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A metric space is compact iff every sequence has a convergent subsequence. Using it show that the product metric space of two compact metric spaces is compact where the product of two metric spaces $(X,d_X),~(Y,d_Y)$ is defined by $$d_{X\times Y}((x_1,y_1),(x_2,y_2))=\max\{d_X(x_1,x_2),d_Y(y_1,y_2)\}$$

I can see the ordered pair sequence of two sequences having convergent subsequences not necessarily posseses convergent subsequence e.g. $$\{(2,1/2),(1/2,2),(3,1/3),(1/3,3),...\}$$

Then how do the given characterization help us?

Shantipriya
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1 Answers1

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Here is the sketch

$z_n=(x_n,y_n)\in X\times Y$ be any sequence $\ni x_n\in X,y_n\in Y$

since $X$ is compact so $x_n$ has a subsequence $x_{n_k}\to x\in X$ and $Y$ is compact so $y_n$ has a subsequence $y_{n_k}\to y$

Then $z_{n_k}=(x_{n_k},y_{n_k})\to (x,y)\in X\times Y$

Myshkin
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