\i lar\i\ i\c cin
\begin{equation*}
\alpha=\upomega^{b_0}\cdot c_0+\upomega^{b_1}\cdot c_1+\dots+\upomega^{b_{n-1}}\cdot c_{n-1}.
\end{equation*}
Bu ifadeye $\alpha$'n\i n \textbf{Cantor normal bi\c cimi} denir.
($0$'\i n Cantor normal bi\c cimi $0$'d\i r.)

\begin{theorem}\label{thm:a+w^m=w^m}
$0<m<\upomega$ ve $\alpha<\upomega^m$ ise $\alpha+\upomega^m=\upomega^m$.
\end{theorem}

\begin{proof}
$\alpha$'n\i n Cantor normal bi\c cimini yaz\i n
ve \Teoremi{thm:w^k+w^m=w^m} kullan\i n.
\end{proof}

\begin{corollary}
$m<\upomega$, $n\in\N$ ve $k\in\N$ ise
\begin{equation*}
(\upomega^m\cdot n+\alpha)\cdot k=\upomega^m\cdot n\cdot k+\alpha.
\end{equation*}
\end{corollary}

\ksk

\"Orne\u gin
\begin{equation*}
(\upomega^5\cdot10+\upomega^3\cdot8+\upomega)\cdot6
=\upomega^5\cdot60+\upomega^3\cdot8+\upomega.
\end{equation*}

Sonucun durumunda a\c sa\u g\i daki e\c sitlik \c c\i kar.
\begin{align*}
(\upomega^m\cdot n+\alpha)\cdot\upomega
&=\upomega^m\cdot n+\underbrace{\alpha+\upomega^m\cdot n}_{\upomega^m\cdot n}+\underbrace{\alpha+\upomega^m\cdot n}_{\upomega^m\cdot n}+\cdots\\
&=\upomega^m\cdot n\cdot\upomega\\
&=\upomega^{m+1}.
\end{align*}
Asl\i nda e\c sitli\u gin ger\c cek kan\i t\i n\i n
\Teoreme{thm:a+w^m=w^m} ihtiyac\i\ yoktur.

\begin{theorem}
$m<\upomega$, $n\in\N$ ve $\alpha<\upomega^m$ ise
\begin{equation*}
(\upomega^m\cdot n+\alpha)\cdot\upomega=\upomega^{m+1},
\end{equation*}
dolay\i s\i yla
$k\in\N$ ise
\begin{equation*}
(\upomega^m\cdot n+\alpha)\cdot\upomega^k=\upomega^{m+k}.
\end{equation*}
\end{theorem}

\begin{proof}
$(\upomega^m\cdot n+\alpha)\cdot k
<\upomega^m\cdot(n+1)\cdot k$
oldu\u gundan
\begin{align*}
	\upomega^{m+1}
	&\leq(\upomega^m\cdot n+\alpha)\cdot\upomega\\
  &=\sup_{x<\upomega}\bigl((\upomega^m\cdot n+\alpha)\cdot x\bigr)\\
  &\leq\sup_{x<\upomega}\bigl(\upomega^m\cdot(n+1)\cdot x\bigr)\\
	&=\upomega^{m+1}.\qedhere
\end{align*}
\end{proof}

\"Orne\u gin
\begin{align*}
	&\phantom{{}={}}(\upomega^3\cdot4+\upomega\cdot6)\cdot(\upomega^2\cdot3+8)\\
	&=(\upomega^3\cdot4+\upomega\cdot6)\cdot\upomega^2\cdot3+(\upomega^3\cdot4+\upomega\cdot6)\cdot8\\
	&=\upomega^5\cdot3+\upomega^3\cdot32+\upomega\cdot6.
\end{align*}

\begin{xca}
$(\upomega^9\cdot9+\upomega^2\cdot9+\upomega\cdot9+9)\cdot(\upomega^2\cdot9+\upomega\cdot9+9)$
\c carp\i m\i n\i n Cantor normal bi\c cimini hesaplay\i n.
\end{xca}

\section{Kardinaller}

\begin{theorem}\label{thm:ab=axb}
  $\alpha\cdot\beta
\approx\alpha\times\beta$.
\end{theorem}

\begin{proof}
  \Teoremde{thm:division}n
	\begin{equation*}
	(\xi,\eta)\mapsto\alpha\cdot\eta+\xi,
	\end{equation*}
$\alpha\times\beta$
kartezyan
\c carp\i m\i ndan
$\alpha\cdot\beta$ ordinal \c carp\i m\i na giden
bir e\c slemedir.
\end{proof}

\begin{theorem}
$\card{\upomega\cdot\upomega}=\upomega$.
\end{theorem}

\ktk

\begin{theorem}
$\{\xi\colon\upomega\leq\xi<\upomega^{\upomega}\}$ k\"umesinin her eleman\i n\i n kardinali
$\upomega$'d\i r.
\end{theorem}

\ktk

\begin{theorem}\label{thm:cntbl-u}
Her $k$ do\u gal say\i s\i\ i\c cin $f_k$
bir $A_k$ k\"umesini $\upomega$'ya g\"oms\"un.
O zaman
\begin{equation*}
\bigcup_{i<\upomega}A_i\preccurlyeq\upomega.
\end{equation*}
\end{theorem}

\begin{proof}
$\bigcup_{i<\upomega}A_i$ bile\c siminde
\begin{equation*}
g(x)=\min\{i\colon x\in A_i\}
\end{equation*}
olsun.  O zaman $x\mapsto\left(g(x),f_{g(x)}(x)\right)$ g\"ondermesi,
bile\c simin $\upomega\times\upomega$ \c carp\i m\i na bir g\"ommesidir.
\end{proof}

\begin{corollary}
$\upomega^{\upomega}\approx\upomega$.
\end{corollary}

\begin{proof}
Her $n$ i\c cin, $\upomega^{n+1}=\upomega^n\cdot\upomega$ oldu\u gundan,
\Teoremin{thm:ab=axb} kan\i t\i n\i ndan
kesin bir $f_n$ i\c cin
\begin{equation*}
f_n\colon\upomega^{n+1}\xrightarrow{\approx}\upomega^n\times\upomega.
\end{equation*}
\c Simdi $g\colon\upomega\times\upomega\xrightarrow{\approx}\upomega$ olsun.
O zaman
\begin{equation*}
  g\circ f_1\colon\upomega^2\xrightarrow{\approx}\upomega.
\end{equation*}
M\"umk\"unse
\begin{equation}\label{eqn:hm}
h_m\colon\upomega^m\xrightarrow{\approx}\upomega
\end{equation}
olsun.
O zaman bir ve tek bir $h_{m+1}$ i\c cin,
\begin{gather*}
h_{m+1}\colon\upomega^{m+1}\to\upomega,\\  
\Forall{\xi}\Forall{\eta}\Forall z
\Bigl(f_n(\xi)=(\eta,z)\lto h_{m+1}(\xi)=g\bigl(h_m(\eta),z\bigr)\Bigr);
\end{gather*}
ve bu durumda
\begin{equation*}
  h_{m+1}\colon\upomega^{m+1}\xrightarrow{\approx}\upomega.
\end{equation*}
T\"umevar\i m ve \"ozyinelemeyle her $m$ sayma say\i s\i\ i\c cin,
bir ve tek bir $h_m$ i\c cin, \eqref{eqn:hm} do\u grudur.
\c Simdi
\begin{equation*}
\upomega^{\upomega}=\sup_{0<x<\upomega}(\upomega^x)=\bigcup_{0<x<\upomega}\upomega^x
\end{equation*}
oldu\u gundan teoremi kullan\i labilir.
\end{proof}

\chapter{Ordinal kuvvet alma}\label{ch:exp}

\section{Tan\i m ve \"ozellikler}

\"Ozyineli tan\i ma g\"ore her $\alpha$ i\c cin
\begin{gather*}
	\alpha^0=1,\\
	\alpha^{\beta'}=\alpha^\beta\cdot\alpha,\\
	\gamma\text{ limit ise }\alpha^\gamma
=\sup_{0<\xi<\gamma}(\alpha^{\xi})
=\limsup_{\xi\to\gamma^-}(\alpha^{\xi}).
\end{gather*}

\begin{theorem}
  $\alpha^1=\alpha$, $1^{\alpha}=1$, ve
  \begin{equation*}
  0^{\alpha}=
  \begin{cases}
    1,&\text{ $\alpha=0$ durumunda},\\
0,&\text{ $\alpha>0$ durumunda}.
  \end{cases}
  \end{equation*}
\end{theorem}

\ktk

\begin{theorem}
$\alpha\geq2$ ise $\xi\mapsto\alpha^{\xi}$ i\c slemi, normaldir.
\end{theorem}

\ktk

\Sekle{fig:w^x} bak\i n.
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  \centering
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  \caption{$\eta=\upomega^{\xi}$ denkleminin grafi\u gi}\label{fig:w^x}  
\end{figure}
\c Sekilde
\begin{equation*}\label{epsilon}
\upvarepsilon_0=\sup\left\{\upomega,\upomega^{\upomega},\upomega^{\upomega^{\upomega}},\dots\right\}.
\end{equation*}

\begin{xca}
$\xi\mapsto\xi^{\xi}$ i\c slemi kesin artan m\i d\i r?
S\"urekli midir?
\end{xca}

\begin{theorem}
$\alpha^{\beta+\gamma}=\alpha^{\beta}\cdot\alpha^{\gamma}$
ve
$\alpha^{\beta\cdot\gamma}=(\alpha^{\beta})^{\gamma}$.
\end{theorem}

