Analysis lecture notes

David Pierce

October 2, 2002

I am preparing these notes while teaching a class in mathematical analysis (METU Math 271, year 2002/3, fall semester). The current table of contents is at the end. Mistakes and inconsistencies are possible here (as in all human productions). I say some things about ℝ which are true—and are later needed— for arbitrary (ordered) fields.

Our main purpose is to understand ℝ.

1 Fields: a definition

A commutative (or abelian) group is a structure

(G, ∗,ˆ , e)

—where ∗ is a binary, ˆ a unary, and e a nullary operation—satisfying:

• ∀x ∀y x ∗ y = y ∗ x [commutativity],

• ∀x ∀y ∀z x ∗ (y ∗ z) = x ∗ y ∗ x [associativity],

• ∀x e ∗ x = x [identity element],

• ∀x x ∗ x̂ = e [inverses].

A field is a structure

(F, +, −, ·, 0, 1)

where F \ {0} is also equipped with ^{−1} such that

• (F, +, −, 0) and (F \ {0}, ·,^{−1}, 1) are commutative groups, and

• ∀x ∀y ∀z x · (y · z) = x · y + x · z [distributivity].

Exercise 1. Prove that a field satisfies:

∀x ∀y (xy = 0 → x = 0 ∨ y = 0) [no zero-divisors].

The set ℝ of real numbers is a field (when equipped with the usual operations).

2 Functions into a field

We shall be interested in functions from arbitrary sets into ℝ. Say S is a set, and f, g : S → ℝ are functions. Then we can use the field-operations to form new functions:

• f + g : x ↦ f(x) + g(x),

• &c.

Exercise 2. Let A be the set of all functions from S to ℝ. Which properties of fields does A have? For example, does it have no zero-divisors?

3 Polynomial and rational functions

Let X be a variable; we define

ℝ[X]

to be the set of polynomials in X with coefficients in ℝ. So a typical element of ℝ[X] is

a_{0} + a_{1}X + a_{2}X^{2} + ⋯ + a_{n}X^{n},

that is,

∑_{k=0}^{n} a_{k}X^{k},

where a_{k} ∈ ℝ. Any member of ℝ[X] determines a function from ℝ to itself. In particular, X is the identity-function on ℝ. Then ℝ[X] is the smallest set of functions from ℝ to itself that contains the identity-function and is closed under the field-operations induced from ℝ.

Note that the identity-function is not the multiplicative identity 1 or the additive identity 0. Also, in this definition, multiplicative inversion is not a fieldoperation.

Say f ∈ ℝ[X]. What can we say about the set

{x ∈ ℝ : f(x) = 0}?

It is a consequence of the field-axioms that this set is finite (and is no bigger than the degree of f). Let 1/f be the function x ↦ f(x)^{−1}. Then 1/f is defined at all but finitely many points of ℝ.

The function 1/f is the multiplicative inverse of f (not the functional inverse, f^{−1}). We can form the function g/f whenever g ∈ ℝ[X]. The set of all such functions is

ℝ(X),

the field of rational functions in X over ℝ.

4 Subfields

A field (F, +, −, ·, 0, 1) is a sub-field of a field (G, ⊕, ⊖, ⊙, 0′, 1′) if F ⊆ G, and

• x ⊕ y = x + y for all x and y in F,

• &c.

Then ℚ is a sub-field of ℝ; in fact, ℚ is the smallest sub-field of ℝ. Also, ℝ is a sub-field of ℂ; but ℤ/pℤ is not a sub-field of ℚ (or ℝ, or ℂ).

Each element a of ℝ can be considered as the function

x ↦ a : ℝ → ℝ.

In this way, ℝ is a sub-field of ℝ(X). Also, ℝ(X) is the smallest field that contains the identity-function and includes ℝ as a sub-field.

