Is this claim about mapping closed discrete sets into open sets true for $l^2$?

64 Views Asked by At

I have found this claim in topological paper:

Let $D$ be closed and discrete and let $D \subseteq \mathbb{R^n}, n\geq 2$. Let $U$ be open and non-bounded with $\overline{U}$ noncompact.

Then there exists a homeomorphism $h: \mathbb{R^n} \mapsto \mathbb{R^n}$ such that $h(D) \subseteq U$.

My question is: Does this claim also apply for $l^2$ spaces? How to prove that? (I am interested in that case, since it would help me prove another topological fact).

The formulation of the claim for $l^2$ would be:

Let $D$ be closed and discrete and let $D \subseteq \mathbb{l^2}$. Let $U$ be open and non-bounded with $\overline{U}$ noncompact.

Then there exists a homeomorphism $h: \mathbb{l^2} \mapsto \mathbb{l^2}$ such that $h(D) \subseteq U$.

Is that true? And why? Thank you for your thoughts.