Show that the image of $T:l^{\infty}\to l^{\infty}$, $(x_n)_n \mapsto \Big(\frac{x_n}{n}\Big)_n$ is not closed in $l^{\infty}$.

Question

Denote the set of all bounded sequences in $\mathbb{R}$ by $l^{\infty}$, endowed with the sup norm $\lVert \rVert _{\infty}$. Define a map $T:l^{\infty} \to l^{\infty}$ as follows:

$$(x_n)_n \mapsto \Big(\frac{x_n}{n}\Big)_n.$$

Then the question is:

Show that the image of $l^{\infty}$ under $T$ is not a closed set in $l^{\infty}$.

My attempts:

As it is enough to show the existence of a sequence $(t_n)$ of bounded sequences from $T(l^{\infty})$ which converges to a sequence outside of it, I first considered the constant sequence $$(1,1, \ldots, 1, \ldots) = t_0$$ $(\notin T(l^{\infty}))$ and tried to look for a sequence $(t_n)$ of sequences in $T(l^{\infty})$ which goes to it. The following are two (failed) attempts to construct such a sequence.

First I thought the sequences $$t_n = (1,1,\ldots,1,\frac{1}{(n+1)^2},\frac{1}{(n+2)^2},\ldots)$$ would work. But they didn't, because $$\lVert t_n-t_0\rVert_{\infty} = \sup_{k\geq n}\Big|1-\frac{1}{k^2}\Big|=1 \nrightarrow 0$$ as $n\to \infty$ and thus $(t_n) \nrightarrow t_0$.

Also I thought about the sequences $$t_n = (1,1,\ldots,1,1+\frac{1}{n+1},1+\frac{1}{n+2},\ldots,1+\frac{1}{n+m},\ldots).$$ In this case $(t_n) \to t_0$ as $n \to \infty$, but none of the $t_n$'s is an element of $T(l^{\infty})$. So this sequence also is of no use.

Any help? Thanks a lot in advance.

Answer 1

Let $t_0=(1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\ldots)$ and $t_m=(1,\frac{1}{\sqrt{2}},\ldots,\frac{1}{\sqrt{m}},0,0,\ldots)$. Clearly $t_m\rightarrow t_0$, $t_m$ is in the image of $T$ and $t_0$ is not in the image of $T$.

Answer 2

Consider $t_0=(\frac {1}{\sqrt{n}})_n$. This too is not in the image of $T$, since its preimage is $(\sqrt{n})_n$. Now define $t_m=(\frac{1}{\sqrt{1}},\frac{1}{\sqrt{2}},\ldots,\frac{1}{\sqrt{m}},\frac{\sqrt{m}}{m+1},\frac{\sqrt{m}}{m+2},\ldots)$, the image of $(\sqrt{1},\sqrt{2},\ldots,\sqrt{m},\sqrt{m},\sqrt{m},\ldots)$. But now $\|t_m-t_0\|_\infty=\sup_{n> m}\{|\frac{1}{\sqrt{n}}-\frac{\sqrt{m}}{n}|\}=\sup_{n> m}\{\frac{1}{\sqrt{n}}|\sqrt{\frac{m}{n}}-1|\}<\frac{1}{m}$, because $0<|\sqrt{\frac{m}{n}}-1|<1$