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?
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asked Jun 19, 2010 at 19:30
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$\begingroup$ It would suffice to know whether H is homoeomorphic to a closed bounded subset of itself. $\endgroup$– Robin ChapmanCommented Jun 19, 2010 at 19:40
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1$\begingroup$ You may want to reprhase the title to actually match the question. $\endgroup$– Andrea FerrettiCommented Jun 19, 2010 at 22:49
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$\begingroup$ 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). $\endgroup$– Garabed GulbenkianCommented Jun 22, 2010 at 18:56
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$\begingroup$ 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. $\endgroup$– Garabed GulbenkianCommented Jun 28, 2010 at 13:47
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1 Answer
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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.
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1$\begingroup$ 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. $\endgroup$ Commented Jun 19, 2010 at 21:46