5 Ordered fields

An ordered field is a field with a binary relation ⩽ satisfying:

• ∀x x ⩽ x,

• ∀x ∀y (x ⩽ y ∧ y ⩽ x → x = y),

• ∀x ∀y ∀z (x ⩽ y ∧ y ⩽ z → x ⩽ z),

• ∀x ∀y (x ⩽ y ∨ y ⩽ x),

• ∀x ∀y ∀z (x ⩽ y → x + z ⩽ y + z),

• ∀x ∀y ∀z (x ⩽ y ∧ 0 ⩽ z → xz ⩽ yz).

Then ℝ is an ordered field.

In an alternative definition, a field is ordered if it has a subset P that is closed under + and · such that the whole field is the disjoint union

{x : −x ∈ P } ⊔ {0} ⊔ P.

Then we write x < y just in case y − x ∈ P. The set P is the set of positive elements of the field.

Exercise 3. Prove the equivalence of the definitions.

6 Convexity and bounds

A set A in space is called convex if, for all points P and Q in A, the line segment joining P and Q lies within A.

A subset A of an ordered field K is convex if K satisfies

∀x ∀y ∀z (x ∈ A ∧ y ∈ A ∧ x ⩽ z ⩽ y → z ∈ A).

An upper bound for A is an element b of K such that

∀x (x ∈ A → x ⩽ b).

Likewise, lower bound. A set is bounded if it has an upper and a lower bound.

Exercise 4. An ordered field has exactly one convex subset that has no upper or lower bound. What is it?

An ordered field has intervals of nine kinds:

\begin{matrix}(−∞, b) & (−∞, b] & (−∞, ∞) \\ (a, b) & (a, b] & (a, ∞) \\ [a, b) & [a, b] & [a, ∞) \end{matrix}

where a < b. Here (−∞, b) = {x : x < b}, &c; a square bracket means the corresponding end-point is contained in the interval.

Exercise 5. Every interval is convex.

7 Completeness

An ordered field is complete if every subset with an upper bound has a least upper bound, also called a supremum. If it exists, the supremum of A is denoted

sup A.

We postulate that ℝ is complete.

An infimum is a greatest lower bound. The infimum of A, if it exists, is denoted

inf A.

Exercise 6. In ℝ, every convex set with at least two members is an interval. (This is not true in ℚ.)

Exercise 7. Here, P is the set of positive elements of ℝ.

(1) If ab ∈ P, what can you conclude about a and b?

(2) Write P as an interval.

(3) Prove that P is closed under x ↦ x^{−1}.

(4) Let P [X] be the set of [non-zero] polynomials in X whose [non-zero] coefficients are from P. If f ∈ P [X], show that P is closed under x ↦ f(x).

(5) If f ∈ ℝ[X], and P is closed under x ↦ f(x), can you conclude that f ∈ P [X]?

Exercise 8. Prove that, in a complete ordered field, all sets with lower bounds have infima.

8 Triangle inequalities

On any ordered field, we define the absolute-value function

x ↦ |x|

by the rule

|x| = \begin{cases}x, & if x ⩾ 0; \\ −x, & if x < 0. \end{cases}

Then

∀x ∀y (|x| ⩽ y ↔ −y ⩽ x ⩽ y).

Hence the triangle-inequality:

|x + y| ⩽ |x| + |y|

and the variants

||x| − |y|| ⩽ |x − y| ⩽ |x| + |y| .

Exercise 9. Prove the triangle-inequality (and variants).

9 The natural numbers

The set ℕ of natural numbers can be defined as the smallest subset of ℝ that contains 0 and that contains x + 1 whenever it contains x. (Any intersection of subsets with this property continues to have this property; so ℕ is the intersection of all subsets of ℝ with this property.)

Therefore, proof by induction works in ℕ: if A ⊆ ℕ, and 0 ∈ A, and x + 1 ∈ A whenever x ∈ A, then A = ℕ.

Exercise 10. Prove that ℕ is well-ordered, that is, every non-empty subset of ℕ has a least element.

