Is it true that if $ E,F$ are two topological vector spaces (or say Banach spaces) over $\mathbb{R}$ such that they have nonempty open subsets $U\subset E, V\subset F$ which are homeomorphic, then the two vector spaces are isomorphic? If false, then what can we say if the two open subsets are $\mathcal{C}^1$-diffeomorphic?

  • $\begingroup$ I would guess that the first question is false, simply on the grounds that it is pretty trivially true for finite dimensional ones, so it's probably false for arbitrary vector spaces. $\endgroup$
    – Simon Rose
    Sep 30, 2015 at 9:49
  • $\begingroup$ My guess is that it is true in all spaces, since it is true for finite-dimentional ones. But I cannot give a proof or a disproof. $\endgroup$
    – usr203050
    Sep 30, 2015 at 9:51
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    $\begingroup$ My guess for the second part is that the differential at some point (or may be at any point) can give an isomorphism between the two spaces. $\endgroup$
    – usr203050
    Sep 30, 2015 at 10:06
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    $\begingroup$ Yes. The interesting question between those two extremes is the case of Lipschitz equivalence (for Banach spaces). Again much work has been done on this case, e.g. by Lindenstrauss et al. $\endgroup$
    – shasta
    Sep 30, 2015 at 10:19

1 Answer 1


This is false. All separable Banach spaces, for example, are homeomorphic.Indeed, there is a considerable body of work on when topological vector spaces are homeomorphic (see Bessaga and Pelczynski "Selected Topics in infinite-dimensional Topology" for starters).

  • $\begingroup$ Ok thank you. So it is far from evident in any case. $\endgroup$
    – usr203050
    Sep 30, 2015 at 9:59
  • $\begingroup$ And what if the subsets are smoothly diffeomorphic? (The second question). $\endgroup$
    – usr203050
    Sep 30, 2015 at 10:03
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    $\begingroup$ Isn't the diffeomorphic part answered by the comment of usr203050? $\endgroup$ Sep 30, 2015 at 10:17
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    $\begingroup$ If diffeomorphism is something that has differential as a bounded invertible operator between tangent spaces, then this very operator is linear isomorphism between spaces. If not, what concretely do you mean? $\endgroup$ Sep 30, 2015 at 10:17
  • $\begingroup$ That is what I mean! :-) Also, the comment of shasta gives something more. $\endgroup$
    – usr203050
    Sep 30, 2015 at 10:29

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