\ktk

\begin{theorem}
$\alpha\geq1$ ise $\xi\mapsto\xi^{\alpha}$ artand\i r.
\end{theorem}

\ktk

\section{Hesaplamalar}

\begin{theorem}[Logaritma alma]
$2\leq\alpha$ ve $1\leq\beta$ ise $(\xi,\eta,\zeta)$ i\c cin
\begin{equation*}
\alpha^{\xi}\cdot\eta+\zeta=\beta\land0<\eta<\alpha\land\zeta<\alpha^{\xi}
\end{equation*}
sisteminin bir ve tek bir \c c\"oz\"um\"u vard\i r.
\end{theorem}

\ktk

Teorem sayesinde $1\leq\alpha$ ise,
baz\i\ $\alpha_0$, $a_0$, ve $\beta_1$ i\c cin
\begin{align*}
	\upomega^{\alpha_0}\cdot a_0+\beta_1&=\alpha,&
	0<a_0&<\upomega,&
	\beta_1&<\upomega^{\alpha_0}.
\end{align*}
E\u ger $1\leq\beta_1$ ise, o zaman
baz\i\ $\alpha_1$, $a_1$, ve $\beta_2$ i\c cin
\begin{align*}
	\upomega^{\alpha_1}\cdot a_1+\beta_2&=\beta_1,&
	0<a_1&<\upomega,&
	\beta_2&<\upomega^{\alpha_1},
\end{align*}
ve saire.  O zaman bir $k$ i\c cin
\begin{gather*}
\alpha_0>\alpha_1>\dots>\alpha_k,\\
\{a_0,\dots,a_k\}\included\N,\\
\alpha=\upomega^{\alpha_0}\cdot a_0+\upomega^{\alpha_1}\cdot a_1+\dots+\upomega^{\alpha_k}\cdot a_k.
\end{gather*}
Son ifade, $\alpha$'n\i n \textbf{Cantor normal bi\c cimidir.}

\begin{theorem}
$\alpha<\upomega^{\beta}$ ise $\alpha+\upomega^{\beta}=\upomega^{\beta}$.
\end{theorem}

\ktk[ (\Teoreme{thm:a+w^m=w^m} bak\i n.)]

\begin{corollary}
$\alpha<\upomega^{\beta}$, $n\in\N$, ve $k\in\N$ ise
\begin{equation*}
(\upomega^{\beta}\cdot n+\alpha)\cdot k=\upomega^{\beta}\cdot n\cdot k.
\end{equation*}
\end{corollary}

\ksk

\begin{theorem}
$\alpha<\upomega^{\beta}$, $n\in\N$, ve $1\leq\gamma$ ise
\begin{equation*}
(\upomega^{\beta}\cdot n+\alpha)\cdot\upomega^{\gamma}=\upomega^{\beta+\gamma}.
\end{equation*}
\end{theorem}

\ktk[ (Bir $\delta$ i\c cin $\gamma=1+\delta$ oldu\u gunu kullanabiliriz.)]

\c Simdi iki Cantor normal bi\c ciminin \c carp\i m\i n\i n 
Cantor normal bi\c cimini hesaplayabiliriz.

\begin{theorem}
  $0<k<\upomega$ ise
  \begin{equation*}
    k^{\upomega^{\xi}}=
    \begin{cases}
      k,&\text{ $\xi=0$ durumunda,}\\
\upomega^{\upomega^{\xi-1}},&\text{ $0<\xi<\upomega$ durumunda,}\\
\upomega^{\upomega^{\xi}},&\text{ $\upomega\leq\xi$ durumunda.}
    \end{cases}
  \end{equation*}
\end{theorem}

\ktk

\begin{theorem}
  $\alpha<\upomega^{\beta}$, $n\in\N$, ve $\gamma$ limit ise
  \begin{equation*}
    (\upomega^{\beta}\cdot n+\alpha)^{\gamma}=\upomega^{\beta\cdot\gamma}.
  \end{equation*}
\end{theorem}

\begin{theorem}
$\upvarepsilon_0$,
\begin{equation*}
\upomega^{\xi}=\xi
\end{equation*}
denkleminin en k\"u\c c\"uk \c c\"oz\"um\"ud\"ur.
\end{theorem}

\ktk

\section{Kardinaller}

Herhangi $\alpha$ ve $\beta$ ordinalleri i\c cin,
$\beta$'dan $\alpha$'ya giden g\"ondermeler,
\begin{equation*}
{}^{\beta}\alpha
\end{equation*}
s\i n\i f\i n\i\ olu\c stursun, ve
\begin{equation*}
\exp(\alpha,\beta)=\bigl\{f\colon f\in{}^{\beta}\alpha\land\{\xi\in\beta\colon f(\xi)>0\}\prec\upomega\bigr\}
\end{equation*}
olsun.

\begin{theorem}
$\alpha^{\beta}\approx\exp(\alpha,\beta)$.
\end{theorem}

\begin{proof}
$\exp(1,\beta)\approx1=1^{\beta}$; ayr\i ca
\begin{equation*}
\exp(0,\beta)\approx
\begin{cases}
	1,&\text{ $\beta=0$ durumunda,}\\
	0,&\text{ $\beta>0$ durumunda,}
\end{cases}
\end{equation*}
dolay\i s\i yla $\exp(0,\beta)\approx 0^{\beta}$.
\c Simdi $\alpha\geq2$ olsun.
E\u ger $\gamma<\alpha^\beta$ ise,
o zaman Cantor normal bi\c cimi gibi,
baz\i\ $n$ do\u gal say\i s\i\ i\c cin,
baz\i\ $\gamma_i$ ve $\delta_i$ i\c cin,
\begin{gather*}
	\beta>\gamma_0>\dots>\gamma_{n-1},\\
	\{\delta_i\colon i<n\}\included\alpha\setminus\{0\},\\
	\gamma=\alpha^{\gamma_0}\cdot c_0+\dots+\alpha^{\gamma_{n-1}}\cdot c_{n-1}.
\end{gather*}
\c Simdi tan\i ma g\"ore
\begin{equation*}
f_{\gamma}(\xi)=
\begin{cases}
	\delta_i,&\text{ $\xi=\gamma_i$ durumunda},\\
0,&\text{ $\xi\in\beta\setminus\{\gamma_i\colon i<n\}$ durumunda}
\end{cases}
\end{equation*}
olsun.
O zaman $f_{\gamma}\in\exp(\alpha,\beta)$.
As\i nda
\begin{equation*}
\xi\mapsto f_{\xi}\colon\alpha^{\beta}\xrightarrow{\approx}\exp(\alpha,\beta).\qedhere
\end{equation*}
\end{proof}

\begin{theorem}
$\upvarepsilon_0\approx\upomega$.
\end{theorem}

\ktk

\chapter{Kardinal kuvvetler}\label{ch:card}

\section{Say\i lamaz k\"umeler}

E\u ger $A\preccurlyeq\upomega$ ise,
o zaman $A$ \textbf{say\i labilir;}
di\u ger durumda $A$ \textbf{say\i lamaz.}
G\"ord\"u\u g\"um\"uz gibi
say\i labilir k\"umelerden
ordinal toplama, \c carpma, ve kuvvet alma ile
say\i lamaz k\"umeler elde edilemez.


Herhangi $\bm A$ s\i n\i f\i\ i\c cin
\begin{equation*}\label{pow}
\pow{\bm A},
\end{equation*}
$\bm A$'n\i n \emph{altk\"umeleri} taraf\i ndan olu\c sturulmu\c s s\i n\i ft\i r.
Yani
\begin{equation*}
\pow{\bm A}=\{X\colon X\included\bm A\}.
\end{equation*}
Buradaki $X$ siyah olmad\i\u g\i ndan \emph{k\"ume} de\u gi\c skenidir.
K\"ume olmayan bir s\i n\i f, bir s\i n\i f\i n eleman\i\ olamaz.
E\u ger $\universe$, t\"um k\"umeler taraf\i ndan olu\c sturulmu\c s s\i n\i f ise,
o zaman
\begin{equation*}
\pow{\universe}=\universe.
\end{equation*}
Ama $n\in\upomega$ ise
\begin{equation*}
n<2^n=\card{\pow n}.
\end{equation*}

\begin{theorem}[Cantor]\label{thm:pow}
Her $A$ k\"umesi i\c cin
\begin{equation*}
A\prec\pow A.
\end{equation*}
\end{theorem}

\begin{proof}
$x\mapsto\{x\}\colon A\xrightarrow{\preccurlyeq}\pow A$.
\c Simdi $f\colon A\xrightarrow{\preccurlyeq}\pow A$ ise
\begin{equation*}
B=\{x\in A\colon x\notin f(x)\}
\end{equation*}
olsun.  O zaman $A$'n\i n her $c$ eleman\i\ i\c cin
\begin{equation*}
c\in B\liff c\notin f(c),
\end{equation*}
dolay\i s\i yla $B\neq f(c)$.
Bu \c sekilde $f$, e\c sleme olamaz.
\end{proof}

\begin{xca}
Cantor Teoreminin kan\i t\i\
$A$'n\i n k\"ume oldu\u gunu nas\i l kullan\i r?
\end{xca}

\begin{axiom}[Kuvvet K\"ume]
Her $A$ k\"umesi i\c cin $\pow A$ s\i n\i f\i\ bir k\"umedir.
\end{axiom}

Herhangi $a$ ve $b$ i\c cin
\begin{equation*}
(a,b)=\bigl\{\{a\},\{a,b\}\bigr\}
\end{equation*}
olsun.

\begin{theorem}
$(a,b)=(c,d)$ ancak ve ancak $a=c$ ve $b=d$.
\end{theorem}

\ktk

\c Simdi $A\times B=\{(x,y)\colon x\in A\land y\in B\}$
tan\i mlanabilir.

\begin{theorem}
$A\times B\included\pow{\pow{A\cup B}}$.
\end{theorem}

\ktk

\begin{theorem}[Hartogs]
Her kardinalin daha b\"uy\"u\u g\"u vard\i r.
\end{theorem}

\begin{proof}
$\bm A=\{\xi\colon\xi\preccurlyeq\kappa\}$ olsun.
O zaman $\bm A\included\on$, ve ayr\i ca $\bm A$ ge\c ci\c slidir,
dolay\i s\i yla $\bm A$ bir k\"umeyse, bir $\alpha$ ordinalidir.
Bu durumda $\alpha\notin\alpha$ oldu\u gundan $\alpha>\kappa$.

