Question

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"?

Answer 1

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.