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Math Help - Continuous and connectivity

  1. #1
    Super Member Showcase_22's Avatar
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    Continuous and connectivity

    Suppose that \{A,B\} is a partition of a topological space X and that f: X \rightarrow Y is a map to another space Y. Prove that if the restrictions f|A and f|B are both continuous then f is continuous.
    Let E \subset X s.t E \cap B \neq \phi and E \cap A \neq \phi.

    I need to show that A and B are connected (since if f is continuous on A and B, then the only place it might be discontinuous is at the boundary, where A meets B).

    A is connected (as is B) since f is a continuous map between any two points.

    This is what I know, but I can't see what I can do from here! Can someone help?
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  2. #2
    Senior Member Tinyboss's Avatar
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    What's the definition of a partition? I'm going to assume it's not just "sets whose union is the space", else a counterexample is X=[0,2],\:A=[0,1),\:B=[1,2] and

    f(x)=\begin{cases}0&x\in A\\1&x\in B.\end{cases}
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  3. #3
    Super Member Showcase_22's Avatar
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    If \{A,B\} is a partition of a set X, then A \cup B=X and A \cap B= \phi.

    I didn't think the second condition helped, so I left it out.
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  4. #4
    Senior Member Tinyboss's Avatar
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    Are you sure there's not another condition, such as A and B are both open or both closed? Else it can fail, as I demonstrated above.
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  5. #5
    Super Member Showcase_22's Avatar
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    Ah yes! Both A,B have to be open.

    Sorry! I forgot the most important condition.
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  6. #6
    Senior Member Tinyboss's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    Ah yes! Both A,B have to be open.

    Sorry! I forgot the most important condition.
    Well, there's a "lemma" that talks about "pasting" continuous functions together. ;-) I don't know if it would be kosher just to cite it, but the proof isn't hard.
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  7. #7
    Super Member Showcase_22's Avatar
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    Okay.

    I know that A and B are open. Since f is continuous then f(A), \ f(B) are also open.

    Im(f)=f(A) \cup f(B). This is the union of two open sets so Im(f) is open.

    Therefore for any E as defined before, f(E) is also going to be open so f maps an open set to an open set. Therefore f is continuous.

    I think that's about right. The odd thing is that this is in the connectedness chapter, that's why I was trying to approach it from a connectedness angle.
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  8. #8
    Senior Member Tinyboss's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    I know that A and B are open. Since f is continuous then f(A), \ f(B) are also open.
    This is false--it only works in the inverse direction. For example, constant functions are continuous, but their images are single points. And the image of f(x)=x^2 is [0,\infty), also not open.

    A function with the property you gave is called an "open map".
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  9. #9
    Super Member Showcase_22's Avatar
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    Sorry I keep getting this wrong! I think i'm getting either flustered or tired.

    let V \in Y be open.

    Then f^{-1}(V) \subset A \cup B.

    If f^{-1}(V) \subset A or f^{-1}(V) \subset B then f^{-1}(V) is open since we know that f is continuous on these intervals.

    If f^{-1}(V) has points in both A and B then partition V into two sets I,J where f^{-1}(I) \subset A and f^{-1}(J) \subset B.

    f^{-1}(I),\ f^{-1}(J) are both open since f is continuous on both A and B.

    Hence f^{-1}(V) is always open. The preimage of an open set is also an open set so f is continuous.
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  10. #10
    Member Focus's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    Sorry I keep getting this wrong! I think i'm getting either flustered or tired.

    let V \in Y be open.

    Then f^{-1}(V) \subset A \cup B.

    If f^{-1}(V) \subset A or f^{-1}(V) \subset B then f^{-1}(V) is open since we know that f is continuous on these intervals.

    If f^{-1}(V) has points in both A and B then partition V into two sets I,J where f^{-1}(I) \subset A and f^{-1}(J) \subset B.

    f^{-1}(I),\ f^{-1}(J) are both open since f is continuous on both A and B.

    Hence f^{-1}(V) is always open. The preimage of an open set is also an open set so f is continuous.
    How do you know that both I and J are open?

    Here is a simple way to do this, take U open in Y, then f^{-1}|_A(U)=f^{-1}(U) \cap A,f^{-1}|_B(U)=f^{-1}(U) \cap B are both open in X. Then f^{-1}(U)=f^{-1}(U) \cap (A\cup B)=(f^{-1}(U) \cap A)\cup(f^{-1}(U) \cap B) a union of two open sets, which is open.
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