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Let $\{A_i\}_{i \in I}$ be a family of connected subsets of a metric space $X$ ($I$ is some set of indices). Show that if the intersection $\bigcap A_i \neq \emptyset$ , then $\bigcup A_i$ is connected.


marked as duplicate by user147263, Ilmari Karonen, Jack Lee, Brian M. Scott general-topology Nov 4 '15 at 1:22

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  • $\begingroup$ Sorry. Don't know how to write math characters on here. $\endgroup$ – user2684794 Nov 3 '15 at 23:52
  • $\begingroup$ There’s a tutorial and quick reference here for writing mathematics on this site. $\endgroup$ – Brian M. Scott Nov 4 '15 at 1:23

Since $\bigcap A_i \ne \emptyset$, let $x \in \bigcap A_i$. Now if $U$ and $V$ are non-empty disjoint open sets separating $\bigcup A_i$ then $x \in U$ or $x \in V$. Without loss of generality, suppose $x \in U$. Now, where does $x$ come from and what can we say about it?

  • $\begingroup$ Not sure what you mean by "separating the union of A_i" $\endgroup$ – user2684794 Nov 4 '15 at 0:06
  • $\begingroup$ Meaning $ A = U\cup V $. $\endgroup$ – Kevin Sheng Nov 4 '15 at 0:14
  • $\begingroup$ I guess you could say x comes from the intersection of A_i but is not an element of V $\endgroup$ – user2684794 Nov 4 '15 at 0:25
  • $\begingroup$ But that must be a contradiction... do you see why? $\endgroup$ – Kevin Sheng Nov 4 '15 at 0:53
  • $\begingroup$ because it's the intersection? so it must be both? $\endgroup$ – user2684794 Nov 4 '15 at 2:26

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