# $L^{\infty}$ and uniform norm compact subset

Consider $$L^{\infty}$$ to be the set of all bounded real sequences and a subset

$$E=$$ $$\{$$ $$(x_n)_n \in L^\infty\,:\,\sup_n|x_n|\leq 1$$ }$I am trying to show that $$E$$ is not a compact subset. Since we are dealing with a metric space, I am in particular trying to show that it is not a sequentially compact subset. With respect to the supremum/ uniform metric. Hence, I must find a sequence with no convergent subsequence. The example I have thought of is the sequence of all 0s with 1 in the nth position. However, I am not quite sure how to formally prove that is sequence does not have a convergent subsequence. Note that its subsequences are either itself or the 0 sequence. • first step: assume limit exists: let it be$e=(e_1,e_2,...)$. What are$e_i$? What is$||e-x_i||$? – Ben Feb 28 '20 at 19:57 • @Ben 1 or 0. Depends – monoidaltransform Feb 28 '20 at 19:59 • What is the pointwise limit of the i-th coordinate of (x_j)? – Ben Feb 28 '20 at 20:00 • You can also use the fact that compact metric spaces are totally bounded, so all of its subsets are totally bounded as well. But the set$\left\{e_n\right\}_n$is not totally bounded (where$e_n=(0,\ldots,0,1,0,\ldots,0,\ldots)$has zeroes everywhere, except for a$1$in the$n$-th position) – user160185 Feb 28 '20 at 20:02 • Any two distinct elements of the sequence you mention are at distance$1.\$ A fortiori there aren't converging subsequences. – Will M. Feb 28 '20 at 20:08

For each $$n\in \mathbb{N}$$, let $$x_n=e_n$$, which is the sequence with all $$0$$ except $$1$$ at the $$n$$-th entry. It is clear that $$(x_n)_n$$ is in $$E$$. Moreover, we have that $$\|x_m-x_n\|_\infty=1$$ for all $$m\not =n$$, which implies that $$(x_n)_n$$ has no subsequence forming a Cauchy sequence in the sup-norm (this can be proved by contradiction). Hence we know that the sequence has no convergent subsequence.