in my Analysis course we saw this statement :

We are given $U,V$ two metric spaces and $k$ the distance

Let $U,V$ compact sets then $\; \longrightarrow \; $$ U \times V $ is also compact

Our teacher told us that we can prove it using compactness by the Bolzano-Weierstrass condition.

So here is what I know , A metric space is compact if it has the B-W Property.

I didn't found any definition in our lecture so I took the following :

Definition: A set S in a metric space has the Bolzano-Weierstrass Property if every sequence in S has a convergent subsequence — i.e., has a subsequence that converges to a point in S.

I don't know how I can apply it to my problem , so thanks in advance for your help.

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    $\begingroup$ Since not every metric space has the Heine-Borel property, the definition for compactness that you are using (closed and bounded) is going to lead to maximal confusion 😅 $\endgroup$ Sep 25 '20 at 17:05
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    $\begingroup$ So you have a sequence in $U\times V$. It looks like $(u_1,v_1),(u_2,v_2),\ldots$. You need to pull out a subsequence where both the first coordinates converge and the second coordinates converge. $\endgroup$ Sep 25 '20 at 17:06
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    $\begingroup$ @MaximilianJanisch J'ai changé merci d'avoir preciser ce que je n'ai pas remarqué / I edited thanks for pointing out what I missed Danke ! $\endgroup$
    – user655132
    Sep 25 '20 at 17:11

In a metric space, a set is compact if and only if it is sequentially compact (meaning every sequence admits a convergent subsequence). (Edit: This is where you invoke B-W).

Take a sequence $(x_n, y_n) \subseteq U \times V$. Taking just the sequence $(x_n) \subseteq U$, we can use sequential compactness to find a subsequence $(x_{n_j})$ so that $x_{n_j} \rightarrow x \in U$. Now examine the sequence $(x_{n_j}, y_{n_j}) \subseteq U \times V$. We can take the sequence $(y_{n_j}) \subseteq V$ and find a subsequence $(y_{n_{j_k}}) \subseteq V$ so that $y_{n_{j_k}} \rightarrow y$. I claim that $x_{n_{j_k}} \rightarrow x$. I then claim that $(x_{n_{j_k}}, y_{n_{j_k}}) \rightarrow (x,y)$. If this is true, then since I took an arbitrary sequence I have sequential compactness.


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