E\u ger $f$ bir $\beta$'y\i\ $\kappa$'ya g\"om\"urse,
o zaman bir
\begin{equation*}
\Bigl\{\bigl(f(\xi),f(\eta)\bigr)\colon\xi\leq\eta<\beta\Bigr\}
\end{equation*}
k\"umesi
elde edilebilir.
Bu \c sekilde elde edilen t\"um k\"umeler,
$\kappa\times\kappa$ \c carp\i m\i n\i n bir $B$ altk\"umesini olu\c sturur.
O halde $B\approx\bm A$ (neden?),
dolay\i s\i yla $\bm A$ da bir k\"umedir.
\end{proof}

Sonu\c c olarak
\begin{equation*}
\kappa^+=\min\{\xi\colon\kappa<\xi\}
\end{equation*}
tan\i mlanabilir, $\kappa^+$, $\kappa$'n\i n \emph{kardinal} ard\i l\i d\i r.

\c Simdi \"ozyineli tan\i ma g\"ore
\begin{gather*}
	\aleph_0=\upomega,\\
	\aleph_{\alpha'}=(\aleph_{\alpha})^+,\\
	\alpha\text{ limit ise }\aleph_{\alpha}=\sup_{\xi<\alpha}\aleph_{\xi}.
\end{gather*}
(Burada $\aleph$, \.Ibrani \emph{alef} harfidir.)

\begin{theorem}
$\xi\mapsto\aleph_{\xi}$ normaldir,
ve her sonsuz kardinal, bir $\alpha$ i\c cin, $\aleph_{\alpha}$'d\i r.
\end{theorem}

\ktk

\begin{lemma}
  Her sonsuz kardinal, $\upomega$'n\i n bir kuvvetidir.
\end{lemma}

\klk

Tan\i ma g\"ore
\begin{align*}
  \kappa\oplus\lambda&=\card{\kappa\sqcup\lambda}=\card{\kappa+\lambda},\\
  \kappa\otimes\lambda&=\card{\kappa\times\lambda}=\card{\kappa\cdot\lambda}
\end{align*}
olsun;
bunlar $\kappa$ ve $\lambda$'n\i n \textbf{kardinal toplam\i}
ve \textbf{kardinal \c carp\i m\i d\i r.}

\begin{theorem}
E\u ger $\kappa$ ve $\lambda$'n\i n biri sonsuz ise
  \begin{equation*}
    \kappa\oplus\lambda=\max(\kappa,\lambda).
  \end{equation*}
E\u ger
$\kappa$ ve $\lambda$'n\i n biri sonsuz ise ve di\u geri s\i f\i r de\u gilse
  \begin{equation*}
    \kappa\otimes\lambda=\max(\kappa,\lambda).
  \end{equation*}
\end{theorem}

\begin{proof}
$\kappa\leq\lambda$ olsun.  O zaman
  \begin{equation*}
   \lambda
\leq\kappa+\lambda
\leq\lambda+\lambda
\preccurlyeq2\cdot\lambda
\leq\lambda\cdot\lambda,
  \end{equation*}
ve $\kappa>0$ ise
\begin{equation*}
  \lambda\leq\kappa\cdot\lambda\leq\lambda\cdot\lambda,
\end{equation*}
dolay\i s\i yla $\lambda\approx\lambda^2$ kan\i tlamak yeter.

Lemmadan bir $\alpha$ i\c cin $\lambda=\upomega^{\alpha}$.
O zaman $\lambda\approx\exp(\upomega,\alpha)$.
\c Simdi $f\colon\upomega\times\upomega\xrightarrow{\approx}\upomega$ olsun.
E\u ger $g$ ve $h$, $\exp(\upomega,\alpha)$ k\"umesinin eleman\i\ ise $g*h$,
\begin{equation*}
  \xi\mapsto f(g(\xi),h(\xi))
\end{equation*}
eleman\i\ olsun.
O zaman
\begin{equation*}
  (g,h)\mapsto g*h\colon\exp(\upomega,\alpha)\xrightarrow{\approx}
\exp(\upomega,\alpha)\times\exp(\upomega,\alpha).\qedhere
\end{equation*}
\end{proof}

Sonu\c c olarak
\begin{equation*}
  \card{\aleph_{\alpha}+\aleph_{\beta}}=\aleph_{\max(\alpha,\beta)}
=\card{\aleph_{\alpha}\cdot\aleph_{\beta}}.
\end{equation*}

\c Simdi herhangi $A$ k\"umesi i\c cin
\begin{equation*}\label{powf}
  \powf A=\{X\in\pow A\colon\card X<\upomega\}
\end{equation*}
olsun.

\begin{theorem}
E\u ger $\kappa$ sonsuz ise $\powf{\kappa}\approx\kappa$.
\end{theorem}

\begin{proof}
Her $m$ i\c cin
  $\{\xi\in\kappa\colon\card{\xi}=m\}\preccurlyeq\kappa^m\approx\kappa$,
dolay\i s\i yla
\begin{equation*}
  \powf{\kappa}=\bigcup_{i\in\upomega}\{\xi\in\kappa\colon\card{\xi}=i\}
\preccurlyeq\upomega\times\kappa\approx\kappa.\qedhere
\end{equation*}
\end{proof}

\begin{theorem}
E\u ger $\beta$ sonsuz ve $2\leq\alpha\leq\beta$ ise
\begin{equation*}
  \card{\alpha^{\beta}}=\card{\beta}.
\end{equation*}
E\u ger $\alpha$ sonsuz ve $1\leq\beta\leq\alpha$ ise
\begin{equation*}
  \card{\alpha^{\beta}}=\card{\alpha}.
\end{equation*}
  \end{theorem}

\ktk

\section{Se\c cme}

\Teoremde{thm:cntbl-u}, 
$\bigcup_{i<\upomega}A_i$ bile\c siminin say\i labilir olmas\i\ i\c cin,
her $A_k$ k\"umesinin say\i labilir olmas\i\ yetmez,
ama $A_k$ k\"umesinin $\upomega$'ya kesin bir g\"ommesi bilinmelidir.
Her $k$ i\c cin,
$A_k$ k\"umesinin $\upomega$'ya g\"ommeleri,
bo\c s olmayan bir $\mathscr B_k$ k\"umesini olu\c sturabilirler;\label{curly-B}
ama g\"ord\"u\u g\"um\"uz aksiyomlarla
\begin{equation*}
\Forall x(x\in\upomega\lto f_x\in\mathscr B_x)
\end{equation*}
ko\c sulunu sa\u glayan 
$x\mapsto f_x$ g\"ondermesinin olup olmad\i\u g\i n\i\ bilmiyoruz.

G\"ord\"u\u g\"um\"uz aksiyomlar, 
\textbf{Zermelo--Fraenkel} veya \textbf{ZF} aksiyomlar\i d\i r.

Her $k$ i\c cin $\mathscr B_k$ k\"umesinden bir $f_k$ se\c cmek isteriz.
\emph{Se\c cim Aksiyomunun} 
bi\c cimlerinin birine g\"ore, bu se\c cme m\"umk\"und\"ur.
Bizim i\c cin, a\c sa\u g\i daki bi\c cim kullan\i l\i\c sl\i\ olacakt\i r.

\begin{axiom}[Se\c cim]
  Her k\"ume iyis\i ralanabilir.
\end{axiom}

\"Orne\u gin $\bigcup_{x\in\upomega}\mathscr B_x$ iyis\i ralan\i rsa,
o zaman istedi\u gimiz $x\mapsto f_x$ g\"ondermesi
\begin{equation*}
  x\mapsto\min\mathscr(B_x)
\end{equation*}
olabilir.

