The Unapologetic Mathematician

Mathematics for the interested outsider

Some compact subspaces

Let’s say we have a compact space X. A subset C\subseteq X may not be itself compact, but there’s one useful case in which it will be. If C is closed, then C is compact.

Let’s take an open cover \{F_i\}_{i\in\mathcal{I}} of C. The sets F_i are open subsets of C, but they may not be open as subsets of X. But by the definition of the subspace topology, each one must be the intersection of C with an open subset of X. Let’s just say that each F_i is an open subset of X to begin with.

Now, we have one more open set floating around. The complement of C is open, since C is closed! So between the collection \{F_i\} and the extra set X\setminus C we’ve got an open cover of X. By compactness of X, this open cover has a finite subcover. We can throw out X\setminus C from the subcover if it’s in there, and we’re left with a finite open cover of C, and so C is compact.

In fact, if we restrict to Hausdorff spaces, C must be closed to be compact. Indeed, we proved that if C is compact and X is Hausdorff then any point x\in X\setminus C can be separated from C by a neighborhood U\subseteq X\setminus C. Since there is such an open neighborhood, x must be an interior point of X\setminus C. And since x was arbitrary, every point of X\setminus C is an interior point, and so X\setminus C must be open.

Putting these two sides together, we can see that if X is compact Hausdorff, then a subset C\subseteq X is compact exactly when it’s closed.

January 15, 2008 - Posted by | Point-Set Topology, Topology


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