Given the theorem for an ordinary series, I need to prove the same for an arbitrary double series. I've seen a very similar question, but I haven't worked with double integrals in real analysis so far and I have to stick to the theorem: Proving double sums are interchangable.

This is a theorem $№\;28$ in the Analysis book written by Svetozar Kurepa and the proof I have to apply in the case of double series:

Instead of $(a_n)_n$ I have $(a_{mn})_{mn}$

Let $(a_n)_n$ be a complex number sequence:


$\text{series}\;\sum|a_n|\;\text{converges}\implies \text{series} \sum a_n\;\text{converges}\;\land\;\underbrace{\Big|\sum_1^{\infty}a_n\Big|\leq\;\sum_1^{\infty}|a_n|}_{\text{triangle inequality}}$


$\text{series}\;\sum|a_n|\;\text{converges}\;\land \exists\; \text{bijection}\; \sigma:\mathbb N\rightarrow \mathbb N\implies \text{series}\; \sum a_{\sigma{(n)}}\;\text{converges}$$ $$\land\;\sum_1^{\infty}a_{\sigma{(n)}}=\sum_1^{\infty}a_n$

Proof of $I.$:

Let $(a_n)_n$ be a real number series and let's define sequences with positive terms: $$a_n^+=\begin{cases} \displaystyle \ a_n,\;a_n\geq 0 \\ \ 0,\;\;a_n<0 \end{cases}\;\;\;\;$$ $$a_n^{-}=\begin{cases} \displaystyle \ 0\;\;a_n\geq 0 \\ \ -a_n\;a_n<0 \end{cases}$$ $$a_n=a_n^+-a_n^-,\;\forall n\in \mathbb N$$ $$\text{series} \sum |a_n|\;\text{converges}\;\land\;a_n^+\leq |a_n|\;\land\;a_n^-\leq |a_n|,\;\forall n\in \mathbb N\implies \sum a_n^+\;\& \sum a_n^-\;\text{converge}$$

Let $A^+$ & $A^-$ be their sums. By the theorem for convergence of linear combinations of $2$ series: $$A^+-A^-=\sum_1^{\infty} a_n^+-\sum_1^{\infty} a_n^-=\sum_1^{\infty} (a_n^+ -a_n^-)=\sum_1^{\infty} a_n\leftarrow \text{converges}$$

Proof of $II.$:

Let $a_n=\alpha_n+i\beta_n\;(\alpha_n,\beta_n\in \mathbb R)$

$$|\alpha_n|\leq\sqrt{\alpha_n^2+\beta_n^2}=|a_n|\implies \sum |\alpha_n|\leq \sum |a_n|\implies \text{series}\;\sum |\alpha_n|\;\text{converges}$$ $$\text{series}\;\sum |\alpha_n|\;\text{converges}\implies\text{series}\sum \alpha_n\;\text{converges}$$ Analogously: $$|\beta_n|\leq|a_n|\implies \text{series}\;\sum |\beta_n|\;\text{converges}$$ $$\text{series}\; \sum \alpha_n\;\&\;\sum \beta_n\;\text{converge}\;\implies \text{series}\;\sum (\alpha_n +\beta_n)=\sum a_n\;\text{converges}$$

Furthermore: $$\sum_1^{\infty} a_n=\sum_1^{\infty} \alpha_n+i\sum_1^{\infty} \beta_n=\sum_1^{\infty}\alpha_{\sigma{(n)}} +i\sum_1^{\infty} \beta_{\sigma{(n)}}=\sum_1^{\infty} (\alpha_{\sigma{(n)}}+i\beta_{\sigma{(n)}})=\sum_1^{\infty} a_{\sigma{(n)}}$$

  • $\begingroup$ "Given the theorem for an ordinary series, I need to prove the same for an arbitrary double series." What theorem are you talking about? $\endgroup$
    – zhw.
    Jan 3 '20 at 18:48
  • $\begingroup$ What is the result about double series that you want to prove? $\endgroup$
    – zhw.
    Jan 4 '20 at 19:36
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    $\begingroup$ The whole point of absolute convergence of series is that we can reorder the terms arbitrarily. Thus, given a double series, pick any convenient bijective function $\,\mathbb{N}\to\mathbb{N}\times\mathbb{N}\,$ to convert any double series into an ordinary series and the same results carry over. $\endgroup$
    – Somos
    Jan 5 '20 at 19:34
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    $\begingroup$ @Somos Well I wouldn't say it's the whole point. We don't tell our beginning calculus students much about reordering. We do tell them that absolute convergence implies convergence. $\endgroup$
    – zhw.
    Jan 5 '20 at 20:38
  • 1
    $\begingroup$ Absolute convergence implies convergence a fortiori. If the convergence is not absolute, then reorderinging the terms can cause the series to converge to another limit or diverge to infinity. $\endgroup$
    – Somos
    Jan 5 '20 at 20:44

Although I am not exactly a specialist in either topology or analysis, this topic that you address I hold quite dear to my heart so may I walk you through a rather general topological approach towards the aim of deriving a far more general form of the result you mention, obtainable by far more general and efficient methods than mere series summation. Please do bear with me, as this is likely to become quite an expound.

