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Let $\mathbb{R}^{\infty}$ be a linear topological space of all sequences $x=(x_{1},x_{2},\ldots,x_{n},\ldots)$ of real numbers with a product topology, or, in other words, let $\mathbb{R}^{\infty}$ be a countable product of real lines.

Is $\mathbb{R}^{\infty}$ homeomorphic to $\mathbb{R}^{\infty}\setminus\{0\}$, where $0=(0,0,\ldots,0,\ldots)$?

If the answer is yes, how to prove this theorem?

Thank you in advance!

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Dear Natalia, the first sentence should be "Let $\mathbb{R}^{\infty}$ be THE linear topological space of all sequences $x=(x_{1},x_{2},\ldots,x_{n},\ldots)$ of real numbers with THE product topology, or, in other words, let $\mathbb{R}^{\infty}$ be THE countable product of real lines." I know that in Russian (your native tongue?) there are no articles and it takes some effort to use them correctly in English. But I am sure you will soon master them : good luck! (and +1 for your question) –  Georges Elencwajg Mar 1 '13 at 9:37
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3 Answers 3

The answer is yes.

In Chapter VI § 2 of Selected Topics in Infinite Dimensional Topology (Bessaga & Pełczyński) is stated the surprising result that $\mathbb{R}^\infty$, the Hilbert space $\ell_2$ and the unit sphere $S \subset\ell_2$ are homeomorphic spaces (${}^*\!$). The background required (developed in the previous chapters) seems to be rather wide. Anyway, proving this result, they exhibit an explicit homeomorphism between $S \smallsetminus \{\ast\}$ and $\ell_2$.

(${^*}\!$) moreover there is a 1966 paper by Bessaga entitled Every infinite-dimensional Hilbert space is diffeomorphic with its unit sphere

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Another reference is R. D. Anderson's paper, Topological properties of the Hilbert cube and the infinite product of open intervals (1967). –  Seirios Dec 15 '13 at 16:45
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In "Infinite-dimensional Topology, Prerequisites and an introduction" by Jan van Mill (table of contents) the author proves a characterization of $R^\infty$, also called $s$ in the book, which allows him to show that for any $\sigma$-compact subset $A$ of $\mathbb{R}^\infty$ we have that $\mathbb{R}^\infty \setminus A$ is homeomorphic to $\mathbb{R}^\infty$. There is considerable machinery involved, but it's a very elegant exposition of these matters, IMHO. So the infinite product of copies of $\mathbb{R}$ behaves quite differently from its finite powers, in which homological/homotopical methods can be used to show e.g. $\mathbb{R}^n$ and $\mathbb{R}^n \setminus \{p\}$ are not homeomorphic.

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Although this is not your original question, $\mathbb{R}^{\infty}\setminus 0$ is locally simply connected.

To prove that, use universal property of product in the category of topological spaces to lift any path from $[0,1] \rightarrow U(n):=\mathbb{R}^{n} \setminus 0$ for $n>2$. Then notice that inverse images of sets of this form under the projection $\mathbb{R}^{\infty} \setminus 0 \to U(n)$ form a basis for the product topology $\mathbb{R}^{\infty} \setminus 0$. This proves that this space is locally simply connected.

I believe this is the simplest answer to your question: the cone of $\mathbb{R}^{\infty}$ and the cone of $\mathbb{R}^{\infty}\setminus 0$ should be homeomorphic. There is a nice description of these two cones as lines in $\mathbb{R}^{\infty}$ which pass through the origin. From this fact you can construct a homeomorphism as the cones minus the vertex deform retract onto their respective bases.

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I am not sure it is enough to prove that it is simply connected. Could you explain a bit more ? –  Selim Ghazouani Mar 1 '13 at 9:11
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It is actually easy to see, that $\mathbb R^\infty \setminus \{ 0 \}$ is contractible and thus homotopic to $\mathbb R^\infty$: Since $S^n$ is compact, any map $S^n \to \mathbb R^\infty \setminus \{ 0 \}$ factors through the inclusion $\mathbb R^N \setminus \{ 0 \} \to \mathbb R^\infty \setminus \{ 0 \}$ for large enough $N$ but any map $S^n \to \mathbb R^N \setminus \{ 0 \}$ is nullhomotopic provided $N > n + 1$. Contractibility now follows from Whiteheads theorem. –  Alexander Thumm Mar 1 '13 at 9:30
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Also note that locally simply connected + connected does not imply simply connected, the sphere $S^1$ being the obvious counterexample. –  Alexander Thumm Mar 1 '13 at 9:33
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@AlexanderThumm: This argument works for the algebraic topologist's $\mathbb{R}^\infty$ (finitely supported sequences with inductive limit topology) which is a CW complex, but $\mathbb{R}^{\mathbb{N}}$ with the product topology is not a CW complex (at least not in an obvious way). –  Martin Mar 1 '13 at 9:53
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@Martin: Yes you are right. This argument only works for $\mathbb R^\infty$ with the limit topology. –  Alexander Thumm Mar 1 '13 at 10:04
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