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Open compacted topology.

  1. Feb 16, 2008 #1
    Let C(X,Y) be the continuous functions space between the topological spaces X,Y, with the open-compact topology. prove that if the sequence {f_n} of C(X,Y) converges to f0 in C(X,Y) then for every point x in X the sequence {f_n(x)} in Y converges to f0(x).

    here's what I did, let x be in X and for every U_x, neighbourhood, of f0(x) i need to prove that there exists N s.t for every n>=N f_n(x) is in U_x.

    but because f_n converges to f_0 in C(X,Y), then for every U open in C(X,Y), s.t f_0 is in U then there exists N s.t for every n>=N f_n is in U, but U is open in C(X,Y) which means that U containes a set W(C,V) which contains f_n, i.e C is compact in X and V is open in Y, s.t f_n(C) is contained in V, but C is compact then so is f_n(C) (cause it's continuous), and thus for every collection of open set in Y which cover f_n(C) also finite union of them covers it.
    Now how to relate those matters, if f0(x) in U_x, then let's take a compact set which contains x then from what is given we know that there exists N s.t for every n>=N s.t f_n in W(C,U_x) then f_n(C) is contained in U_x which means that f_n(x) is in U_x.

    but here's the catch how do i know that there exists a compact set in X which contains x, unless X is a union of compact sets i don't see how to do this question, any hints?

    thanks in advance.
     
  2. jcsd
  3. Feb 16, 2008 #2

    morphism

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    I haven't really read your entire post, but in regards to the last line, surely {x} is a compact set that contains x!
     
  4. Feb 17, 2008 #3
    yes i know, but can you or someone else check my work or suggest another way to prove this statement?

    thanks in advance.
     
  5. Feb 17, 2008 #4

    morphism

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    I think you're making it much more harder than it is.

    If U is an open set containing f0(x) in Y, then W({x},U) is an open set containing f0 in C(X,Y).
     
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