G\"odel'in kan\i tlad\i\u g\i\ teoreme g\"ore, 
ZF aksiyomlar\i n\i n bir \emph{modelinde,}
Se\c cim Aksiyomu do\u grudur.
Cohen'in kan\i tlad\i\u g\i\ teoreme g\"ore, 
ZF aksiyomlar\i n\i n bir modelinde,
Se\c cim Aksiyomu yanl\i\c st\i r.
K\i saca Se\c cim Aksiyomu, ZF'den ba\u g\i ms\i zd\i r.

\begin{sloppypar}
  Se\c cim Aksiyomunu varsay\i yoruz.
  Bununla ZF, \textbf{ZFC}'dir.
  \c Simdi her k\"umenin kardinali vard\i r.
  Tan\i ma g\"ore
  \begin{equation*}
    \kappa^{\lambda}=\card{{}^{\kappa}\lambda}.
  \end{equation*}
  Bu kuvvet, ordinal k\"uvvet de\u gil,
  \textbf{kardinal kuvvettir.}
  Orne\u gin
  $\aleph_0=\upomega$ oldu\u gu halde
  $2^{\aleph_0}$, kardinal kuvvet olarak anla\c s\i l\i r,
  ve bu kuvvet, $2^{\upomega}$ ordinal kuvvetinden farkl\i d\i r.
  Asl\i nda $2^{\upomega}=\upomega$, ama sonraki teoreme g\"ore
  $2^{\aleph_0}>\aleph_0$.
\end{sloppypar}

\begin{theorem}
  $2^{\kappa}=\card{\pow{\kappa}}$.
\end{theorem}

\ktk

A\c sa\u g\i daki kurallar kolayd\i r.
\begin{align*}
&\begin{gathered}
	0<\lambda\lto 0^{\lambda}=0,\\
	\kappa^0=1,\\
	1^{\lambda}=1,\\
	\kappa^1=\kappa,
	\end{gathered}&
&\begin{gathered}
	\kappa^{\lambda\cardsum\mu}=\kappa^{\lambda}\cardprod\kappa^{\mu},\\
	\kappa^{\lambda\cardprod\mu}=(\kappa^{\lambda})^{\mu},\\
	\kappa\leq\mu\land\lambda\leq\nu\lto\kappa^{\lambda}\leq\mu^{\nu}.	
\end{gathered}
\end{align*}

\begin{theorem}
$2\leq\kappa$, $1\leq\lambda$, ve $\aleph_0\leq\max\{\kappa,\lambda\}$ olsun.  
O zaman
\begin{gather}\label{eqn:kl}
\kappa\leq 2^{\lambda}\lto\kappa^{\lambda}=2^{\lambda},\\\label{eqn:lk}
\lambda\leq\kappa\lto
%\kappa\leq
\kappa^{\lambda}\leq2^{\kappa}.
\end{gather}
\end{theorem}

\begin{proof}
Hipoteze g\"ore $\kappa\leq2^{\lambda}$ ise 
$2\leq\kappa\leq 2^{\lambda}$ ve $\lambda$ sonsuzdur, dolay\i s\i yla
\begin{equation*}
2^{\lambda}\leq\kappa^{\lambda}\leq(2^{\lambda})^{\lambda}
=2^{\lambda\cardprod\lambda}=2^{\lambda}.
\end{equation*}
Ayr\i ca $\lambda\leq\kappa$ ise $\kappa$ sonsuzdur, dolay\i s\i yla
\begin{equation*}
\kappa\leq\kappa^{\lambda}
\leq(2^{\kappa})^{\lambda}
=2^{\kappa\cardprod\lambda}
=2^{\kappa}.\qedhere
\end{equation*}
\end{proof}

\begin{sloppypar}
  Tekrar $\kappa$ ve $\lambda$'n\ in biri sonsuz olsun.
  E\u ger $\lambda\leq\kappa\leq2^{\lambda}$ ise, 
  o zaman \eqref{eqn:kl} gerektirmesine g\"ore
  \begin{equation*}
    \kappa^{\lambda}=2^{\lambda}\leq2^{\kappa};
  \end{equation*}
  burada \eqref{eqn:lk} gerekmez.
  Bir durumda, 
  e\u ger \eqref{eqn:kl} gerektirmesinin hipotezi do\u gru de\u gilse,
  o zaman $2^{\lambda}<\kappa$, dolay\i s\i yla $\lambda\leq\kappa$,
  ve \eqref{eqn:lk} kullan\i labilir.
  Bu \c sekilde teoremin yerine
  \begin{gather*}%\label{eqn:kl}
    \kappa\leq 2^{\lambda}\lto\kappa^{\lambda}=2^{\lambda},\\%\label{eqn:lk}
    2^{\lambda}<\kappa\lto
    %\kappa\leq
    \kappa^{\lambda}\leq2^{\kappa}.
  \end{gather*}
  kurallar\i\ kullan\i labilir.
  (Tekrar
  $2\leq\kappa$, $1\leq\lambda$, 
  ve $\aleph_0\leq\max\{\kappa,\lambda\}$ olmal\i d\i r.)
  \"Orne\u gin
  \begin{gather*}
    2\leq\kappa\leq2^{\aleph_0}\lto\kappa^{\aleph_0}=2^{\aleph_0},\\
    2^{\aleph_0}<\kappa\lto%\kappa\leq
    \kappa^{\aleph_0}\leq2^{\kappa}.
  \end{gather*}
\end{sloppypar}

\c Simdi a\c sa\u g\i daki tan\i m yapabiliriz.
\begin{gather*}
  \beth_0=\aleph_0,\\
\beth_{\alpha'}=\card{\pow{\beth_{\alpha}}}=2^{\beth_{\alpha}},\\
\alpha\text{ limit ise }\beth_{\alpha}=\sup_{\xi<\alpha}\beth_{\xi}.
\end{gather*}
(Burada $\beth$, \.Ibrani \emph{beth} harfidir.)
O zaman $\xi\mapsto\beth_{\xi}$ normaldir, ve
\begin{equation*}
  \aleph_{\alpha}\leq\beth_{\alpha}.
\end{equation*}

\begin{theorem}
T\"um $\kappa$ ve $\lambda$ i\c cin
\begin{gather*}
2\leq\kappa\leq\beth_{\alpha+1}\lto\kappa^{\beth_{\alpha}}=\beth_{\alpha+1},\\
1\leq\lambda\leq\beth_{\alpha}\lto{\beth_{\alpha+1}}^{\lambda}=\beth_{\alpha+1}.
\end{gather*}
\end{theorem}

\ktk

\textbf{Kontinuum Hipotezi} veya \textbf{KH,}
$\aleph_1=\beth_1$ \"onermesidir.
\textbf{Genelle\c stirilmi\c s Kontinuum Hipotezi} veya \textbf{GKH,}
$\Forall{\xi}\aleph_{\xi}=\beth_{\xi}$ \"onermesidir.
G\"odel'in kan\i tlad\i\u g\i\ teoreme g\"ore, 
ZFC aksiyomlar\i n\i n bir modelinde,
GKH do\u grudur.
Cohen'in kan\i tlad\i\u g\i\ teoreme g\"ore, 
ZFC aksiyomlar\i n\i n bir modelinde,
KH yanl\i\c st\i r.
Bu \c sekilde KH, ZFC'den ba\u g\i ms\i zd\i r.

\appendix

\chapter{Harfler}\label{ch:let}

Metinde simge olarak kullan\i l\i rken harfler a\c sa\u g\i daki anlamlara gelir.

\begin{description}
\item[``Tahta siyah\i'' harfleri]\mbox{}
\begin{itemize}[]
\item
$\R$\letsep ger\c cel say\i lar k\"umesi \letterrefs{\sayfada{R}}
\item
$\Q$\letsep kesirli say\i lar k\"umesi \letterrefs{\sayfada{Q}}
\item
$\Z$\letsep tamsay\i lar k\"umesi \letterrefs{\sayfada{Z}}
\item
$\N$\letsep $\{1,2,3,\dots\}$ sayma say\i lar k\"umesi \letterrefs{\sayfada{N}}
\end{itemize}
\item[K\"u\c c\"uk Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$a$, $b$, $c$, $d$, $e$\letsep say\i lar veya k\"umeler
\item
$f$, $g$, $h$\letsep k\"umede tan\i mlanmi\c s g\"ondermeler
\item
$i$, $j$\letsep do\u gal say\i\ de\u gi\c skenler
\item
$k$, $\ell$, $m$, $n$\letsep do\u gal say\i lar
\item
$p$\letsep asal say\i\ \letterrefs{\ref{p} ve \sayfada{p2}}
\item
$u$, $x$, $y$, $z$\letsep say\i\ veya k\"ume de\u gi\c skenleri
\end{itemize}
\item[Dikey k\"u\c c\"uk Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$\sup$\letsep supremum
\item
$\min$\letsep minimum (en k\"u\c c\"uk)
\item
$\max$\letsep maksimum (en b\"uy\"uk)
\end{itemize}
\item[B\"uy\"uk Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$A$, $B$, $C$, $D$\letsep k\"umeler \letterrefs{\sayfada{A}}
\item
$X$, $Y$, $Z$\letsep k\"ume de\u gi\c skenleri
\end{itemize}
\item[K\i v\i rc\i k Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$\mathscr A$, $\mathscr B$, $\mathscr C$\letsep elemanlar\i\ k\"ume veya g\"onderme olan k\"umeler
\letterrefs{\pageref{curly-C} ve \sayfanumarada{curly-B}}
\item
$\pow A$\letsep$\{X\colon X\included A\}$ \letterrefs{\pageref{pow} ve \sayfanumarada{thm:pow}}
\item
$\powf A$\letsep$\{X\in\pow A\colon\card X<\upomega\}$ \letterrefs{\sayfada{powf}}
\end{itemize}
\item[B\"uy\"uk siyah Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$\bm A$, $\bm B$, $\bm C$\letsep s\i n\i flar \letterrefs{\S \ref{sect:sets-classes}, \sayfada{sect:sets-classes}}
\item
$\bm F$, $\bm G$, $\bm H$\letsep s\i n\i fta tan\i mlanm\i\c s g\"ondermeler
\letterrefs{\Teorem{thm:ord-rec}, \sayfada{thm:ord-rec}}
\end{itemize}
\item[Dikey b\"uy\"uk siyah Latin harfleri]\mbox{}
\begin{itemize}[]
\item
$\universe$\letsep evrensel s\i n\i f
\item
$\on$\letsep ordinaller s\i n\i f\i
\item
$\cn$\letsep kardinaller s\i n\i f\i
\end{itemize}
\item[Yunan harfleri]\mbox{}
\begin{itemize}[]
\item
$\alpha$, $\beta$, $\gamma$, $\delta$, $\theta$\letsep ordinaller \letterrefs{\sayfada{minus-gr}}
\item
$\xi$, $\eta$, $\zeta$\letsep ordinal de\u gi\c skenler \letterrefs{\sayfada{minus-gr}}
\item
$\kappa$, $\lambda$, $\mu$, $\nu$\letsep kardinaller \letterrefs{\sayfada{card}}
\item
$\phi$, $\psi$, $\chi$\letsep form\"uller \letterrefs{\sayfada{phi}}
\end{itemize}
\item[Dikey Yunan harfi]\mbox{}
\begin{itemize}[]
\item
$\upvarepsilon_0$\letsep$\sup\{\upomega,\upomega^{\upomega},\upomega^{\upomega^{\upomega}},\dots\}$
\letterrefs{\sayfada{epsilon}}
\item
$\upomega$\letsep $\{0,1,2,\dots\}$ do\u gal say\i lar\i\ k\"umesi \letterrefs{\sayfada{eqn:omega=}}
\end{itemize}
\item[Harflerden t\"ureyen simgeler]\mbox{}
\begin{itemize}[]
\item
$\in$\letsep eleman olma ba\u g\i nt\i s\i\ 
(``$a$ \gr{>est`i} $B$'' demek ``$a$, bir $B$'dir'')
\item
$\forall$\letsep her \lips i\c cin (\emph{for \textbf All})
\item
$\exists$\letsep baz\i\lips i\c cin (\emph{there \textbf Exists})
\item
$\cup$, $\bigcup$\letsep bile\c sim (\emph{\textbf Union})
\end{itemize}
\end{description}

