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Thread: An open set is disconnected iff it is the union of two non-empty disjoint open sets

  1. #1
    Senior Member Pinkk's Avatar
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    An open set is disconnected iff it is the union of two non-empty disjoint open sets

    How do I proof this is the definition of disconnectedness that I am using that $\displaystyle S$ is disconnected if there exists two sets $\displaystyle S_{1}, S_{2}$ such that they are both non-emtpy, $\displaystyle S_{1} \cup S_{2} = S$ and $\displaystyle cl(S_{1})\cap S_{2} = cl(S_{2})\cap S_{1} = \emptyset$

    Attempting to prove the first condition, that disconnected implies the open set union condition, I tried to show that if $\displaystyle (S_{1},S_{2})$ form a disconnection, then so does $\displaystyle (S_{1} - \partial S_{1}, S_{2}-\partial S_{2})$ as well, but I am not even sure if this is the correct approach because I am not nsure how to show that those two sets would be non-empty and that their union would be $\displaystyle S$. And I am not sure at all how to show the other direction of the proof (that the open set union condition implies disconnectedness). Any help would be appreciated.
    Last edited by Pinkk; Sep 26th 2010 at 03:25 PM.
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  2. #2
    MHF Contributor

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    This is an ideal case for using RL Moore’s way of defining connected sets.
    Two non-empty sets are separated if neither contains a point or a limit point of the other.
    Then A set is connected if and only if it is not the union of two separated sets.
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  3. #3
    Senior Member Pinkk's Avatar
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    Then would I use that to assume to the contrary that a set is disconnected yet it is not the union of two non-empty disjoint open sets, so the disconnection must be of closed sets...? I am not sure how that definition follows to the proposition that I want to prove.

    Edit: So here's my attempt at a proof:

    Suppose $\displaystyle S_{1}, S_{2}$ are two non-empty disjoin open sets such that $\displaystyle S = S_{1} \cup S_{2}$ but suppose to the contrary that they do not form a disconnection of $\displaystyle S$. Then, without loss of generality, $\displaystyle cl(S_{1}) \cap S_{2} \ne \emptyset$. Since $\displaystyle S_{1} \cap S_{2} = \emptyset$, it follows that there exists $\displaystyle x\in \partial S_{1} \cap S_{2}$. So there exists $\displaystyle \delta > 0$ such that $\displaystyle B(\delta, x) \subset S_{2}$ and this ball contains elements in $\displaystyle S_{1}$ and $\displaystyle (S_{1})^{c}$ since x is a boundary point of $\displaystyle S_{1}$. This means there exists $\displaystyle y \in S_{1} \cap S_{2}$, which is a contradiction, so $\displaystyle cl(S_{1}) \cap S_{2} = \emptyset$ and since the index of the sets was arbitrarily fixed, it follows $\displaystyle (S_{1}, S_{2})$ is in fact a disconnection.

    But now I'm still having trouble proving the converse...
    Last edited by Pinkk; Sep 26th 2010 at 07:04 PM.
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