Basic topology

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Chapter 2: Basic topology Contents
Contents
1.  Finite, countable, and uncountable sets
2.  Metric spaces
3.  Compact sets
4.  Perfect sets
5.  Connected sets

1. Finite, countable, and uncountable sets

Theorem 1. Every infinite subset of a countable set

A

is countable.

The theorem shows that, roughly speaking, countable sets represent the ``smallest" infinity: No uncountable set can be a subset of a countable set.

Theorem 2. The union of a sequence of countable sets is countable.

Theorem 3. Let

A

be a countable set, and let

B_n

be the set of all

n

-tuples

(a_1, \dots , a_n)

, where

a_k\in A\, (k = 1, \dots , n)

, and the elements

a_1, \dots , a_n

need not be distinct. Then

B_n

is countable.

Corollary 4.

\mathbb{Q}

is countable.

In fact, even the set of all algebraic numbers is countable.

Theorem 5. Let

A

be the set of all sequences whose elements are the digits

0

and

1

. This set

A

is uncountable.

2. Metric spaces

Definition 6. A set

X

, whose elements we shall call points, is said to be a metric space if with any two points

p

and

q

X

there is associated a real number

d(p, q)

, called the distance from

p

q

, such that

$d(p,q)>0$ if $p\neq q$ ; $d(p,p)=0$ ;
$d(p,q)=d(q,p)$ ;
$d(p,q)\leqslant d(p,r)+d(r,q)$ , for any $r\in X$ .

Any function with these three properties is called a distance function, or a metric.

Definition 7.

E

is open if every point of

E

is an interior point of

E

E

is closed if every limit point of

E

is a point of

E

$E$ is perfect if $E$ is closed and if every point of $E$ is a limit point of $E$ .

Theorem 8. Every neighborhood is an open set.

Theorem 9. If

p

is a limit point of a set

E

, then every neighborhood of

p

contains infinitely many points of

E

Corollary 10. A finite point set has no limit points.

Theorem 11. Let

\{E_\alpha\}

be a (finite or infinite) collection of sets

E_\alpha

. Then

(\bigcup_\alpha E_\alpha)^c=\bigcap_\alpha E_\alpha^c

Theorem 12. A set

E

is open if and only if its complement is closed.

Corollary 13. A set

F

is closed if and only if its complement is open.

Theorem 14. For any collection

\{G_\alpha\}

of open sets,

\bigcup _ \alpha G_\alpha

is open. For any finite collection

G_1,\dots,G_n

of open sets,

\bigcap_{i=1}^n G_i

is open.

For any collection $\{F_\alpha\}$ of closed sets, $\bigcap _ \alpha F_\alpha$ is closed. For any finite collection $F_1,\dots,F_n$ of closed sets, $\bigcup_{i=1}^n F_i$ is closed.

Theorem 15. If

X

is a metric space and

E \subset X

, then

$\bar{E}$ is closed,
$E=\bar{E}$ if and only if $E$ is closed,
$\bar{E} \subset F$ for every closed set $F \subset X$ such that $E \subset F$ .

By (1) and (3), $E$ is the smallest closed subset of $X$ that contains $E$ .

Theorem 16. Let

E

be a nonempty set of real numbers which is bounded above. Let

y = \sup E

. Then

y \in \bar E

. Hence

y \in E

E

is closed.

Suppose $E \subset Y \subset X$ , where $X$ is a metric space. To say that $E$ is an open subset of $X$ means that to each point $p \in E$ there is associated a positive number $r$ such that the conditions $d(p, q)<r$ , $q \in X$ imply that $q \in E$ . But we have already observed that $Y$ is also a metric space, so that our definitions may equally well be made within $Y$ . To be quite explicit, let us say that $E$ is open relative to $Y$ if to each $p \in E$ there is associated an $r > 0$ such that $q \in E$ whenever $d(p, q)<r$ and $q \in Y$ . A set may be open relative to $Y$ without being an open subset of $X$ . However, there is a simple relation between these concepts, which we now state.

Theorem 17. Suppose

Y \subset X

. A subset

E

Y

is open relative to

Y

if and only if

E = Y \cap G

for some open subset

G

X

3. Compact sets

Definition 18. A subset

K

of a metric space

X

is said to be {compact} if every open cover of

K

contains a finite subcover.

The notion of compactness is of great importance in analysis, especially in connection with continuity.

It is clear that every finite set is compact. The existence of a large class of infinite compact sets in $\mathbb R^k$ will follow from Theorem 2..