\chapter{Mant\i k}\label{mantik}

\section{Form\"uller}

Form\"ullerde kulland\i\u g\i m\i z simgelerin birka\c c tane t\"ur\"u vard\i r:
\begin{compactenum}[1)]
\item
\textbf{de\u gi\c skenler}\index{de\u gi\c sken} (\eng{variables}):
$z$, $y$, $x$, \dots; $x_0$, $x_1$, $x_2$, \dots;%
\glossary{$x$, $y$, $z$, \dots}
\item
\textbf{sabitler}\index{sabit} (\eng{constants}): $a$, $b$, $c$,
\dots; $a_0$, $a_1$, $a_2$, \dots;%
\glossary{$a$, $b$, $c$, \dots}
\item
\textbf{iki-konumlu ba\u glay\i c\i lar}\index{ba\u glay\i c\i} (\eng{binary connectives}): $\land$, $\lor$, $\lto$, $\liff$;\footnote{Bazen $\lto$ ile $\liff$ oklar\i n\i n yerine $\to$ ile $\leftrightarrow$ i\c saretleri yaz\i l\i r.}
\item
bir \textbf{tek-konumlu ba\u glay\i c\i} (\eng{singulary connective}): $\lnot$;
\item
\textbf{niceleyiciler}\index{niceleyici} (\eng{quantifiers}): $\exists$, $\forall$;
\item
\textbf{ayra\c clar}\index{ayra\c c} (\eng{parentheses, brackets}): $($, $)$;
\item
bir \textbf{y\"uklem}\index{y\"uklem} (\eng{predicate}): $\in$ (epsilon).
\end{compactenum}
Bir \textbf{terim}%
\index{terim}\label{terim}
(\eng{term}), ya de\u gi\c sken ya da sabittir.  E\u ger $t$ ile $u$, iki terim ise, o zaman
\begin{equation*}
t\in u
\end{equation*}
ifadesi, bir \textbf{b\"ol\"unemeyen form\"uld\"ur}%
\index{form\"ul} (\eng{atomic formula}).  
Genelde \textbf{form\"ullerin} tan\i m\i, \"ozyinelidir:
\begin{compactenum}
\item
B\"ol\"unemeyen bir form\"ul, bir form\"uld\"ur.
\item
E\u ger $\phi$ bir form\"ul ise, o zaman
\begin{equation*}
\lnot\phi
\end{equation*}
ifadesi de bir form\"uld\"ur.
\item
E\u ger $\phi$ ile $\psi$ iki form\"ul ise, o zaman
\begin{align*}
&(\phi\land\psi),&
&(\phi\lor\psi),&
&(\phi\lto\psi),&
&(\phi\liff\psi)
\end{align*}
 ifadeleri de form\"uld\"ur.
\item
E\u ger $\phi$ bir form\"ul ise, ve $x$ bir de\u gi\c sken ise, o zaman
\begin{align*}
&\Exists x\phi,&\Forall x\phi
\end{align*}
ifadeleri de form\"uld\"ur.
\end{compactenum}
Form\"ullerin her t\"ur\"un\"un ad\i\ vard\i r:
\begin{compactenum}
\item
$\lnot\phi$ form\"ul\"u, bir 
\textbf{de\u gillemedir}%
\index{de\u gilleme}
(\eng{negation}).
\item
$(\phi\land\psi)$ form\"ul\"u, bir 
\textbf{birle\c sme}%
\index{birle\c sme}
veya \textbf{t\"umel evetlemedir}
\index{t\"umel evetleme}%
\index{evetleme}
(\eng{conjunction}).
\item
$(\phi\lor\psi)$ form\"ul\"u, bir 
\textbf{ayr\i lma}%
\index{ayr\i lma}
veya \textbf{tikel evetlemedir} 
(\eng{disjunction}). 
\item
$(\phi\lto\psi)$ form\"ul\"u, bir 
\textbf{gerektirme}%
%\textbf{kar\i\c st\i rmad\i r}
\index{gerektirme}
(\eng{implication}). 
\item
$(\phi\liff\psi)$ form\"ul\"u, bir 
\textbf{denkliktir}%
\index{denklik}
(\eng{equivalence}). 
\item
$\Exists x\phi$ form\"ul\"u, bir 
\textbf{\"orneklemedir}%
\index{\"ornekleme}
(\eng{instantiation}). 
\item
$\Forall x\phi$ form\"ul\"u, bir 
\textbf{genelle\c stirmedir}%
\index{genelle\c stirme}
(\eng{generalization}). 
\end{compactenum}
Bu t\"urlerin adlar\i, \c cok \"onemli de\u gildir.  Fakat a\c sa\u
g\i daki teorem \c cok \"onemlidir. 

\begin{theorem}
Her form\"ul\"un tek bir \c sekilde tek bir t\"ur\"u vard\i r.
\end{theorem}

Mesela ayn\i\ form\"ul, hem gerektirme, hem \"ornekleme olamaz:
$\Exists x(\phi\lto\psi)$ form\"ul\"u, gerektirme de\u gil,
\"orneklemedir; $(\Exists x\phi\lto\psi)$ form\"ul\"u, \"ornekleme de\u gil, gerektirmedir.

Ayr\i ca $(\phi\land(\psi\land\theta))$ form\"ul\"u, tek bir \c
sekilde birle\c smedir.  Asl\i nda sadece $\phi$ ile
$(\psi\land\theta)$ form\"ullerinin birle\c smesidir.  E\u ger $A$
harf\/i, $\phi\land(\psi$ ifadesini g\"osterirse ve $B$ harf\/i,
$\theta)$ ifadesini g\"osterirse, o zaman $(A\land B)$ ifadesi,
$(\phi\land(\psi\land\theta))$ form\"ul\"un\"u g\"osterir; ama tan\i
ma g\"ore bu form\"ul, $A$ ile $B$ ifadelerinin birle\c smesi de\u
gildir, \c c\"unk\"u $A$ ile $B$ ifadeleri (yani $A$ ile $B$ taraf\i
ndan g\"osterilen ifadeler), form\"ul de\u gildir. 

Teoremi kan\i tlamayaca\u g\i z.  
Fakat teoremi kullanarak a\c sa\u g\i daki \"ozyineli tan\i m\i\ yapabiliriz.  
Bir de\u g\i\c skenin bir form\"ulde birka\c c tane 
\textbf{ge\c ci\c si}%
\index{ge\c cis}
(\eng{occurrence})
olabilir.  Mesela $\Forall x(x\in y\liff x\in z)$ form\"ul\"unde $x$ de\u gi\c skeninin \"u\c c tane ge\c ci\c si vard\i r (ve $y$ ile $z$ de\u gi\c skenlerinin birer ge\c ci\c si vard\i r).
\begin{compactenum}
\item
B\"ol\"unemeyen bir form\"ulde bir de\u gi\c skenin her ge\c ci\c si,
\textbf{serbest} bir ge\c ci\c stir.
\item
Bir de\u gi\c skenin $\phi$ form\"ul\"undeki her serbest ge\c ci\c si,
$\lnot\phi$, $(\phi*\psi)$, ve $(\psi*\phi)$ form\"ullerinde de
serbesttir.  (Burada $*$ i\c sareti, herhangi bir iki-konumlu ba\u glay\i c\i
d\i r.) 
\item
E\u ger $x$ ile $y$, iki \emph{farkl\i} de\u gi\c sken ise, o zaman $x$ de\u
gi\c skeninin $\phi$ form\"ul\"unde her serbest ge\c ci\c si, $\Exists
y\phi$ ile $\Forall y\phi$ form\"ullerinde de serbesttir. 
\item
$\Exists x\phi$ ile $\Forall x\phi$ form\"ullerinde $x$ de\u gi\c
  skeninin hi\c c serbest ge\c ci\c si yoktur.
\end{compactenum}
Bir form\"ulde bir de\u gi\c skenin serbest ge\c ci\c si varsa, bu
de\u gi\c sken, form\"ul\"un bir \textbf{serbest de\u gi\c skenidir.}
Serbest de\u gi\c skeni olmayan bir form\"ul, bir
\textbf{c\"umledir.}\index{c\"umle}  C\"umleler i\c cin $\sigma$, $\tau$,
ve $\rho$ gibi Yunan harflerini kullanaca\u g\i z. 

