Is the closure of the span equal to the span of union closed spaces?

Question

Let $V$ be a infinite dimensional Banach space over the complex plane $\mathbb{C}$

Let $\{v_n\}_{n \in \mathbb{N}} \subset V$ be a sequence of linear independent vectors in $V$

Let $m \in \mathbb{N}$ be a fixed positive integer

I would like to know if it is true that:

$$ \overline { \operatorname{span} (\{v_n\}_{n \geq 1}) } = \operatorname{span} \left( \{v_n\}_{n \leq m} \bigcup \overline { \operatorname{span} (\{v_n\}_{n \geq m+1}) } \right) $$

Thanks.

Answer 1

Let $E_m = \operatorname{span}(\{v_n\}_{n \leq m})$ and $E'_m = \operatorname{span}(\{v_n\}_{n \geq m+1})$. We obviously have $$\operatorname{span}(\{v_n\}_{n \geq 1}) = E_m + E'_m \subset E_m + \overline{E'_m} = \operatorname{span}(E_m \cup \overline{E'_m}) \\ = \operatorname{span}(\{v_n\}_{n \leq m}) \cup \overline{E'_m}) \subset \overline{\operatorname{span}(\{v_n\}_{n \geq 1})} .$$ It therefore suffices to show that $E_m + \overline{E'_m}$ is closed. But this is well-known because $E_m$ is finite-dimensional and $\overline{E'_m}$ is closed. See Sum of closed subspaces of normed linear space.