$\mathbb{R}^n\not\cong\mathbb{R}^n\setminus\{0\}$ are not homeomorphic is a triviality from Algebraic Topology. On the other hand, if $X$ is an infinite dimensional Banach space, then $X \cong X\setminus\{0\}$.

Question. In the $\infty$-dimensional spaces, which sets are "removable" like $\{0\}$ in $X$? Can you give an explicit homeomorphic map in such a case?

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    $\begingroup$ Wow, $X \simeq X \setminus \{0\}$, really? Do you have a reference for that? Or is getting the reference part of the question? $\endgroup$ – Squark Feb 3 '17 at 21:26
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    $\begingroup$ V. Klee, Two topological properties of topological linear spaces, Israel J. Math. 4 (1964), 211-220. $\endgroup$ – T. Amdeberhan Feb 3 '17 at 21:28
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    $\begingroup$ I suggest that the title should be changed to better reflect the question $\endgroup$ – Yemon Choi Feb 3 '17 at 22:05
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    $\begingroup$ There is a huge literature on this. Start with MR0250342 Henderson, David W. Infinite-dimensional manifolds are open subsets of Hilbert space. Topology 9 1970 25–33. $\endgroup$ – Bill Johnson Feb 3 '17 at 22:56
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    $\begingroup$ See also mathoverflow.net/questions/254521/… and references therein $\endgroup$ – Thomas Rot Feb 4 '17 at 1:50

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