I have to prove that the range $\mathcal{R}(T)$ of bounded linear operator $T:X\rightarrow Y$; $X,Y$ normed spaces need not be closed in $Y$. As a hint I'm given that I could consider $T:\mathcal{l^{\infty}}\rightarrow \mathcal{l^{\infty}}$ where $y=Tx$ is such that $y_n=\frac{x_n}{n}$.
Now my attempt is as follows:
We take the sequence $y^{(n)}=(1,\frac{1}{\sqrt{2}},..., \frac{1}{\sqrt{n}}, 0,0,...)$. Then clearly we can take for each $y^{(n)}$ a $x^{(n)}=(1, \frac{2}{\sqrt{2}},...,\frac{n}{\sqrt{n}},0,0,...)\in \mathcal{l^{\infty}}$ so that $Tx^{(n)}=y^{(n)}$, and thus $y^{(n)}\in \mathcal{R}(T)$. Consider, the limit $y=\lim_{n}y^{(n)}=(1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},...)\in \mathcal{l^{\infty}}$. If this were to be in $\mathcal{R}(T)$, then we must have $x=(1, \frac{2}{\sqrt{2}},\frac{3}{\sqrt{3}},...)\in \mathcal{l^{\infty}}$, but this is clearly not the case, since $x$ is a sequence increasing to $\infty$. Thus we have a converging sequence in $\mathcal{R}(T)$, whose limit is not in $\mathcal{R}(T)$, and thus $\mathcal{R}(T)$ cannot be closed.
I wonder if this is correct, and even if it is I'm still curious about other proofs of this statements, with or without use of the suggested operator.
Thanks.
