2
$\begingroup$

Let $X$ be a CW complex. Prove that a set $A\subseteq X$ is compact if and only if $A$ is closed and has a nonempty intersection with finitely many open cells of $X$ only.

What I know: compact means every open cover has a finite subcover. (I'm not sure if there's another definition of compact which would be more helpful here)

I think most of my confusion comes from working with CW complexes and how that affects the way I should approach this problem. Please explain as much as possible!

$\endgroup$
3
$\begingroup$

There are some important theorems about compact sets that can simplify your work:

  1. If $A$ is closed, $A\subseteq B$, and $B$ is compact, then $A$ is compact

  2. If $A$ is compact and $f:A\to B$ is continuous, then $f(A)$ is compact.

  3. If $A$ is compact, $A\subseteq B$, and $B$ is Hausdorff, then $A$ is closed.

The complete proof can be given using these things, but I will just prove one direction.

A closed cell is compact. To see this, note that the map used for attachment goes from $D^n$ to $X$ is continuous. Since images of compact sets under continuous maps are compact, and $D^n$ is compact, we get that a closed cell is compact.

Any finite subcomplex of $X$ is compact. Since closed cells are closed and compact, the disjoint union of finitely-many of them is closed and compact. This means that any finite subcomplex of $X$ is comapct. Since any closed subset of a compact set is comapct$-$and $A$ is closed$-$it suffices to show that $A$ is contained in a finite subcomplex of $X$.

This is immediate from the hypotheses that $A$ intersects only finitely-many open cells. Simply take the subcomplex formed by using the closures of these open cells, of which there are finitely-many.

For the other direction, you can try to prove $X$ is Hausdorff by showing each of its $n-$skeletons is Hausdorff (use induction) and then use the fact that $A$ is compact to show $A$ is closed. To show that $A$ intersects finitely-many open cells, the argument is less intuitive, but you can pick a point in each open cell that $A$ intersects, and show the collection of all such points will be closed, and that any subset of this collection is closed. This makes the collection of these points discrete and, since it is closed, it will be compact because $A$ is assumed to be compact in this case. But any discrete compact space must be finite by the open cover definition you give above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.