Math4201 Topology I (Lecture 26)

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 $\bigcup_{i=1}^k U_{x_i}\times W\subseteq N$.

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 $\bigcup_{i=1}^k X\times W_{y_i}=X\times Y$.

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 $y$. $y<c\in C$.

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.

Continue next time.