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

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.