  1. First of all, let us enumerate a series of elementary definitions and notation conventions. Given a family of sets $A$ indexed by index set $I$ (if you wish to know more we can into the details of what a ''family of objects'' actually is), we define its direct product $$\prod_{i \in I} A_i$$ as the set of all families $x$ of objects indexed by $I$ and such that at every index $i \in I$ the corresponding component $x_i \in A_i$ is an element in $A_i$. In case all the sets in the given family are equal to one other and to a certain fixed set $B$, then we simplify the notation:

$$\prod_{i \in I}B=B^I$$

This is nothing else than the set of all families of elements of $B$ indexed by $I$. For arbitrary set $A$ we write $$\mathscr{P}(A)=\{X|\ X \subseteq A\}$$

for its powerset (set of all subsets of $A$). Therefore, combining the two notations introduced, saying that $M \in \mathscr{P}(A)^I$ means that $M$ is a family of subsets of $A$ indexed by $I$, so for every $i \in I$ we will have $M_i \subseteq A$.

On a set $A$ a binary relation $R$ will simply be a subset $R \subseteq A \times A$; we say that $x$ and $y$ are related by $R$ (are in the relation $R$) when $(x, y) \in R$ and we express this alternatively as $xRy$. We remark a distinguished relation on $A$, called the diagonal of $A$ and defined by $\Delta_A=\{(x,x)\}_{x \in A}$; in plain terms it is the relation of equality on $A$.

For arbitrary relation $R$ we define its inverse (or dual or reciprocal) $R^{-1}$ by the property $xR^{-1}y \Leftrightarrow yRx$.

  1. Let us recall notions from order theory: given a set $A$ we say $R$ is an order relation on $A$ if $R$ is a binary relation on $A$ satisfying the axioms of reflexiveness ($\Delta_A:=\{(x,x)\}_{x \in A} \subseteq R$), antisymmetry (for any $x, y$ if simultaneously $xRy$ and $yRx$ then $x=y$; or symbolically $R \cap R^{-1} \subseteq \Delta_A$) and transitivity (for any $x, y, z$ if $xRy$ and $yRz$ then $xRz$; in concise terms $R \circ R \subseteq R$).

An ordered set will be an ordered pair $(A, R)$ such that $R$ is an order relation on $A$; we will express the relation $xRy$ as $x \leqslant_R y$, since it is quite suggestive to employ the $\leqslant$ symbol for orders. For arbitrary set $A$ let us write $\mathscr{Ord}(A)$ for the set of all order relations on $A$.

Given a family $A$ of sets indexed by $I$ and a family of orders $$R \in \prod_{i \in I} \mathscr{Ord}(A_i)$$

which means that for each index $i \in I$ the relation $R_i$ is an order on $A_i$, we can define a natural order relation on the cartesian product

$$\prod_{i \in I} A_i$$

structure which will be called the direct product of the orders $R_i$ with $i \in I$, denoted by $S$ and defined such that

$$x \leqslant_S y \Leftrightarrow (\forall i)(i \in I \Rightarrow x_i \leqslant_{R_i} y_i)$$

This definition amounts to saying that two families $x, y \in \prod_{i \in I} A_i$ are comparable through $S$ when they are comparable componentwise, through $R_i$ at each index $i$. The ordered set $$(\prod_{i \in I} A_i, S)$$

is called the direct product of the family of ordered sets $(A_i, R_i)_{i \in I}$.

A map $f: A \to B$ between two ordered sets $(A, R)$ and $(B, S)$ is called increasing or isotonic when $$(\forall x, y)(x \leqslant_R y \Rightarrow f(x) \leqslant_S f(y))$$

and it is called an isomorphism of ordered sets (order isomorphism for short) if it is isotonic and invertible such that its inverse also is isotonic.

By an upward directed set we mean an ordered set $(A, R)$ such that for any $x, y \in A$ there will exist an element $z \in A$ with the property $x, y \leqslant_R z$.

A typical example of such objects can be found in the instance of the set of all finite subsets of a given set $A$, collection which we are going to denote by $\mathscr{F}(A)=\{X \subseteq A|\ |X|< \aleph_0\}$ and which we will order by inclusion: since for any finite subsets $X, Y \subseteq A$ the union $X \cup Y \subseteq A$ is also finite and obviously $X, Y \subseteq X \cup Y$, the property of upward-directedness is indeed satisfied. As an ordered set, $\mathscr{F}(A)$ will have a maximum if and only if $A$ is finite (in which case the maximum is $A$ itself).

Before concluding let us make the observation that a direct product of upward directed (ordered) sets is itself an upward directed set.

