1
$\begingroup$

Problem. Let $(X,d_X)$ and $(Y,d_Y)$ be two compact metric spaces (see the definition here). Then show that the product metric space $(X\times Y,d_{X\times Y})$ is also compact.

Now this can be done easily using sequences but the proof using the notion of sequences doesn't seem to be general enough. So I was looking for a "purely topological" proof but can't find one.

Can anyone help?

$\endgroup$
  • $\begingroup$ Maybe Tychonoff's theorem? $\endgroup$ – Henricus V. Apr 8 '16 at 13:44
  • $\begingroup$ @HenryW.: I don't know anything about it. Our professor gave us this exercise immediately after introducing the open cover definition of compactness. $\endgroup$ – user 170039 Apr 8 '16 at 13:47
  • 1
    $\begingroup$ I think Tychonoff's theorem is overkill for just a product of two spaces. $\endgroup$ – Aidan Sims Apr 8 '16 at 14:52
  • $\begingroup$ See also here: math.stackexchange.com/questions/567335/… (Perhaps it can be considered a duplicate.) $\endgroup$ – Martin Sleziak Apr 8 '16 at 14:57
2
$\begingroup$

Take a cover by open sets $U_\alpha$.

First fix $x \in X$. The set $\{x\} \times Y$ is homeomorphic to $Y$ and so compact. It is covered by the $U_\alpha$ and so there is a finite subcover $U^x_{\alpha_1}, \dots, U^x_{\alpha_{n_x}}$ of $\{x\} \times Y$. Let $U_x := \bigcup^{n_x}_{i=1} U^x_{\alpha_i}$. Since $U_x$ is open, for each $y \in Y$ there is a basic open set $V^x_y \times W^x_y$ contained in $U_x$ and containing $(x,y)$. Since $Y$ is compact and the sets $\{W^x_y : y \in Y\}$ cover $Y$, there is a finite set $F^x \subseteq Y$ such that $\{W^x_y : y \in F^y\}$ covers $Y$. Let $V^x := \bigcap_{y \in F^x} V^x_y$. Then the set $V^x \times Y$ is contained in $U_x$.

Now since $X$ is compact, the cover $\{V^x : x \in X\}$ of $X$ by open sets has a finite subcover $V^{x_1}, \dots, V^{x_k}$. It follows that the sets $V^{x_1} \times Y, \dots, V^{x_k} \times Y$ cover $X \times Y$ and so the sets $U_{x_1}, \dots, U_{x_k}$ cover $X \times Y$. Hence $$ \{U^{x_j}_{\alpha_i} : j \le k, i \le n_{x_j}\} $$ is a finite subcover of the original cover $U_\alpha$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.