# Heine-Borel - Finite Intersection equivalence in subspaces

I'm doing a course in Calculus/Real Analysis, and we're covering really basic topological ideas (metric spaces, open/closed sets etc.). From my notes it looks like we proved the following in class (Heine-Borel):

(1) A set K is compact iff (2) for every open cover of K there is a finite subcover.

And then we claimed without proof that the right hand side is clearly equivalent to the following property: (3) for every set of closed subsets from K that have the finite intersection property, the intersection of said sets is non-empty.

It looks pretty easy to prove this equivalence when K is an entire space. Like this proof here. But from my notes it appears to be a theorem that holds true with any subset of the metric space.

My question is how that proof holds up when K (as mentioned in class) is not the whole space but a subset of the space. The proof relies on the complement of the sets, which means that given an open cover, the complement of the sets are closed but are still in the space (obviously, since the complement is with respect to the space). However when we take the complement w.r.t. the space and K is no longer the whole space, the sets we have are no longer from K necessarily. It also appears of no help to intersect them with K, since we cannot assume K is closed (that's part of what we're trying to prove in the sense that compactness is closed+bounded).

The difference seems petty but it's bugging me to no end.

Apologies about the wordiness, my latex skills are about as poor as my topology skills. Thanks.

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1
Note that (2) is the usual definition of compactness. I don’t know what definition your course is using, but it may be something that’s actually not equivalent to compactness in all topological spaces. – Brian M. Scott Sep 22 '11 at 18:44
@BrianM.Scott, we defined compactness of K as: every sequence in K has a subsequence that converges to an element in K, which I think is pretty general. – davin Sep 23 '11 at 0:27
@davin: It is actually a property called sequential compactness, which in metric spaces is equivalent to compactness. – Asaf Karagila Sep 23 '11 at 6:38

## 1 Answer

Every set with a topology is a space.

In particular, if $X$ is any space (metric, in this case) and $K\subseteq X$ is a compact subset of $X$ then $K$ is a space equipped with the relative topology.

$U\subseteq K$ is open if and only if there exists some $V\subseteq X$ which is open in the original topology and $K\cap V=U$.

Now if $K$ is a compact subset of $X$ then $K$ forms a compact space with the relative topology. This is since every cover of $K$ by open sets can be extended to an open cover in $X$, where we know $K$ is compact.

From here we can just assume that $K$ is now the entire space, and prove as in the link.

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I didn't really get your answer at first, because it sounds like you're saying "if we know K is compact then we can consider K a space", which only works to prove one direction of the theorem (still better than what I had). And after some thought I realised that that direction is the only problematic one. Thanks. – davin Sep 23 '11 at 0:33
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