  1. Next, let us recall the notion of generalized sequence of elements of a set: given set $A$, index set $I$ and order relation $R$ such that $(I, R)$ is upward directed, for any family of elements $x \in A^I$ we call $(x, R)$ a generalized sequence of elements of $A$ (with respect to order $R$, if we are to be very pedantic; it is the specific choice of an order that determines the ''sequence'' of elements in the family $x$). Although an abuse of language, it is quite customary however to refer to a generalized sequence $(x, R)$ as merely the sequence $x$, omitting any mention of the underlying order.

  2. Given a topological space $(X, \mathscr{T})$ (I hope you were at least introduced to the notion), an index set $I$ upward directed by order $R$ and family $x \in X^I$ we say that the generalized sequence $(x, R)$ converges to element $t \in X$ (the syntagm ''sequence $(x, R)$ has $t$ as a limit" is also acceptable) if:

for any neighbourhood $V \in \mathscr{V}(t)$ there will exist an index $i \in I$ such that for all $j \geqslant_R i$ it holds that $x_j \in V$

which we express symbolically as

$$x_i \xrightarrow[R]{i \in I} t\ (\mathscr{T})$$

or more simply

$$x \xrightarrow[R]\ t\ (\mathscr{T})$$

this being the most general notation to be used in the case of not necessarily Hausdorff spaces (when limits might not be unique whenever they exist). For reasons of rigour we make explicit in brackets the topology with respect to which the convergence takes place; however, when there is no risk of ambiguity (and thus in the vast majority of cases) we shall omit any reference to the background topology.

In the particular case of a Hausdorff space, we also adopt notation

$$\mathrm{lim}_{i \in I\\ R}x_i=t$$

or in a simplified version


We shall say a generalized sequence converges (or is convergent) in $X$ if there exists a $t \in X$ such that $x$ converges to $t$. Let us notice that if the ordered set $(I, R)$ has a maximum $\alpha$ then any generalized sequence $x$ indexed by it will be convergent, having $x_{\alpha}$ as a limit.

This section we conclude with a couple of remarks that will play a definite role later on:

  • If $(I, R), (J, S)$ are two upward directed sets, $\varphi: J \to I$ an isomorphism between them and $x \in X^I$ such that $x \xrightarrow[R]{i \in I} t \in X$ then $((x_{\varphi(j)})_{j \in J}, S)$ is also a generalized sequence and $x_{\varphi(j)} \xrightarrow[S]{j \in J} t$; in particular, in a Hausdorff space we can write:

$$\mathrm{lim}_{j \in J\\ S} x_{\varphi(j)}=\mathrm{lim}_{i \in I\\ R}x_i$$

(the change of variable for generalized sequences).

  • Consider an index set $J$, a family of topological spaces $(X_j, \mathscr{T}_j)_{j \in J}$, a family of points

$$t \in \prod_{j \in J} X_j$$

a family $(I_j, R_j)_{j \in J}$ of upward directed sets and a family of families (!) $$u \in \prod_{j \in J} X_j^{I_j}$$

(explicitly, for each $j \in J$ the object $u_j$ is itself a family of points in $X_j$ indexed by $I_j$) such that for each $j \in J$ we have

$$u_j \xrightarrow[R_j]\ t_j\ (\mathscr{T}_j)$$

If we denote by $(Y, \mathscr{S})$ the direct product of the family of topological spaces $(X_j, \mathscr{T}_j)$ (so in particular $Y=\prod_{j \in J} X_j$), $(K, S)$ the direct product of the family of ordered sets $(I_j, R_j)$ (so that $K=\prod_{j \in J}I_j$) and if we introduce the family $$y \in Y^K\\ y_k=((u_j)_{k_j})_{j \in J}$$

(family which is itself informally speaking the direct product of the family of families $u$), then we have that $(y, S)$ is a generalized sequence (informally, the direct product of the family of generalized sequences $(u_j, R_j)_{j \in J}$) and that

$$y \xrightarrow[S]\ t\ (\mathscr{S})$$

(compatibility of direct products with convergence of generalized sequences).

  1. The concept coming up next is going to be crucial for the appropriate generalization of (double/multiple etc) series convergence. Consider an arbitrary commutative topological monoid (which we are going to handle in additive notation) $(M, +, \mathscr{T})$: this means that the subjacent structure $(M, +)$ is a commutative monoid and $\mathscr{T}$ is a topology on $M$ compatible with the monoid structure. Further consider an arbitrary index set $I$ and a family $x \in M^I$: from it we can naturally define the generalized sequence of (finite) partial sums of $x$ as the family $$s \in M^{\mathscr{F}(I)}\\ s_F=\sum_{i \in F}x_i$$

where we implicitly understand the upward directed order on $\mathscr{F}(I)$ to be inclusion (cnf. paragraph 1) and we shall not mention it explicitly.