The definition of ℕ works for any ordered field. (Why not a non-ordered field?)

10 Archimedean ordered fields

An ordered field is called Archimedean if ℕ has no upper bound in the field.

Theorem. ℝ is Archimedean.

Proof. Suppose ℕ has an upper bound. Then ℕ has a supremum, say a. Then a − 1 is not an upper bound of ℕ, so some n in ℕ satisfies

a − 1 < n.

Hence a < n + 1, but n + 1 ∈ ℕ, contradicting that a is an upper bound for ℕ.

Exercise 11. Make ℝ(X) into a non-Archimedean ordered field such that

a ⩽ X

for all a in ℝ.

11 The integers

We can define the set ℤ of integers to be the set

{x ∈ ℝ : −x ∈ ℕ} ∪ ℕ.

Note that 0 is in both sets. The set ℤ is closed under + and − and ·.

Because ℝ is Archimedean, we can define the greatest-integer function

x ↦ ⌊x⌋

by the rule

⌊x⌋ = min{n ∈ ℤ : x < n + 1}.

Then ⌊x⌋ ⩽ x < ⌊x⌋ + 1.

12 The rational numbers

The set ℚ of rational numbers is

{x ∈ ℝ : xy ∈ ℤ for some positive integer y}.

This is a field. Rational numbers exist in any ordered field.

Lemma. In an Archimedean ordered field, if

0 ⩽ x ⩽ r

for all positive rational numbers r, then x = 0.

Proof. Under the assumption, we have 0 ⩽ 1/n for all positive integers n. If x ≠ 0, then 1/x is an upper bound for ℕ.

13 Open and closed sets

A subset A of ℝ is called open if, for every x in A, there is a positive real number ε such that

(x − ε, x + ε) ⊆ A.

A subset of ℝ is closed if its complement is open.

Exercise 12. The intervals (−∞, ∞), (−∞, b), (a, b) and (a, ∞) are open. The intervals (−∞, ∞), (−∞, b], [a, b] and [a, ∞) are closed. The set {a} is closed.

Lemma. If A is a closed subset of ℝ with an upper bound, then

sup A ∈ A.

Proof. Suppose b is an upper bound of A, but b ∉ A. Since A is closed, the set ℝ \ A is open. Therefore

(b − ε, b + ε) ⊆ ℝ \ A

for some positive ε. Hence b − ε is an upper bound for A. (Why?

A ⊆ (−∞, b],

but also

A ⊆ ℝ \ (b − ε, b + ε),

so

A ⊆ (−∞, b] \ (b − ε, b + ε) = (−∞, b − ε],

which means b − ε is an upper bound for A.) Thus b is not sup A. That is, no upper bound of A that is not in A is sup A; but sup A is an upper bound of A; therefore sup A is in A.

14 Sequences of sets

A sequence is a function on ℕ. A function n ↦ a_{n} : ℕ → S may also be written (a_{n} : n ∈ ℕ) or (a_{n})_{n} or just (a_{n}).

Theorem. Suppose (F_{n} : n ∈ ℕ) is a sequence of non-empty bounded closed subsets of ℝ such that

F_{n+1} ⊆ F_{n}

for all n in ℕ. Then

⋂_{n∈ℕ} F_{n} ≠ ∅.

Proof. Since F_{n} is bounded and empty, sup F_{n} exists; and sup F_{n} ∈ F_{n} by the Lemma, since F_{n} is closed. Then the set

{sup F_{n} : n ∈ ℕ}

is bounded below by any lower bound of F_{0}. Hence this set has an infimum, say L. I claim that L ∈ F_{n} for each n in ℕ. To prove this, since F_{n} is closed, it is enough to show that

F_{n} ∩ (L − ε, L + ε) ≠ ∅

whenever ε > 0. But by definition of L, for each positive ε, there is m in ℕ such that

L ⩽ sup F_{m} < L + ε.

