Generalized associate law for unordered sums Let


*

*$I'$ be a nonempty set

*$I_m$ be a nonempty set for $m\in I'$

*$I:=\left\{(m,n):m\in I'\text{ and }n\in I_m\right\}$

*$E$ be a normed $\mathbb R$-vector space

*$(x_{mn})_{(m,\:n)\in I}\subseteq E$ be summable



I want to show that
  
  
*
  
*$(x_{mn})_{n\in I_m}$ is summable for all $m\in I'$
  
*$\left(\sum_{n\in I_m}x_{mn}\right)_{m\in I'}$ is summable and $$\sum_{m\in I'}\sum_{n\in I_m}x_{mn}=\sum_{(m,\:n)\in I}x_{mn}\tag1$$
  

My attempt for (1.) is as follows: Let $\varepsilon>0$. Since $(x_{mn})_{(m,\:n)\in I}$ is summable, it is Cauchy and hence there is a finite $J\subseteq I$ with $$\left\|\sum_{(m,\:n)\in K}x_{mn}\right\|_E<\varepsilon\;\;\;\text{for all finite }K\subseteq I\setminus J\tag2\;.$$ Let $m\in I'$ and $$J_m:=\left\{n:(m,n)\in J\right\}\;.$$ By definition, $$\left\{(m,n):n\in I_m\setminus J_m\right\}\subseteq I\setminus J\tag3$$ and hence $$\left\|\sum_{n\in K_m}x_{mn}\right\|_E<\varepsilon\;\;\;\text{for all finite }K_m\subseteq I_m\setminus J_m\;.\tag4$$

If $E$ is complete, we obtain (1.) by $(4)$. I would like to know, if we're able to prove this even without the completeness assumption on $E$. In any case, I wasn't able to find the right approach for a proof of $(2.)$. So, how can we prove it?

 A: *

*You have showed that every subfamily of a Cauchy family is Cauchy. We really need completeness in the last step of the whole proof. For example $∑_{n ∈ ℕ} e_n/n = (1, \frac12, \frac13, …)$ and $∑_{n ∈ 2ℕ} e_n/n = (0, \frac12, 0, \frac14, 0, …)$ in $\ell_∞$, but if we take $E = \operatorname{span}\{e_n/n: n ∈ ℕ,\ (1, \frac12, \frac13, …)\}$, then the subfamily $∑_{n ∈ 2ℕ} e_n/n$ is not convergent in $E$.


*Let us write $x =_ε y$ for $\lVert x - y\rVert_E < ε$. Let us suppose that $I$ is a disjoint union of the sets $I_m$ for $m ∈ I'$.
We are given $ε > 0$. There is finite $F ⊆ I$ such that $∑_{i ∈ F'} x_i =_ε x$ for every finite $F' ⊇ F$. Let $G$ be the finite set $\{m ∈ I': F ∩ I_m ≠ ∅\}$. And let $G' ⊇ G$ be finite. We will show that $∑_{m ∈ G'} ∑_{i ∈ I_m} x_i =_{2ε} x$.

Let $f\colon G' \to \{1, …, \lvert G'\rvert\}$ be a bijection. For every $m ∈ G'$ there is finite $F ∩ I_m ⊆ F_m ⊆ I_m$ such that $∑_{i ∈ F_m} x_i =_{ε/2^{f(m)}} ∑_{i ∈ I_m} x_i$. Let $F' = ⋃_{m ∈ G'} F_m$.
Since $∑_{m ∈ G'} ε/2^{f(m)} ≤ ε$, we have $∑_{m ∈ G'} ∑_{i ∈ I_m} x_i =_ε ∑_{m ∈ G'} ∑_{i ∈ F_m} x_i = ∑_{i ∈ F'} x_i =_ε x$.

