In class, I have seen the claim below, but for $\mathbb{R^n}$. The question is, whether it still applies for $\ell^2$ and how to prove it.
The claim for $\ell^2$
Let $E, F$ be closed, disjoint sets in $l^2$. Then there extists $n \in \mathbb{N}$ and $E_1, E_2, ... E_n$, $F_1, F_2, ... F_n$ closed sets such that $E = \bigcup_{i=1}^{n} E_{i}$. $F = \bigcup_{j=1}^{n} F_{j}$. Also, for each $E_i$ and $F_j$ and for each U, V open and non-empty in $l^2$, there is a homeomorphism $h: l^2 \rightarrow l^2$ such that $h(E_i) \subseteq U$ and $h(F_j) \subseteq V$.
My idea:
- Take arbitrary $U$ and $V$ and suppose for all homeomorphisms $h: l^2 \rightarrow l^2$, there is a point $x \in U\setminus h(E_i)$.
- Prove by contradiction that such homeomorphism cannot exist?