Let $\mathbb{R}^\omega$ be endowed with the product topology. Is there a nonempty open set $U\subseteq \mathbb{R}^\omega$ such that $\mathbb{R}^\omega\cong \mathbb{R}^\omega\setminus \text{cl}(U)$? 


(By $\text{cl}(\cdot)$ we denote the topological closure.)


**Edit.** Apologies for omitting the word "open" in the question above.



Punching a hole into $\mathbb{R}^\omega$
If $U= (-\infty,0) \times \mathbb R \times \mathbb R \times \dots$, then 
$\mathrm{cl}(U) = (-\infty,0] \times \mathbb R \times \mathbb R \times \dots$, and the complement of this is $(0,+\infty) \times \mathbb R \times \mathbb R \times \dots$, which is homeomorphic to the whole space.


