Let $X$ and $Y$ be two subsets of $\ell^2$ space over $\mathbb{C}$ such that: $X \cup Y$ is linearly independent, $X \cap Y = \emptyset$ and $\inf\_{x \in X, y \in Y} \| x-y \|>0$ and such that each of them is: infinite, closed and bounded.


My question is if is it true that
$$
\overline{
\operatorname{span}
X
}
\cap
\overline{
\operatorname{span}
Y
}
=
\{0\}
$$



The intersection of closure of span of infinite, linearly independent, closed, bounded, separated subsets of $\ell^2$
No. Here's a simple example.


In $\ell\_2$ with orthonormal basis $(e\_n)\_{n\ge 1}$ choose $X=\{e\_n:n\ge 1\}$ (so $X$ is closed and bounded). Then choose a sequence $v\_n$ of vectors with $\|v\_n\|\le 1/2$ and define $f\_n=e\_{3n-2}+e\_{3n-1}+v\_n$, $Y=\{f\_n:n\ge 1\}$. Then $Y$ is also closed (discrete) and bounded, and $d(f\_n,e\_m)\ge 1/2$ for all $n,m$. Since $\mathrm{Span}(X)$ is dense, the intersection 0 condition does not hold.


Then it is enough to find $(v\_n)$ such that $X\cup Y$ is linearly independent. We choose $v\_n=\sum\_{k}u\_{n,k}e\_{3k}$. This already ensures that $(f\_n)$ is linearly independent. Write $P\_n=\{k:u\_{n,k}\neq 0\}$. We require that $P\_n$ is infinite and $P\_n\cap P\_m=\emptyset$ for all $n\neq m$. Then this ensures that $X\cup Y$ is linearly independent: indeed, write a linear relation as $\sum\_{i=1}^n\lambda\_ie\_i=\sum\_{i=1}^m\mu\_if\_i$, with $\mu\_m\neq 0$. Then there exists $k\in P\_m$ with $k>n$. Projecting orthogonally on $e\_k$ then yields $0=\mu\_mu\_{m,k}$, a contradiction.