\section{Do\u gruluk ve Yanl\i\c sl\i k}

Bir $\phi$ form\"ul\"un\"un tek serbest de\u gi\c skeni $x$ ise, o
zaman form\"ul 
\begin{equation*}
\phi(x)
\end{equation*}
olarak yaz\i labilir.  O halde $a$ bir sabit ise, ve $x$ de\u gi\c
skeninin $\phi$ form\"ul\"undeki her \emph{serbest} ge\c ci\c sinin yerine
$a$ konulursa, \c c\i kan c\"umle 
\begin{equation*}
\phi(a)
\end{equation*}
olarak yaz\i labilir.  \c Simdi 
\textbf{do\u grulu\u gu}%
\index{do\u gruluk}\label{truth}
(\eng{truth}) ve
\textbf{yanl\i\c sl\i\u g\i}%
\index{yanl\i\c sl\i k}
(\eng{falsehood})
tan\i mlayabiliriz: 
\begin{compactenum}
\item
E\u ger $b$ k\"umesi, $a$ k\"umesini i\c cerirse, o zaman $a\in b$
c\"umlesi do\u grudur; i\c cermezse, yanl\i\c st\i r. 
\item
E\u ger $\sigma$ c\"umlesi do\u gruysa, o zaman $\lnot\sigma$ de\u
gillemesi yanl\i\c st\i r; $\sigma$ yanl\i\c s ise, $\lnot\sigma$ do\u
grudur. 
\item
E\u ger hem $\sigma$ hem $\tau$ do\u gruysa, o zaman
$(\sigma\land\tau)$ birle\c smesi de do\u grudur; $\sigma$ ile $\tau$
c\"umlelerinin biri yanl\i\c s ise, birle\c smesi de yanl\i\c st\i r. 
\item
E\u ger bir $a$ k\"umesi i\c cin $\phi(a)$ c\"umlesi do\u gruysa, o
zaman $\Exists x\phi(x)$ \"orneklemesi de do\u grudur; hi\c c \"oyle
bir $a$ yoksa, \"ornekleme yanl\i\c st\i r. 
\item
$(\sigma\lor\tau)$ c\"umlesi, $\lnot(\lnot\sigma\land\lnot\tau)$
  c\"umlesinin anlam\i na gelir, yani bu iki c\"umle ayn\i\ zamanda ya
  do\u grudur, ya da yanl\i\c st\i r. 
\item
$(\sigma\lto\tau)$ c\"umlesi, $(\lnot\sigma\lor\tau)$ c\"umlesinin anlam\i na gelir.
\item
$(\sigma\liff\tau)$ c\"umlesi, $\bigl((\sigma\lto\tau)\land(\tau\lto\sigma)\bigr)$ c\"umlesinin anlam\i na gelir.
\item
$\Forall x\phi(x)$ c\"umlesi, $\lnot\Exists x\lnot\phi(x)$ c\"umlesinin anlam\i na gelir.
\end{compactenum}
\"Ozel olarak form\"ullerde $\lor$, $\lto$, $\liff$, ve $\forall$
simgeleri gerekmez; sadece kolayl\i k i\c cin kullanaca\u g\i z.  Ama
$(\sigma\lto\tau)$ c\"umlesi do\u grudur ancak ve ancak $\tau$ do\u
gru veya $\sigma$ yanl\i\c st\i r; ve $(\sigma\liff\tau)$ c\"umlesi
do\u grudur ancak ve ancak hem $\sigma$ hem $\tau$ ya do\u gru ya
yanl\i\c st\i r.  Ayr\i ca $\Forall x\phi(x)$ do\u grudur ancak ve
ancak her $a$ k\"umesi i\c cin $\phi(a)$ do\u grudur. 

Birka\c c tane k\i saltma daha kullan\i r\i z:
\begin{compactenum}
\item
$\lnot\; t\in u$ form\"ul\"un\"un yerine $t\notin u$ ifadesini yazar\i z;
\item
Bir $(\phi*\psi)$ form\"ul\"un\"un en d\i\c staki ayra\c clar\i
n\i\ yazmay\i z.
\item
$\lto$ ile $\liff$ ba\u glay\i c\i lar\i na g\"ore $\land$ ile
$\lor$ ba\u glay\i c\i lar\i na \"onceli\u gi veririz:  Mesela
$\phi\land\psi\lto\chi$ ifadesi, $(\phi\land\psi)\lto\chi$
form\"ul\"un\"un anlam\i na gelir.   
\item
$\phi\lto\psi\lto\chi$ ifadesi, $\phi\lto(\psi\lto\chi)$
form\"ul\"un\"un anlam\i na gelir. 
\end{compactenum}
Bir $\phi$ form\"ul\"un\"un serbest de\u gi\c skenleri $x$ ile $y$
ise, o zaman form\"ul 
\begin{equation*}
\phi(x,y)
\end{equation*}
olarak yaz\i labilir.  O halde $a$ ile $b$, iki sabit ise, ve $x$ de\u
gi\c skeninin $\phi$ form\"ul\"undeki her serbest ge\c ci\c sinin
yerine $a$ konulursa, ve benzer \c sekilde $y$ de\u gi\c skeninin her
serbest ge\c ci\c sinin yerine $b$ konulursa, \c c\i kan c\"umle
\begin{equation*}
\phi(a,b)
\end{equation*}
olarak yaz\i labilir.  

Genelde $\phi$ form\"ul\"un\"un serbest de\u gi\c skenleri, bir $\vec
x$ listesini olu\c sturursa, o zaman form\"ul 
\begin{equation*}
\phi(\vec x)
\end{equation*}
olarak yaz\i labilir; ayr\i ca
\begin{align*}
\Forall{\vec x}&\phi(\vec x),&
\Exists{\vec x}&\phi(\vec x)
\end{align*}
c\"umleleri yaz\i labilir.  E\u ger $\vec a$, uzunlu\u gun $\vec x$
listesinin uzunlu\u gu olan bir sabit listesiyse, o zaman 
\begin{equation*}
\phi(\vec a)
\end{equation*}
c\"umlesi de \c c\i kar.
E\u ger $\phi(\vec x)$ ile $\psi(\vec x)$, iki form\"ul ise, ve \emph{sadece
do\u grulu\u gun tan\i m\i n\i\ kullanarak}
\begin{equation*}
\Forall{\vec x}\bigl(\phi(\vec x)\liff\psi(\vec x)\bigr)
\end{equation*}
c\"umlesinin do\u grulu\u gu kan\i tlanabilirse, o zaman $\phi$ ile
$\psi$ birbirine 
\textbf{(mant\i\u ga g\"ore) denktir}\index{denk} (\eng{logically
  equivalent}):
k\i saca
\begin{equation*}
  \phi\denk\psi.
\end{equation*}
\"Oyleyse $\phi$ ile $\psi$ birbirine denktir, ancak
ve ancak her $\vec a$ sabit listesi i\c cin, \emph{do\u grulu\u gun tan\i
m\i na g\"ore}
\begin{equation*}
\phi(\vec a)\liff\psi(\vec a)
\end{equation*}
c\"umlesi do\u grudur.  \"Orne\u gin, yukar\i daki tan\i mlara g\"ore
\begin{gather*}
	\phi\lor\psi\denk\lnot(\lnot\phi\land\lnot\psi),\\
	\phi\lto\psi\denk\lnot\phi\lor\psi,\\
	\phi\liff\psi\denk(\phi\lto\psi)\land(\psi\lto\phi),\\
	\Forall x\phi\denk\lnot\Exists x\lnot\phi.
\end{gather*}
Ama $\Exists y\Forall x\bigl(\phi(x)\lto x\in y\bigr)$ ile
$\Exists y\Forall x\bigl(\phi(x)\liff x\in y\bigr)$, denk de\u gildir.

\begin{theorem}\label{thm:denklik}
\mbox{}
\begin{compactenum}
\item
Her form\"ul, kendisine denktir.
\item
E\u ger $\phi$ ile $\psi$ denk ise, o zaman $\psi$ ile $\phi$ denktir.
\item
E\u ger $\phi$ ile $\psi$ denk ise, ve $\psi$ ile $\chi$ denk ise, o zaman $\phi$ ile $\chi$ denktir.
\end{compactenum}
\begin{comment}
  

Yani
\begin{gather*}
	\phi\denk\phi,\\
	\phi\denk\psi\lto\psi\denk\phi,\\
	\phi\denk\psi\land\psi\denk\chi\lto\phi\denk\chi.
\end{gather*}



