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Topology: Hausdorff Spaces

  1. Nov 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Let X be a topological space, Y a Hausdorff space, and let f:X -> Y and g:X -> Y be continuous. Show that {x [tex]\in[/tex] X : f(x) = g(x)} is closed. Hence if f(x) = g(x) for all x in a dense subset of X, then f = g.


    2. Relevant equations
    Y is Hausdorff => for every x, y in Y with x != y, there exist disjoint open sets U, V with x in U and y in V.
    f continuous iff f-1(V) is open in X whenever V is open in Y, iff f-1(F) is closed whenever F is closed.

    3. The attempt at a solution
    I could show the set is closed by proving that its complement is open, but do I want to take that route? The complement is {x in X : f(x) != g(x)}. So is this itself a Hausdorff space? I'm not sure if this is the right way to go with this, or even if it's correct. Thanks, as always, for any help.
     
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  3. Nov 6, 2008 #2

    HallsofIvy

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    The question "is this itself a Hausdorff space" is meaningless. It is not a space at all- it is a subset of a topological space. (You can make it a topological space by using the "inherited" topology but surely that's not what you want to do. You want to show that it is closed in X but every set is closed in its own inherited topology.) In any case, the set of all {x| f(x)= g(x)} or {x| f(x)!= g(x)} is a subset of X, not Y, and you are not given that X is hausdorff.
     
  4. Nov 6, 2008 #3

    Dick

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    Yes. Show {x:f(x)!=g(x)} is open. Start by using that f(x) and g(x) are in Y and Y is Hausdorff.
     
  5. Nov 6, 2008 #4
    Ok, if I use that fact, then if f(x) != g(x), there exist subsets U, V disjoint and open with f(x) in U and g(x) in V. Since f is continuous, f-1(U) is open in X (because U is open in Y). Similarly, g-1(V) is open in X (because V is open in Y). Thus {x in X | f(x) != g(x)} is open, and it's complement, {x in X | f(x) = g(x)} is closed.

    Is this correct, or am I missing something?
     
  6. Nov 6, 2008 #5

    Dick

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    The conclusion is correct. You didn't state exactly why this means that x has a open neighborhood where f!=g. I understand it and I think you understand it. You might just want to be a little more explicit.
     
  7. Nov 6, 2008 #6
    Is it basically by definition? Since f is continuous, f-1(N) is a neighborhood of x for every neighborhood N of f(x). Maybe I don't understand it, heh.
     
  8. Nov 6, 2008 #7

    Dick

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    Maybe you don't quite understand it. I was giving you the benefit of the doubt for the gap in the proof. You have that f^(-1)(U) and f^(-1)(V) are both open neighborhoods of x. What about their intersection? Is it also an open neighborhood of x? If so, then can you show that if z is an element of the intersection, then f(z)!=g(z)? That would mean f!=g is open.
     
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