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The space of continuous functions.

  1. Feb 19, 2008 #1
    Let X be a compact space, (Y,p) a compact metric space, let F be a closed subset of C(X,Y) (the continuous functions space) (i guess it obviously means in the open-compact topology, although it's not mentioned there) which satisifes:
    for every e>0 and every x in X there exists a neighbourhood U of x s.t for every f in F:
    f(U) is a subset of B_e(f(x)) which is an open ball with radius e.
    prove that F is compact.

    well i want to prove that C(X,Y) is compact and then it will follow that F is compact.

    now let U(W(C,U))=C(X,Y) which is an arbitrary covering of C(X,Y), i.e for every C compact in X and every U open in Y s.t f(C) is contained in U.
    now i need to find a finite covering, now because Y is compact and metric then it is covered by a finite union of open balls, because f in F is continuous then f^-1(B_e(f(x))
    is open in X, so finite union of them cover the entire set X, so because every compact set in X is actually covered by a finite union of f^-1(B_e(f(x))) and also every every open set in Y is covered by a finite union of open balls, then C(X,Y) is covered by a finite union of W(C_i,B_e_j(f(x))) s.t C_i is compact in some f^-1(B_e_j(f(x)) i.e f(C_i) is contianed in B_e_j(f(x))), so C(X,Y) is compact and then also F is compact as beiing a closed subset of C(X,Y).

    am i right or way off?

    thanks in advance.
     
  2. jcsd
  3. Feb 19, 2008 #2

    JasonRox

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    How does that follow?
     
  4. Feb 19, 2008 #3
    well there's a theorem that if X is a compact topological space and A a closed subset of it then A is compact in X.
     
  5. Feb 19, 2008 #4

    JasonRox

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    What is the continuous functions space? Which one? Define an open set or subbasis element of it?

    I've heard of compact-open topology but not open-compact topology. Maybe it's the reverse, but we would have to prove it's a topology first, which I don't think it is.

    http://en.wikipedia.org/wiki/Metric_space

    What can this lead to?
     
  6. Feb 19, 2008 #5

    gel

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    C(X,Y) isn't compact.
     
  7. Feb 20, 2008 #6
    instead of telling me I'm wrong, you could have also pointed me to a possible solution, could you not?

    anyway, jason, open-compact is the same as compact-open, i just changed the order.
     
  8. Feb 20, 2008 #7
    okay, i see it now, thanks jason.
     
  9. Feb 20, 2008 #8
    wait a minute gel, if C(X,Y) is metric, and because Y is complete it's also complete, then by the condition given on F, C(X,Y) is absolutely bounded, then by a thoeorem we learned in class, C(X,Y) must be compact, cause a metric space is compact iff it's complete and absolutely bounded (absolutely bounded, or totally, is that there's the metric space has a finite e>0 net.
     
  10. Feb 20, 2008 #9

    JasonRox

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  11. Feb 20, 2008 #10
    So what i said is wrong?
    just got the wrong terminology on absolutely bounded (should be totally bounded), it's a problem that you learn the terms in hebrew and after you look at the definitions in munkres with the theorems. (the lecturer is awfull, an entire lecture today was only with figures, with no formalism, and a lot of handwaving of the worst kind, we learnt on manifolds, a topic which he didn't prove anything in it rigorously).
     
  12. Feb 20, 2008 #11

    morphism

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    It is wrong. You can't conclude that C(X,Y) is totally bounded based on what you're given. What you could try to do is prove that F is totally bounded, although I don't think you will find this very fruitful. Instead, maybe you could try the Arzela-Ascoli theorem.
     
  13. Feb 20, 2008 #12

    gel

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    Yes, but I was in a rush, and I think it's useful to know when you're going about something the wrong way.

    yes C(X,Y) is complete, so F is too. C(X,Y) isn't absolutely bounded though. Maybe you meant to say that F is absolutely bounded, which I think it is (if I understand your definition correctly - what is an e>0 net? is it just a cover by open balls of radius e?) Then compactness of F follows.
     
    Last edited: Feb 20, 2008
  14. Feb 20, 2008 #13

    morphism

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    Maybe I'm missing something, but the condition on F doesn't really look like it implies total boundedness easily. It's saying that F is equicontinuous, hence my inclination to try Arzela-Ascoli.
     
  15. Feb 20, 2008 #14

    gel

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    Well looking at wikipedia statement of Arzela-Ascoli theorem, it does imply that F is compact

    In the original question, it said that F is closed and equicontinuous. Also, Y is compact => F is pointwise compact. So the original question was almost a restatement of the Arzela-Ascoli theorem anyway.

    I can also see that F will be compact from Tychonoff's theorem. The space of all functions X->Y (not just continuous ones) is compact under the product topology, and F is closed with this topology.
     
    Last edited: Feb 20, 2008
  16. Feb 21, 2008 #15

    morphism

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    That's an interesting observation, but is it enough here? The topology C(X,Y) gets from the compact-open subbase here is the topology of uniform convergence, i.e. the topology generated by the uniform metric, whereas the topology it gets as a subspace of the product space Y^X is the topology of pointwise convergence, which in general is strictly weaker than the former. So certainly F is going to be closed in the product topology if it's to be closed in the compact-open topology, but for the converse we need something stronger. Or did you have something else in mind?

    Edit:
    Evidently the equicontinuity of F can be our "something stronger", because it reduces the compact-open topology on F to the product topology. <Proof removed.> So I guess it is enough to prove that F is closed in the product topology.
     
    Last edited: Feb 21, 2008
  17. Feb 21, 2008 #16
    eventually i realised before handing in the assignment that I should show that F is complete and totally bounded, i hope my proof is rigorous enough, well the fact that F is complete follows from the fact that Y is complete then C(X,Y) is complete then F is complete, and the condition that F satisfies assures us it's totally bounded and it's metric then it's compact.
     
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