TITLE: Bounded linear operator maps norm-bounded, closed sets to closed sets. Implies closed range?
QUESTION [8 upvotes]: Question
Suppose $T:X\rightarrow Y$ is a continuous, injective linear operator between Banach spaces. Suppose, in addition, that $T$ maps norm bounded closed sets in $X$ to closed sets in $Y$. Then the range of $T$ is closed in $Y$.
This is a problem related to one given An Invitation to Operator Theory by Abramovich and Aliprantis and I'd just like to verify my proof.
Attempt
We assume that $T$ is as above, and we shall prove that it has closed range. If $y_n = T x_n$ and $y_n\rightarrow y$, we want to show that $y=Tx$ for some $x\in X$. First, suppose $\{x_n\}_{n\geq 1}$ is unbounded. Then
$$\lim_n\, T(x_n/\|x_n\|) = \lim_n\, y_n/\|x_n\| = 0.$$
But the set $B=\{ x\in X: \| x\|=1\}$ is closed and norm-bounded, so its image under $T$ is closed. In particular, we must have $Tz=0$ for some $z\in B$. This contradicts the fact that $T$ is injective. So the sequence $\{x_n\}_{n\geq 1}$ is bounded in $X$. Since $\{x_n\}_{n\geq 1}$ is bounded, the set
$$ A=\mathrm{cl} \{ x_1, x_2, \ldots, x_n, \ldots \}$$
is closed and norm bounded. Hence $T(A)$ is closed in $Y$. In particular, $y=\lim_n\, Tx_n = Tx$ for some $x\in A \subset X$. So the range of $T$ is closed.
Thanks in advance!

REPLY [4 votes]: You are assuming that $(x_{n})$ unbounded implies $\left(\frac{1}{\|x_{n}\|}\right)$ goes to $0$, so that you can then conclude $\lim_{n\to\infty}\frac{y_{n}}{\|x_{n}\|}=0$.
This is not necessarily true.
For example, the sequence $(x_{n})$ in $\mathbb{R}$ given by $x_{n}=\begin{cases}n & \mbox{if }n\mbox{ is even}\\1 & \mbox{if }n\mbox{ is odd}\end{cases}$ is unbounded.
However, $\frac{1}{\|x_{n}\|}=\begin{cases}\frac{1}{n} & \mbox{if }n\mbox{ is even}\\1 & \mbox{if }n\mbox{ is odd}\end{cases}$ does not converge.
Fortunately, $\left(\frac{1}{\|x_{n}\|}\right)$ does contain a subsequence converging to $0$. So you can get around the problem with a minor change to your proof:
Assume $(x_{n})$ is not bounded. Then there is a subsequence $(x_{n_{k}})$ of $(x_{n})$ such that $x_{n_{k}}\not=0$ for all $k$ and $\lim_{k\to\infty}\|x_{n_{k}}\|=\infty$.
Then $\lim_{k\to\infty}\frac{y_{n_{k}}}{\|x_{n_{k}}\|}=0$.