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Thread: Disjoint connected

  1. #1
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    Disjoint connected

    Suppose $\displaystyle A,B $ are connected subsets of some metric space $\displaystyle X $. Prove that if $\displaystyle A \cap B \neq \emptyset $ the $\displaystyle A \cup B $ is connected.

    So suppose $\displaystyle A \cup B $ is not connected.

    Then $\displaystyle A \cup B = C \cup D $ where $\displaystyle C $ and $\displaystyle D $ are disjoint and clopen. So $\displaystyle C = E_{O} \cap X $ and $\displaystyle C = E_{C} \cap X $. Also $\displaystyle D = F_{O} \cap X $ and $\displaystyle D = F_{C} \cap X $.

    So $\displaystyle A \cup B = (E_{O} \cap X) \cup (F_{O} \cap X) $. Now what? How do you conclude that $\displaystyle A \cap B = \emptyset $?
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    Quote Originally Posted by heathrowjohnny View Post
    Suppose $\displaystyle A,B $ are connected subsets of some metric space $\displaystyle X $. Prove that if $\displaystyle A \cap B \neq \emptyset $ the $\displaystyle A \cup B $ is connected.
    So suppose $\displaystyle A \cup B $ is not connected.
    Then $\displaystyle A \cup B = C \cup D $ where $\displaystyle C $ and $\displaystyle D $ are disjoint and clopen.
    At that point, show that A must be a subset of either C or D.
    Then B must be a subset of the other. There is your contradiction.
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    Quote Originally Posted by Plato View Post
    At that point, show that A must be a subset of either C or D.
    Then B must be a subset of the other. There is your contradiction.
    Isn't this also contropostition? But this all depends on relative openess/closeness right?

    Writing $\displaystyle C = E_{O} \cap X $ is what you should do? E.g. $\displaystyle E_O $ is an open set set. We have to show $\displaystyle C $ is open relative to $\displaystyle A \cup B $ right?

    So want to show that $\displaystyle A \subseteq E_{O} \cap X $ and so on....?
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    Quote Originally Posted by heathrowjohnny View Post
    We have to show $\displaystyle C $ is open relative to $\displaystyle A \cup B $ right?
    I have no idea how your textbook/instructor requires you to prove connectivity.
    The usual definition simply says that X is connected if and only if there do no exist two sets C & D such that $\displaystyle \begin{gathered} X \cap C \ne \emptyset \;,\;X \cap D \ne \emptyset \;\& \;X \subseteq C \cup D \hfill \\ \overline C \cap D = \emptyset \;\& \;\overline D \cap C = \emptyset \hfill \\ \end{gathered}$ .
    Put simply: X is not a subset of the union of two sets, each having a nonempty intersection with X, and neither contains a point or a limit point of the other. That is known as a separation of X.
    It seems that you are to use some other approach to connectivity.
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