With this in place we say that the family $x$ is summable in $M$ if the associated generalized sequence of partial sums $s$ converges in $M$; any limit of $s$ will be called a sum of family $x$. If the topology considered is Hausdorff, then a summable family will have only one sum, and we agree to write:

$$\sum_{i \in I}x_i=\mathrm{lim}_{F \in \mathscr{F}(I)} \sum_{i \in F}x_i$$

Written explicitly, the claim that $t$ is a sum of family $x$ means that:

for every neighbourhood $V \in \mathscr{V}(t)$ there exists a finite subset $K \subseteq I$ such that for every finite subset $H$ such that $K \subseteq H \subseteq I$ one has

$$\sum_{i \in H}x_i \in V$$

Let us end this paragraph with a couple of important remarks:

  • a) If the index set $I$ is finite, then by some of the remarks of paragraphs 2 and 4, any family $x$ indexed by $I$ will be summable having $\sum_{i \in I}x_i$ as a sum! Therefore, in a Hausdorff commutative topological monoid the sum of a finite family understood in the ''topological sense'' we just introduced in this section is nothing else than its algebraic sum in the ordinary algebraic sense.

  • b) Given an arbitrary family of objects $x$ indexed by $I$ and arbitrary subset $J$ allow me to denote by $x_{|J}=(x_i)_{i \in J}$ the restriction of $x$ to $J$; With this specification, if $x \in M^I$ and $J, J' \subseteq I$ are disjoint ($J \cap J'=\varnothing$) and such that both restrictions $x_{|J}, x_{|J'}$ are summable of sums say $s$ and respectively $t$, then the ''joint'' restriction $x_{|J \cup J'}$ is also summable, having $s+t$ as a sum; this observation easily extends to a finite collection of subsets of $I$ that are pairwise disjoint and such that the restriction of $x$ to each of them is summable.

  • c) Consider a permutation $\sigma \in \Sigma(I)$ of the set $I$ and the ''permutated'' family $y=(x_{\sigma(i)})_{i \in I}$. It is not difficult to show that if $x$ is summable of sum $s$ then $y$ is also summable admitting $s$ as a sum.

  • d) Given a sequence $x \in M^{\mathbb{N}}$, if $x$ is summable as a family with sum $s$ then the series $x$ (of general term $x_n, n \in \mathbb{N}$) also converges to sum $s$. Symbolically, if

$$\sum_{n \in H} x_n \xrightarrow{H \in \mathscr{F}(\mathbb{N})}\ s$$

then we also have

$$\sum_{k=0}^n x_k \xrightarrow {n \to \infty}\ s$$

In the case of a Hausdorff topology, if sequence $x$ is summable then

$$\sum_{n \in \mathbb{N}}x_n=\sum_{n=0}^{\infty}x_n$$

This result is also valid for double (in general multiple) sequences/series and it is this observation which makes the entire theory I am expounding on here applicable to your particular case of summating series!

We are finally prepared to give a very general formulation of the result known as:

Generalized Associativity Theorem: Let $(G, +, \mathscr{T})$ be a complete Hausdorff abelian topological group, $I, K$ two index sets, a family of subsets $J \in \mathscr{P}(I)^K$ yielding a partition of $I$, in other words such that the properties:

$$\bigcup_{k \in K}J_k=I$$ $$(\forall k,l)(k,l \in K \wedge k \neq l \Rightarrow J_k \cap J_l = \varnothing)$$

are fullfilled and a family $x \in G^I$ that is summable in $G$. We then have that

  • for any $k \in K$, the restriction $x_{|J_k}$ is summable (this allows us to introduce $y_k:=\sum_{i \in J_k}x_i$)
  • the family of partial sums $(y_k)_{k \in K}$ is also summable and:

$$\sum_{i \in I}x_i=\sum_{k \in K}y_k$$

Written in developed fashion the relation above becomes:

$$\sum_{i \in I} x_i=\sum_{k \in K} \sum_{i \in J_k}x_i$$

which is indeed a relation of general associativity. I am not going to prove this very general statement however, as it relates to abstract notions of uniform structures (and I do not think it yet helpful for your understanding to take matters to such a level of abstraction), and will instead concentrate on a more familiar and particular version of it. Do however write back if you would like to learn more about the general setting.

As the situation stands right now, my feeble compiler is no longer able to efficiently handle further re-edits of this answer, so I see myself compelled to continue the intended presentation in a separate answer. I apologise for the highly unconventional approach, but I ask all those interested to bear with me.

  • $\begingroup$ thank you for your patience, precision, and understanding! I'm not familiar with all the expressions and notation since I'm at my first year, but I did read on this topic. If I got it right, $\leq_{R}$ is something like $+_{n}$ in $\mathbb Z_{/n}$, so it denotes the algebraic structure? Have I understood this right:$$|\mathscr{F}(A)|<|\mathcal P(\mathbb R)|=2^{\aleph_0}?$$ I ask to keep pace with your mind. And $\mathscr{V}$ is a $\text{family of neighbourhoods}$? And is $F=I$? $$$$ Then $I$ is a quotient set generated by $\mathscr{P}(I)^K$? $\endgroup$
    – Invisible
    Jan 10 '20 at 7:07
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    $\begingroup$ @VerkhovtsevaKatya: I'm glad you found it helpful and I will be back later on with details (for the moment I had to leave my desk and attend some matters outside). $\endgroup$
    – ΑΘΩ
    Jan 10 '20 at 7:15
  • $\begingroup$ @ ΑΘΩ, no problem, I enjoy thinking further about what you wrote. $\endgroup$
    – Invisible
    Jan 10 '20 at 7:26
  • $\begingroup$ @VerkhostsevaKatya: I am back at my desk now. Rather than replying in the comment section to your previous questions, let me implement a first paragraph (numbered 0, preceding 1) of general conventions and syntax. Also, once I finish writing the full answer, do make sure to read through the already posted portion once again, since I estimate I will be adding many new details in between. $\endgroup$
    – ΑΘΩ
    Jan 10 '20 at 10:30
  • $\begingroup$ thank you for the effort! $\endgroup$
    – Invisible
    Jan 10 '20 at 10:55

