1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Topology - compactness

  1. May 31, 2010 #1
    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. jcsd
  3. Jun 1, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    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, [itex][0,\infty)[/itex] is a closed set but is not compact.

     
  4. Jun 1, 2010 #3
    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
  5. Jun 1, 2010 #4
    My bad, I forgot to mention that X is Hausdorff and compact.
     
  6. Jun 1, 2010 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  7. Jun 1, 2010 #6
    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.
     
  8. Jun 1, 2010 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  9. Jun 1, 2010 #8
    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
  10. Jun 2, 2010 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook