How to show the intersection of arbitrary compact sets is compact in a general metric space?

I understand that if you are working in $\mathbb{R}^n$, then the intersection of an arbitrary collection of compact sets is compact because it is closed and bounded. But what if you are not in $\mathbb{R}^n$? How can you show that the intersection of arbitrary compact sets in a general metric space is compact?

This is homework so I only need a hint. Thank you!

Solution 1:

Hint: A closed subset of a compact set is compact. What do you know about compact sets in Hausdorff spaces?

Solution 2:

Metric Spaces are Hausdorff, so compact sets are closed. Now, arbitrary intersection of closed sets are closed. So for every open cover of the intersection, we can get an extension to a cover for the whole metric space. Now just use the definition.

Solution 3:

Are you using the sequential compactness definition? Any sequence in the intersection will be a sequence in any one of those particular compact sets, say $A$, in the intersection. This has a convergent subsequence in $A$ and since the intersection of closed sets is closed (and compact sets are closed in metric spaces), this same subsequence must converge in the intersection.

Solution 4:

Let $\{x(n)\}$ be any sequence in the intersection of compact subspaces of $X$. Then $\{x(n)\}$ is a member of every compact subspace individually. Compact spaces are closed in general hence every subspace contains the limit of the sequence. Now since arbitrary intersection of closed sets is closed this implies that the intersection is also closed. Hence $n$ contains the limit of $\{x(n)\}$ and hence converges. Since $\{x(n)\}$ is any general sequence in the intersection which converges (hence has a convergent subsequence) makes it compact by the sequential criteria. This approach seems good for your homework.