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This is from a 2011 qualifier in functional analysis:

Let $T: E\rightarrow F $ be a linear continuous surjective map between Banach spaces. If $F$ is separable, prove there is a separable closed subspace $E_o$ of $E$ such that $T(E_o)=F$.

Of course se we are in the hypothesis of the open mapping Theorem and we have:

$$B_F(0,c)\subset T(B_E)$$

Because $F$ is separable, there is a dense countable set $\{y_n\}$ in $B_F(0,c)$. To each one of these elements we may take $x_n\in B_E$ such that $T(x_n)=y_n$. I am tempted do define $E_o=\overline{span(\{x_n\:| n \in \mathbb{N}\})}$. However I am having trouble proving $T(E_o)=F$. If we take $y\in F$ a non zero element, there is a sequence of $y_{n_j}$ such that:

$$T(x_{n_j})=y_{n_j}\rightarrow \frac{y}{\lVert y \rVert}c$$

However I have no guarantee the $x_{n_j}$ converge to a point in $E_o$:

Kadmos
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    I think $T|{E_0}$ is injective by construction and therefore you can use that it has an inverse to prove that $x{n_j} = (T|{E_0})^{-1}(y{n_j})$ converges. However, I'm not 100% sure about it being injective... – Levie B Feb 09 '24 at 13:02
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    Isn't the image of $E_0$ complete by the open property? – André Caldas Feb 09 '24 at 13:06

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