We observe that if $E \subset Y \subset X$ , then $E$ may be open relative to $Y$ without being open relative to $X$ . The property of being open thus depends on the space in which $E$ is embedded.

Compactness, however, behaves better, as we shall now see. To formulate the next theorem, let us say, temporarily, that $K$ is compact relative to $X$ if the requirements of Definition def23 are met.

Theorem 19. Suppose

K \subset Y \subset X

. Then

K

is compact relative to

X

if and only if

K

is compact relative to

Y

By virtue of this theorem we are able, in many situations, to regard compact sets as metric spaces in their own right, without paying any attention to any embedding space. In particular, although it makes little sense to talk of open spaces, or of closed spaces (every metric space $X$ is an open subset of itself, and is a closed subset of itself), it does make sense to talk of compact metric spaces.

Theorem 20. Compact subsets of metric spaces are closed.

Theorem 21. Closed subsets of compact sets are compact.

Corollary 22. If

F

is closed and

K

is compact, then

F \cap K

is compact.

Theorem 23. If

\{K_\alpha\}

is a collection of compact subsets of a metric space

X

such that the intersection of every finite subcollection of

\{K_\alpha\}

is nonempty, then

\bigcap K_\alpha

is nonempty.

Corollary 24. If

\{K_n\}

is a sequence of nonempty compact sets such that

K_n\supset K_{n+1}\, (n = 1, 2, 3,\dots)

, then

\bigcap_{n=1}^\infty K_n

is not empty.

Theorem 25. If

E

is an infinite subset of a compact set

K

, then

E

has a limit point in

K

Theorem 26. If

\{I_n\}

is a sequence of intervals in

\mathbb R^1

such that

I_n\supset I_{n+1}\, (n = 1, 2, 3,\dots)

, then

\bigcap_{n=1}^\infty I_n

is not empty.

Theorem 27. Let

k

be a positive integer. If

\{I_n\}

is a sequence of

k

-cells such that

I_n\supset I_{n+1}\, (n = 1, 2, 3,\dots)

, then

\bigcap_{n=1}^\infty I_n

is not empty.

Theorem 28. Every

k

-cell is compact.

The equivalence of (1) and (2) in the next theorem is known as the Heine-Borel theorem.

Theorem 29. If a set

E

\mathbb R^k

has one of the following three properties, then it has the other two.

$E$ is closed and bounded.
$E$ is compact.
Every infinite subset of $E$ has a limit point in $E$ .

We should remark, at this point, that (2) and (3) are equivalent in any metric space but that (1) does not, in general, imply (2) and (3).

Theorem 30 (Weierstrass). Every bounded infinite subset of

\mathbb R^k

has a limit point in

\mathbb R^k

Proof. Theorem theorem233. Suppose $F \subset K \subset X$ , $F$ is closed (relative to $X$ ), and $K$ is compact. Let $\{V_\alpha\}$ be an open cover of $F$ . If $F^c$ is adjoined to $\{V_\alpha\}$ , we obtain an open cover $\Omega$ of $K$ . Since $K$ is compact, there is a finite subcollection $\Phi$ of $\Omega$ which covers $K$ , and hence $F$ . If $F^c$ is a member of $\Phi$ , we may remove it from $\Phi$ and still retain an open cover of $F$ . We have thus shown that a finite subcollection of $\{V_\alpha\}$ covers $F$ .

Theorem theorem234. Fix a member $K_1$ of $\{K_\alpha\}$ and put $G_\alpha=K_\alpha^c$ . Assume that no point of $K_1$ belongs to every $K_\alpha$ . Then the sets $G_\alpha$ form an open cover of $K_1$ ; and since $K_1$ is compact, there are finitely many indices $\alpha_1,\dots,\alpha_n$ such that $K_1 \subset G_{\alpha_1}\cup\dots\cup G_{\alpha_n}$ . But this means that $K_1\cap K_{\alpha_1}\cap\dots\cap K_{\alpha_n}$ is empty, in contradiction to our hypothesis.

Theorem theorem235. If no point of $K$ were a limit point of $E$ , then each $q \in K$ would have a neighborhood $V_q$ which contains at most one point of $E$ (namely, $q$ , if $q \in E$ ). It is clear that no finite subcollection of $\{V_q\}$ can cover $E$ ; and the same is true of $K$ , since $E \subset K$ . This contradicts the compactness of $K$ .

