Let $X$ be a separable Banach space of dimensional $>2$. When does there exist a sequence positive integers $\{N_n\}_{n \in \mathbb{n}}$ such that
- For any sequence of distinct finite-dimensional subspaces $\{L_n\}_{n \in \mathbb{N}}$ of $X$ we have: $$ N_n\leq \dim(L_n) (\forall n \in \mathbb{N})\,\Rightarrow \,\overline{\bigcup_{n \in \mathbb{N}} L_n} =X. $$ By distict, I simply mean that $L_n\neq L_m$ if $n\neq m$ but we can have containement.
- There exists a sequence of subspaces $\{\tilde{L}_n\}$ of $X$ for which $$ Dim(\tilde{L}_n)<N_n \mbox{ and } \overline{\bigcup_{n \in \mathbb{N}} \tilde{L}_n} \neq X. $$ Let's call such $\{N_n\}$ critical for $X$.
Finite-Dimensional Intuition:
If $X$ is finite-dimensional with dimension at-least $2$, then WLOG assume $X=\mathbb{R}^D$ for some positive integer $D>2$. Then $N_n=\min\{n,D\}$ works. But if $\tilde{L}_n$ are rotations of $\mathbb{R}^{D-1}\subseteq \mathbb{R}^D$ by $\pi_n$ about the origin, then $\tilde{L}_n$ satisfy the second condition. So $N_n=\min\{n,D\}$ is critical in the above sense.
Note: We can always construct a dense linearly independent subset of $X$ by from a topological basis (since X is infinite-dimensional). So we can find $L_n$ with $1$-dimension each. However, here, I'm looking for a "critical" integer $N$ above which any collection of $\{L_n\}$ must satisfy $\overline{\bigcup_{n \in \mathbb{N}} L_n} =X$.
**Edit: ** I removed the requirement of each $L_n$ to be distinct so that I can give my (updated) intuition in the finite-dimensional setting.
