Show a set is compact

**billa** · January 22nd 2011, 08:46 AM

If A and B are compact subsets of some metric space M with metric d, then I think that the cross product $A\times B$ is a compact subset of the metric space $M^2$ where the metric in this space is defined by

$p((x,y),(z,w))=\sqrt{(d(x,z)^2+d(y,w)^2}$

What is the best way to show that AxB is indeed compact? Is there a particularly good way to do this? I think I have a proof showing that every sequence in AxB has a convergent subsequence.

Thanks.

**roninpro** · January 22nd 2011, 08:51 AM

Showing that every sequence has a convergent subsequence is probably the easiest way to do it. You would need to use the fact that a sequence $\{(x_n,y_n)\}$ in $A\times B$ converges if and only if $\{x_n\}$ converges in $A$ and $\{y_n\}$ converges in $B$ .

On the other hand, it might be possible to use the definition directly: take an open cover of $A\times B$ and try to show that it has a finite subcover. You can relate this to the compactness of $A$ and $B$ by using the projection maps: $\pi_1: A\times B\to A$ given by $\pi_1(x,y)=x$ and $\pi_2: A\times B\to B$ given by $\pi_2(x,y)=y$ .

**DrSteve** · January 22nd 2011, 09:02 AM

By the way, I think you mean "cartesian product" and not "cross product"

I think that both methods suggested by roninpro should work.

**Tinyboss** · January 22nd 2011, 10:32 AM

Show that the given metric induces the product topology, and invoke the theorem that finite (easy proof) or arbitrary (Tychonoff) products of compact spaces are compact.

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