\end{comment}
\end{theorem}

\begin{proof}
\begin{asparaenum}
\item
$\sigma\liff\sigma$ her zaman do\u grudur.
\item
$\sigma\liff\tau$ do\u gru olsun.  O zaman hem $\sigma$ hem $\tau$ ya do\u gru ya yanl\i\c st\i r.  \"Oyleyse hem $\tau$ hem $\sigma$ ya do\u gru ya yanl\i\c st\i r; yani $\tau\liff\sigma$ do\u grudur.
\item
$\sigma\liff\tau$ ve $\tau\liff\rho$ do\u gru olsun.  E\u ger $\sigma$ do\u gruysa, o zaman $\tau$ do\u gru olmal\i, ve sonu\c c olarak $\rho$ do\u gru olmal\i, dolay\i s\i yla $\sigma\liff\rho$ do\u grudur.  Benzer \c sekilde $\sigma$ yanl\i\c s ise $\sigma\liff\rho$ tekrar do\u grudur.\qedhere
\end{asparaenum}
\end{proof}

\begin{theorem}\mbox{}\label{thm:lto}
\begin{compactenum}
\item
$\phi\lto\psi\lto\chi$ ile $\phi\land\psi\lto\chi$ denktir.
\item
E\u ger $x$ de\u gi\c skeni, $\phi$ form\"ul\"unde serbest de\u gilse, 
o zaman
\begin{equation*}
\Forall x(\phi\lto\psi)\denk\phi\lto\Forall x\psi.
\end{equation*}
\end{compactenum}
\end{theorem}

\begin{proof}
\begin{asparaenum}
\item
$\sigma\lto\tau\lto\rho$ do\u gru olsun.  E\u ger $\sigma\land\tau$
  c\"umlesi de do\u gruysa, o zaman hem $\sigma$ hem $\tau$ do\u
  grudur, ve sonu\c c olarak $\tau\lto\rho$ do\u grudur, ve $\rho$
  do\u grudur.  Yani $\sigma\land\tau\lto\rho$ do\u grudur. 

Tersi i\c cin $\sigma\land\tau\lto\rho$ do\u gru olsun.  O zaman
$\sigma\land\tau$ yanl\i\c s veya $\rho$ do\u grudur.  Yani $\sigma$
yanl\i\c s, veya $\tau$ yanl\i\c s, veya $\rho$ do\u grudur.  E\u ger
$\sigma$ do\u gruysa, o zaman $\tau$ yanl\i\c s, veya $\rho$ do\u
grudur, yani $\tau\lto\rho$ do\u grudur.  Sonu\c c olarak
$\sigma\lto\tau\lto\rho$ do\u grudur. 

\item
$\Forall x(\sigma\lto\phi(x))$ do\u gru olsun.  
O zaman her $a$ i\c cin $\sigma\lto\phi(a)$ do\u grudur.  
Sonu\c c olarak $\sigma$ do\u gruysa, 
o zaman her $a$ i\c cin $\phi(a)$ do\u gru\-dur.  
Yani $\sigma\lto\Forall x\phi(x)$ do\u grudur.

Benzer \c sekilde $\sigma\lto\Forall x\phi(x)$ do\u gruysa 
$\Forall x(\sigma\lto\phi(x))$ do\u grudur.\qedhere
\end{asparaenum}
\end{proof}

\chapter{Kof\/inallik}\label{ch:cof}

\section{Tan\i m ve \"ozellikler}

\begin{sloppypar}
  Sonsuz bir $\kappa$ kardinali limit ordinali oldu\u gundan
  \begin{equation*}
    \kappa=\sup\{\xi\colon\xi<\kappa\}=\bigcup_{\xi<\kappa}\xi.
  \end{equation*}
  Bazen bir kardinal, 
  kendisinden k\"u\c c\"uk bir altk\"umenin supremumu\-dur.  
  \"Orne\u gin $\upomega<\aleph_{\upomega}$, ama
  \begin{equation*}
    \aleph_{\upomega}=\sup\{\aleph_x\colon x\in\upomega\}.
  \end{equation*}
  Genelde $\alpha$ limit, $b\included\alpha$, ve
  \begin{equation*}
    \Forall{\xi}\bigl(\xi<\alpha\lto\Exists{\eta}(\eta\in b\land\xi<\eta)\bigr)
  \end{equation*}
  ise,
  $b$ altk\"umesi, $\alpha$ ordinalinin \textbf{s\i n\i rs\i z}%
  \index{s\i n\i rs\i z}
  (\eng{unbounded})
  altk\"umesidir.
  Bu durumda
  \begin{equation*}
    \alpha=\sup(b).
  \end{equation*}
  \"Orne\u gin her limit ordinali, kendisinde s\i n\i rs\i zd\i r.  
  Ayr\i ca $\{\aleph_x\colon x\in\upomega\}$, 
  $\aleph_{\upomega}$ ordinalinde s\i n\i rs\i zd\i r.
  Bir limit ordinalinin s\i n\i rs\i z alt\-k\"umelerinin 
  en k\"u\c c\"uk kardinaline, 
  ordinalin
  \textbf{kof\/inalli\u gi}%
  \index{kof{}inallik}
  (\eng{cofinality})
  denir, ve bu kardinal,
  $\cof{\alpha}$
  \glossary{$\cof{\alpha}$}%
  olarak yaz\i labilir.  Yani
  \begin{equation*}
    \cof{\alpha}=\min\{\cardinal(x)\colon x\included\alpha\land\sup(x)=\alpha\}.
  \end{equation*}
  Ayr\i ca, tan\i ma g\"ore,
  \begin{align*}
    \cof0&=0,&\cof{\alpha+1}&=1
  \end{align*}
  denebilir,
  ama bu durumlar\i\ kullanmayaca\u g\i z.
\end{sloppypar}

\begin{theorem}
Her $\alpha$ limit ordinali i\c cin,
tan\i m k\"umesi $\cof{\alpha}$ olan,
de\u ger k\"umesi $\alpha$ ordinalinin s\i n\i rs\i z bir altk\"umesi olan,
kesin artan bir g\"onderme vard\i r.
\end{theorem}

\begin{proof}
$f\colon\cof{\alpha}\to\alpha$ olsun, 
ve $f[\alpha]$, $\alpha$ ordinalinin s\i n\i rs\i z bir altk\"umesi olsun.  
\"Ozyinelemeyle, tan\i m k\"umesi $\cof{\alpha}$ olan,
\begin{equation*}
g(\beta)=\max\Bigl(f(\beta),\sup\bigl(g[\beta]\bigr)\Bigr)
\end{equation*}
ko\c sulunu sa\u glayan bir $g$ g\"ondermesi vard\i r.
E\u ger $\beta<\cof{\alpha}$ ve $g[\beta]\included\alpha$ ise, 
o zaman $g[\beta]$, $\alpha$ ordinalinin s\i n\i rs\i z altk\"umesi de\u gil,
dolay\i s\i yla $g(\beta)\in\alpha$; ayr\i ca $f(\beta)\leq g(\beta)$.  
\"Oyleyse $g$, istedi\u gimiz gibidir.
\end{proof}

\begin{theorem}
$\alpha$ ve $\beta$ limit ordinalleri olsun.
E\u ger $f\colon\alpha\to\beta$ ve kesin artan ise,
ve $\beta=\bigcup f[\alpha]$ ise,
o zaman
\begin{equation*}
\cof{\alpha}=\cof{\beta}.
\end{equation*}
\end{theorem}

\begin{proof}
$\cof{\beta}\leq\cof{\alpha}$ ve $\cof{\alpha}\leq\cof{\beta}$
e\c sitsizliklerini kan\i tlayaca\u g\i z.
\begin{asparaenum}
\item
$g\colon\cof{\alpha}\to\alpha$
ve $\bigcup g[\cof{\alpha}]=\alpha$ olsun.  
%$(f\circ g)[\gamma]$ g\"or\"unt\"us\"un\"un $\beta$ ordinalinde s\i n\i rs\i z oldu\u gunu kan\i tlayaca\u g\i z.  
$\delta<\beta$ ise,
hipoteze g\"ore $\alpha$ ordinalinin bir $\theta$ eleman\i\ i\c cin
\begin{equation*}
\delta<f(\theta).
\end{equation*}
O zaman $\cof{\alpha}$ kardinalinin bir $\iota$ eleman\i\ i\c cin
\begin{align*}
\theta&<g(\iota),&
\delta<f(\theta)&<f\bigl(g(\iota)\bigr).
\end{align*}
\"Oyleyse $\bigcup(f\circ g)[\cof{\alpha}]=\beta$,
dolay\i s\i yla $\cof{\beta}\leq\cof{\alpha}$.
\item
$h\colon\cof{\beta}\to\beta$ ve $\bigcup h[\cof{\beta}]=\beta$ olsun.
$\delta<\cof{\beta}$ ise
\begin{equation*}
k(\delta)=\min\{\xi\in\alpha\colon h(\delta)<f(\xi)\}
\end{equation*}
olsun.  O zaman $k\colon\cof{\beta}\to\alpha$.  
E\u ger $\theta\in\alpha$ ise, o zaman
$\cof{\beta}$ kardinalinin
\begin{equation*}
  f(\theta)<h(\delta)
\end{equation*}
ko\c sulunu sa\u glayan bir $\delta$ eleman\i\ vard\i r.
O zaman
\begin{equation*}
f(\theta)<h(\delta)<f(k(\delta)),  
\end{equation*}\sloppy
dolay\i s\i yla $\theta<k(\delta)$,
\c c\"unk\"u $f$ kesin artand\i r.
\"Oyleyse $\bigcup k[\cof{\beta}]=\alpha$,
dolay\i s\i yla
$\cof{\alpha}\leq\cof{\beta}$ 
ve asl\i nda $\cof{\alpha}=\cof{\beta}$.\qedhere
\end{asparaenum}
\end{proof}

