Let $Y=${$(x_j/j):(x_j)\in l_\infty$}. Is Y closed in $l_\infty$?
(I could not find any sequence in Y tending to some point outside Y. Please give some hint for such kind of sequence or any other hints/ways.)
Let $Y=${$(x_j/j):(x_j)\in l_\infty$}. Is Y closed in $l_\infty$?
(I could not find any sequence in Y tending to some point outside Y. Please give some hint for such kind of sequence or any other hints/ways.)
Copyright © 2021 JogjaFile Inc.
No, it is not closed.
Consider $$ y_j=\frac1{\sqrt{j}}, \quad (y_j)\in\ell^\infty. $$
It is easy to see that $(y_j)\not\in Y$.
Now define $(x_j^n)$ via $$ x_j^n = y_j \cdot \mathbb 1_{j\leq n} $$
Then $(x_j^n)\in Y$, and $\| (x_j^n)- (y_j)\|_{\ell^\infty} \leq \frac1{\sqrt{n}}$.