(continuation-middle part)

In order to state and actually prove the more familiar version of the main theorem in question, a few more notions need to be introduced:

  1. Normed vector spaces (and in particular Banach spaces). We call complex normed vector space a structure $(V, +, \cdot, \lVert \cdot \rVert)$ satisfying the following groups of axioms:

    • $V$ is a set and $+: V \times V$ a binary operation on it such that the subjacent structure $(V, +)$ is an abelian group. In detail this means that the addition $+$ is associative, admits a neutral element (which will be unique with this property and we will distinguish under the notation $0_V$), that every element $x \in V$ admits an opposite $y \in V$ such that $x+y=y+x=0_V$ (this opposite will also be unique and denoted by $-x$) and finally that the addition is also commutative.
    • The second operation is an external action $\cdot: \mathbb{C} \times V \to V$ of $\mathbb{C}$ on $V$ by group endomorphisms, which explicitly means that

$$(\forall \lambda, x, y)(\lambda \in \mathbb{C} \wedge x, y \in V \Rightarrow \lambda (x+y)=\lambda x+\lambda y)$$

together with the fulfillment of the following additional relations: $$(\forall \lambda, \mu, x)(\lambda, \mu \in \mathbb{C} \wedge x \in V \Rightarrow (\lambda+\mu)x=\lambda x+\mu x)\\ (\forall \lambda, \mu, x)(\lambda, \mu \in \mathbb{C} \wedge x \in V \Rightarrow (\lambda \mu)x=\lambda (\mu x))\\ (\forall x)(x \in V \Rightarrow 1x=x)$$

So far these axioms tell us that the subjacent structure $(V, +, \cdot)$ is a complex vector space.

  • The map $\lVert \cdot \rVert: V \to [0, \infty)$ is a norm on the subjacent vector space structure, which explicitly means that $$(\forall x)(x \in V \Rightarrow (\lVert x \rVert=0 \Leftrightarrow x=0_V))$$


$$(\forall x, y)(x, y \in V \Rightarrow \lVert x+y \rVert \leqslant \lVert x \rVert+ \lVert y \rVert)$$


$$(\forall \lambda, x)(\lambda \in \mathbb{C} \wedge x \in V \Rightarrow \lVert \lambda x \rVert=|\lambda| \lVert x \rVert)$$

(absolute homogeneity).

An absolutely analogous definition is given to real normed vector spaces (vector spaces over $\mathbb{R}$ satisfying the analogous axioms).

Any norm on a vector space induces a metric given by $$d: V \to V\\ d(x,y)=\lVert x-y \rVert$$

If the metric space thus obtained is complete, then the normed space is said to be a (real or complex) Banach space. I want to point out an important aspect: according to standard textbook definitions a sequence $x \in V^{\mathbb{N}}$ is said to be Cauchy if $$(\forall \rho)(\rho >0 \Rightarrow (\exists k)(k \in \mathbb{N} \wedge (\forall m, n)(m, n \geqslant k \Rightarrow \lVert x_m-x_n \rVert < \rho)))$$

and the normed space $(V, +, \cdot, \lVert \cdot \rVert)$ is called complete if any Cauchy sequence of vectors converges.

Given an arbitrary upward directed set $(I, R)$ and family $x \in V^I$ there is the more general definition of the generalized sequence $(x, R)$ being Cauchy when:

$$(\forall \rho)(\rho >0 \Rightarrow (\exists i)(i \in I \wedge (\forall j, k)(j, k \geqslant_R i \Rightarrow \lVert x_j-x_k \rVert < \rho)))$$

In particular, considering an arbitrary family $x \in V^I$ let us notice that claiming its associated generalized sequence of partial sums is Cauchy is equivalent to claiming that $$(\forall \rho)(\rho >0 \Rightarrow (\exists F)(F \in \mathscr{F}(I) \wedge (\forall H)(H \in \mathscr{F}(I) \wedge H \cap F=\varnothing \Rightarrow \left\lVert \sum_{i \in H}x_i \right\rVert <\rho))) \tag{Ca}$$

We have the following

Proposition: if the generalized sequence $(x, R)$ converges in the normed space $V$ then it is a Cauchy generalized sequence.

