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Define the space of finite sequence $c_{00}:=\{\{x_k\}_{k=1}^{\infty}: x_k^{(n)}=0 \mbox{ eventually} \}$ with sup norm, $x_k^{(n)}$ is the $n$-th coordinate of $x_k$. I try to show that $c_{00}$ is not complete without using the property "Let $X$ be a Banach space with norm $\|\cdot\|_X$. Let $Y$ be a subspace of $X$. Then $Y$ is Banach space iff $Y$ is closed in $X$."

I try to prove it by definition but I am stuck the last step.

I take a sequence $x_1=(1,0,\dots,), x_2=(1,1/2, 0,\dots), \dots, x_k=(1,\frac{1}{2}, \frac{1}{3},\dots, \frac{1}{k},\dots)$. Clearly, for $m>n$, $$ \|x_m-x_n\|_{\infty}=\frac{1}{n+1}\to 0 $$ So it is a Cauchy sequence in $c_{00}$.

However, to prove $\{x_n\}$ does not converges in $c_{00}$, I try to prove it by contradiction. I am stuck on this step.

If I assume that $\{x_n\}$ converges to a limit point $x\in c_{00}$, then as $n\to \infty$, $$ \|x_n-x\|_{\infty}\to 0. $$

How do we get the contradiction?

Hermi
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    An overkill approach: a countably-infinite dimensional normed space is never complete by the Baire category theorem – Lukas Heger Oct 07 '23 at 23:11
  • $|x_n-x|{\infty} \geq |x{n,i}-x_i|$ for each $i$. Can you say what $x_i$ has to be? – Kavi Rama Murthy Oct 07 '23 at 23:11
  • It's easy to guess what the limit of your sequence is and it is indeed not in $c_{00}$. Show the norm-convergence takes place and you are done (normed spaces are Hausdorff, hence uniqueness of limits). – Oskar Vavtar Oct 07 '23 at 23:17
  • @geetha290krm Yes! So it means $x_{ni}=x_i$ for $n$ large enough. Then for $n\ge N$ for some constant $N>0$, $x=(1, 1/2, \dots, \frac{1}{N+1},\frac{1}{N+2},\dots)$, so $x$ cannot be in $c_{00}$? – Hermi Oct 07 '23 at 23:30
  • Yes, that is it! – Kavi Rama Murthy Oct 07 '23 at 23:38
  • @LukasHeger Your argument is an over over overkill. For something that follows easily from definition of norm you don't use Baire Category Theorem. – Kavi Rama Murthy Oct 07 '23 at 23:59

2 Answers2

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The incompleteness can be shown from within.

Let $x\in c_{00}$. Then $$ x=(a_1,\ldots,a_m,0,0,\ldots). $$ For any $n\geq m+1$ you have $$ d(x_n,x)=\|x_n-x\|\geq\frac1{m+1} $$ since the $(m+1)^{\rm th}$ entry of $x$ is $0$. So the sequence $\{x_n\}$ is at a positive distance from any element in $c_{00}$ and hence it is not convergent, though it is Cauchy.

Martin Argerami
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Assume $(x_{i}) \rightarrow x \in c_{00}$. Then $x$ is eventually $0$, let's say starting at index $n$, so $\forall k \geq n, x^{(k)}=0 \space$. Then $\forall m \geq n, \|x_{m}-x\| \geq \frac{1}{m} \implies (x_{i}) \nrightarrow x \in c_{00} \implies c_{00}$ is incomplete ($(x_{i})$ converges but not to a point in $c_{00}$, that is, to $x=(1,\frac{1}{2},\frac{1}{3},...)$).

It's worth pointing out that any Cauchy sequence converges, just not necessarily to a limit inside the given space. Here's an interesting answer: Does every Cauchy sequence converge to *something*, just possibly in a different space?.

ssv
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