\"Ozel durum olarak $\bm F$ normal ve $\alpha$ limit ise
\begin{equation*}
  \cof{\bm F(\alpha)}=\cof{\alpha}.
\end{equation*}

\begin{theorem}
  $\alpha$ limit ise $\cof{\aleph_{\alpha}}=\cof{\alpha}$.
\end{theorem}

\begin{proof}
$\xi\mapsto\aleph_{\xi}$ normaldir.
\end{proof}

\begin{theorem}
  Cantor normal bi\c ciminde
\begin{equation*}
\alpha=\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_n}\cdot a_n
\end{equation*}
ve $\alpha_n>0$ ise, o zaman
\begin{equation*}
\cof{\alpha}
=\begin{cases}
	\upomega,&\text{ e\u ger $\alpha_n$ bir ard\i lsa},\\
	\cof{\alpha_n},&\text{ e\u ger $\alpha_n$ bir limitse}.
\end{cases}
\end{equation*}
\end{theorem}

\begin{proof}
Son teoreme g\"ore
$\alpha$ limit, $\gamma\geq1$, ve $\delta\geq2$ ise
\begin{equation*}
  \cof{\alpha} 
=\cof{\beta+\alpha} 
=\cof{\gamma\cdot\alpha}
=\cof{\delta^{\alpha}}.\qedhere
\end{equation*}
\end{proof}


Bazen bu hesaplama bize yard\i m etmez.  Mesela $f(0)=0$ ve $f(n+1)=\upomega^{f(n)}$ ve $\alpha=\sup(f[\upomega])$ ise, yani
\begin{equation*}
\alpha =\sup\{0,1,\upomega,\upomega^{\upomega},\upomega^{\upomega^{\upomega}},\dots\}
\end{equation*}
ise, o zaman $\cof{\alpha}=\upomega$, ama $\alpha=\upomega^{\alpha}$.

\begin{theorem}\label{thm:succ-cof}
Her $\alpha$ ordinali i\c cin
\begin{equation*}
\cof{\aleph_{\alpha+1}}=\aleph_{\alpha+1}.  
\end{equation*}
\end{theorem}

\begin{proof}
$\beta<\aleph_{\alpha+1}$ ve $f\colon\beta\to\aleph_{\alpha+1}$ olsun.  O zaman
\begin{equation*}
\sup(f[\beta])=\bigcup_{\xi<\beta}f(\xi).
\end{equation*}
Bu bile\c simden $\aleph_{\alpha}\times\aleph_{\alpha}$ \c carp\i m\i na giden bir $h$ g\"ommesini tan\i mlayaca\u g\i z.
Se\c cim Aksiyomu sayesinde $\bigcup\{{}^{\xi}\aleph_{\alpha}\colon\xi<\aleph_{\alpha+1}\}$ k\"umesi iyi\-s\i ralanabilir.  Bu s\i ralamaya g\"ore $\delta<\aleph_{\alpha+1}$ ise  ${}^\delta\aleph_{\alpha}$ k\"umesinin en k\"u\c c\"uk \emph{g\"ommesi,} $g_{\delta}$ olsun.  O zaman $\gamma<\sup(f[\beta])$ ise
\begin{align*}
\delta&=\min\{z\in\beta\colon\gamma<f(z)\},& h(\gamma)&=\bigl(g_{\beta}(\delta),g_{\delta}(\gamma)\bigr)
\end{align*}
olsun.
B\"oylece
\begin{equation*}
\cardinal\bigl(\sup(f[\beta])\bigr)\leq\cardinal(\aleph_{\alpha}\times\aleph_{\alpha})=\aleph_{\alpha},
\end{equation*}
dolay\i s\i yla $\sup(f[\beta])<\aleph_{\alpha+1}$.  Sonu\c c olarak $\cof{\aleph_{\alpha+1}}=\aleph_{\alpha+1}$.
\end{proof}

\section{Hesaplamalar}

\begin{theorem}
$2\leq\kappa$, $1\leq\lambda$, ve $\aleph_0\leq\max\{\kappa,\lambda\}$ olsun.  O zaman
\begin{align*}
	\lambda\geq\cof{\kappa}&\lto\kappa<\kappa^{\lambda},\\
	\gch\land\lambda<\cof{\kappa}&\lto\kappa=\kappa^{\lambda}.
\end{align*}
\end{theorem}

\begin{proof}
$\cof{\kappa}\leq\lambda$ ise ${}^{\lambda}\kappa$ k\"umesinin
\begin{equation*}
\kappa=\bigcup_{\xi<\lambda}f(\xi)
\end{equation*}
ko\c sulunu sa\u glayan bir $f$ eleman\i\ vard\i r.  
\c Simdi $\xi\mapsto g_{\xi}\colon\kappa\to{}^{\lambda}\kappa$ olsun.  
O zaman ${}^{\lambda}\kappa$ k\"umesinin 
$\{g_{\xi}\colon\xi<\kappa\}$ k\"umesinde olmayan bir
\begin{equation*}
\eta\mapsto\min\Bigl(\kappa\setminus\bigl\{g_{\xi}(\eta)\colon\xi<f(\eta)\bigr\}\Bigr)
\end{equation*}
eleman\i\ vard\i r.  

\c Simdi $\lambda<\cof{\kappa}$ olsun.  
O zaman \Teoremin{thm:succ-cof} kan\i t\i ndaki gibi
\begin{multline*}
{}^{\lambda}\kappa
=\bigcup_{\xi<\kappa}{}^{\lambda}\xi
=\bigcup_{\lambda\leq\xi<\kappa}{}^{\lambda}\xi\\
\preccurlyeq\bigcup_{\lambda\leq\xi<\kappa}{}^{\lambda}(\card{\xi})
=\bigcup_{\substack{\lambda\leq\xi<\kappa\\\xi\in\cn}}{}^{\lambda}\xi
\preccurlyeq\bigcup_{\substack{\lambda\leq\xi<\kappa\\\xi\in\cn}}{}^{\xi}2.
\end{multline*}
E\u ger $\gch$ do\u gruysa $\mu<\kappa\lto2^{\mu}\leq\kappa$, dolay\i s\i yla $\kappa^{\lambda}\leq\kappa$.
\end{proof}

\c Simdi, g\"osterdiklerimize g\"ore, e\u ger $\kappa+\lambda$
sonsuzsa, o zaman
\begin{align*}
2\leq\kappa\leq2^{\lambda}&\lto\kappa^{\lambda}=2^{\lambda},\\
\cof{\kappa}\leq\lambda\leq\kappa&\lto\kappa<\kappa^{\lambda}\leq2^{\kappa},\\
1\leq\lambda<\cof{\kappa}&\lto\kappa\leq\kappa^{\lambda}\leq2^{\kappa}.
\end{align*}
Ayr\i ca%, e\u ger $\gch$ do\u gruysa,
\begin{equation*}
\gch\lto
\kappa^{\lambda}=
\begin{cases}
  \lambda^+,&\text{ e\u ger $2\leq\kappa<\lambda$ ise},\\
\kappa^+,&\text{ e\u ger $\cof{\kappa}\leq\lambda\leq\kappa$ ise},\\
\kappa,&\text{ e\u ger $1\leq\lambda<\cof{\kappa}$ ise}.
\end{cases}
\end{equation*}
\"Ozel olarak% e\u ger $\gch$ do\u gruysa,
\begin{equation*}
\gch\lto
{\aleph_{\alpha}}^{\aleph_{\beta}}=
\begin{cases}
	\aleph_{\beta+1},&\text{ e\u ger $\alpha<\beta$ ise},\\
	\aleph_{\alpha+1},&\text{ e\u ger $\cof{\alpha}\leq\aleph_{\beta}\leq\aleph_{\alpha}$ ise},\\
	\aleph_{\alpha},&\text{ e\u ger $\aleph_{\beta}<\cof{\alpha}$ ise}.	
\end{cases}
\end{equation*}

\end{document}
\chapter{Old figures}

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  \caption{$\eta=\upomega+\xi$ denkleminin grafi\u gi}\label{fig:y=omega+x}
  
\end{figure}

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  \caption{$\eta=\xi+\upomega$ denkleminin grafi\u gi}\label{fig:y=x+omega}  
\end{figure}


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  \caption{$\eta=\xi+\upomega$ denkleminin grafi\u gi}%\label{fig:x+w}  
\end{figure}


\begin{figure}
  \centering
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  \caption{$\eta=\upomega\cdot\xi$ denkleminin grafi\u gi}\label{fig:y=omega.x}
  
\end{figure}

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\psline(0,1.25)(-0.03,1.25)
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\psline(! 0 2 0.75 10 exp sub)(! -0.03 2 0.75 10 exp sub)
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\psline(! 0 2 0.75 12 exp sub)(! -0.03 2 0.75 12 exp sub)
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%%%%%%%%%%%%%%%%%%%%%%%%%
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\psline(1,0)(1,-0.03)
\psline(1.25,0)(1.25,-0.03)
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\psline(! 2 0.75 16 exp sub 0)(! 2 0.75 16 exp sub -0.03)
\psline(2,0)(2,-0.03)
%%%%%%%%%%%%%%%%%%%%%%
\psset{linestyle=dotted}
\psline(1,0)(1,3.1)
\psline(2,0)(2,3.1)
\psline(0,1)(2.1,1)
\psline(0,2)(2.1,2)
\psline(0,3)(2.1,3)
\psdot[dotstyle=o](1,1)
\psdot[dotstyle=o](2,2)
  \end{pspicture}

  \caption{$\eta=\xi\cdot\upomega$ denkleminin grafi\u gi}\label{fig:y=x.omega}  
\end{figure}

%\end{document}


