# A question about open subsets of Hilbert space

If H is (a separable and infinite dimensional) Hilbert space and if U is a non-empty open subset of H that is not connected, does the boundary B of U always have at least one component that is not a singleton?

-
Just a few questions: Why do you put so strict hypothesis (i.e. separability and Hilbert)? What about the finite-dimesional case (dim>1)? What about Banach spaces? And if we remove completeness? – Michele Triestino Sep 26 '10 at 17:50
Evidently, I was thinking of a real vector space.. Are you interested in complex Hilbert spaces? – Michele Triestino Sep 26 '10 at 17:56
Michele, every Banach space is homeomorphic to a Hilbert space, and every complex Hilbert space is homeomorphic (actually real-linear homeomorphic to) a real Hilbert space. – Bill Johnson Sep 27 '10 at 0:09
Can't you simplify the question to does every connected (bounded) subset of a Hilbert space have boundary consisting of more than one point. – Owen Sizemore Sep 27 '10 at 3:36
@Bill Johnson: I didn't know that result. Can you give a reference? – Nate Eldredge Sep 27 '10 at 13:03

If $\dim >2$, choose a plane such that its intersection with U is not connected and use the above statement.
I did not assume that $\dim<\infty$. – Anton Petrunin Sep 29 '10 at 19:47