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We know that a product of two (or finitely many) compact topological spaces is compact. And we also know that in a metric space, compactness is equivalent to sequential compactness. So a product of two sequentially compact metric spaces is sequentially compact.

My question is this:

Let $(X, d_X)$, $(Y,d_Y)$ be two sequentially compact metric spaces, and let $\{x_n \times y_n \}$ be an arbitrary sequence in the product $X \times Y$. This implies that $\{x_n\}$ is a sequence in $X$ and $\{y_n\}$ is a sequence in $Y$.

Now since $X$ is sequentially compact, there exists a subsequence $\{x_{k_n}\}$, say, of $\{x_n\}$ which is convergent; similarly there is a subsequence $\{y_{r_n}\}$ of $\{y_n\}$ which is convergent, $\{k_n\}$ and $\{r_n\}$ being strictly increasing sequences of natural numbers.

If $k_n = r_n$ for infinitely many $n$, then we can explicitly determine a convergent subsequence of the sequence $\{x_n \times y_n \}$.

How do we explicitly determine a convergent subsequence of the sequence $\{x_n \times y_n \}$ --- whose existence is gauranteed by the sequential compactness of the product $X \times Y$ --- if $k_n = r_n$ for at most finitely many $n$?

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Jasper’s answer contains the key idea, but it also contains a fair bit of unnecessary material. Start with your sequence $\big\langle\langle x_n,y_n\rangle:n\in\Bbb N\big\rangle$ in $X\times Y$. You know that $\langle x_n:n\in\Bbb N\rangle$ has a convergent subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ in $X$, say with limit $x$. Now consider the sequence $\langle y_{n_k}:k\in\Bbb N\rangle$ in $Y$: it has a convergent subsequence $\langle y_{n_{k_j}}:j\in\Bbb N\rangle$ in $Y$, say with limit $y$. Now show that $$\Big\langle\left\langle x_{n_{k_j}},y_{n_{k_j}}\right\rangle:j\in\Bbb N\Big\rangle$$ converges to $\langle x,y\rangle$ in $X\times Y$.

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  • $\begingroup$ Absolutely! Thank you so much! $\endgroup$ – Saaqib Mahmood Aug 18 '13 at 5:32
  • $\begingroup$ @Saaqib: You’re very welcome! $\endgroup$ – Brian M. Scott Aug 18 '13 at 5:34

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