Homework Help: Separating Sets in a Completely Regular Space

1. Jul 20, 2009

e(ho0n3

The problem statement, all variables and given/known data
Let X by a completely regular space and let A and B be closed, disjoint subsets of X. Prove that if A is compact, then there is a continuous function f : X --> [0,1] such that f(A) = {0} and f(B) = {1}.

The attempt at a solution
Let {U} be an open covering of A, U_1, ..., U_n be the corresponding finite open covering afforded by compactness of A. Let a_i in U_i and let f_i : X --> [0,1] be the continuous function separating a_i and B afforded by the complete regularity of X. Define f by f_1 * ... * f_n. This is all I have at the moment. Unfortunately, my f is only 0 at a_i and I don't know how to extend it to be 0 on all of A without breaking continuity.

I believe that the open covering I choose for A must somehow play a role. For example, if the open covering is just {A}, then, proceeding as I did above, compactness plays no role. Any tips?

2. Jul 20, 2009

HallsofIvy

Have you used the fact that X is completely regular? What is the definition of "completely regular"?

3. Jul 20, 2009

e(ho0n3

Yes I did: The f_i's in my attempt are from complete regularity.

The definition is: For any a in X and any closed subset B of X, there exists a continuous function f : X --> [0,1] such that f(a) = 0 and f(B) = {1}.

4. Jul 20, 2009

Billy Bob

Obtain f_a for each a in A. Now pick any number between 0 and 1, say 1/2. Let U_a be the open set (containing a) where f_a is less than 1/2. Then invoke compactness.

Note: If you subtract 1/2 from f_a, you'll get a continuous function that is less than 0 on U_a. So, consider the max of 0 and f_a, which is also a continuous function.

Now use scaling and/or addition of these functions to get your desired function.

5. Jul 20, 2009

e(ho0n3

That is ingenious. What led you to devise the U_a's the way you did?

6. Jul 20, 2009

Billy Bob

"Cheating." (ashamed...)

7. Jul 20, 2009

e(ho0n3

In hindsight, it seems to me that there are two viable approaches to this problem:

1. Invoke compactness first and then complete regularity.
2. Invoke complete regularity first and then compactness.

The first approach was my approach, and that clearly lead nowhere. The second I did not try, but I was already heading in that direction with last comment I made in my original post. In any case, thanks for the help.

8. Jul 20, 2009

Billy Bob

Before "cheating," I was considering a third approach, using a theorem instead of first principles. I prefer first principles. But how about saying since X is completely regular, then by a theorem, X is (homeomorphic to) a subspace of a normal space X'. Now A is cpt in X, so cpt in X', so closed in X'. Can we get B' closed in X', disjoint from A, such that B= B' intersect X. Then apply Urysohn's lemma to A and B'.

9. Jul 20, 2009

e(ho0n3

I like that approach. However, I'm not aware of any embedding theorems for completely regular spaces. Is there such a theorem?

10. Jul 20, 2009

Billy Bob

There are several.

If X is completely regular, then X can be imbedded in a parallelotope (i.e. cartesian product of unit intevals [0,1]).

X is c.r. iff X is (homeomorphic to) a subspace of a cpt Hausdorff space iff X is (homeomorphic to) a subspace of a normal space.

Also, there is the Stone-Čech compactification of a c.r. space.

11. Jul 20, 2009

e(ho0n3

Neat. Looks like my forays in topology have just begun.