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Math Help - 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 S is disconnected if there exists two sets S_{1}, S_{2} such that they are both non-emtpy, S_{1} \cup S_{2} = S and 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 (S_{1},S_{2}) form a disconnection, then so does (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 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; September 26th 2010 at 04:25 PM.
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  2. #2
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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 S_{1}, S_{2} are two non-empty disjoin open sets such that S = S_{1} \cup S_{2} but suppose to the contrary that they do not form a disconnection of S. Then, without loss of generality, cl(S_{1}) \cap S_{2} \ne \emptyset. Since S_{1} \cap S_{2} = \emptyset, it follows that there exists x\in \partial S_{1} \cap S_{2}. So there exists \delta > 0 such that B(\delta, x) \subset S_{2} and this ball contains elements in S_{1} and (S_{1})^{c} since x is a boundary point of S_{1}. This means there exists y \in S_{1} \cap S_{2}, which is a contradiction, so cl(S_{1}) \cap S_{2} = \emptyset and since the index of the sets was arbitrarily fixed, it follows (S_{1}, S_{2}) is in fact a disconnection.

    But now I'm still having trouble proving the converse...
    Last edited by Pinkk; September 26th 2010 at 08:04 PM.
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