Let k = max{m, n}; then

L ⩽ sup F_{k} ⩽ sup F_{m} < L + ε

and sup F_{k} ∈ F_{k} ⊆ F_{n}. Therefore sup F_{k} ∈ F_{n} ∩ (L − ε, L + ε).

The sets F_{n} of the Theorem can be called nested.

Example. Let F_{n} = {x ∈ ℝ : |x^{2} − 2| ⩽ 1/n}. Then each F_{n} is closed, and F_{n+1} ⊆ F_{n}, so

⋂_{n∈ℕ} F_{n} ≠ ∅.

In fact, the intersection is {±√2}.

Exercise 13. The Theorem puts four conditions on the sequence of nested sets F_{n}: each F_{n} must

(1) be non-empty,

(2) be closed,

(3) have an upper bound, and

(4) have a lower bound.

Show that each of these conditions is necessary (that is, if any one of them is removed, then the theorem cannot be proved).

Note that the theorem is true in any complete ordered field. The next theorem gives an alternative definition for complete.

Theorem. Suppose K is an Archimedean ordered field, and suppose that

⋂_{n∈ℕ}[a_{n}, b_{n}] ≠ ∅

whenever ([a_{n}, b_{n}] : n ∈ ℕ) is a sequence of nested closed intervals of K. Then K is a complete ordered field.

Proof. Let C be a subset of K with element a and upper bound b. We shall recursively define a nested sequence ([a_{n}, b_{n}] : n ∈ ℕ) in the following way. First,

[a_{0}, b_{0}] = [a, b].

Then [a_{0}, b_{0}] contains a member of C and an upper bound of C. Suppose [a_{k}, b_{k}] has been defined and contains a point of C and an upper bound of C. Then one of the intervals

[a_{k}, \frac{a_{k} + b_{k}}{2} ] , [\frac{a_{k} + b_{k}}{2} , b_{k}]

has the same property (why?). Let [a_{k+1}, b_{k+1}] be this interval. If both intervals have this property, then we can just define

[a_{k+1}, b_{k+1}] = [a_{k}, \frac{a_{k} + b_{k}}{2} ] .

By assumption, there is some c in ⋂_{n∈ℕ}[a_{n}, b_{n}]. Then c is an upper bound of C (why?), and no upper bound of C is less than c (why?). Therefore c = sup C.

Exercise 14. Supply the missing details in the proof of the last theorem.

15 Different treatments of the real numbers

We have treated ℝ axiomatically. That is, we have written down

• axioms for fields,

• additional axioms for ordered fields, and finally

• the completeness axiom for fields.

Then we have declared that ℝ is a complete ordered field.

The alternative approach is constructive. Here, one starts with ℕ—as given by axioms, namely

• 0 ≠ n + 1 for any n in ℕ,

• ∀x ∀y (x + 1 = y + 1 → x = y), and

• for all subsets A of ℕ, if 0 ∈ A, and n + 1 ∈ A whenever n ∈ A, then A = ℕ.

Then, using ℕ, one defines ℤ and the Archimedean ordered field ℚ. Finally, it is possible to define an ordered field-structure on the set of all convex open subsets of ℚ that have upper but not lower bounds. The result is a complete ordered field, and ℝ can be defined to be this field.

Again, we are taking the axiomatic approach to ℝ, and not the constructive approach. But we shall see presently why the constructive approach works.

A subset A of an ordered field K is called dense in K if every interval of K contains an element of A.

Lemma. ℚ is dense in ℝ.

Proof. It is enough to show that if a, b ∈ ℝ and a < b, then a < c < b for some c in ℚ. But since a < b, we have 0 < b − a. From an earlier lemma, since ℝ is Archimedean, we conclude that some positive rational number r satisfies r < b − a. We may assume (why?) that r = 2/n for some natural number n. Then nb − na > 2. Therefore

na < ⌊nb⌋ − 1 < ⌊nb⌋ ⩽ nb

(why?). Finally, a < (⌊nb⌋ − 1)/n < b, and (⌊nb⌋ − 1)/n ∈ ℚ.

