# Topology - compactness

1. May 31, 2010

### economist13

1. The problem statement, all variables and given/known data

Let X be a set and t & T be two topologies on X. Prove that if (X,t) is Hausdorff and (X, T) is Compact with t a subset of T, then t=T. (i.e., T is a subset of t).

3. The attempt at a solution

potentially useful theorem: (X,t) Hausdorff and X compact implies that each subset F is compact iff it is closed.

I don't really know which direction to go from here...

Last edited: Jun 1, 2010
2. Jun 1, 2010

### HallsofIvy

This is certainly not true. Every compact subset of a Hausdorff space is closed but the converse is not true. For example, in the real numbers with the usual topology, $[0,\infty)$ is a closed set but is not compact.

3. Jun 1, 2010

### VeeEight

There is a nice theorem regarding homeomorphisms from compact to hausdorff spaces

You should see the proof of the theorem HallsofIvy posted - compact Hausdorff spaces come up often in point set topology.

Last edited: Jun 1, 2010
4. Jun 1, 2010

### economist13

My bad, I forgot to mention that X is Hausdorff and compact.

5. Jun 1, 2010

### Dick

You are given that X is Hausdorff under t and compact under T. You don't know it's Hausdorff under T or that it's compact under t until you prove it by proving t=T.

6. Jun 1, 2010

### economist13

actually it is enough to know that if t is a subset of T and X is compact with T, then it is compact with t.

Choose any open covering for (X, t), {Ui} for i in some index I, then it is also and open of (X,T), since t a subset of T. Since (X,T) compact, there is a finite open cover, {Ui} i=1,...,n. But each of those Ui is also in (X,t) thus (X,t) is compact.

It is also not too difficult to show that (X,T) is Hausdorff. Choose x and y not equal in X. Then since (X, t) is Hausdorff, there exist M and N, both open, such that they are disjoint and contain x and y, respectively. Since t a subset of T, then M and N are in T, thus (X,T) also has the Hausdorff property.

So in neither case do we need T=t.

additionally, my edit was simply w.r.t the theorem that X compact and Hausdorff then subset F is compact iff F is closed.

7. Jun 1, 2010

### Dick

Now you are getting someplace. You are actually proving things. But what you are supposed to prove is that t=T. How do you prove that? You want to show an open set O in T is also in t.

8. Jun 1, 2010

### economist13

I know what I need to show to prove that T=t. i.e., O in T => O in t, since we are given t a subset of T. I just don't see how to do this with the information given.

I guess I should have been more clear with what I need help with. I just am missing whatever key observation I need to make to achieve any useful progress.

EDIT: How's this, since (X,t) compact & Hausdorff (shown above) and (X,T) compact & Hausdorff, then i:(X,T)->(X,t), where i(x)=x, is a homeomorphism. Thus for each open set O in T, i(O) is in t. But i(O)=O since i is the identity function. Thus T=t.

Thanks VeeEight, Dick and HallsofIvy

Last edited: Jun 2, 2010
9. Jun 2, 2010

### Dick

You could also just take the direct approach. If O is open in T then C=complement(O) is closed in T. Therefore C is compact. Since (C,T) is compact can't you prove (C,t) is also compact (and therefore closed) in t the same way you did with X? Sorry to underestimate the level you were coming from!