Redundance in $l^p$ space.

Question

I am covering the following problem:

Let $A \subset l^p$, with $p \in [1,\infty)$, then is equivalent:

i)A is relatively compact

ii)A is bounded and we have $$\lim_{n \rightarrow \infty} \sup_{x \in A} ( \sum_{i=n}^{\infty}|x(i)|^p)^{\frac{1}{p}} = 0$$

My question is: As far as I see is the second message is somehow redundant, cause if we know that $A$ is bounded, then we have $\forall x \in A: ||x|| \le C$ for some $C>0$ and this surely implies the thing with the limit as $sup ||x|| \le C$ still holds? CAUSE: We surely know that this limit is zero for all elements in A, since they have a bounded norm. So I am asking myself. Where is the point in this limit. What does it tell me that I did not know by just having that $A$ is bounded?

Answer 1

The second part of (ii) gives information about the remainder of the series defining the norm. This is not contained in the fact that the set is bounded: consider indeed $A:=\{e^{(k)},k\in\mathbb N\}$ where $e^{(k)}_j=1$ if $k=j$ and $0$ otherwise. This set is bounded but the other property does not hold.