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Show a metric space X is connected iff \forall non-empty proper subset A \subset X the boundary (set of points in X whose neighborhoods contain points from both A and the complement) is non-empty.

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  • $\begingroup$ What definition of connected are you using? $\endgroup$ – Tim Raczkowski Nov 4 '15 at 16:36
  • $\begingroup$ if it cannot be written as the union of two disjoint, non-empty closed (or open) subsets $\endgroup$ – user2684794 Nov 4 '15 at 16:52
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If $X$ is disconnected, so $X = A \cup B$ where $A$ and $B$ are disjoint non-empty closed (and open) subsets. What is $\operatorname{Bd}(A)$?

Suppose $A$ is non-empty and suppose $\operatorname{Bd}(A) = \emptyset$; as the boundary of $A$ is the difference between closure and interior of $A$, we have that these coincide and so $A$ is clopen. So $X = A \cup \ldots $ proves $X$ is disconnected.

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  • $\begingroup$ A union A complement? $\endgroup$ – user2684794 Nov 4 '15 at 17:46
  • $\begingroup$ Indeed. And $X \setminus A$ is also... $\endgroup$ – Henno Brandsma Nov 4 '15 at 17:47
  • $\begingroup$ X \ A is also A complement? or B? Are you allowed to say both of those? $\endgroup$ – user2684794 Nov 4 '15 at 17:48
  • $\begingroup$ $B = X \setminus A$, one could say. But closed and open is what I meant. $\endgroup$ – Henno Brandsma Nov 4 '15 at 17:49
  • $\begingroup$ I'm trying to let it sink in. $\endgroup$ – user2684794 Nov 4 '15 at 17:53

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