Let $X$ be a closed subspace of $\ell^2$ over $\mathbb{C}$


Let $\{x\_m\}\_{m \in \mathbb{N}}$ a basis for $X$ equivalent to canonical basis $\{e\_m\}\_{m \in \mathbb{N}}$ for $\ell^2$


I Would like to know if it is true that:
$$
\bigcap\_{p=1}^\infty \overline{ \operatorname{span} } \{x\_m\}\_{m \geq p}
=
\{0\}
$$


Thanks.


P.S.


Let $(x\_n)$ be a basis for $X$ and $(y\_n)$ be a basis for $Y$. We say that $(x\_n)$ and $(y\_n)$ are equivalent if the convergence of $\sum a\_n x\_n$ is equivalent to that of $\sum a\_n y\_n$ (Sequence and series in Banach spaces. Joseph Diestel)



Intersection of Hilbert spaces with basis equivalent to canonical basis for $\ell^2$
I assume "basis" means "Schauder basis"? If so, then you don't need any additional assumption to conclude that this intersection is zero. $X$ can be any Banach space with a Schauder basis. For any $v \in X$ we can uniquely write $v = \sum a\_nx\_n$; define $P\_n(v) = a\_n$. This is a bounded linear functional, as explained on the [Wikipedia page for Schauder bases](https://en.wikipedia.org/wiki/Schauder_basis). So if $v \in {\rm span}\{x\_m\}\_{m \geq p}$ then $P\_n(v) = 0$ for all $n < p$, and by continuity the same is true for $v \in \overline{\rm span}\{x\_m\}\_{m \geq p}$. Thus any $v \in \bigcap\_{p = 1}^\infty \overline{\rm span}\{x\_m\}\_{m \geq p}$ satisfies $P\_n(v) =0$ for all $n$, and therefore $v = 0$.


