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.)


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.