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Let H be a separable and infinite-dimensional Hilbert space. Is every closed subset of H homeomorphic to some closed and bounded subset of H?

flag	asked Jun 19 2010 at 19:30 Garabed Gulbenkian 3,011●6●13

It would suffice to know whether H is homoeomorphic to a closed bounded subset of itself. – Robin Chapman Jun 19 2010 at 19:40

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You may want to reprhase the title to actually match the question. – Andrea Ferretti Jun 19 2010 at 22:49

to Robin Chapman: Actually, as you can see from the response below, your question has an affirmative answer-since it has been proved that H is homeomorphic to its own closed unit ball. But I still do not see how I can use this result to prove that my question has an affirmative answer (or that it has a negative answer). – Garabed Gulbenkian Jun 22 2010 at 18:56

I will never understand why it took me so long to see that your question is equivalent to mine but I finally do see it. Thanks for teaching me something about homeomorphisms that I once learned but had somehow forgotten. – Garabed Gulbenkian Jun 28 2010 at 13:47

Ady · Answer 1 · 2010-06-19 20:23:03Z UTC

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It is an old result of Klee saying that the infinite-dimensional Hilbert space is homeomorphic with both its unit sphere and its closed unit ball. See e.g. http://www.ams.org/journals/bull/1961-67-03/S0002-9904-1961-10589-2/S0002-9904-1961-10589-2.pdf , and the references therein.

answered Jun 19 2010 at 20:23

Ady
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Also, H is homeomorphic with any closed half-space of it, which is the reason why a "C<sup>0</sup> Hilbert manifold with boundary" is in fact a C<sup>0</sup> Hilbert manifold without boundary. – Pietro Majer Jun 19 2010 at 21:46

When can boundedness be characterized topologically in Metric spaces?

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