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Math Help - topology; connectedness

  1. #1
    Senior Member abhishekkgp's Avatar
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    topology; connectedness

    Let S be a disconnected space, say S=U \cup V, where U \cap V = \Phi and U, V are both open and non-empty. If x \in U and y \in V, prove that there can be no continuous function f:[0,1] \rightarrow S such that f(0)=x and f(1)=y.

    I want to know HOW important is the "such that" thing.
    Here is my attempt:

    Let f be continuous.
    Then f^{-1}(U), f^{-1}(V) are both open in [0,1].
    Also S=U \cup V \Rightarrow f^{-1}(U) \cup f^{-1}(V)=[0,1].
    Again U \cap V = \Phi \Rightarrow f^{-1}(U) \cap f^{-1}(V)=\Phi. Also 0 \in f^{-1}(U), 1 \in f^{-1}(V) hence these are non empty.
    The above suggests that [0,1] is disconnected which is wrong hence we have a contradiction.
    Have i missed something or is the proof correct?
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  2. #2
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    Re: topology; connectedness

    Quote Originally Posted by abhishekkgp View Post
    Let S be a disconnected space, say S=U \cup V, where U \cap V = \Phi and U, V are both open and non-empty. If x \in U and y \in V, prove that there can be no continuous function f:[0,1] \rightarrow S such that f(0)=x and f(1)=y.
    Let f be continuous.
    Then f^{-1}(U), f^{-1}(V) are both open in [0,1].
    Also S=U \cup V \Rightarrow f^{-1}(U) \cup f^{-1}(V)=[0,1].
    Again U \cap V = \Phi \Rightarrow f^{-1}(U) \cap f^{-1}(V)=\Phi. Also 0 \in f^{-1}(U), 1 \in f^{-1}(V) hence these are non empty.
    The above suggests that [0,1] is disconnected which is wrong hence we have a contradiction.
    Have i missed something or is the proof correct?
    Well that is close to a complete proof.
    We know that the continuous image of a connected set is connected.
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Re: topology; connectedness

    Quote Originally Posted by Plato View Post
    Well that is close to a complete proof.
    We know that the continuous image of a connected set is connected.
    I was worried about my proof because x and y are images of the END POINTS of [0,1] while my proof would work even if x and y were images of ANY two distinct points in [0,1].
    My proof's right, right?
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