Prove some set in $(l^\infty,d_\infty)$ is closed.

Asked Mar 15 '21 at 08:07

Active Mar 15 '21 at 08:13

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Can someone verify if my reasoning is ok?

After parsing the definitions, I realized that I need to show that if I let $x \in \bar A$, the closure, then $x \in A$ to show closed, which is same as proving $n\vert x_n\vert \leq 1$ for all $n\geq 1$.

So after thinking for a bit, I didn't get anywhere with direct proof so I did it by contradiction. Suppose $\sup n \vert x_n \vert > 1$, then we have a $N \in \mathbb{N}$ and $k > 1$ so that $N\vert x_N\vert = k$. Problem is now if we let $\epsilon = \frac{k-1}{N+1}$ then since $x \in \bar A$, for all $r > 0$: $$ B_r(x) \cap A \neq \emptyset \implies \exists y \in A \text{ so that } \sup\vert x_n - y_n \vert < \epsilon $$ And by some triangle inequality it follows: $$ N\vert x_N\vert\leq N\vert y_N \vert + N\epsilon $$ Which I guess is a contradiction. Is this correct?

edited Mar 15 '21 at 08:13

asked Mar 15 '21 at 08:07

nabu1227

1

It is correct if you change the last inequality to a strict inequality. – Kavi Rama Murthy Mar 15 '21 at 08:15
I think I edited $\epsilon = (k-1)/N$ into $\epsilon = (k-1)/(N+1)$, now it works right? I don't need strict inequality – nabu1227 Mar 15 '21 at 08:16
1

Fine, but you can also work with the original $\epsilon$ since $|x_N|<|y_N|+\epsilon$. – Kavi Rama Murthy Mar 15 '21 at 08:17
oh right thats true,thanks – nabu1227 Mar 15 '21 at 08:18

Prove some set in $(l^\infty,d_\infty)$ is closed.

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