A question on linearly independent vectors in a Banach space Given a list of linearly independent vectors $\{x_1,...,x_n\}$ in a Banach space. If for each $1 \leq i\leq n$, there is a sequence of vectors $\{y_m^{(i)}\}_{m=1}^{\infty}$ converges to $x_i$. Then when $m$ is big enough, whether $\{y_m^{(i)}:1 \leq i\leq n\}$ is also linearly independent?
Thank you for all helps!
 A: Yes. Suppose for the sake of convenience that the $y$'s are dependent for every $m$. Choose $c_m^i$ with $$\sum_{i=1}^nc_m^iy_m^i=0,$$and, say, $$\sum_{i=1}^n|c_m^i|=1.$$Now the vectors $c_m\in\Bbb R^n$ have a non-zero accumulation point...
A: If this Banach space has finite dimension, you can form the matrix $X$ using expansion of $x_i$'s as columns. Since $\lbrace x_i \rbrace$ is a linearly independent system of vectors, ${\rm det}\, X \neq 0$. And you can also form the matrix $Y_m$ using expansions of $y_m^{(i)}$ in the same basis. Then ${\rm det} \; Y_m$ is also non-zero from some moment — this is just a consequence of continuity of ${\rm det}$.
A: This is not true, even in the simplest case of $\mathbb{R}^2$.  Take $x_1, x_2$ as the standard basis, $y^{(1)}_m=(1,1/m)$, $y^{(2)}_m=(1/m,1)$.
A: The definition of linear independence is that if $\sum q_n x_n = 0$, then $q_n = 0\  \forall n$.
So let's assume by contradiction that for all m, $\{y^{(m)}\}$ are linearly dependant.
Then $\sum q_n y_n^{(m)} = 0$, but not all $q_n =0$.
But $y_n^{(m)} \to x_n$, and therefore we have $\sum q_n x_n = 0$, but not all $q_n =0$.
This contradicts out hypothesis, and proves the assertion.
Edit: nevermind, I'm assuming that the coefficients stay the same.
