Suppose $X$ is a connected complete metric space with more than one point. Must $X$ contain a non-singleton non-empty connected proper open subset?
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Thank you to Daniel and David. – user156619 Jun 27 '14 at 23:01
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1Can a small open ball be disconnected? – mfl Jun 27 '14 at 23:01
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1The Theorem in the first answer here implies that it does. – David Mitra Jun 27 '14 at 23:31
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@DavidMitra I don't see how. We can get a nontrivial proper connected subset from the theorem, but does it have to be open? – Jun 28 '14 at 04:23
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@Thisismuchhealthier. Oops. I forgot about the open condition. – David Mitra Jun 28 '14 at 09:22
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Probably yes.
Theorem. If $X$ is a connected separable metric space or compact Hausdorff space with more than one point, then $X$ has a proper open connected subset.
proof. Clearly $X$ is infinite. Thus if $X\setminus \{x\}$ is connected for any $x\in X$, then you have an open set as desired. If $X$ is compact Hausdorff, then $X$ has a non-cut point, and so we're done.
Otherwise, every point of $X$ is a cut-point. In Topology. Vol. II by K. Kuratowski [Theorem 1, page 160], it is shown that for a connected separable metric space $Z$, the set $Z\setminus \{z\}$ is connected or is the union of two connected sets for every $z\in Z$ except for a countable set of points of $Z$. As $X$ is uncountable, for some $x\in X$ we have that $X\setminus \{x\}$ is the union of two disjoint connected open sets. Either one of these works for the desired open set.
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