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?

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    $\begingroup$ 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? $\endgroup$ – Michele Triestino Sep 26 '10 at 17:50
  • $\begingroup$ Evidently, I was thinking of a real vector space.. Are you interested in complex Hilbert spaces? $\endgroup$ – Michele Triestino Sep 26 '10 at 17:56
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    $\begingroup$ 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. $\endgroup$ – Bill Johnson Sep 27 '10 at 0:09
  • $\begingroup$ Can't you simplify the question to does every connected (bounded) subset of a Hilbert space have boundary consisting of more than one point. $\endgroup$ – Owen Sizemore Sep 27 '10 at 3:36
  • $\begingroup$ @Bill Johnson: I didn't know that result. Can you give a reference? $\endgroup$ – Nate Eldredge Sep 27 '10 at 13:03

Clearly it is true in the plane.

If $\dim >2$, choose a plane such that its intersection with U is not connected and use the above statement.

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  • $\begingroup$ Thanks, Anton, for showing how one could go about proving that my question has a "yes" answer when restricted to a finite dimensional Euclidean space having dimension not less than 2. This answer follows from the fact that closed and totally disconnected subsets of such Euclidean spaces are zero dimensional. I asked my question about Hilbert space because (1) I do not know if it has been answered in this case (2)Hilbert space is closest to Euclidean spaces and homeomorphic to separable Banach spaces as Bill Johnson has pointed out. $\endgroup$ – Garabed Gulbenkian Sep 29 '10 at 19:43
  • $\begingroup$ I did not assume that $\dim<\infty$. $\endgroup$ – Anton Petrunin Sep 29 '10 at 19:47
  • $\begingroup$ @Anton Petrunin: Applying your strategy to Hilbert space, suppose That P denotes the plane in H which we choose to intersect with U. This intersection will be an open subset ,V, of P. Then we need to prove (1)We can always choose P so that, if U is not connected then V is not connected. (2) the boundary ,F, of V is a subset of the boundary ,B, of U. When we consider the humunguously complicated shapes that open subsets of Hilbert space (such as U) can have, I wonder how difficult it might be to actually prove (1) and (2). $\endgroup$ – Garabed Gulbenkian Oct 2 '10 at 15:10
  • $\begingroup$ My mistake! It is not at all difficult to prove (1) and (2), so your method works for the infinite-dimensional case as well. Bravo. $\endgroup$ – Garabed Gulbenkian Oct 5 '10 at 21:54

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