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Prove that if two sets $A$ and $B$ are compact then so is their Cartesian product $A \times B = \{(a,b): a \in A, b\in B\}$.

The hint is to use Bolzano Weiertrass theorem and an argument of sequence to proof the statement.

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    $\begingroup$ Yes, the hint is good. Where are you stuck at? $\endgroup$ – egreg Nov 14 '13 at 21:40
  • $\begingroup$ Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need. $\endgroup$ – Brett Frankel Nov 14 '13 at 21:54
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    $\begingroup$ To avoid the axiom of countable choice or other (mostly stronger) choice principles, this is usually done using the tube lemma. $\endgroup$ – dfeuer Nov 15 '13 at 0:22
  • $\begingroup$ In the general case, this is known as Tychonoff's Theorem. $\endgroup$ – jpv Jul 12 '14 at 16:53
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    $\begingroup$ The OP does not say whether they are interested in a proof of this fact for metric spaces or (more generally) for topological spaces. But since Bolzano-Weierstrass is mentioned in the question, it is possible that the question is supposed to be only about $\mathbb R^n$? $\endgroup$ – Martin Sleziak Apr 8 '16 at 14:59
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A set $S$ is compact if from any sequence of elements in $S$ you can extract a sub-sequence with a limit in $S$.

If we are given a sequence $(u_n)$ of $A \times B$, then you can write $u_n=(a_n,b_n)$. Since $A$ is compact, you can find a sub-sequence $(a_{f(n)})$ with a limit in $A$. Then, since B is also compact, you can extract a sub-sequence $(b_{f(g(n))})$ of $(b_{f(n)})$ with a limit in B. Thus, the sub-sequence $(u_{f(g(n))})$ of $(u_n)$ has its limit in $A \times B$. This proves that $A \times B$ is compact.

I hope you can understand my explanation, I know my english is approximative...

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  • $\begingroup$ Is there a proof using the open-cover definition of compactness? $\endgroup$ – Fang Jing Jun 8 '14 at 20:19
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There is indeed a topological proof using the open covers definition of compactness.

Let $A$ and $B$ be compact sets and $\{O_\lambda\}_{\lambda\in\Lambda}$ be an open cover of $A \times B$. For each $(a,b) \in A \times B$, we can choose some $\lambda = \lambda(a,b)$ such that $(a,b) \in O_{\lambda(a,b)}$. By construction, $O_{\lambda(a,b)}$ is open, hence the point $(a,b)$ is contained in some open box $X \subset O_{\lambda(a,b)}$ where $X = U_{(a,b)} \times V_{(a,b)}$, where $U_{(a,b)} \subset A$ and $V_{(a,b)} \subset B$.

Suppose we fix $a$ and vary $b$. Then for every point $(a,b)$ we find that the point is contained in an open box in the product $A \times B$, and that box is then itself the product of a subset of $A$ with a subset of $B$. Proceeding in this manner, we observe that the collection of sets $\{V_{(a,b)}\}_{b\in B}$ is an open cover of $B$. Since by assumption $B$ is compact, we can find a finite cover $\{V_{(a,b_j(a))}\}$ of $B$ that consists of finitely many open sets containing points $\{(a,b_j(a))\}$.

Now let $U_a = \bigcap_j U_{(a,b_j(a))}$. Since $U_a$ is the intersection of finitely many open sets, it is itself open. Since $A$ is compact, there are finitely many $a_i$ such that $\{U_{a_i}\}$ forms an open cover of $A$. Then it follows that the collection of sets $\{O_{(a_i,b_j(a_i))}\}$ (for all combinations of $i,j$) is a finite cover of $A \times B$, hence $A \times B$ is compact.

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  • $\begingroup$ How does it follow that $\{U_{a_i}\}$ covers $A$? Couldn't $U_a$ be an empty set? $\endgroup$ – Björn Lindqvist Jul 7 '18 at 20:16
  • $\begingroup$ @BjörnLindqvist $a\in U_a$. $\endgroup$ – Noah Schweber Feb 26 at 18:34

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