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Thread: topology; connectedness

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    Senior Member abhishekkgp's Avatar
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    topology; connectedness

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

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

    Let $\displaystyle f$ be continuous.
    Then $\displaystyle f^{-1}(U), f^{-1}(V)$ are both open in $\displaystyle [0,1]$.
    Also $\displaystyle S=U \cup V \Rightarrow f^{-1}(U) \cup f^{-1}(V)=[0,1]$.
    Again $\displaystyle U \cap V = \Phi \Rightarrow f^{-1}(U) \cap f^{-1}(V)=\Phi$. Also $\displaystyle 0 \in f^{-1}(U), 1 \in f^{-1}(V)$ hence these are non empty.
    The above suggests that $\displaystyle [0,1]$ 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 $\displaystyle S$ be a disconnected space, say $\displaystyle S=U \cup V$, where $\displaystyle U \cap V = \Phi$ and $\displaystyle U, V$ are both open and non-empty. If$\displaystyle x \in U$ and $\displaystyle y \in V$, prove that there can be no continuous function $\displaystyle f:[0,1] \rightarrow S$ such that $\displaystyle f(0)=x$ and $\displaystyle f(1)=y$.
    Let $\displaystyle f$ be continuous.
    Then $\displaystyle f^{-1}(U), f^{-1}(V)$ are both open in $\displaystyle [0,1]$.
    Also $\displaystyle S=U \cup V \Rightarrow f^{-1}(U) \cup f^{-1}(V)=[0,1]$.
    Again $\displaystyle U \cap V = \Phi \Rightarrow f^{-1}(U) \cap f^{-1}(V)=\Phi$. Also $\displaystyle 0 \in f^{-1}(U), 1 \in f^{-1}(V)$ hence these are non empty.
    The above suggests that $\displaystyle [0,1]$ 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 $\displaystyle x$ and $\displaystyle y$ 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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