# Proof that $U \times V$ is a compact

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.

• 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 😅 Sep 25 '20 at 17:05
• 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. Sep 25 '20 at 17:06
• @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 !
– user655132
Sep 25 '20 at 17:11

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.