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Math Help - locally constant

  1. #1
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    locally constant

    Hello,

    I want to show the following :
    Let U be a open subset of IR^2.
    If f is constant on each connected component, then f:U->\IR is locally constant.

    My Idea was this:
    Let x be a arbitrary point of U. We have to show that there is a (open) nbh. V of x, s.t.
    f|V is constant.
    We know x is an elm. of a conn. component V' of U and f is constant on V'.

    Now i couldn't show that there exist an (open) nbh. V of x, s.t. V is a subset of V'.

    Is this the right way? how can i show this property of conn. components.


    Thank you for your help
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  2. #2
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    You do have the correct idea. Just recall that a component is a maximally connected subset. So there is no problem getting a neighborhood as a subset.
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  3. #3
    Senior Member Tinyboss's Avatar
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    Are connected components in this space open or closed? Or...?
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  4. #4
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    Connectedness components in any topological space are both open and closed.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Connectedness components in any topological space are both open and closed.
    This is not always true. It's true that if X is a topological space and C\subseteq X is a component that C must be closed since \overline{C} is a connected set containing C and so by maximality we may conclude that C=\overline{C}. That said, the following theorem is true

    Theorem: Let X be a topological space. Then, every connected component of X is open if and only if X is locally connected.
    [
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