If X,Y Banach spaces and T: X $\to$ Y bounded linear operator with T(X) dense and not equal to Y. Prove that there exists y in Y such that $\left\lVert{x_n}\right\lVert \to \infty$ for every sequence $x_n$ for which $Tx_n\to y$.
I know I am supposed to use the open mapping theorem for this but I honestly can't see how. I also read somewhere on here that T(X) must be a first category set but that didn't help much either. A hint would be appreciated.
Hint:
Let $p$ be an integer. Let $C_p$ be the subset of $Y$ such that for every $y\in C_p$, there exists a sequence $y_n$ where $x_n$ is in the closed ball $B(0,p)$ and $lim_nT(x_n)=y$. Show that $C_p$ is closed.
You can write $T(X)=\cup_n V_n$ where $V_n$ is a closed meagre set and $n$ is an integer. $C_p\cap V_n$ is closed and meager, its complementary $U_{n,p}$ is dense.
Baire category theorem implies that $U=\cap U_{n,p}$ is not empty. Let $y\in U$ every for every sequence $x_n$ which satisfies $limT(x_n)=y$, we have $lim_n\|x_n\|=+\infty$.
If $X$ is finite dimensional there is a very short proof:
Suppose $y\in Y\setminus T(X)$. Because $X$ and $Y$ are Banach, if $T(x_n)\rightarrow y$, then $x_n$ (or any subsequence) must not converge in $X$ because of the closed graph theorem (if a subsequence converged to $x$, then $T(x)=y$, which contradicts our choice of $y$). In a finite dimensional space, balls are compact, so for $(x_n)_n$ not to have any convergent subsequence, it must not be contained in any ball.
Thus, the natural numbers $n$ that satisfy $x_n\in B_{M}$ in our sequence are such that $\lVert T(x_n)-y\rVert\geq \varepsilon$ for a specified $\varepsilon$ (if there weren't such an $\varepsilon$ there would exist a bounded sequence in $X$ whose image converges to $y$, which is absurd). For $n\geq {n_o}(\varepsilon)$, we have $\lVert T(x_n)-y\rVert< \varepsilon$, so $\lVert x_n \rVert\geq M$ and $\lVert x_n \rVert\rightarrow \infty$. $\square$