One remarkable thing in this context is that claiming:

any generalized Cauchy sequence converges in $V$

seems to be a stronger statement than:

any (ordinary) Cauchy sequence converges in $V$ (this being standard textbook completeness)

however the two statements are actually equivalent! In other words, Banach spaces are precisely those spaces for which the converse of the above proposition also holds.

  1. The monoid of extended positive real numbers: the set $[0, \infty]=\overline{\mathbb{R}}_{+}$ comes equipped with a natural order relation (for which $\infty$ is the maximum); this order relation (which we leave unnamed) is total (any two elements are comparable) and complete (any subset admits a supremum) and induces a topology (any total order induces the so-called order topology, having the open intervals with respect to that order as a basis) which we also leave unnamed; finally, this set comes equipped with a natural extension of the standard addition of (finite) real numbers, achieved by defining $\infty+x=x+\infty=\infty$ (in other words setting $\infty$ to be what is known as an absorptive element in abstract algebra).

Under this addition, the interval $[0, \infty]$ is a monoid (i.e. the addition is associative and admits a neutral element, namely $0$); since the order and topology defined above are both compatible with this addition, this monoid structure becomes a totally ordered topological monoid when equipped with the respective order and topology.

The important thing that must be pointed out here is that

any family of extended positive real numbers is summable!

Indeed, if $x \in [0, \infty]^I$ then it is easy to show that

$$\sum_{i \in I}x_i=\sup_{F \in \mathscr{F}(I)} \sum_{i \in F}x_i$$

As an immediate consequence of this, a family $x \in [0, \infty)^I$ will be summable in $\mathbb{R}$ (actually in $[0, \infty)$) if and only if $\sum_{i \in I}x_i < \infty$ if and only if the set of finite partial sums $$\left\{\sum_{i \in F}x_i \right\}_{F \in \mathscr{F}(I)}$$ is bounded.

Before we move on, let us state an important proposition that establishes a connection between paragraphs 6 and 7

Proposition. Let $x \in V^I$ be a family of vectors in the Banach space $V$ (by abuse of language, we omit any reference to the auxiliary structure on the support set $V$). Then if the family $(\lVert x_i \rVert)_{i \in I}$ is summable (in $\mathbb{R}$), then $x$ is summable in $V$ (absolute summability entails summability).

Proof: Let us introduce the family of values of norms

$$\theta=(\lVert x_i \rVert)_{i \in I}$$

which is by hypothesis summable in $[0, \infty)$ and therefore in $\mathbb{R}$ which we consider equipped with its canonical Banach space structure. Let us also consider the generalized sequences of partial sums of $x$ and $\theta$, defined as follows

$$u=\left(\sum_{i \in F}x_i \right)_{F \in \mathscr{F}(I)}\\ \mu=\left(\sum_{i \in F} \theta_i \right)_{F \in \mathscr{F}(I)}$$

The generalized sequence $\mu$ being convergent is Cauchy and therefore satisfies formulation (Ca) (paragraph 6), and it will suffice to prove that $u$ also satisfies the same formulation (for then it will be Cauchy and thus convergent, by completenes of $V$).

The subadditivity of the norm ensures that for any finite $F \subseteq I$ we have $\lVert u_F \rVert \leqslant \mu_F$; considering an arbitrary $\rho$, there exists $F \in \mathscr{F}(I)$ such that for any $H \in \mathscr{F}(I)$ with $H \cap F=\varnothing$ we have $\lVert u_H \rVert \leqslant \mu_H < \rho$, since $\mu$ satisfies (Ca). This means that $u$ also satisfies (Ca), as desired. $\Box$

Let us finally prove the generalized associativity theorem in the particular case of a real Banach space (explicitly, the abelian group $(G, +)$ will be the subjacent additive group $(V, +)$, equipped with the metric topology induced by the norm).

Preliminary lemma: Let $x \in V^I$ be summable in the Banach space $V$ and $J \subseteq I$ be arbitrary. Then the restriction $x_{|J}$ is also summable (subfamilies of summable families remain summable in complete spaces).

Proof: We abbreviate $y:=x_{|J}$ and introduce the generalized sequences of partial sums of these two families $$u=\left(\sum_{i \in F}x_i\right)_{F \in \mathscr{F}(I)}\\ v=\left(\sum_{i \in F}y_i\right)_{F \in \mathscr{F}(J)}$$

Bearing in mind that $y_i=x_i$ for any $i \in J$ and that $\mathscr{F}(J) \subseteq \mathscr{F}(I)$, it is easy to see that $v=u_{| \mathscr{F}(J)}$.

The family $x$ being summable, $u$ is by definition convergent and therefore Cauchy by the proposition of paragraph 6. It suffices to show that $v$ is also Cauchy, and this we achieve with the help of formulation (Ca) from the same paragraph: given arbitrary $\rho >0$, there must exist $F \in \mathscr{F}(I)$ such that for any $H \in \mathscr{F}(I)$ with $H \cap F= \varnothing$ we have $\lVert u_H \rVert < \rho$.

