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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.2012-10-31
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    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.2012-10-31
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    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.2012-10-31
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    @ZhenLin: am sorry, I should have clarified. Yes, I am talking about sequential compactness.2012-11-01

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