Ytukyg

Show that the space $S$ of all real sequences that are $0$ after a certain rank is not complete with the uniform norm. Find a metric space $X$ such that $bar{S} = X.$

I took the sequence $x_{k}^{n}=(1,1/2,1/3,cdots ,1/n,0,0,cdots)$ then for $ngeq m$
$$||x_{k}^{n}-x_{k}^{m}||_{infty}leq frac{1}{m+1}to 0$$ and so the sequence ${x^{n}_{k}}$ is Cauchy. However it is not convergent since the limit $x=(1,1/2,1/3,cdots)$ does not belong to $S.$ So $S$ is not complete. Next, I am guessing that $X= l^{infty},$ but I am not sure how to prove this. Any hints will be much appreicated.

If $X=ell^{infty}$, then you would be able to approximate the element
$$bar{x}=(1,1,1,dots) $$
with eventually vanishing sequences. This is clearly not true. So $X$ is a smaller space.

More generally the problem is represent by elements containing infinitely many elements larger than $varepsilon$ for some $varepsilon>0$. Indeed, in this case, if $left{x^kright}_{kin mathbb{N}}subset S$, we have
$$|x^k-bar{x}|_{infty}geq varepsilon $$
and so $x^knot to bar{x}$. But these are precisely the elements $bar{x}=(bar{x}_n)_{nin mathbb{N}}$ such that $lim_{nto +infty}bar{x}_nneq 0$. So we might guess that
$$X=left{x=(x_n)_{nin mathbb{N}}in ell^{infty}:lim_{nto +infty}x_n=0right} $$
it remains to verify that indeed, each element $xin X$ can be approximated by a sequence $left{x^kright}_{kin mathbb{N}}$ elements in $S$, and this is easily done by truncating the sequence $(x_n)_{nin mathbb{N}}$ at the $k$-th index.