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Math Help - Continuity in metric spaces

  1. #1
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    Continuity in metric spaces

    Let X be a metric space, s.t. X = C \cup D where C and D are closed sets.

    Suppose f: X -> Y is continuous on C and D.

    Show f is continuous on X

    My proof:

    x \in X implies x \in C, or x \in D or x \in C \cap D

    (I am just going to treat the elements in the intersection as belonging to C or D only.)

    Let V \subset Y be closed in Y.

    By continuity of f:

    f^{-1}(V) is closed in C
    f^{-1}(V) is closed in D

    I want to say this is enough to prove f is continuous on X.

    But I think I am lacking something. I don't really use the fact that C and D are closed.

    This theorem could say open and I could literally do the same thing. But I am 100% sure this is not true for open sets.

    I'm really not sure what to do. I don't think considering complaints here helps at all.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Continuity in metric spaces

    Quote Originally Posted by Jame View Post
    Let X be a metric space, s.t. X = C \cup D where C and D are closed sets.

    Suppose f: X -> Y is continuous on C and D.

    Show f is continuous on X

    My proof:

    x \in X implies x \in C, or x \in D or x \in C \cap D

    (I am just going to treat the elements in the intersection as belonging to C or D only.)

    Let V \subset Y be closed in Y.

    By continuity of f:

    f^{-1}(V) is closed in C
    f^{-1}(V) is closed in D

    I want to say this is enough to prove f is continuous on X.

    But I think I am lacking something. I don't really use the fact that C and D are closed.

    This theorem could say open and I could literally do the same thing. But I am 100% sure this is not true for open sets.

    I'm really not sure what to do. I don't think considering complaints here helps at all.
    Well, they need to agree on C\cap D...but anwaysys, so you know that f^{-1}(V)=C\cap A and f^{-1}(V)=D\cap B for closes sets A,B\subseteq X but since C,D are closed this implies that C\cap A,D\cap B are closed in X and since f^{-1}(V)=(C\cap A)\cup (D\cap B) the conclusion follows.


    Also, this is true for any finitely many closed sets X=C_1\cup\cdots\cup C_n and true for arbitrary open sets \displaystyle X=\bigcup_{\alpha\in\mathcal{A}}D_\alpha.
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    Re: Continuity in metric spaces

    Cool. Makes sense.

    I have a question. Why must we union C \cap A and D \cap B? Is knowing f^{-1}(V) = C \cap A enough to say f is continuous on X? C \cap A is a closed set in X.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Continuity in metric spaces

    Quote Originally Posted by Jame View Post
    Cool. Makes sense.

    I have a question. Why must we union C \cap A and D \cap B? Is knowing f^{-1}(V) = C \cap A enough to say f is continuous on X? C \cap A is a closed set in X.
    Because technically what is meant is that (f_{\mid C})^{-1}(V)=C\cap A where f_{\mid C} is the restriction of f to C, but I suppressed that.
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