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Separating Sets in a Completely Regular Space

  1. Jul 20, 2009 #1
    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. jcsd
  3. Jul 20, 2009 #2

    HallsofIvy

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    Have you used the fact that X is completely regular? What is the definition of "completely regular"?
     
  4. Jul 20, 2009 #3
    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}.
     
  5. Jul 20, 2009 #4
    Don't start with the open covering.

    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.
     
  6. Jul 20, 2009 #5
    That is ingenious. What led you to devise the U_a's the way you did?
     
  7. Jul 20, 2009 #6
    "Cheating." (ashamed...)
     
  8. Jul 20, 2009 #7
    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.
     
  9. Jul 20, 2009 #8
    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'.
     
  10. Jul 20, 2009 #9
    I like that approach. However, I'm not aware of any embedding theorems for completely regular spaces. Is there such a theorem?
     
  11. Jul 20, 2009 #10
    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.
     
  12. Jul 20, 2009 #11
    Neat. Looks like my forays in topology have just begun.
     
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