Surjective operator to separable space Let $X,Y$ be Banach spaces and $T:X\to Y$ be a surjective bounded linear operator. If $Y$ is separable then there exists a separable subspace $Z$ of $X$ such that $T(Z)=Y$. I tried the following: 
$Y = \overline{\{y_1, \dots, y_n, \dots\}}$. Since $T$ is surjective $y_i = Tx_i$ for some $x_i \in X$. Let $Z = \overline{\text{span}(x_1,\dots,x_n,\dots)}$. I know that $Z$ is separable so if I could prove that $T(Z)=Y$ the proof would be complete. However I could only prove that $\overline{T(Z)}=Y$. Any help would be appreciated. Thanks.
 A: By open mapping theorem, there is $c>0$ such that $$B(0,1)\subseteq T(B(0,c))$$ let $S=\{s_j\}_j\cap B(0,1)$, where $\{s_j\}_j$ is the dense subset of $Y$. Select an element $x_{j}\in B(0,c)$ such that $f(x_{j})={s_j}$ and let $$Z=\overline{\text{span}(x_{j})_j}$$ Take $y\in B(0,1/2)$, then there is $s_{j_1}\in S$ such that $$\|y-s_{j_1}\|<1/4\rightarrow \|2y-2s_{j_1}\|<\frac{1}{2}$$ then $2y-2s_{j_1}\in B(0,1/2)$, so there is $s_{j_2}\in S$ such that $$\|2y-2s_{j_1}-s_{j_2}\|<\frac{1}{4}\rightarrow \|y-s_{j_1}-\frac{s_{j_2}}{2}\|<\frac{1}{8}$$ so we create a sequence $\{s_{j_n}\}_n$ such that $$\|y-s_{j_1}-\cdots-\frac{s_{j_n}}{2^n}\|<\frac{1}{2^{n+1}}$$ consider now the sequence $$t_n=x_{j_1}+\frac{x_{j_2}}{2}+\cdots+\frac{x_{j_n}}{2^n}$$ then $\|y-T(t_n)\|<2^{-n-1}$ and the sequence belongs to $Z$ and, for every pair $n<m$ $$\|t_m-t_n\|=\|\frac{x_{j_{n+1}}}{2^{n+1}}+\cdots+\frac{x_{j_m}}{2^m}\|\leqslant\|\frac{x_{j_{n+1}}}{2^{n+1}}\|+\cdots+\|\frac{x_{j_m}}{2^m}\|<\frac{c}{2^n}$$ (where I used that $\|x_{j_n}\|<c$ for all $n$). We can draw the following conclusions:

*

*$Z$, being a closed subspace of a Banach space, it's Banach


*The sequence $\{t_n\}_n$ is a sequence in $Z$ and is Cauchy, therefore there it has a limit $t\in Z$


*$T(t_n)$ converges to $y$
So, we conclude that $$T(t)=T(\lim_nt_n)=\lim_n T(t_n)=y$$ therefore $y\in T(Z)\rightarrow B(0,1/2)\subseteq T(Z)$, which implies that $Y\subseteq T(Z)$ (since $Z$ is a vector subspace).
