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It is renowned that removing a point from an $\infty-$dimensional vector space $H$ preserves its contractibility (by Kuiper's theorem). Is there any hope that $H\setminus A$ is not contractible, where $A$ is a disjoint union of points (possibly infinitely many, but still countable)?

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  • $\begingroup$ I wonder if it is correct to think of $H \setminus A$ as some sort of connected sum of infinite dimensional spheres. I would think that this notion would be contractible on every addition. $\endgroup$ Nov 12, 2016 at 16:32
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    $\begingroup$ You mean wedge sum? I also tought about this inconvenient. Somehow it seems to me that, having infinite dimensions, we can always get rid of "finite dimensional" obstacles when contracting spheres to a point. But what if we remove an orthonormal base in a Hilbert space? It seems reasonable to me that all moves in any direction would meet an obstacle. $\endgroup$ Nov 12, 2016 at 17:05
  • $\begingroup$ The subject of infinite dimensional topology deals with topological vector spaces satisfying certain assumptions (e.g. Frechet spaces or absorbing spaces). Without extra assumptions there is no theory. The existing theory tends to work well for spaces that arise in analysis and geometry. What space $H$ you are interested in? $\endgroup$ Nov 13, 2016 at 14:32
  • $\begingroup$ I'm interested in the space of Riemannian metrics over a compact manifold $M$, endowed with the $\mathcal C^k$-topology. Sorry for not being specific. $\endgroup$ Nov 14, 2016 at 8:38

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In comments it is stated that the space of most interest to the OP is the space of $C^k$ Riemannian metrics on a compact manifold.

If you look at the space of $C^k$ Riemannian metrics with $C^k$ topology, where $k$ is finite or infinite, then the space is Fréchet and separable. In any separable Fréchet space $X$ the complement of any countable union of compact sets is homeomorphic to $X$ (this is proved in Bessaga-Pełczyński book, see definition IV.5.1 and section V.6).

Perhaps, the OP is interested in the space of $C^\infty$ Riemannian metrics with $C^k$ topology, where $k$ is finite. Theorem 1.1 in https://arxiv.org/abs/1510.07269 determines the homeomorphism type of the space. Namely, if $\Sigma$ denotes the linear span of the Hilbert cube in $l^2$, then the space of the $C^\infty$ Riemannian metrics with $C^k$ compact-open topology is homeomorphic to $\Sigma^\omega$, the product of countably many copies of $\Sigma$. I do not know whether any countable subset of $\Sigma^\omega$ has a contractible complement.

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No, at least if $H$ is Banach. The argument is similar to this MO post that proves $\Bbb R^3 \setminus S$, where $S$ is some countable set, is simply connected.

Let $S \subset H$ be a countable set. Fix a map $i: S^n \to H \setminus S$; let $X = F(D^n,\partial S^n; H)$ be the space of null-homotopies of $i$: the space of continuous maps $D^n \to H$ that restrict to $i$ on the boundary, equipped with the compact-open topology. As long as $H$ is at least an F-space, $X$ is completely metrizable, so the Baire category theorem applies. Enumerating $S$, let $X_k$ be the space of null-homotopies whose image does not contain $k$. if $H$ is Banach, we can use smooth approximation and the Sard-Smale transversality theorem to see that the $X_k$ are open, dense, nonempty sets, and because Baire applies to $X$ the intersection $\bigcap_k X_k$ is nonempty, providing a null-homotopy of $i$. More generally we can delete the union of countably many finite-dimensional submanifolds of $X$ and preserve (weak!) contractibility.

It seems like an interesting question to ask which topological vector spaces always have $H \setminus S$ weakly contractible.

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Such things are discussed in

C. Bessaga and A. Pełczyński Selected topics in infinite-dimensional topology.

I don't have access to the book now, but if I remember correctly the principle is that removing a compact set of a Banach manifold does not affect the homotopy --and even the diffeomorphism-- type of the manifold. You should have a look at this book for precise statements.

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  • $\begingroup$ And by the above argument you can remove a countable union of compact sets without changing the homotopy type (and thus diffeomorphism type if that union is a closed set). How do they show that you can remove eg a wild arc? $\endgroup$
    – mme
    Nov 13, 2016 at 16:32
  • $\begingroup$ @MikeMiller: see e.g. "Negligible Subsets of Infinite-Dimensional Fréchet Manifolds" by Cutler, jstor.org/stable/2036608?seq=1#page_scan_tab_contents. $\endgroup$ Nov 13, 2016 at 18:55

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