1. Suppose $(x\_n)$ and $(y\_n)$ are two basic sequences in a separable Banach space $X$ such that $\overline{span}\{(x\_n), (y\_n)\}=X$. Can we always pass to subsequences $(x\_{n\_k})$ and $(y\_{n\_k})$ such that $\overline{span}\{(x\_{n\_k}), (y\_{n\_k})\}\neq X$?


I had the impression that this must be "obviously true", and perhaps still is embarrassingly easy, but I just can't see the argument. 


2. A somewhat related question, to which I do not know the answer either: Can any separable Banach space be written as the closed span of finitely many basic sequences?



Two questions about basic sequences
I would suggest to get a positive answer to the second question as follows: Let $\{x\_i\}$ be a basic sequence in $X$ (existing by Mazur's result) and let $\{z\_i\}$ be a rapidly converging to zero sequence in $X$ with dense linear span. Let $\{y\_i\}$ be given by $y\_i=x\_i+z\_i$. The sequence $\{y\_i\}$ is basic by the result of Krein-Milman-Rutman. The linear span of the union is dense because it contains $\{z\_i\}$.


The answer to the first question is negative. It is done by using the construction for the second question, and in addition picking $\{z\_i\}$ in such a way that each of it subsequences has a dense linear span in $X$. Such constructions are known, one of them: Consider a sequence $\{u\_k\}$ of normalized vectors with dense linear span and let $z\_i=\sum\_{k=1}^\infty\left(\tau\_i\right)^ku\_k$, where $\{\tau\_i\}$ is a rapdly converging to $0$ sequence of real numbers. Any subsequence of $\{z\_i\}$ has a dense linear span because otherwise there is $x^\* \in X^\*$, $x^\*\ne 0$, such that the analytic (in $\lambda$) function $\sum\_{k=1}^\infty\lambda^kx^\*(u\_k)$ would be a nonzero function with infinitely many zeros in $[0,1]$.