Theorem. There is a one-to-one correspondence between:

• real numbers, and

• convex open subsets of ℚ that are bounded above, but not below.

Proof. The correspondence is given by the function

x ↦ {r ∈ ℚ : r < x}

(why?).

Exercise 15. Supply the details in the proof of the last theorem.

(If A is a convex subset of ℚ with an upper but not a lower bound, then the pair (A, ℚ \ A) is called a cut. It was Dedekind who first defined the real numbers in terms of cuts.)

16 Topology

Topology is the study of open and closed sets as such.

Theorem. ℝ and ∅ are open subsets of ℝ. If A and B are open subsets of ℝ, then so is A ∩ B. If {U_{i} : i ∈ I} is a family of open subsets of ℝ, then the union

⋃_{i∈I} U_{i}

is open.

Proof. It is trivial that ℝ and ∅ are open. If A and B are open, and x ∈ A ∩ B, then x ∈ A and x ∈ B, so there are positive reals ε_{A} and ε_{B} such that

(x − ε_{S}, x + ε_{S}) ⊆ S

when S ∈ {A, B}. Let ε = min{ε_{A}, ε_{B}}; then

(x − ε, x + ε) ⊆ A ∩ B.

Thus A ∩ B is open. Finally, if U_{i} is open for each i in the index-set I, and x ∈ ⋃_{i∈I} U_{i}, then x ∈ U_{i} for some i in I, so (x − ε, x + ε) ⊆ U_{i}, and therefore

(x − ε, x + ε) ⊆ ⋃_{i∈I} U_{i}.

Therefore ⋃_{i∈I} U_{i} is open.

Exercise 16. Prove that:

• ℝ and ∅ are closed subsets of ℝ; 10 David Pierce, October 2, 2002

• if A and B are closed subsets of ℝ, then so is A ∪ B;

• if {F_{i} : i ∈ I} is a family of open subsets of ℝ, then the union

⋂_{i∈I} F_{i}

is open. An arbitrary intersection of open sets may not be open:

Example. ⋂_{n∈ℕ} (−\frac{1}{n + 1}, \frac{1}{n + 1}) = \{0\}.

Exercise 17. Show that the only subsets of ℝ that are both open and closed are ∅ and ℝ itself. (The exercise shows that ℝ is connected.)

17 The Cantor set

By definition, a non-empty open subset of ℝ includes an open interval. Does a non-empty closed set include a closed interval? No. We can obtain a closed set that has no subsets that are intervals in the following way. Start with a closed interval. Remove an open interval, so two closed intervals remain. Remove an open interval from each of these, and continue. In the end, no interval can remain. More precisely, we can proceed thus. If a and b are integers, we write

a ≡ b (mod 3)

if a − b is a multiple of 3. For n in ℕ, let U_{n} be the set

{x ∈ ℝ : ⌊3^{n}x⌋ < 3^{n}x ∧ ⌊3^{n}x⌋ ≡ 1 (mod 3)}.

Then let F be the set

[0, 1] \ ⋃_{n∈ℕ} U_{n}.

This set F is called the Cantor set. Exercise 18. The Cantor set is closed, but none of its subsets is an interval.

Contents

1 Fields: a definition 1 2 Functions into a field 1 3 Polynomial and rational functions 2 4 Subfields 2 5 Ordered fields 3 6 Convexity and bounds 3 Analysis notes 11

7 Completeness 4 8 Triangle inequalities 4 9 The natural numbers 5 10 Archimedean ordered fields 5 11 The integers 5 12 The rational numbers 6 13 Open and closed sets 6 14 Sequences of sets 6 15 Different treatments of the real numbers 8 16 Topology 9 17 The Cantor set 10