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We know that if $A$ and $B$ are compact (assuming A and B are non-empty), then the Cartesian product $A \text{x} B$ is compact. But how do you go the other way round.

We have to show that any sequence $(a_k)$ in A and $(b_k)$ in B have subsequences that converges in A and B respectively. We are given that any subseqence of the sequence $(a_k, b_k)$ is convergent in $A \text{ x } B$. I have at loss at how to I use this information to claim that subsequences of $(a_k) \text{and} (b_k)$ are convergent in $A$ and $B$ respectively.

Should I claim that $(a_k, b_k)$ is convergent iff each $a_k$ and $b_k$ is convergent, and be done with it?

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An easier strategy might be to pick a point $a\in A$ and show that $\{a\}\times B$ is closed in $A\times B$. Since a closed subspace of a compact space is compact, you're done. (Noting that $B$ is clearly homeomorphic to $\{a\}\times B$.) A similar argument gives you that $A$ is compact. – Shawn Henry Oct 31 '12 at 19:55
Shawn, can we make the same argument without stating the last point (B is homeomorphic to {a} x B)? I am just trying to keep it really basic. – user43901 Oct 31 '12 at 20:40
Do you mean compact metric spaces? In general sequential compactness is not the same as compactness. Shawn's suggestion is the easiest proof for general compact Hausdorff spaces. The easiest proof for general compact topological spaces is to use projections and the fact that the image of a compact space is again compact. – Zhen Lin Oct 31 '12 at 20:41
@ZhenLin: am sorry, I should have clarified. Yes, I am talking about sequential compactness. – user43901 Nov 1 '12 at 0:21

HINT: Let $\langle a_k:k\in\Bbb N\rangle$ be a sequence in $A$. Let $b$ be any point of $B$, and consider the sequence $$\big\langle\langle a_k,b\rangle:k\in\Bbb N\big\rangle$$ in $A\times B$. The same basic idea works to handle sequences in $B$.

Added: Here’s a rough sketch that may help a bit.

enter image description here

I’ve drawn this as if $A$ and $B$ were $[0,1]$, because that’s easy to visualize. The red dots in $A$ (down at the bottom) are four terms of the sequence $\langle a_k:k\in\Bbb N\rangle$; the red dots in $A\times B$ (the square) are all on the line representing $A\times\{b\}$: they are the corresponding terms of the sequence $$\big\langle\langle a_k,b\rangle:k\in\Bbb N\big\rangle\tag{1}$$ in $A\times B$.

You know that every sequence in $A\times B$ has a convergent subsequence, so $(1)$ converges to some point $\langle x,y\rangle\in A\times B$. In fact, $y=b$; why? Thus, $\langle x,y\rangle=\langle x,b\rangle$ is on the same ‘line’ as the sequence $(1)$, the set $A\times\{b\}$. Now show, using the definition of the product topology, that $\langle a_k:k\in\Bbb N\rangle$ converges to $x$.

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@BrianMScott: Try fleshing that hint out for me. I was thinking of how to use the fact that the sequence $(a_k,b_k)$ in AxB can be used to arrive that the conclusion we need. I think I have a hard time thinking about the sequence $(a_k,b_k)$, ie, ordered pairs forming a sequence. I would like some help in envisioning this, and an explication will help. I have found a proof to my question, but I rather understand it before I look at it. – user43901 Oct 31 '12 at 19:07
@user43901: You aren’t using a sequence $\big\langle\langle a_k,b_k\rangle:k\in\Bbb N\big\rangle$; you’re using a sequence $\big\langle\langle a_k,b\rangle:k\in\Bbb N\big\rangle$ whose terms all have the same second component. I’m not sure that I can say much more without just doing the proof; let me think about that while I produce a sketch that may help you to visualize what’s going on. – Brian M. Scott Oct 31 '12 at 19:11
@user43901: I added both a sketch and a considerable further hint; see if they’re enough for you to finish the job. – Brian M. Scott Oct 31 '12 at 19:36
@BrianMScott: thank you so much for adding this. I am working through it now, but I apologize for not understanding when you say "use the definition of product topology" since I am not aware of that. So if there is a way we can use the basic facts of compactness and convergences instead of using other terms, I would be grateful. – user43901 Oct 31 '12 at 20:35
@user43901: Do you know that if $V\subseteq A$ and $W\subseteq B$ are open sets, then $V\times W$ is an open set in $A\times B$? And that if $U$ is an open neighborhood of a point $p\in A\times B$, there are open sets $V\subseteq A$ and $W\subseteq B$ such that $p\in V\times W\subseteq U$? I’m hard-pressed to see how you can be expected to prove the result if you don’t have at least that much information. – Brian M. Scott Oct 31 '12 at 20:41

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