\section{Metric spaces}

\news[Euclidean spaces]

If $n\in\N$, we can define $\R^n$ to be the set of all \defn{ordered
  $n$-tuples}
$$(a_0,\dots,a_{n-1}),$$
where $a_i\in\R$ in each case.  Otherwise, $\R^n$ can be described as
the Cartesian product
$$\underbrace{\R\times\dots\times\R}_n.$$
The $n$-tuple $(a_0,\dots,a_{n-1})$ can be abbreviated
$\tuple a$; it might be called a \defn{point} of $\R^n$, or a
\defn{vector} in $\R^n$.  Then the structure  
$$(\R^n,+,-,\tuple 0)$$ 
is a commutative group, where:
\begin{itemize}
  \item
$\tuple a+\tuple b=(a_0+b_0,\dots,a_{n-1}+b_{n-1})$;
\item
$-\tuple a=(-a_0,\dots,-a_{n-1})$;
\item
$\tuple 0=(0,\dots,0)$.
\end{itemize}
(Note that $\R^1=\R$, and $\R^0=\{\emptyset\}$.)
Each vector $\tuple a$ in $\R^n$ has a \defn{norm} or \defn{length},
namely
$$\radix\sum_{i<n}a_i^2.$$
Since a sum of squares of real numbers is always non-negative, the
norm is well-defined.  The norm of $\tuple a$ is denoted
$$\norm{\tuple a}.$$
The \defn{distance} from point $\tuple a$ to point $\tuple b$ is the
norm
$$\norm{\tuple b-\tuple a}.$$
This terminology is reasonable, since:
\begin{itemize}
  \item
the distance from $\tuple a$ to $\tuple b$ is the same as the
distance from $\tuple b$ to $\tuple a$ (so we can speak of the
distance \emph{between} two points);
\item
the distance between \emph{distinct} points is positive, while the distance
between a point and itself is $0$;
\item
the distance from $\tuple a$ to $\tuple b$ is no greater than the
the distance from $\tuple a$ to $\tuple c$ plus the distance
from $\tuple c$ to $\tuple b$.
\end{itemize}
The first two of these facts are clear from the definition.  To prove
the last fact, we first define the \defn{dot-product} on $\R^n$ by the
rule
$$\tuple a\cdot\tuple b=\frac{\norm{\tuple a+\tuple b}^2 - \norm{\tuple
    a}^2 - \norm{\tuple b}^2}2.$$
Also, any $r$ in $\R$ determines an operation $\tuple x\mapsto r\tuple
x:\R^n\to\R^n$, where
$$r\tuple x=(rx_0,\dots,rx_{n-1}).$$
In this context, elements of $\R$ are \defn{scalars}, and $r\tuple x$
can be called a \defn{scalar multiple} of $\tuple x$.

\begin{lemma*}
Suppose $\tuple x,\tuple y\in\R^n$ and $r\in \R$.  Then:
\begin{enumerate}
  \item  
  $\norm{r\tuple x}=\abs r\norm{\tuple x}$;
\item
$\tuple x\cdot\tuple y=\sum_{i<n}x_iy_i$;
\item
$(r\tuple x)\cdot \tuple y=r(\tuple x\cdot\tuple y)$.
\end{enumerate}
\end{lemma*}

\begin{exercise}
  Prove the lemma.
\end{exercise}

\begin{lemma*}[Cauchy--Bunyakovski--Schwartz inequality]
  For all $\tuple a$ and $\tuple b$ in $\R^n$ we have
$$\abs{\tuple a\cdot\tuple b}\leq\norm{\tuple a}\norm{\tuple b},$$
with equality holding just in case one of $\tuple a$ and $\tuple b$ is
a scalar multiple of the other.
\end{lemma*}

\begin{proof}
The equation 
\begin{equation}\label{eqn:vectors}
\norm{x\tuple a+\tuple b}^2=0
\end{equation}
 has at most one solution
  $x$ (why?).
  For all scalars $x$ we have
  \begin{equation*}
\norm{x\tuple a+\tuple b}^2=   x^2\norm{\tuple a}^2 + 2x\tuple
a\cdot\tuple b + \norm{\tuple b}^2. 
  \end{equation*}
The right member of this equation is a quadratic polynomial in $x$; its
discriminant, $D$, is not positive, so
$$(\tuple a\cdot\tuple b)^2\leq
\norm{\tuple a}^2\norm{\tuple b}^2$$
and therefore
$$\abs{\tuple a\cdot\tuple b}\leq
\norm{\tuple a}\norm{\tuple b}.$$
If equation (\ref{eqn:vectors}) does have a solution, then $D=0$, so
the last inequality is an equality; if (\ref{eqn:vectors}) has no
solution, then the inequality is strict.
\end{proof}

