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Let $S$ be the space of all eventually zero sequences. I need to prove that $(S,\|.\|_\infty)$ is not complete. I've already shown that $S$ is a dense subspace of the Banach space $(C_0,\|.\|_\infty),$ where $C_0$ is the space of all sequences converging to $0$. My aim is to prove that $(S,\|.\|_\infty)$ is not closed. Here is my attempt.

Consider the sequence $$(x^n)=\Big((0,0,0,\cdots),(1,0,0,\cdots),(1,1/2,0,\cdots),\cdots,(1,1/2,\cdots,1/n,0,0,\cdots),\cdots\Big)$$ in $S$.

Now for all $n,$ $\|x^n-x\|_\infty=1/n$, where $x=(1/n)_{n\in\mathbb{Z^+}}$.

Hence $\|x^n-x\|_\infty\to0$ as $n\to\infty$ or rather $(x^n)$ converges to $x$.

But $x\notin S$ while $x\in C_0=\overline S$. Therefore $(S,\|.\|_\infty)$ is not closed and hence not complete.

Could someone please tell me if this argument is ok? Thanks.

Janitha357
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  • Yep, it's correct – orangeskid Aug 23 '17 at 16:18
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    It's fine. But if you've already shown density, all you have to say is it can't be closed as $x\notin S$ ($S$ closure has to be all of $C_0$). – David Mitra Aug 23 '17 at 16:24
  • This seems slightly confusing. What is $x^n$? It seems like if $x^n=(1,1/2,\dots,1/n,0,0,\dots)$, then $|x-x_n|_{\infty}=\frac{1}{n+1}$. It still works, but you need to be clearer. – Thomas Andrews Aug 23 '17 at 16:28
  • @ThomasAndrews I prepended the sequence $(0,0,\cdots)$ to $(x^n)$ to get $1/n$ instead of $1/(n+1)$. I thought it would be nice. – Janitha357 Aug 23 '17 at 16:40

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