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:
\bigcap\limits_{i\in\mathcal{I}}F_i=\varnothing
Then the complement of this intersection is all of X. We can rewrite this as the union of the complements of the F_i:
X=\bigcup\limits_{i\in\mathcal{I}}F_i^c
Since we’re assuming X to be compact, we can find some finite subcollection \mathcal{A}\subseteq\mathcal{I} so that
X=\bigcup\limits_{i\in\mathcal{A}}F_i^c
which, taking complements again, implies that
\bigcap\limits_{i\in\mathcal{A}}F_i=\varnothing
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.

January 14, 2008 Posted by | Point-Set Topology, Topology | 6 Comments

   

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