\begin{exercise}
  Supply the missing detail(s) in the proof.
\end{exercise}

\begin{theorem*}[triangle inequality]
$\norm{\tuple a+\tuple b}\leq\norm{\tuple a}+\norm{\tuple b}$
for all vectors $\tuple a$ and $\tuple b$ of $\R^n$.
\end{theorem*}

\begin{proof}
By the C--B--S inequality, we get
$$\norm{\tuple a+\tuple b}^2= 
   \norm{\tuple a}^2 + 2\tuple a\cdot\tuple b + \norm{\tuple b}^2
\leq\norm{\tuple a}^2+ 2\norm{\tuple a}\norm{\tuple b} +\norm{\tuple
   b}^2 =(\norm{\tuple a}+\norm{\tuple b})^2,$$ 
hence the claim.
\end{proof}

In the theorem, if we simultaneously replace $\tuple a$ with $\tuple
a-\tuple c$, and $\tuple b$ with $\tuple c-\tuple b$, we get
$$\norm{\tuple a-\tuple b}\leq\norm{\tuple a-\tuple c}+\norm{\tuple
  c-\tuple b}.$$
So we have a good notion of distance in $\R^n$.  We may refer to
  $\R^n$ as \defn{Euclidean $n$-space}, because of the following.

\begin{exercise}
Two vectors $\tuple a$ and $\tuple b$ are called \defn{orthogonal} if
$\tuple a\cdot\tuple b=0$; in this case we may write
$$\tuple a\perp\tuple b.$$ 
Prove the `Pythagorean Theorem' for $\R^n$, namely: 
$$\tuple a\perp\tuple b\iff \norm{\tuple a+\tuple b}^2=\norm{\tuple
  a}^2+\norm{\tuple b}^2.$$
\end{exercise}

\news[The Euclidean topology]

The \defn{open ball} at a point $\tuple a$ of $\R^n$ with radius $\epsilon$
in $(0,\infty)$ is the set
$$\{\tuple x\in\R^n:\norm{\tuple x-\tuple a}<\epsilon\},$$
which may be denoted
$$B(\tuple a;\epsilon).$$
\begin{example*}
If $a\in\R$, then $B(a,\epsilon)=(a-\epsilon,a+\epsilon)$.
\end{example*}

Suppose $\tuple a\in E\included \R^n$.  Then $E$ is called a
\defn{neighborhood} of $\tuple a$ if
$$B(\tuple a;\epsilon)\included E$$
for some positive $\epsilon$.  In this case, $\tuple a$ is
called an \defn{interior point} of $E$.  A subset of $\R^n$ is
\defn{open} if it is a neighborhood of each of its points, that is,
each of its points is an interior point.  The 
complement of an open set is \defn{closed}.

Our terminology is not ambiguous:

\begin{lemma*}
  Open balls are open.
\end{lemma*}

\begin{proof}
  Suppose $\tuple x\in B(\tuple a;\epsilon)$.  Let
  $\epsilon'=\epsilon-\norm{\tuple x-\tuple a}$; this is positive.  If
  $\tuple u\in B(\tuple x;\epsilon')$, then
$$\norm{\tuple u-\tuple a}\leq\norm{\tuple u-\tuple x}+\norm{\tuple
    x-\tuple a}\leq\epsilon'+\norm{\tuple
    x-\tuple a}=\epsilon$$
by the triangle inequality, so $\tuple u\in B(\tuple a;\epsilon)$.
Thus $B(\tuple x;\epsilon')\included B(\tuple
    a;\epsilon)$.  Therefore an open ball is a neighborhood of all of its
    points. 
\end{proof}

Our earlier general statements about open and closed sets in $\R$ are
true in $\R^n$:

\begin{theorem*}
  In $\R^n$, both $\emptyset$ and $\R^n$ are open; finite
  intersections of open subsets are open; arbitrary unions of open
  sets are open.
\end{theorem*}

\begin{exercise}
  Prove the theorem.
\end{exercise}

A point $\tuple a$ is a \defn{cluster point} (or \defn{accumulation
  point}) of a set $E$ if every neighborhood of $\tuple a$ contains a 
  point of $E\setminus\{\tuple a\}$.