Then obviously $G:=F \cap J \in \mathscr{F}(J)$ and for any $H \in \mathscr{F}(J)$ such that $H \cap G=\varnothing$ it follows at once that $H \in \mathscr{F}(I), H=H \cap J$ (since $H \subseteq J$) and that $H \cap F=(H \cap J) \cap F=H \cap (J \cap F)=H \cap G=\varnothing$ so that we have $v_H=u_H$ and hence $\lVert v_H \rVert=\lVert u_H \rVert < \rho$. This means that $v$ also satisfies the (Ca) formulation and is indeed Cauchy. $\Box$

This second part also turned out to be quite long and I am afraid the capacities of my compiler will be exceeded once again. Allow me to continue by opening up a third part below, which I trust will be the last one in this grand exposé.

  • $\begingroup$ @Katya: I've resumed working on this second part of the answer and I estimate I will be done a few hours later (at the most). Once it is all done, it would be a good idea to go through everything top to bottom once again, since I believe I will have rearranged some passages by the time it is done. $\endgroup$
    – ΑΘΩ
    Jan 11 '20 at 7:32

(continuation-last part)

Proof of the main Theorem of Associativity:

Consider the family $x \in V^I$ summable in the Banach space $V$, together with a partition $J \in \mathscr{P}(I)^K$ of the index set $I$. By the preliminary lemma at the end of part 2, we know that the subfamilies $x_{|J_k}$ are summable for each $k \in K$ so let us introduce the generalized sequences of finite partial sums

$$u=\left(\sum_{i \in F}x_i\right)_{F \in \mathscr{F}(I)}\\ v_k=\left(\sum_{i \in F}x_i\right)_{F \in \mathscr{F}(J_k)},\ k \in K$$

together with the sums

$$s=\sum_{i \in I}x_i\\ y_k=\sum_{i \in J_k}x_i,\ k \in K$$

Our objective is to show that the family of partial sums $(y_k)_{k \in K}$ is summable with the same sum $s$ as family $x$, so to this end let us also introduce the corresponding generalized sequence of finite partial sums

$$w=\left(\sum_{k \in L}y_k\right)_{L \in \mathscr{F}(K)}$$

Let us consider an arbitrary $\rho>0$; the generalized sequence $u$ by hypothesis converges to $s$ and is thus Cauchy, satisfying formulation (Ca), facts from which we gather that at the same time there exist

  • a finite $F' \subseteq I$ such that for any $H \in \mathscr{F}(I)$ with $F' \subseteq H$ we have $\lVert u_H-s \rVert <\frac{\rho}{3}$ (this from convergence)
  • a finite $F'' \subseteq I$ such that for any $H \in \mathscr{F}(I)$ with $H \cap F''=\varnothing$ we have $\lVert u_H \rVert < \frac{\rho}{3}$ (this from formulation (Ca)).

Thus by introducing $F=F' \cup F'' \in \mathscr{F}(I)$ we realize that at the same time

  • a) for any $H \in \mathscr{F}(I)$ such that $F \subseteq H$ we have $\lVert u_H-s \rVert < \frac{\rho}{3}$ (since then $H \supseteq F \supseteq F'$)

  • b) for any $H \in \mathscr{F}(I)$ such that $H \cap F=\varnothing$ we have $\lVert u_H \rVert <\frac{\rho}{3}$ (since then $H \cap F'' \subseteq H \cap F=\varnothing$).

Fixing such an $F$ enjoying the above properties, let us introduce

$$L=\{k \in K|\ F \cap J_k \neq \varnothing \}$$

It is then clear that we have $$F=F \cap I=F \cap \left( \bigcup_{k \in K} J_k\right)=\bigcup_{k \in K}(F \cap J_k)=\bigcup_{k \in L} (F \cap J_k)=F \cap \left(\bigcup_{k \in L} J_k \right) \tag{*}$$

and since the sets $J_k, k \in K$ are pairwise disjoint (when corresponding to distinct indices) we can write

$$|F|=\left| \bigcup_{k \in L}(F \cap J_k) \right|=\sum_{k \in L} |F \cap J_k| \geqslant \sum_{k \in L} 1=|L|$$

This means $|L|< \aleph_0$ and therefore that $L \in \mathscr{F}(K)$. We now make the claim that in order to prove that $w$ also converges to $s$ it suffices to establish the following:

  1. that $\lVert w_L-s \rVert \leqslant \frac{\rho}{3}$
  2. that for any $M \in \mathscr{F}(K)$ such that $M \cap L=\varnothing$ we have $\lVert w_M \rVert \leqslant \frac{\rho}{3}$.

