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Math Help - The union of non-disjoint connected is connected

  1. #1
    Senior Member Pinkk's Avatar
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    The union of non-disjoint connected is connected

    Suppose S_{1},S_{2} are connected sets in \mathbb{R}^{n} and S_{1}\cap S_{2} \ne \emptyset. Show that S_{1}\cup S_{2} is connected. Does this hold true for S_{1} \cap S_{2}?

    Okay, so I have seen proofs that rely on topological properties that we have not really discussed or proven in my class (like for instance, if U,V are open in A connected, and U \cup V = A, then U \cap V \ne \emptyset. Assuming that, I understand the proof for this theorem pretty well, but how do I prove that assumption? Or is there another way to solve this question without invoking open sets/topology in general? Thanks.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Pinkk View Post
    Suppose S_{1},S_{2} are connected sets in \mathbb{R}^{n} and S_{1}\cap S_{2} \ne \emptyset. Show that S_{1}\cup S_{2} is connected. Does this hold true for S_{1} \cap S_{2}?

    Okay, so I have seen proofs that rely on topological properties that we have not really discussed or proven in my class (like for instance, if U,V are open in A connected, and U \cup V = A, then U \cap V \ne \emptyset. Assuming that, I understand the proof for this theorem pretty well, but how do I prove that assumption? Or is there another way to solve this question without invoking open sets/topology in general? Thanks.
    Since "being connected" is defined in terms of open sets, I don't think there is a way around talking of open sets and the like in a proof of the above proposition.

    To show that the union S_1\cup S_2 of connected sets S_{1,2} is connected, if S_1\cap S_2\neq \emptyset, it suffices to show that given any two open sets U,V with S_1\cup S_2\subseteq U\cup V, and U\cap V=\emptyset, it follows that either S_1\cup S_2\subseteq U or S_1\cup S_2\subseteq V.

    And why is that? Well, suppose that x\in S_1\cap S_2, then we have either x\in U or x\in V (but not both, because U and V are disjoint).
    If x\in U, then, because S_{1,2} are both connected, it follows that S_1\subseteq U and S_2\subseteq U, and therefore S_1\cup S_2\subseteq U.
    Similarly, if x\in V it follows that S_1\cup S_2\subseteq V.

    Finally, to the question as to whether S_1\cap S_2 is connected if S_{1,2} are connected and S_1\cap S_2\neq \emptyset. The answer is no, for consider the intersection of a line and a circle in \mathbb{R}^2 that consists of two isolated points. In such a case, the intersection is clearly not connected.
    Last edited by Failure; September 15th 2010 at 11:21 PM.
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  3. #3
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    There is another characterization of connect sets.
    A set is connected if it is not the union of two separated sets
    Two sets are separated if each is non-empty and neither contains a point nor a limit point of the other.

    Suppose that U~\&~V are separated sets such S_1\cup S_2=U\cup V.
    We know that \left( {\exists t \in S_1  \cap S_2 } \right) so t\in U\text{ or }t\in V.
    In either case, because each of S_1~\&~S_2 is connected there follows a contradiction.
    So S_1\cup S_2=U\cup V must be connected.
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  4. #4
    Senior Member Pinkk's Avatar
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    I don't see how the contradiction automatically follows from t being in either U or V. That is, I don't see why if t\in S_{1} and S_{1} means S_{1} \subset U.
    Last edited by Pinkk; September 16th 2010 at 09:03 PM.
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  5. #5
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    Otherwise S_1\cap U~\&~ S_1\cap V would be a separation of  S_1 which is connected.

    The basic idea underlying this proof is this theorem:
    If a connected set is a subset of the union of two separated sets then the set is a subset of one of the two.
    Last edited by Plato; September 17th 2010 at 03:33 AM.
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