  \begin{example*}
    The set $\{1/(n+1):n\in\N\}$ has exactly one cluster point, $0$. 
  \end{example*}

  \begin{lemma*}
    A subset of $\R^n$ is closed if and only if it contains all of its
    cluster points. 
  \end{lemma*}

  \begin{proof}
Let $A$ be a subset of $\R^n$.  The following statements are
equivalent:
\begin{itemize}
  \item
The set $A$ is closed. 
\item
The complement $\R^n\setminus
    A$ is a neighborhood of each of its points.
\item
No point of the complement of $A$ is a cluster point of $A$.
\item
Every cluster point of $A$ is in $A$.
\end{itemize}
This proves the claim.
  \end{proof}

\news[Bolzano--Weierstra\ss\ Theorem]

An \defn{open interval} in $\R^n$ is a Cartesian product
$$I_0\times\dots\times I_{n-1},$$
where the $I_i$ are open intervals of $\R$.  If each $I_i$ is
$(a_i,b_i)$, then the product $I_0\times\dots\times I_{n-1}$ might be
denoted
$$(\tuple a,\tuple b).$$  
\defn{Closed  intervals} of $\R^n$ are defined similarly.

\begin{lemma*}
  Open intervals in $\R$ are open sets; closed intervals are closed.
  Every open set is a union of open balls.  Every open set is a union
  of open intervals.
\end{lemma*}

\begin{exercise}
  Prove the lemma.
\end{exercise}

\begin{lemma*}
  For each $i$ less than $n$, suppose $(I^{(i)}_m:m\in\N)$ is a
  sequence of bounded closed intervals in $\R$.  Then
$$\bigcap_{m\in\N}I_m^{(0)}\times\dots\times
  I_m^{(n-1)}\neq\emptyset.$$ 
\end{lemma*}

\begin{proof}
  Each intersection $\bigcap_{m\in\N}I_m^{(i)}$ contains an element
  $a_i$ by an earlier theorem; hence the intersection of the Cartesian
  products contains $(a_0,\dots,a_{n-1})$.
\end{proof}

A set is \defn{bounded} if it is included in some open ball.

\begin{theorem*}[Bolzano--Weierstra\ss]
  Every bounded infinite subset of $\R^n$ has an cluster point.
\end{theorem*}

\begin{proof}
  Suppose $A$ is an infinite subset of $B(\tuple a;r)$.  Let
  $\tuple b$ be the tuple
$$(a_0-r,\dots,a_{n-1}-r),$$
and let $\tuple r=(r,\dots,r)$.
Then $A\included[\tuple b,\tuple b+2\tuple r]$.

Let us also write $I_0$ for the interval $[\tuple b,\tuple b+2\tuple
  r]$.  Then $I_0$ is the product of closed intervals in $\R$ of equal
  length.  By halving these intervals of $\R$, we can subdivide $I_0$
  into $2^n$ intervals, thus: 
$$I_0=\bigcup\{I^{(e_0,\dots,e_{n-1})}:e_0,\dots,e_{n-1}\in\{0,1\}\},$$
where
$$I^{(e_0,\dots,e_{n-1})}=[b_0+e_0r,b_0+(e_0+1)r]\times\dots
\times[b_{n-1}+e_{n-1}r, b_{n-1}+ (e_{n-1}+1)r].$$
One of these must contain infinitely many elements of $A$.  Call this
interval $I_1$, and continue the process.

We get a sequence $(I_k:k\in\N)$ of closed bounded intervals of
$\R^n$, each containing infinitely many points of $A$.  Also $I_k$ is
the product of intervals of length $2r/2^k$.  The intersection of the
intervals $I_k$ contains a point $\tuple c$ by the last lemma.

