By a "totally disconnected" point set I mean one whose only connected subsets are singletons. Can a finite dimensional Euclidean space whose dimension is at least two, be separated by any subset that is "totally disconnected"? Such a subset could not be closed in the space, for then it would be locally compact and therefore zero-dimensional. If we move beyond locally compact spaces, can a separable and infinite dimensional Hilbert space be separated by any subset that is "totally disconnected"?
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asked Aug 26, 2010 at 17:55
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1 Answer
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Assume the complement of $S$ in $\mathbb{R}^n$ is not connected, say $A$ and $B$ are relatively closed and disjoint in $\mathbb{R}^n\setminus S$ (and nonempty of course); let $O$ be the complement of the closure of $B$ and $U$ the complement of the closure of $A$, then $O$ and $U$ are disjoint nonempty open subsets of $\mathbb{R}^n$ and the complement of their union, $F$, is closed in $\mathbb{R}^n$, a subset of $S$ and it separates $\mathbb{R}^n$. In short: $S$ contains a closed set that also separates; as you noted that set is zero-dimensional and hence the answer is `no' for Euclidean spaces. I don't know (yet) about Hilbert space.
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$\begingroup$ Many thanks for your response and its very nice proof. It looks as if the same line of argument could be used for Hilbert space up to the point where you have to show whether or not a closed and totally disconnected subset of Hilbert space can diconnect it. $\endgroup$ Commented Aug 27, 2010 at 20:16
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$\begingroup$ (To be pedantic one should mention that under the given assumptions $S$ has empty interior). $\endgroup$ Commented May 10, 2013 at 23:04