Before actually proving claims 1 and 2, let us see how they enable us to draw the desired conclusion: for arbitrary $M \in \mathscr{F}(K)$ such that $L \subseteq M$ we will have $$\lVert w_M-s \rVert=\lVert w_L+w_{M \setminus L}-s \rVert \leqslant \lVert w_L-s \rVert+ \lVert w_{M \setminus L} \rVert \leqslant \frac{2 \rho}{3}<\rho$$

as obviously $(M \setminus L) \cap L=\varnothing$; the arbitrariness of $\rho >0$ allows us to infer that

$$w_M \xrightarrow{M \in \mathscr{F}(K)}\ s$$

On then to establishing the two claims, in the given order:

  1. Let us define $$I'=\bigcup_{k \in L} J_k$$

such that $(J_k)_{k \in L}$ becomes a finite partition of $I'$ and that $F \subseteq I'$ by virtue of relation (*). Then by the remark b) of paragraph 5, the subfamily $x':=x_{|I'}$ has the property that $$\sum_{i \in I'} x_i=\sum_{k \in L} y_k=w_L$$

It is clear that $u':=u_{\mathscr{F}(I')}$ is the generalized sequence of finite partial sums of the subfamily $x'$; since for any $H \in \mathscr{F}(I')\subseteq \mathscr{F}(I)$ such that $F \subseteq H$ we know via relation a) that $\lVert u'_H -s \rVert=\lVert u_H-s \rVert < \frac{\rho}{3}$, bearing in mind the continuity of the norm $\lVert \cdot \rVert$ we can pass to the limit in this inequality (family of inequalities) and infer that $$\mathrm{lim}_{H \in \mathscr{F}(I')} \lVert u_H-s \rVert=\left\lVert \mathrm{lim}_{H \in \mathscr{F}(I')} u'_H-s \right\rVert=\left\lVert \sum_{i \in I'} x_i -s \right\rVert=\lVert w_L-s \rVert \leqslant \frac{\rho}{3}$$

  1. We keep the notations of the previous section and consider an arbitrary $M \in \mathscr{F}(K)$ such that $M \cap L=\varnothing$ and define this time $$I''= \bigcup_{k \in M} J_k$$

such that $(J_k)_{k \in M}$ is a finite partition of $I''$ and we also deduce

$$F \cap I'' \subseteq I' \cap I''=\bigcup_{k \in L\\ l \in M} (J_k \cap J_l)=\bigcup_{k \in L\\ l \in M} \varnothing=\varnothing$$

as $L$ and $M$ being disjoint entails that $k \neq l$ for every pair $k \in L, l \in M$.

In fashion entirely analogous to what we have just seen in the previous section, the subfamily $x'':=x_{|I''}$ will have associated generalized sequence of finite partial sums given by $u''=u_{|\mathscr{F}(I'')}$ and by another application of remark b) from paragraph 5 we know that

$$\sum_{i \in I''}x_i=\sum_{k \in M} y_k=w_M$$

This time, since for any $H \in \mathscr{F}(I'')$ it follows that $H \cap F \subseteq I'' \cap F=\varnothing$ we have by relation b) that $\lVert u''_H \rVert=\lVert u_H \rVert < \frac{\rho}{3}$ so by passing to the limit in the norm we infer that

$$\mathrm{lim}_{H \in \mathscr{F}(I'')} \lVert u''_H \rVert=\left\lVert \mathrm{lim}_{H \in \mathscr{F}(I'')} u''_H \right\rVert=\left\lVert \sum_{i \in I''} x_i \right\rVert=\lVert w_M \rVert \leqslant \frac{\rho}{3}$$

This concludes our long-awaited proof. $\Box$

Before indicating how the generalized associativity theorem settles the result you were inquiring about, let me state one more remarkable result (without proof):

Theorem. In a finite dimensional real Banach space, a family of vectors is summable if and only if it is absolutely summable.

(the family $x \in V^I$ is absolutely summable if the family of absolute values $(\lVert x_i \rVert)_{i \in I}$ is summable in $\mathbb{R}$).

That absolute summability is a sufficient condition for summability in Banach spaces we have already seen, however the remarkable aspect is that in finite dimensional Banach spaces it is also necessary.

In your particular case we are considering a double sequence $z \in \mathbb{C}^{\mathbb{N} \times \mathbb{N}}$ such that the double series $$\sum_{m, n=0}^{\infty} z_{mn}$$

is absolutely convergent, which means that $$\sum_{m,n=0}^{\infty} |z_{mn}|=\sum_{r \in \mathbb{N} \times \mathbb{N}} |z_r|< \infty$$

where we have used remark d) of paragraph 5.

Thus, $z$ is absolutely summable and hence summable as a family, allowing you to apply the associativity theorem to derive

$$\sum_{r \in \mathbb{N} \times \mathbb{N}} z_r=\sum_{m \in \mathbb{N}} \sum_{n \in \mathbb{N}} z_{mn}=\sum_{n \in \mathbb{N}} \sum_{m \in \mathbb{N}} z_{mn} \tag{**}$$

Bearing in mind that the summability of (double) sequences entails equality between the sum as a family and the sum of the associated series you deduce that $$\sum_{m \in \mathbb{N}} \sum_{n \in \mathbb{N}} z_{mn}=\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} z_{mn}$$

etc, which brings the equality (**) to the desired form

$$\sum_{m,n=0}^{\infty} z_{mn}=\sum_{m=0}^{\infty} \sum_{n=0}^{\infty}z_{mn}=\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} z_{mn}$$


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