Is $\mathbb{R}^{\infty}$ homeomorphic to $\mathbb{R}^{\infty}\setminus\{0\}$?

Question

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!

Answer 1

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

Answer 2

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.

Answer 3

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.