Is $\mathbb{R}^\omega \cong (\mathbb{R}^\omega \setminus \{x\})$, where $\mathbb{R}^\omega$ is given the product topology, and $x\in\mathbb{R}^\omega $?
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asked Jun 7, 2018 at 15:46
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1$\begingroup$ This question also came up in mathoverflow.net/questions/301892/… $\endgroup$– R.P.Commented Jun 7, 2018 at 15:53
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11$\begingroup$ Yes, any two homotopy equivalent Hilbert manifolds are homeomorphic, see en.wikipedia.org/wiki/Hilbert_manifold. The countable product of lines is homeomorphic to the separable Hilbert space, see Bessaga C., Pelczynski A. "Selected Topics in Infinite-Dimensional topology" for much more information. $\endgroup$– Igor BelegradekCommented Jun 7, 2018 at 16:11
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1$\begingroup$ In fact, the complement in $\mathbb R^\omega$ of the union of any countable family of compact subsets is homeomorphic to $\mathbb R^\omega$, see e.g. Theorem V.6.4 in Bessaga-Pelczynski's book. $\endgroup$– Igor BelegradekCommented Jun 7, 2018 at 20:12
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$\begingroup$ Great @IgorBelegradek, thanks! Can you put this into an answer so we can close this thread? $\endgroup$– Dominic van der ZypenCommented Jun 7, 2018 at 20:42
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1 Answer
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The answer is yes. Any two homotopy equivalent Hilbert manifolds are homeomorphic. The countable product of lines is homeomorphic to the separable Hilbert space, see p.174 of [Bessaga C., Pelczynski A., Selected Topics in Infinite-Dimensional topology] where much more information is given.
More generally, the complement in $\mathbb R^\omega$ of the union of any countable family of compact subsets is homeomorphic to $\mathbb R^\omega$, see e.g. Theorem V.6.4 in Bessaga-Pelczynski's book.
answered Jun 7, 2018 at 21:36
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$\begingroup$ How is $\mathbb R^\omega$ homotopy equivalent to itself with a point removed? $\endgroup$– WojowuCommented Jun 7, 2018 at 21:49
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1$\begingroup$ The proof of the second statement is also in van Mill's book "infinite-dimensional topology, prerequisites and an introduction", in a self-contained way. $\endgroup$ Commented Jun 7, 2018 at 22:09
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2$\begingroup$ @Wojowu ... $\mathbb R^\omega \setminus \{0\}$ is contractible and thus homotopic to a point ... unlike $\mathbb R^n \setminus \{0\}$ and thus contrary to our intuition. $\endgroup$ Commented Jun 7, 2018 at 22:17
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1$\begingroup$ @GeraldEdgar ...how is it homotopic to a point? Sorry for such silly questions, I have never studied those topics. $\endgroup$– WojowuCommented Jun 7, 2018 at 22:19
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