Is $\mathbb{R}^\omega \cong (\mathbb{R}^\omega \setminus \{x\})$, where $\mathbb{R}^\omega$ is given the product topology, and $x\in\mathbb{R}^\omega $?



Is $\mathbb{R}^\omega \cong (\mathbb{R}^\omega \setminus \{x\})$?
The answer is yes. Any two homotopy equivalent [Hilbert manifolds](https://en.wikipedia.org/wiki/Hilbert_manifold) are homeomorphic. The countable product of lines is homeomorphic to the separable Hilbert space, see p.174 of [Bessaga C., Pelczynski A., Selected Topics in Infinite-Dimensional topology] where much more information is given. 


More generally, the complement in $\mathbb R^\omega$ of the union of any countable family of compact subsets is homeomorphic to $\mathbb R^\omega$, see e.g. Theorem V.6.4 in Bessaga-Pelczynski's book.


