The intersection of closure of span of infinite, linearly independent, closed, bounded, connected and disjoint subsets of $\ell^2$

Question

Let $X$ and $Y$ be two subsets of $\ell^2$ space over $\mathbb{C}$ such that each of them is: infinite, linearly independent, closed, bounded, connected and $X \cap Y = \emptyset$

I would like to know if is it true that $$ \overline{ \operatorname{span} X } \cap \overline{ \operatorname{span} Y } = \{0\} $$

thanks.

ok @PhoemueX i understand, the answer is 'no', is it the same if $X$ and $Y$ are infinite? — Matey Math, Dec 16 '17 at 19:32
I have no proof yet, but I think that $X$ and $Y$ can just contain one element since they are linearly independent and connected. — Mundron Schmidt, Dec 16 '17 at 19:37

Mundron Schmidt · Accepted Answer · 2017-12-17T06:17:57.340

Assume there exists an infinite, connected, closed, bounded and linearly independent set $X$ and $\alpha\in\mathbb C\setminus\{0,1\}$, then you can define $Y=\{\alpha x~:~x\in X\}$. Then $Y$ is also infinite, connected, closed, bounded and linearly independent and $X\cap Y=\emptyset$ but $\overline{\operatorname{span}X}=\overline{\operatorname{span}Y}$.

So this argument work for $\ell^2$ over $\mathbb C$ and over $\mathbb R$.

But I suppose there cannot exist such a set. I am not sure, but somehow I think that a connected and linearly independent set cannot contain more than one element. But I have no proof yet.

The intersection of closure of span of infinite, linearly independent, closed, bounded, connected and disjoint subsets of $\ell^2$

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