Let $(X,\|.\|)$ be a reflexive Banach space and $(A_i)_{i\geq 1}$ a sequence of separable subset of $X$.\
Show that the linear subspace $Y$ of $X$ generated by : ${\displaystyle\bigcup_{i} A_i}$ is closed.
An idea please.
2026-02-23 04:41:56.1771821716
Show that the linear subspace $Y$ of $X$ generated by : ${\displaystyle\bigcup_{i} A_i}$ is closed.
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This is surely false. Take $X=\ell^{2}$. Let $(e_i)$ be the usual basis and take $A_i$ to be the one dimensional subspace generated by $e_i$, $i=1,2...$. Then the span of $\cup_i A_i$ is not closed.
For example $\sum \frac 1 i e_i$ belongs to the closure of this span but it does not belong to the span.