The point $\tuple c$ is a cluster point of $A$.  Indeed, if $\epsilon>0$, let
$k$ be large enough that $2r\radix n/2^k<\epsilon$.  (Why is this
possible?)  Then 
$$I_k\included B(\tuple c;\epsilon)$$
(why?), so the latter interval contains a point of $A$. 
\end{proof}

\begin{exercise}
  Supply the details.
\end{exercise}

\news[Cantor Intersection Theorem]

\begin{lemma*}
  If $\tuple a$ is a cluster point of $E$, then every neighborhood of
  $\tuple a$ contains infinitely many points of $E$.
\end{lemma*}

\begin{proof}
  Suppose $U$ is a neighborhood of $\tuple a$, and
$$U\cap (E\setminus\{\tuple a\})=\{\tuple b_0,\dots,\tuple b_{n-1}\}.$$
Let $\epsilon=\min\{\norm{\tuple b_0-\tuple a},\dots,\norm{\tuple
  b_{n-1}-\tuple a}\}$.  Then $U\cap B(\tuple a;\epsilon)$ is a
neighborhood of $\tuple a$ that contains no points of $E$ distinct
from $\tuple a$; so $\tuple a$ is not a cluster point of $E$.
\end{proof}

An application of the Bolzano--Weierstra\ss\ theorem is:

\begin{theorem*}[Cantor Intersection]
  Suppose $(F_n:n\in\N)$ is a sequence of bounded non-empty closed
  subsets of $\R^n$ such that 
$$F_{n+1}\included F_n$$ 
in each case.  Then $\bigcap_{n\in\N}F_n\neq\emptyset$.
\end{theorem*}

\begin{proof}
  If some $F_n$ is finite, then  $\bigcap_{n\in\N}F_n\neq\emptyset$
  (why?).  So suppose that each $F_n$ is infinite.  Then there is a
  sequence $(\tuple x_n:n\in\N)$ of distinct points such that $\tuple
  x_n\in F_n$ for each $n$.  (If $\tuple x_n$ have been chosen for all
  $n$ less than $k$, then choose $\tuple x_k$ from
  $F_k\setminus\{\tuple x_n:n<k\}$.)  As a bounded infinite set,
  $\{\tuple x_n:n\in \N\}$ has a cluster point, $\tuple x$.  Also,
  each $F_k$ contains all but finitely many points of $\{\tuple
  x_n:n\in \N\}$.  Hence, by 
  the Lemma, each neighborhood of $\tuple x$ contains infinitely many
  points of $F_k$, so $\tuple x\in F_k$ since $F_k$ is closed.
  Therefore $\tuple x\in\bigcap_{n\in\N}F_n$.
\end{proof}

\news[Compactness]

An \defn{open covering} of a subset $E$ of $\R^n$ is a set
$\{U_i:i\in I\}$ of open subsets of $\R^n$ such that
$$E\included\bigcup_{i\in I}U_i.$$
A \defn{sub-cover} of an open covering of $E$ is a subset of the
covering that is also 
an open covering of $E$.  The set $E$ is called \defn{compact} if
every open covering of $E$ has a finite sub-cover.

It is generally straightforward to show that a set is \emph{not} compact:

\begin{example*}
  Let $F$ be the set
$$\{(1/(n+1),2):n\in\N\}$$
of intervals in $\R$.  Then $F$ is an open covering of $(0,1]$, and
  any infinite subset of $F$ is a sub-cover; but no finite subset is a
  sub-cover.  Therefore $(0,1]$ is not compact.  The set $F$ does not
    cover $[0,1]$.  The set 
  $F\cup\{(-\epsilon,\epsilon)\}$ does cover $[0,1]$, and so does the
  finite subset
$$\{(-\epsilon,\epsilon),(1/(n+1),2)\},$$
provided $n+1>1/\epsilon$.
\end{example*}

\begin{lemma*}
  All compact subsets of $\R^n$ are closed and bounded.
\end{lemma*}

\begin{proof}
  If $A$ is an unbounded subset of $\R^n$, then $\{B(\tuple
  0;k+1):k\in\N\}$ is an open covering of $A$ with no finite
  sub-cover.  If $C$ is a non-closed subset of $\R^n$, then $C$ has a
  cluster point $\tuple b$ that is not in $C$.  Hence $\{\{\tuple
  x\in\R^n: \norm{\tuple x-\tuple b}\geq 1/(n+1)\}:n\in\N\}$ is an
  open covering of $C$ with no finite sub-cover.
\end{proof}

\begin{theorem*}[Heine--Borel]
  Closed and bounded subsets of $\R^n$ are compact.
\end{theorem*}

