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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.)

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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 $\| (x_j^n)- (y_j)\|_{\ell^\infty} \leq \frac1{\sqrt{n}}$.

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  • $\begingroup$ What is $1_{j\leq} n$ $\endgroup$ – Infinity Aug 17 '17 at 10:34
  • $\begingroup$ $\mathbb 1_{j\leq n}$ is $1$ if $j\leq n$ but $0$ if $j>n$. $\endgroup$ – supinf Aug 17 '17 at 10:39

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