Let (X,d) be a metric space and A,B $\subset$ X be two compact subsets. Show $A\cap B$ is also compact.

I attempted this question by showing the intersection is bounded and closed.

But I stated that Bounded and Closed $\Rightarrow$ Compact (Heine-Borel) but I didn't realise this only holds for $\mathbb R^n$.

Most of the other similar problems on here were dealing with $\mathbb R^n$, so how would you go about showing this for a general space X?

Any help would be greatly appreciated!

  • $\begingroup$ Refer to "Topology of Metric Spaces" by S Kumaresan. $\endgroup$ – Shiva Prakash Dec 6 '13 at 12:22

In metric space, any compact subset is closed. In particular, this means $X \setminus B$ is open. For any open cover $\mathscr{O}$ of $A \cap B$, $\mathscr{O} \cup \{ X \setminus B \}$ will be an open cover of $A$. Since $A$ is a compact, $\mathscr{O} \cup \{ X \setminus B \}$ has a finite sub-cover $\mathscr{F}$. It is easy to see $\mathscr{F} \setminus \{ X \setminus B \}$ is a finite sub-cover of $\mathscr{O}$ for $A \cap B$. Since the open cover $\mathscr{O}$ is arbitrary, $A \cap B$ is compact.

  • $\begingroup$ This makes a lot of sense, thanks! So just to clarify the definition of compactness, in $R^n$ you can say $A\ Compact \Leftrightarrow A\ bounded \wedge A\ closed$ but for a general metric space X $\ A \ Compact \Rightarrow A\ bounded \wedge A\ closed$ $\endgroup$ – user113867 Dec 8 '13 at 12:19
  • $\begingroup$ @user113867 Compactness is a pure topological concept (compactness = every open cover has finite sub-cover). What you say is not the definition of compactness but they are true. $\endgroup$ – achille hui Dec 8 '13 at 14:30

This is true for arbitrary Hausdorff spaces, not only for metric spaces.

Try to prove the following slight generalisation: any closed set in a compact space is compact. This should be easy with the usual definition of compactness (any cover admits a finite subcover). If you insist on working with metric spaces, it's even easier, using the definition that a set in a metric space is compact iff every sequence has a convergent subsequence.

  • $\begingroup$ Actually, this is not necessarily true for non-Hausdorff spaces. Take the line with two origins, and the two "intervals" corresponding to $[-1,1]$. $\endgroup$ – ronno Dec 6 '13 at 12:24
  • $\begingroup$ @ronno: Touche. I thought about a different formulation: intersection of a closed set and a compact set is compact, which is true in general. But of course in non-Hausdorff spaces a compact set need not be closed... $\endgroup$ – tomasz Dec 6 '13 at 12:37

$A$ is closed in $X$, so $A \cap B$ is closed in $B$. Now, a closed subset of a compact space is compact.


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