I need to prove that

$$A=\{a\in \ell^1:\sum_{i=1}^{\infty}|a_n| \le 1\}$$

is closed, bounded and not a compact subset in $\ell^1$. Boundedness is trivial, but I get stuck in the other two. Proving subsets of $l^1$ are not closed seems easy, because one sequence whose limit is not in the subsets does it, but I’m stuck in proving that any sequence converges to a point in $A$. Thanks in advance!


It is closed because it is $f^{-1}\bigl([0,1]\bigr)$, with $f(\sum_{n=1}^\infty a_n)=\sum_{n=1}^\infty|a_n|$.

And it is not compact because, if $e_n$ is the series whose $n$th term is $1$ and all others are $0$, then $(e_n)_{n\in\mathbb N}$ has no convergent subsequence (since the distance between any two distinct terms is $1$).

  • $\begingroup$ Thank you so much for the answer. Wouldn’t it be easier to define $f(a)=\sum_{i=1}^\infty|a_n|$ in order to prove continuity of $f$? $\endgroup$ – Firage Mar 5 '18 at 1:31
  • 1
    $\begingroup$ @Firage Well, that's what I did, right?! $\endgroup$ – José Carlos Santos Mar 5 '18 at 7:16

PART 1. In any normed linear space, if $\lim_{n\to \infty}\|v_n-v\|=0 $ then $\lim_{n\to \infty}\|v_n\|=\|v\|.$ PROOF:

(i). $\|v_n\|=\|(v_n-v)+v\|\leq \|v_n-v\|+\|v\|. \;$.... So $\|v_n\|-\|v\|\leq \|v_n-v\|.$

(ii). $\|v\|=\|(v-v_n)+v_n\|\leq \|v-v_n\|+\|v_n\|.\; $.... So $\|v\|-\|v_n\|\leq \|v-v_n\|.$

(iii). From (i) and (ii) the absolute value of $\|v_n\|-\|v\|$ cannot exceed $\|v_n-v\|.$

So in your Q, if $(a_n)_{n\in \Bbb N}$ is a sequence of members of $A$ converging to $a\in l^1$ then $\|a\|=\lim_{n\to \infty}\|a_n\|\leq \sup_{n\in \Bbb N}\|a_n\|\leq 1, $ so $a\in A.$ Therefoe $A$ is closed.

It $is$ true that $A$ with the metric $d(u,v)=\|u-v\|$ is a $complete$ metric space, but that is another matter.

PART 2. The Kronecker delta. Definition: $\delta (a,b)=1 $ if $a=b. $ And $\delta (a,b)=0 $ if $a \ne b.$

For $n\in \Bbb N$ let $v_n=(\delta (j,n))_{j\in \Bbb N}.$ Let $V=\{v_n:n\in \Bbb N\}.$ Then $V$ is an infinite subset of $A.$

Let $C=\{B(v, 1):v\in A\}$ where $B(v,1)=\{u\in l^1: \|v-u\|<1\}. $ Then $C$ is an open cover of $A.$ We have $\|v_n-v_m\|=2$ when $n\ne m,$ so the triangle inequality implies that any member of $C$ contains at most $1$ member of $V.$

So if $D$ is a finite subset of $C$ then $\cup D$ contains only finitely many members of $V, $ so $D$ is not a cover of $A.$ Therefore $A$ is not compact.


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