An amazingly useful property for a space is that it be “compact”. We define this term by saying that if is any collection of open subsets of indexed by any (possibly infinite) set so that their union is the whole of — the sexy words are “open cover” — then there is some finite collection of the index set so that the union of this finite number of open sets still contains all of — the sexy words are “has a finite subcover”.
So why does this matter? Well, let’s consider a Hausdorff space , a point , and a finite collection of points . Given any point , we can separate and by open neighborhoods and , precisely because is Hausdorff. Then we can take the intersection and the union . The set is a neighborhood of , since it’s a finite intersection of neighborhoods, while the set is a neighborhood of . These two sets can’t intersect, and so we have separated and by neighborhoods.
But what if is an infinite set? Then the infinite intersection may not be a neighborhood of ! Infinite operations sometimes cause problems in topology, but compactness can make them finite. If is a compact subset of , then we can proceed as before. For each we have open neighborhoods and , and so — the open sets form a cover of . Then compactness tells us that we can pick a finite collection so that the union of that finite collection of sets still covers — we only need a finite number of the to cover . The finite intersection will then be a neighborhood of which doesn’t touch , and so we can separate any point and any compact set by neighborhoods.
As an exercise, do the exact same thing again to show that in a Hausdorff space we can separate any two compact sets and 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 of closed subsets of so that the intersection of any finite collection of the is nonempty. Then I assert that the intersection of all of the will be nonempty as well if is compact. To see this, assume that the intersection is empty:
Then the complement of this intersection is all of . We can rewrite this as the union of the complements of the :
Since we’re assuming to be compact, we can find some finite subcollection 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 are nonempty, then the intersection of all the are nonempty — then we can derive the first definition of compactness from it.