Not seeing the justification for an absolutely summable series satisfying an inequation. I'm reading in Dieudonne:  

In order that a denumerable family $(x_\alpha)_{\alpha\in A}$ of elements of a Banach
  space $E$ be absolutely summable, a necessary and sufficient condition
  is that the finite sums $\sum_{\alpha\in J}||x_\alpha||$ ($J \subset A$ and finite) be bounded. Then,
  for any $\varepsilon > 0$, there exists a finite subset $H$ of $A$ such that, for any
  finite subset $K$ of $A$ for which $H\cap K=\emptyset$, $\sum_{\alpha\in K}||x_\alpha||\leq\varepsilon$, and for any finite subset $L\supset H$ of $A$, $||\sum_{\alpha\in A}x_\alpha - \sum_{\alpha\in H}x_\alpha||\leq 2\varepsilon$.

He says the first two statements are obvious and begins his proof with the claim about $L$.  

Then, for any finite subset $L\supset H$, we can write $L = H \cup K$ with
  $H\cap K=\emptyset$, hence $||\sum_{\alpha\in L}x_\alpha-\sum_{\alpha\in H}x_\alpha||\leq \varepsilon$; from the definition of the sum $\sum_{\alpha\in A}x_\alpha$ it follows (after ordering $A$ by an arbitrary bijection of $\bf N$ onto $A$) that $||\sum_{\alpha\in A}x_\alpha - \sum_{\alpha\in H}x_\alpha||\leq \varepsilon$ ...

I've cut off the proof where I can't figure out the justification.  The next few words that he has in the proof make sense to me, but I'm not sure what justifies this last inequation.  The definition of the sum is that the series under a bijection with $\bf N$ be absolutely convergent.  So there should be some $n$ such that for $m>n$ under the map $\phi$ we have $\sum_{i\geq m} ||x_{\phi(i)}||\leq \varepsilon$ but this seems to assume that $H$ is exactly the set of terms with index less than $n$.  What we know about $H$ is just this fact about disjoint finite sets ... I can kind of see the intuition here but the rigorous justification is eluding me.
 A: Let $\phi \colon \mathbf{N} \to A$ be a bijection. Then $S := \phi^{-1}(H)$ is a finite subset of $\mathbf{N}$. If $H = \varnothing$, let $m = -1$, else let $m = \max S$. Then, for $n \geqslant m$, we have $H \subset \phi([0,n])$. Let $K_n = \phi([0,n]) \setminus H$, and
$$s_n := \sum_{i = 0}^n x_{\phi(i)}.$$
By the first mentioned property of $H$, we find
$$\biggl\lVert s_n - \sum_{\alpha \in H} x_{\alpha}\biggr\rVert = \biggl\lVert\sum_{i = 0}^n x_{\phi(i)} - \sum_{\alpha \in H} x_{\alpha}\biggr\rVert = \biggl\lVert\sum_{\alpha \in K_n} x_{\alpha}\biggr\rVert \leqslant \sum_{\alpha \in K_n} \lVert x_{\alpha}\rVert \leqslant \varepsilon,$$
and by the continuity of the norm it follows that
$$\biggl\lVert \lim_{n\to \infty} s_n - \sum_{\alpha \in H} x_{\alpha}\biggr\rVert = \biggl\lVert \sum_{i = 0}^{\infty} x_{\phi(i)} - \sum_{\alpha \in H} x_{\alpha}\biggr\rVert \leqslant \varepsilon.$$
By what you have quoted, it seems he defined $\sum\limits_{\alpha \in A} x_{\alpha}$ as $\sum\limits_{i = 0}^{\infty} x_{\phi(i)}$ for a bijection between $\mathbf{N}$ and $A$. If that is the case, it needs of course be proved that the value doesn't depend on the chosen bijection.
