Cartesian product of two compact metric spaces is again compact, proof using sequential compactness

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.

• Well of course you can. Use the definition of the product topology / product metric. Oct 27, 2019 at 11:10
• I am not aware of any such definition at this point. My instructor has added this anyway and gave no hints to solve Oct 27, 2019 at 11:12
• 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. Oct 27, 2019 at 11:14
• Your proof works. Oct 27, 2019 at 11:15
• @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. Oct 27, 2019 at 11:42

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).