# 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

Staff Emeritus
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.