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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
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  3. Jun 1, 2010 #2

    HallsofIvy

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    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

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    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

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    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

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    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!
     
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