Theorem theorem239. If (1) holds, then $E \subset I$ for some $k$ -cell $I$ , and (2) follows from Theorems theorem238 and theorem233. Theorem theorem235 shows that (2) implies (3). It remains to be shown that (3) implies (1). If $E$ is not bounded, then $E$ contains points $\mathbf{x}_n$ with $|\mathbf{x}_n|>n$ . $\{\mathbf{x}_n\}$ is infinite and clearly has no limit point in $\mathbb{R}^k$ , hence has none in $E$ . Thus (3) implies that $E$ is bounded. If $E$ is not closed, then there is a point $\mathbf{x}_0 \in \mathbb{R}^k$ which is a limit point of $E$ but not a point of $E$ . For $n = 1, 2, 3, \dots$ , there are points $\mathbf{x}_n \in E$ such that $|\mathbf{x}_n-\mathbf{x}_0|<1/n$ . Let $S$ be the set of these points $\mathbf{x}_n$ . Then $S$ is infinite, $S$ has $\mathbf{x}_0$ as a limit point, and $S$ has no other limit point in $\mathbb{R}^k$ . For if $\mathbf{y}\in\mathbb{R}^k,\, \mathbf{y}\neq\mathbf{x}_0$ , then $|\mathbf{x}_n-\mathbf{y}|\geqslant|\mathbf{x}_0-\mathbf{y}|-|\mathbf{x}_n-\mathbf{x}_0|\geqslant|\mathbf{x}_0-\mathbf{y}|-1/n \geqslant\frac{1}{2}|\mathbf{x}_0-\mathbf{y}|$ for large $n$ ; this shows that $\mathbf{y}$ is not a limit point of $S$ . Thus $S$ has no limit point in $E$ , in contradiction to (3); hence $E$ must be closed if (3) holds.

Theorem theorem2310. Being bounded, the set $E$ is a subset of a $k$ -cell $I \subset \mathbb{R}^k$ . By Theorem theorem238, $I$ is compact, and so $E$ has a limit point in $I$ , by Theorem theorem235.

4. Perfect sets

Theorem 31. Let

P

be a nonempty perfect set in

\mathbb R^k

. Then

P

is uncountable.

Corollary 32. Every interval

[a, b] (a<b)

is uncountable. In particular, the set of all real numbers is uncountable.

Proof. Since $P$ has limit points, $P$ must be infinite. Suppose $P$ is countable, and denote the points of $P$ by $\mathbf{x}_1 , \mathbf{x}_2 , \mathbf{x}_3 , \dots$ . We shall construct a sequence $\{V_n\}$ of neighborhoods, as follows.

Let $V_1$ be any neighborhood of $\mathbf{x} _ 1$ . If $V_1$ consists of all $\mathbf{y}\in\mathbb{R}^k$ such that $|\mathbf{y}-\mathbf{x} _ 1|<r$ , the closure $\bar V_1$ of $V_1$ is the set of all $\mathbf{y} \in \mathbb{R}^k$ such that $|\mathbf{y}-\mathbf{x} _ 1| \leqslant r$ . Suppose $V_n$ has been constructed, so that $V_n \cap P$ is not empty. Since every point of $P$ is a limit point of $P$ , there is a neighbor hood $V_{n + 1}$ such that (i) $\bar V_{n+1}\subset V_n$ , (ii) $\mathbf{x} _ n\notin\bar V_{n+1}$ , (iii) $V_{n + 1}\cap P$ is not empty. By (iii), $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed.

Put $K_n = \bar V_{n}\cap P$ . Since $\bar V_n$ is closed and bounded, $\bar V_n$ is compact. Since $\mathbf{x} _ n \notin K_{n+1}$ , no point of $P$ lies in $\bigcap_{n=1}^\infty K_n$ . Since $K_n \subset P$ , this implies that $\bigcap_{n=1}^\infty K_n$ is empty. But each $K_n$ is nonempty, by (iii), and $K_n \supset K_{n+1}$ , by (i); this contradicts the Corollary tuilun232.

{The Cantor set} One of the most interesting properties of the Cantor set is that it provides us with an example of an uncountable set of measure zero.
5. Connected sets

Definition 33. Two subsets

A

and

B

of a metric space

X

are said to be separated if both

A \cap \bar B

and

\bar A \cap B

are empty, i.e., if no point of

A

lies in the closure of

B

and no point of

B

lies in the closure of

A

A set $E \subset X$ is said to be connected if $E$ is not a union of two nonempty separated sets.

The connected subsets of the line have a particularly simple structure:

Theorem 34. A subset

E

of the real line

\mathbb{R}^1

is connected if and only if it has the following property: If

x \in E, y \in E

, and

x<z<y

, then

z \in E

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Basic topology

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