4.1 KiB
Math4201 Topology I (Lecture 26)
Continue on compact spaces
Compact spaces
Tube lemma
Let X be a compact topological space and y be a topological space. Let N\subseteq X\times Y be an open set contains X\times \{y_0\} for y_0\in Y. Then there exists an open set W\subseteq Y is open containing y_0 such that N contains X\times W.
Proof
For any x\in X, there are open sets x\in U_x\subseteq X and y_0\in V_x\subseteq Y such that U_x\times V_x\subseteq N.
In particular, \{U_x\}_{x\in X} is an open cover of X. Since \forall x\in X, x\in U_x, so there exists a finite subcover \bigcup_{i=1}^k U_{x_i}=X.
Take W=\bigcap_{i=1}^k V_{x_i}. This is intersection of finitely many open sets, so it is open.
y_0\in V_{x_i} for all x_i\in X, so y_0\in W.
So U_{x_i}\times W\subseteq U_{x_i}\times V_{x_i}\subseteq N for all i=1, ..., k.
So \bigcup_{i=1}^k U_{x_i}\times W\subseteq N.
Product of compact space is compact
Let X and Y be compact spaces, then X\times Y is compact.
Proof
Let \{N_\alpha\}_{\alpha\in I} be an open covering of X\times Y.
For any y_0\in Y, X\times \{y_0\} is a compact subspace of X\times Y.
So there are finitely many $N_\alpha$'s whose union M_{y_0} is open containing X\times \{y_0\}.
Using the tube lemma, M_{y_0} contains X\times W_{y_0} for some open neighborhood W_{y_0}\subseteq Y of y_0.
Now note that \{W_{y_0}\}_{y_0\in Y} is an open cover of Y.
In particular, there are finitely many y_1, ..., y_n\in Y such that Y=\bigcup_{i=1}^n W_{y_i} by compactness of Y.
So \forall i, X\times W_{y_i}\subseteq M_{y_i}.
This is a union of finitely many $N_\alpha$'s, so it is open.
This implies that taking the union of all such $N_\alpha$'s, for all 1\leq i\leq k, which is finite, covers \bigcup_{i=1}^k X\times W_{y_i}=X\times Y.
Closed intervals in real numbers are compact
[a,b] is compact in \mathbb{R}.
Proof
Let \{U_\alpha\}_{\alpha\in I} be an open cover of [a,b].
Define:
C=\{c\in [a,b]\mid [a,c]\text{ is covered by finitely many } U_\alpha's\}
Our goal is to show that b\in C.
Clearly a is covered by one U_\alpha, so [a,a]\in C.
Take y=\sup C\in [a,b].
Since y\in [a,b], there is U_\alpha such that y\in U_\alpha.
Since U_\alpha is open, there exists an open interval (y-\epsilon, y+\epsilon)\subseteq U_\alpha. So there is some z\in C such that z\in (y-\epsilon, y+\epsilon). Otherwise y-\epsilon is an upper bound of C, which contradicts the definition of y.
So [a,z] can be covered by finitely many $U_\alpha$'s. U_{\alpha_1}, ..., U_{\alpha_k} and \bigcup_{i=1}^k U_{\alpha_i}\supseteq [a,z].
So \bigcup_{i=1}^k U_{\alpha_i}\supseteq [a,z]\cup (y-\epsilon, y+\epsilon)=[a,y+\epsilon].
If y\neq b, then there is an element c\in [a,b] that belongs to (y,y+\epsilon) and [a,c]\subseteq [a,y+\epsilon) can be covered by finitely many $U_\alpha$'s, so c\in C. This contradicts the definition of y. y<c\in C.
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.
Suppose K\subseteq \mathbb{R}^n is closed and bounded.
Continue next time.