\begin{proof}
  If possible, say $E$ is a closed bounded subset of $\R^n$, and $F$
  is an infinite open
  covering of $E$ with \emph{no} finite sub-cover.  As in the proof of
  the Bolzano--Weierstra\ss\ Theorem, $E$ is included in a closed
  bounded interval $I_0$ that is a product of intervals of length
  $2r$ for some positive $r$.
  Subdivide $I_0$ into $2^n$ sub-intervals, each a product of
  intervals of length $r$.  Then $F$ covers the intersection of $A$
  with each of these, so for at least
  one of them---say $I_1$---the intersection $A\cap I_1$ is not
  covered by a finite subset of $F$.  Now sub-divide $I_1$ in the same
  way, and continue.  We get a nested sequence $(I_k:k\in\N)$ of
  closed bounded intervals.  Each intersection $A\cap I_k$ is
  non-empty and closed, so there is a point $\tuple b$ in
  $A\cap\bigcap_{k\in \N}I_k$ by the Cantor Intersection Theorem.

  Now, $\tuple b\in U$ for some $U$ in $F$.  Since $U$ is open, we
  have $B(\tuple b;\epsilon)\included U$ for some positive $\epsilon$.
  Hence $I_k\included U$ if $k$ is large enough.  Then $\{U\}$ is a
  finite covering of $A\cap I_k$, contrary to our choice of $I_k$.
  Therefore, if $E$ is closed and bounded, it must be compact.  
\end{proof}

A collection $\{F_i:i\in I\}$ of subsets of $\R^n$ has the
\defn{finite intersection property} (or \defn{f.i.p.}) if 
$$\bigcap_{i\in I_0}F_i\neq\emptyset$$
whenever $I_0$ is a finite subset of $I$.
\begin{theorem*}
  Let $A$ be a subset of $\R^n$.  The following statements are
  equivalent.
  \begin{enumerate}
    \item
$A$ is compact.
\item
$A$ is closed and bounded.
\item
Every collection of closed subsets of $A$ with the finite intersection
property has non-empty intersection.
\item
Every infinite subset of $A$ has a cluster point in $A$.
  \end{enumerate}
\end{theorem*}

\begin{exercise}
  Prove the theorem.
\end{exercise}

\news[Metric spaces]

Some of our definitions are made, and some of our theorems are true,
in a more general setting.  A set $M$ is called a \defn{metric space}
if it is equipped with a function $d:M\times M\to [0,\infty)$ such
  that, for all $a$, $b$ and $c$ in $M$, we have:
\begin{enumerate}
  \item
$d(a,b)=d(b,a)$;
\item
$d(a,b)=0\iff a=b$;
\item
$d(a,c)\leq d(a,b)+d(b,c)$.
\end{enumerate}
The function $d$ is then called a \defn{metric}, and $d(a,b)$ is the
\defn{distance} between $a$ and $b$ (with respect to $d$).  

More than one metric can be defined on the same set:

\begin{exercise}
  Let $d:\R^2\times \R^2\to[0,\infty)$ be the function defined by the
  rule 
$$d((x,y),(x',y'))=\abs{x-x'}+\abs{y-y'}.$$
Prove that $d$ is a metric on $\R^2$.  (This is called the
\tech{taxi-cab metric} by people who are thinking of cities like New
York, whose streets form a rectangular grid.)
\end{exercise}

Then \tech{open balls}, and
therefore \tech{neighborhoods}, \tech{interior points}, \tech{open}
and \tech{closed sets} and \tech{cluster points}, can be defined just
as before. 

\begin{theorem*}
  Let $(M,d)$ be a metric space.  Then $\emptyset$ and $M$ are open;
  intersections of finitely many open sets are open; and unions of
  arbitrarily many open sets are open.  A subset of $M$ is closed if
  and only if it contains all of its cluster points.
\end{theorem*}

\begin{exercise}
  Prove the theorem.
\end{exercise}

The definition of a \tech{bounded} set also makes sense in an arbitrary
metric space.  However,
the proof of the Bolzano--Weierstra\ss\ Theorem requires \emph{more}
than that $\R^n$ is a metric space:

\begin{exercise}
  $(\Q,d)$ is obviously a metric space when $d$ is the usual Euclidean
  distance ($d(a,b)=\abs{a-b}$).  Show that $\{x\in\Q:x^2<2\}$ is a
  bounded infinite subset of $\Q$ that does not have a cluster point.
\end{exercise}

