a) Give an example of a bounded closed subset of $$ A = \{(x_n) \in \ell^1: \sum_{n\geq1} x_n = 1\}$$ which is not compact. The metric we consider on A is induced by the normal norm on $\ell^1$.

b) Show that the set $$B = \{(x_n)\in \ell^1: \sum_{n\geq1} n|x_n| = 1\}$$ is compact in $\ell^1$.

My try: a)The sequences in $\hat{A} = \{e_i = (0,\ldots 0 , 1 ,0 \ldots, ),\; \forall i \in \mathbb{N}\}$ has no converging subsequences. Is it a closed subset? It seems like it contains all its boundary points (A is not vector space). b) If we take a sequence $(x^k) \in B$ we also have $(x^k) \in \ell^\infty$. But since all elements in $\ell^\infty$ has subsequences that converges, this is true for $(x^k)$?

Edited: Davids method seems to work, both Im trying to get it to work with the following hint.

Hint: You can use without proof the diagonalization process to conclude that every bounded sequence $(x_n) \in\ell^\infty$ has a subsequence $x^{n_k}$ that converges in each component. Moreover, sequences in $\ell^1$ are obviously bounded in their $\ell^1$ norm.

So what I need to show is that the convergence of a sequence does not end up outside B?


(a) The attempt works. To see that $S:=\{e_i,i\in\Bbb N\}$ is closed for the $\ell^1$ norm, let $x\notin S$. There are two index $i$ and $j$ such that $x_ix_j\neq 0$. Let $r:=\min\{|x_i|,|x_j|\}$. Then the open ball of center $x$ and radius $r$ is contained in the complement of $S$.

(b) The problem is that we have to check that we have convergence in $\ell^1$ for the subsequence.

As $\ell^1$ is complete, we can check that $B$ is precompact, i.e. given $\delta>0$, we can cover $B$ by finitely many balls of radius $<\delta$. It's equivalent to show both properties hold:

  1. $B$ is bounded in the $\ell^1$ norm;
  2. $\lim_{N\to +\infty}\sup_{x\in B}\sum_{k=N}^{+\infty}|x_k|=0$.

Indeed, if a set $S$ is precompact, with $\delta=1$ we get that it's bounded, and $2.$ is a $2\delta$ argument (I almost behaves as a finite set).

Conversely, assume that $1.$ and $2.$ hold and fix $\delta$. Use this $\delta$ in the definition of the limit to get an integer $N$ such that $\sup_{x\in B}\sum_{n=N+1}^{+\infty}|x_n|<\delta$. Then use precompactness of $[-M,M]^N$, where $M=\sup_{x\in B}\lVert x\rVert_1$.

Note that this criterion works for $\ell^p$, $1\leqslant p<\infty$.

In our case, each element of $B$ has a norm $\leqslant 1$, and for all $x\in V$, $$\sum_{k=N}^{+\infty}|x_k|\leqslant \frac 1N\sum_{k= N}^{+\infty}k|x_k|\leqslant \frac 1N.$$

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  • $\begingroup$ very nice! Can you please expand a little why 1 and 2 are equivalent to B being pre compact? $\endgroup$ – Johan Dec 8 '12 at 15:03
  • $\begingroup$ thanks! So we can not use that every subsequence in $\ell^\infty$ converges in each component? and show that this limit is in B? $\endgroup$ – Johan Dec 8 '12 at 15:44
  • $\begingroup$ Reading this again, Dont we also need to check that B is closed also? I was thinking about this: math.stackexchange.com/questions/218428/… $\endgroup$ – Johan Jan 9 '13 at 7:53
  • $\begingroup$ Yes, the properties 1. and 2. give that the set has a compact closure. So we also have to check closedness. $\endgroup$ – Davide Giraudo Jan 9 '13 at 9:47

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