Proving that if $A$ and $B$ are compact spaces, then $A\times B$ is compact

Question

I am trying to prove that if A and B are compact sets, then AxB is compact as well.

I mostly need feedback on my proof and if it is correct.

I approach with Bolzano-Weierstrauss theorem.

Let Sn be a sequence in AxB, where Sn = (an, bn) for every ordered-pair in AxB. We need to show that for an arbitrary sequence (defined above as Sn) there exists a convergent subsequence. That is, Snk is a converging subsequence such that Snk = (ank, bnk).

Since A is compact, then there exists a sequence an with a convergent subsequence ank.

Since B is compact, then there exists a sequence bn with a convergent subsequence bnk.

Therefore, since ank and bnk converge, Snk = (ank, bnk) must also converge*.

* I'm not really sure how to prove this fact. I guess in other words I want to show that if J and K are convergent sequences then JxK is a convergent sequence? I could probably approach this proof as a lemma and prove it using delta-epsilon, but I feel I am overcomplicating my proof at this point...

Otherwise, does my proof look reasonable? How can it be improved? Or did I muck up the definitions?

Answer 1

Let it be that every sequence in $A$ has a convergent subsequence and let it be that every sequence in $B$ has a convergent subsequence.

Let it be that $(a_n,b_n)_n$ is a sequence in $A\times B$.

It is our aim to prove that this sequence has a convergent subsequence.

First we note that $(a_n)_n$ has a convergent subsequence $(a_{n_k})_k$.

However we are not ready yet, because there is no guarantee that $(b_{n_k})_k$ is a convergent sequence

But we do observe that sequence $(b_{n_k})_k$ has a convergent subsequence $(b_{n_{k_i}})_i$.

Note that $(a_{n_{k_i}})_i$ is a subsequence of the convergent sequence $(a_{n_k})_k$, hence is also convergent.

From this we conclude that $(a_{n_{k_i}},b_{n_{k_i}})_i$ is a convergent subsequence of $(a_n,b_n)$ and we are done.