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.

  • 1
    $\begingroup$ first step: assume limit exists: let it be $e=(e_1,e_2,...)$. What are $e_i$? What is $||e-x_i||$ ? $\endgroup$ – Ben Feb 28 at 19:57
  • $\begingroup$ @Ben 1 or 0. Depends $\endgroup$ – orientablesurface Feb 28 at 19:59
  • $\begingroup$ What is the pointwise limit of the i-th coordinate of (x_j)? $\endgroup$ – Ben Feb 28 at 20:00
  • $\begingroup$ 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) $\endgroup$ – user160185 Feb 28 at 20:02
  • $\begingroup$ Any two distinct elements of the sequence you mention are at distance $1.$ A fortiori there aren't converging subsequences. $\endgroup$ – Will M. Feb 28 at 20:08

I think you have almost figured out the correct argument.

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.

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