Every continuous linear map from a Banach space to a NVS with a countable algebraic basis must have finite rank.

Question

A proof I used for a homework problem (which was an intermediary step to prove some unrelated result), that if $T:X \to Y$ is a continuous linear map, where $X$ is a Banach space, and $Y$ has a countable algebraic basis, then $T$ has finite rank, is as follows:

Let $(e_{n})_{n \geq 1}$ be a basis for $Y$, and let $E_{n} = \langle e_{1}, \ldots, e_{n} \rangle$. Then $E_{n}$ is closed, since it is finite-dimensional, and $T^{-1}(E_{n}) := C_{n}$ is closed since $T$ is continuous. Then, $\bigcup_{n} C_{n} = T^{-1} \left(\bigcup_{n} E_{n}\right) = T^{-1}(Y) = X$, whence by Baire $B(x, \delta) \subset C_{N}$, some $N$. Then if $y \in X$, $||y|| < \delta$, $Ty = T(x + y) - Tx \in E_{N}$, hence $\text{Im}(T) \in E_{N}$ and so $T$ has finite rank.

I've been thinking a bit about my proof. I was wondering whether it is possible to prove the same result without explicitly using Baire, but instead the result that a Banach space cannot have a countable algebraic basis. I know this is actually also proved using Baire, but I was pursuing a more linear-algebraic argument such as:

Suppose $T$ had infinite rank, then $\text{Im}(T)$ is an infinite-dimensional subspace of $Y$, hence has a countable algebraic basis $(f_{n})_{n \geq 1}$ since $Y$ does. I tried to pursue a contradiction by showing that maybe you can form a countable algebraic basis for $X$ by considering preimages? But this did not seem to work.

Any ideas would be appreciated.

Edit: assume $X$ is infinite-dimensional, and so is $Y$ (otherwise the result is trivial).

Answer 1

Note that $K=\ker T$ is a closed subspace of $X$. Consider the quotient space $Z=X/K$ with the canonical projection $\pi\colon X\rightarrow Z$. Then $Z$ is a Banach space, and there exists an injective bounded linear map $S\colon Z\rightarrow Y$ such that $T=S\circ\pi$.

Let $\mathcal{B}=\{v_n\}_{n\in\mathbb{N}}$ be a countable basis for $S(Z)=T(X)$. For each $n$, choose the unique $x_n\in Z$ satisfying $S(x_n)=v_n$. Then $\{(x_n)\}_{n\in\mathbb{N}}$ is a countable basis for $Z$. Since $Z$ is Banach, it must be finite-dimensional; hence $T(X)=S(Z)$ is also finite-dimensional.