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More specifically, letting $\mathbb S$ be the set $\{\kern.5mm\exp({\rm i}\,t):0\le t<2\pi\kern.6mm\}$ with the induced topology from the complex plane, let $\mathbb T^{\kern.6mm\infty}$ be the product topological space of denumerably many of copies of $\mathbb S$. E.g. from page 15 in this paper we know that all separable infinite-dimensional Fréchet spaces are homeomorphic. So the question is whether $\mathbb T^{\kern.6mm\infty}$ is homeomorphic to an open subset in some separable infinite-dimensional Fréchet space.

Edit. After reading Igor Belegradek's comment, I now realize what I should have realized already before posting the question. Namely, the motivation behind the question was whether $\mathbb T^{\kern.6mm\infty}$ could be made a manifold modelled on Fréchet spaces with charts defined on open subsets. These sets are $\sigma$−compact, and hence cannot be homeomorphic to open subsets in any infinite-dimensional Baire topological vector space. So there is no hope to turn $\mathbb T^{\kern.6mm\infty}$ into a manifold modelled on Fréchet spaces. The question can be closed.

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  • $\begingroup$ Your notation is rather nonstandard: $S^\infty$ usually denotes an infinite-dimensional sphere, rather then an infinite product of circles. $\endgroup$ Sep 25, 2013 at 18:25
  • $\begingroup$ Would perhaps $\mathbb T^{\kern.5mm\infty}$ be better? $\endgroup$
    – TaQ
    Sep 25, 2013 at 18:28
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    $\begingroup$ Sure, either that or $(S^1)^\infty$ would be great. $\endgroup$ Sep 25, 2013 at 18:30
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    $\begingroup$ The space $T^\infty$ is compact (by Tychonov), so if it is homeomorphic to an subset of a metric space, that subset is also closed. Thus $T^\infty$ cannot be homeomorphic to an open subset of a Frechet space. $\endgroup$ Sep 25, 2013 at 19:11
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    $\begingroup$ Yet another argument: $\mathbb{T}^\infty$ is not locally homeomorphic to an open subset of any topological vector space because it is not locally contractible (no open set is simply connected). $\endgroup$ Sep 25, 2013 at 20:19

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