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

    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 $\displaystyle U,V$ are open in $\displaystyle A$ connected, and $\displaystyle U \cup V = A$, then $\displaystyle U \cap V \ne \emptyset$. 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 $\displaystyle S_{1},S_{2}$ are connected sets in $\displaystyle \mathbb{R}^{n}$ and $\displaystyle S_{1}\cap S_{2} \ne \emptyset$. Show that $\displaystyle S_{1}\cup S_{2}$ is connected. Does this hold true for $\displaystyle S_{1} \cap S_{2}$?

    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 $\displaystyle U,V$ are open in $\displaystyle A$ connected, and $\displaystyle U \cup V = A$, then $\displaystyle U \cap V \ne \emptyset$. 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 $\displaystyle S_1\cup S_2$ of connected sets $\displaystyle S_{1,2}$ is connected, if $\displaystyle S_1\cap S_2\neq \emptyset$, it suffices to show that given any two open sets $\displaystyle U,V$ with $\displaystyle S_1\cup S_2\subseteq U\cup V$, and $\displaystyle U\cap V=\emptyset$, it follows that either $\displaystyle S_1\cup S_2\subseteq U$ or $\displaystyle S_1\cup S_2\subseteq V$.

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

    Finally, to the question as to whether $\displaystyle S_1\cap S_2$ is connected if $\displaystyle S_{1,2}$ are connected and $\displaystyle S_1\cap S_2\neq \emptyset$. The answer is no, for consider the intersection of a line and a circle in $\displaystyle \mathbb{R}^2$ that consists of two isolated points. In such a case, the intersection is clearly not connected.
    Last edited by Failure; Sep 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 $\displaystyle U~\&~V$ are separated sets such $\displaystyle S_1\cup S_2=U\cup V$.
    We know that $\displaystyle \left( {\exists t \in S_1 \cap S_2 } \right)$ so $\displaystyle t\in U\text{ or }t\in V$.
    In either case, because each of $\displaystyle S_1~\&~S_2$ is connected there follows a contradiction.
    So $\displaystyle S_1\cup S_2=U\cup V$ 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 $\displaystyle t$ being in either $\displaystyle U$ or $\displaystyle V$. That is, I don't see why if $\displaystyle t\in S_{1}$ and $\displaystyle S_{1}$ means $\displaystyle S_{1} \subset U$.
    Last edited by Pinkk; Sep 16th 2010 at 09:03 PM.
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  5. #5
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    Otherwise $\displaystyle S_1\cap U~\&~ S_1\cap V $ would be a separation of $\displaystyle S_1$ 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; Sep 17th 2010 at 03:33 AM.
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