Let H be a separable and infinite dimensional Hilbert space. Let C be a closed connected subset of H which contains more than one point. Can C ever be a countable union of disjoint closed subsets of H? Sierpinski proved that the answer is negative if H is locally compact. But what happens if the condition of local compactness is not imposed?
A question about closed and connected subsets of Hilbert space
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general-topology
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0I don't know the answer, but I wonder what Hilbert space has to do with anything. Is Sierpinkski's result true for general topological spaces, and is there an obvious example of a non-locally compact topological space with this property that isn't a Hilbert space? – 2017-02-20
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0I mentioned Hilbert Space because it one of the most familiar metric spaces that is not locally compact. Sierpinski's theorem is often stated in books about topology for finite dimensional Euclidean spaces. – 2017-02-20