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Let consider first the space $$V=x_i.$$

This is a complete metric space (refer to my course). Now I was wondering if the set $$mathcal A=xin ell^infty mid ,$$
is compact. I know that in a metric space, a set is compact $iff$ its sequentially compact (i.e. every sequence has a convergent subsequence). I consider the sequence $$x_n=x^i_n,$$
where $x_n^i=(0,0...,0,1,0,...)$ where the $1$ is at the $i^th$ position. In other words, $$x_1=(1,0,0,...)$$
$$x_2=(0,1,0,0,...)$$
$$x_3=(0,0,1,0,...)$$

We have that $$|x_n^i|_ell^infty =1$$
for all $n$ and thus $(x_n)_n$ is a bounded sequence. What could be a convergent subsequence ? I guess that there is no convergent subsequence. To me, if a subsequence converge, it must converge to $0$ (I really don't know how to prove this), but in the same time the norm of the limit must be 1 I guess. So it can't have convergent subsequence. Am I right ? So $mathcal A$ is not compact ?

According to Riesz's lemma a normed vector space is of finite dimension if an only if its closed unit ball is compact. As your space $V$ is of infinite dimension, $mathcal A$ can't be compact.

Hint: if the sequence $(x^i)subset ell^infty$ converges to $x$ in $ell^infty$, then it converges component-wise, i.e. $lim_ito +infty x^i_n = x_n$ for every $n$.

Note that for $i neq j$, $||x_i - x_j||_infty = max, = 1$, so your sequence cannot have a Cauchy subsequence.

There is a fairly common observation here. $[-1,1]^omega = mathcalA$ and the set of that "cube"'s corners, $-1,1^omega$, is a countable set of points with pairwise distance bounded below. With the open cover definition of compact, the collection of open balls of radius $2$ centered on the set of corners is an open cover with no finite subcover (since each center is contained in only one ball in the collection and there are infinitely many corners).