The Unapologetic Mathematician

Mathematics for the interested outsider

Compact Spaces

An amazingly useful property for a space X is that it be “compact”. We define this term by saying that if \{U_i\}_{i\in\mathcal{I}} is any collection of open subsets of X indexed by any (possibly infinite) set \mathcal{I} so that their union \bigcup\limits_{i\in\mathcal{I}}U_i is the whole of X — the sexy words are “open cover” — then there is some finite collection of the index set \mathcal{A}\subseteq\mathcal{I} so that the union of this finite number of open sets \bigcup\limits_{i\in\mathcal{A}}U_i still contains all of X — the sexy words are “has a finite subcover”.

So why does this matter? Well, let’s consider a Hausdorff space X, a point x\in X, and a finite collection of points A\subseteq X. Given any point a\in A, we can separate x and a by open neighborhoods x\in U_a and a\in V_a, precisely because X is Hausdorff. Then we can take the intersection U=\bigcap\limits_{a\in A}U_a and the union V=\bigcup\limits_{a\in A}V_a. The set U is a neighborhood of X, since it’s a finite intersection of neighborhoods, while the set V is a neighborhood of A. These two sets can’t intersect, and so we have separated x and A by neighborhoods.

But what if A is an infinite set? Then the infinite intersection \bigcap\limits_{a\in A}U_a may not be a neighborhood of x! Infinite operations sometimes cause problems in topology, but compactness can make them finite. If A is a compact subset of X, then we can proceed as before. For each a\in A we have open neighborhoods x\in U_a and a\in V_a, and so A\subseteq\bigcup\limits_{a\in A}V_a — the open sets V_a form a cover of A. Then compactness tells us that we can pick a finite collection A'\subseteq A so that the union V=\bigcup\limits_{a\in A'}V_a of that finite collection of sets still covers A — we only need a finite number of the V_a to cover A. The finite intersection U=\bigcap\limits_{a\in A'}U_a will then be a neighborhood of x which doesn’t touch V, and so we can separate any point x\in X and any compact set A\subseteq X by neighborhoods.

As an exercise, do the exact same thing again to show that in a Hausdorff space X we can separate any two compact sets A\subseteq X and B\subseteq X by neighborhoods.

In a sense, this shows that while compact spaces may be infinite, they sometimes behave as nicely as finite sets. This can make a lot of things simpler in the long run. And just like we saw for connectivity, we are often interested in things behaving nicely near a point. We thus define a space to be “locally compact” if every point has a neighborhood which is compact (in the subspace topology).

There’s an equivalent definition in terms of closed sets, which is dual to this one. Let’s say we have a collection \{F_i\}_{i\in\mathcal{I}} of closed subsets of X so that the intersection of any finite collection of the F_i is nonempty. Then I assert that the intersection of all of the F_i will be nonempty as well if X is compact. To see this, assume that the intersection is empty:
Then the complement of this intersection is all of X. We can rewrite this as the union of the complements of the F_i:
Since we’re assuming X to be compact, we can find some finite subcollection \mathcal{A}\subseteq\mathcal{I} so that
which, taking complements again, implies that
but we assumed that all of the finite intersections were nonempty!

Now turn this around and show that if we assume this “finite intersection property” — that if all finite intersections of a collection of closed sets F_i are nonempty, then the intersection of all the F_i are nonempty — then we can derive the first definition of compactness from it.

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January 14, 2008 - Posted by | Point-Set Topology, Topology


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