The definition of a \tech{compact} set makes sense in any metric
space.  
We can prove this much:

\begin{theorem*}
  A metric space is compact if and only if every
  infinite subset has a cluster point.
\end{theorem*}

\begin{proof}
{}  [I didn't do this in class.]  Let $(M,d)$ be a metric space.
  Suppose first that $M$ has an
  infinite subset $A$ with no cluster point.  Then every element $x$
  of $M$ has an open neighborhood $U_x$ that contains only finitely
  many points of $A$.  Therefore $\{U_x:x\in A\}$ is an open covering
  of $M$, but the union of any finite subset of this contains only
  finitely many points of $A$, and this is not $M$.  Therefore $(M,d)$
  is not compact.

Suppose conversely that every infinite subset of $A$ does have a
cluster point.  Then for each positive $\epsilon$ there is a finite
set $\{x_0,\dots,x_{n-1}\}$ such that
$A=\bigcup_{i<n}B(x_i;\epsilon)$.  (Otherwise, for some positive
$\epsilon$, an infinite set can be constructed such that
$d(x,y)\geq\epsilon$ for all distinct members $x$ and $y$; this set
has no cluster point.)

Suppose $\mathcal E$ is a collection of open subsets of $A$ such that
no finite subset of $\mathcal E$ covers $A$.  We shall show
$\bigcup\mathcal E\neq A$.

Now, $A$ is covered by finitely many balls $B(x;1)$; so one of
these---say $B(a_0;1)$---is not covered by any finite subset of
$\mathcal E$.  By recursion, we get a sequence $(a_n:n\in\N)$ of
elements of $M$ such that
$$a_{n+1}\in B(a_n;2^{-n}),$$
and $B(a_n;2^{-n})$ is not covered by any finite subset of $\mathcal
E$.  If the set $\{a_n:n\in\N\}$ is infinite, then it has a cluster point
$a$.  (If the set is finite, let $a$ be such that $a=a_n$ for infinitely
many values of $n$.)  If $U$ is an open neighborhood of $a$, then
$a\in B(a;\epsilon)\included U$ for some positive $\epsilon$.  Let $N$
be large enough that $2^{-N}\leq\epsilon/2$.  Now,
$B(a;\epsilon/2)$ contains infinitely many points of $\{a_n:n\in\N\}$, if
this set is infinite; in any case, for some $n$ in $\N$, we have
$n\geq N$ and $B(a_n;2^{-n})\included U$.  Therefore $U\notin\mathcal
E$.  That is, $\mathcal E$ contains no neighborhood of $a$.  Therefore
$a\notin\bigcup\mathcal E$.  
\end{proof}

\news[Supplement]

A subset of a metric space is \defn{dense} (in that space) if every
open set contains a point of the subset.  Recall that $\Q$ is dense in
$\R$. 

\begin{exercise}
  Show that $\Q^n$ is dense in $\R^n$.
\end{exercise}

\begin{exercise}
  Let $\tuple x\in\R^n$, and let $U$ be a neighborhood of $\tuple x$.
  Then there is a point $\tuple u$ of $\Q^n$, and there is a positive
  rational number $r$, such that $\tuple x\in B(\tuple u;r)\included
  U$. 
\end{exercise}

A consequence is:

\begin{theorem*}[Lindel\"of Covering]
  If $A\included \R^n$. and $\{U_i:i\in I\}$ is an open covering of
  $A$, then the index-set $I$ has a countable subset $J$ such that
  $\{U_i:i\in I\}$ covers $A$.
\end{theorem*}

\begin{exercise}
  Suppose $S\included\R^n$, and for every $\tuple x$ in $S$ there is a
  positive $\delta$ such that $S\cap B(\tuple x;\delta)$ is a
  countable set.  Prove that $S$ is countable.
\end{exercise}

\begin{exercise}
  Suppose $A\included\R^n$.  A point $\tuple x$ of $\R^n$ is called a
  \defn{condensation point} of $A$ if $A\cap B(\tuple x;\epsilon)$ is
  uncountable for every positive $\epsilon$.  Prove that if $A$ is
  uncountable, then $A$ has a condensation point.  (You can use the
  preceeding exercise, or you can mimic the proof of the
  Bolzano--Weierstra\ss\ Theorem.)
\end{exercise}