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Cartesian product of two compact metric spaces $(X,d_X)$ and $(Y,d_Y)$ is again compact.

I will prove using sequential compactness.

Let $(x_n,y_n)$ be any sequence in $X\times Y$.

because $X$ is compact and $(x_n)\in X$ so that $\exists(x_{n_k})$ subsequence such that $(x_{n_k})\to x_0 \in X$

now, $(y_{n_k})\in Y$, and $Y$ is compact, so $\exists (y_{n_{k_l}})$ subsequence of $(y_{n_k})$ such that

$$ (y_{n_{k_l}}) \to y_0\in Y$$

Now can we say that $$(x_{n_{k_l}},y_{n_{k_l}}) \to (x_0, y_0) \in X\times Y$$?

because if it were to be true then we have produced a convergent subsequence and we're done.

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  • $\begingroup$ Well of course you can. Use the definition of the product topology / product metric. $\endgroup$ – астон вілла олоф мэллбэрг Oct 27 '19 at 11:10
  • $\begingroup$ I am not aware of any such definition at this point. My instructor has added this anyway and gave no hints to solve $\endgroup$ – Abhay Oct 27 '19 at 11:12
  • $\begingroup$ Then you must be content with the fact that it can be proved, and is very easy if you know the definition. If you do not know the definition of the topology, then your instructor cannot expect you to work in that space! It's like asking you to play a game without telling you the rules, there's no point if you don't know when you've won and when you've lost. Anyway, the answer below addresses the product concern. $\endgroup$ – астон вілла олоф мэллбэрг Oct 27 '19 at 11:14
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    $\begingroup$ Your proof works. $\endgroup$ – Prototank Oct 27 '19 at 11:15
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    $\begingroup$ @Abhay $d((x,y),(x',y')) = \max(d_X(x,x'),d_Y(y,y'))$ or $d((x,y),(x',y')) = d_X(x,x') + d_Y(y,y')$ are the most common choices; both work for this problem. $\endgroup$ – Henno Brandsma Oct 27 '19 at 11:42
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Yes, you can say that because the sub-subsequence $x_{n_{k_l}}$ of $x_{n_k}$ also converges to $x_0$ and a product sequence converges iff both component sequences converge (this is a general property of the product topology).

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