NoteNextra-origin/content/Math4201/Math4201_L27.md

Math4201 Topology I (Lecture 27)

Continue on compact spaces

Compact spaces

Heine-Borel theorem

A subset K\subseteq \mathbb{R}^n is compact if and only if it is closed and bounded with respect to the standard metric on \mathbb{R}^n.

Definition of bounded

A\subseteq \mathbb{R}^n is bounded if there exists c\in \mathbb{R}^{>0} such that d(x,y)<c for all x,y\in A.

Proof for Heine-Borel theorem

Suppose k\subseteq \mathbb{R}^n is compact.

Since \mathbb{R}^n is Hausdorff, K\subseteq \mathbb{R}^n is compact, so K is closed subspace of \mathbb{R}^n. by Proposition of compact subspaces with Hausdorff property.

To show that K is bounded, consider the open cover with the following balls:

B_1(0), B_2(0), ..., B_n(0), ...

Since K is compact, there are n_1, ..., n_k\in \mathbb{N} such that K\subseteq \bigcup_{i=1}^k B_{n_i}(0). Note that B_{n_i}(0) is bounded, so K is bounded. \forall x,y\in B_{n_i}(0), d(x,y)<2n_i. So K is bounded.

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Suppose K\subseteq \mathbb{R}^n is closed and bounded.

First let M=[a_1,b_1]\times [a_2,b_2]\times \cdots \times [a_n,b_n].

This is compact because it is a product of compact spaces.

Since K is bounded, we can find $[a_i,b_i]$s such that K\subseteq M.

Since K is closed subspace of \mathbb{R}^n, K is closed in M.

Since any closed subspace of a compact space is compact, K is compact.

Warning

This theorem is not true for general topological spaces.

For example, take X=B_1(0) with the standard topology on \mathbb{R}^n.

Take K=B_1(0), this is not compact because it is not closed in \mathbb{R}^n.

Extreme Value Theorem

If f:X\to \mathbb{R} is continuous map with X being compact. Then f attains its minimum and maximum.

Proof

Let M=\sup\{f(x)\mid x\in X\} and m=\inf\{f(x)\mid x\in X\}.

We want to show that there are x_m,x_M\in X such that f(x_m)=m and f(x_M)=M.

Consider the open covering of X given as

\{U_\alpha\coloneqq f^{-1}((-\infty, \alpha))\}_{\alpha\in \mathbb{R}}

If X doesn't attain its maximum, then this is an open covering of X:

1. U_\alpha is open because f is continuous and (-\infty, \alpha) is open in \mathbb{R}.
2. \bigcup_{\alpha\in \mathbb{R}} U_\alpha = X because for any x\in X, by the assumption there is x'\in X with f(x)<f(x') (otherwise f(x) is the maximum value). Then x\in U_{f(x')}.

So there is an open covering of X and hence it's got a finite subcover \{U_{\alpha_i}\}_{i=1}^n.

X=\bigcup_{i=1}^n U_{\alpha_i}=\bigcup_{i=1}^n f^{-1}((-\infty, \alpha_i))=f^{-1}(-\infty, \alpha_k)

and \alpha_1\leq \alpha_2\leq \cdots \leq \alpha_n. There is x_i such that \alpha_i=f(x_i).

Note that x_k\notin U_{\alpha_k} because f(x_k)>\alpha_k. So x_k\notin X. This contradicts the assumption that X doesn't attain its maximum.

Theorem of uniform continuity

Let f:(X,d)\to (X',d') be a continuous map between two metric spaces. Let X be compact, then for any \epsilon > 0, there exists \delta > 0 such that for any x_1,x_2\in X, if d(x_1,x_2)<\delta, then d'(f(x_1),f(x_2))<\epsilon.

Definition of uniform continuous function

f is uniformly continuous if for any \epsilon > 0, there exists \delta > 0 such that for any x_1,x_2\in X, if d(x_1,x_2)<\delta, then d'(f(x_1),f(x_2))<\epsilon.

Example of uniform continuous function

Let f(x)=x^2 on \mathbb{R}.

This is not uniformly continuous because for fixed \epsilon > 0, the interval \delta will converge to zero as x_1,x_2 goes to infinity.

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However, if we take f\mid_{[0,1]}, this is uniformly continuous because for fixed \epsilon > 0, we can choose \delta = \epsilon.

Lebesgue number lemma

Let X be a compact metric space and \{U_\alpha\}_{\alpha\in I} be an open cover of X. Then there is \delta>0 such that for any two points x_1,x_2\in X with d(x_1,x_2)<\delta, there is \alpha\in I such that x_1,x_2\in U_\alpha.

Proof of uniform continuity theorem

Let \epsilon > 0. be given and consider

\{f^{-1}(B_{\epsilon/2}^{d'}((x'))\}_{x'\in X'}

We claim that there is an open covering of X.

1. f^{-1}(B_{\epsilon/2}^{d'}((x'))) is open because f is continuous and B_{\epsilon/2}^{d'}((x')) is open in X'.
2. X=\bigcup_{x'\in X'} f^{-1}(B_{\epsilon/2}^{d'}((x'))) because for any x\in X, x\in f^{-1}(B_{\epsilon/2}^{d'}((f(x))).

Since X is compact, there is a finite subcover \{f^{-1}(B_{\epsilon/2}^{d'}((x')))\}_{i=1}^n.

By Lebesgue number lemma, there is \delta>0 such that for any two points x_1,x_2\in X with d(x_1,x_2)<\delta, there is x'\in X' such that x_1,x_2\in f^{-1}(B_{\epsilon/2}^{d'}((x'))).

So f(x_1),f(x_2)\in B_{\epsilon/2}^{d'}((x')).

Apply the triangle inequality with d'(x_1,x') and d'(x_2,x'), we have d'(f(x_1),f(x_2))<2\epsilon/2=\epsilon.