Proof of Sums of Sequences in [0,∞]: Real Series, Linearity, Comparison, Rearrangement, Double Series and Monotone Convergence

Throughout, the order of $[-\infty,\infty]$ is used through Order and Suprema in the Extended Real Line (i), and sums over finite sets of maps into $[0,\infty]$ lie in $[0,\infty]$ (Iterated Operations and Operations over Finite Sets Take Values in the Underlying Set (ii)). For $n\in\mathbb{N}$, every element of $[n]$ is a natural number by Elements of a Segment Are Natural Numbers, and Mixed Transitivity of the Order on the Natural Numbers (i), so $[n]\subseteq\mathbb{N}$, and $[n]$ is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vii). Write $\Sigma a=\sum_{k\in\mathbb{N}}a_{k}$ and $\Sigma b=\sum_{k\in\mathbb{N}}b_{k}$.

Step 0: three tools.

(S1) For a sequence $(u_{k})$ in $[0,\infty]$ and $n\in\mathbb{N}$, $\sum_{k=1}^{n}u_{k}=\sum_{k\in[n]}u_{k}$ by Natural Numbers as Indices: ℕ ⊆ ℕ₀, 1 ∈ ℕ₀, and the Two Readings of an Operation from 1 to n (iii), $+$ being associative and commutative by Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i). In particular $s_{n}=\sum_{k\in[n]}a_{k}$.

(S2) For every finite subset $F$ of $\mathbb{N}$ there is $N\in\mathbb{N}$ with $F\subseteq[N]$. If $F=\emptyset$, take $N=1$. Otherwise, the order of $\mathbb{N}_{0}$ is a well-order, hence a total order (The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion (v)), and $F\subseteq\mathbb{N}\subseteq\mathbb{N}_{0}$; so $F$ has a greatest element $N$ by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (viii), and $N\in F\subseteq\mathbb{N}$. Every $k\in F$ satisfies $k\le N$ and $1\le k$, the latter by Arithmetic and Order of the Natural Numbers (x); so $k\in[N]$ by Intervals of Natural Numbers (i) and Intervals of Natural Numbers (ii).

(S3) Let $D$ and $D'$ be sets, $\rho:D\to D'$ an injective map, $F\subseteq D$ finite and $f:D'\to[0,\infty]$. Then $\rho(F)$ is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vi), and $\rho|_{F}:F\to\rho(F)$ is injective, as $\rho$ is, and surjective by The Image and the Preimage of a Class under a Class (i), hence a bijection. The map $F\to[0,\infty]$, $k\mapsto f(\rho(k))$, exists (maps $u\mapsto t(u)$ given by an expression, here and below, exist by Maps and Relations Given by Formulas (i)). By Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (iii), $\sum_{k\in F}f(\rho(k))=\sum_{j\in\rho(F)}f(j)$.

Proof of (iv). Let $F\subseteq\mathbb{N}$ be finite. By (S2), $F\subseteq[N]$ for some $N\in\mathbb{N}$, so by Suprema and Finite Sums in [0,∞] (v) and (S1), $\sum_{k\in F}a_{k}\le\sum_{k\in[N]}a_{k}=s_{N}\le\Sigma a$, the last as $\Sigma a$ is an upper bound of the partial sums (Sum of a Sequence in [0,∞] (ii)).

Proof of (v). Let $t\in[0,\infty]$ bound all sums over finite subsets of $\mathbb{N}$. Each $[n]$ is a finite subset of $\mathbb{N}$, so $s_{n}=\sum_{k\in[n]}a_{k}\le t$ for every $n$ by (S1); thus $t$ is an upper bound of the partial sums, and $\Sigma a\le t$, as $\Sigma a$ is their least upper bound (Sum of a Sequence in [0,∞] (ii)).

Proof of (i). By Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (i), applied to the restriction of $(a_{k})$ to $[n+1]$, $s_{n+1}=s_{n}+a_{n+1}$, so $s_{n}\le s_{n+1}$ by Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (v).

Proof of (ii). $\Sigma a$ is an upper bound of the partial sums by Sum of a Sequence in [0,∞] (ii), so $s_{n}\le\Sigma a$.

Proof of (iii). For $k\in\mathbb{N}$, $\{k\}$ is a finite subset of $\mathbb{N}$ (Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iii)) and $\sum_{j\in\{k\}}a_{j}=a_{k}$ by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (i); so $a_{k}\le\Sigma a$ by (iv).

Proof of (vi). Let $a_{k}\in\mathbb{R}$ for every $k$; then $0\le a_{k}$ in $\mathbb{R}$, as $a_{k}\in[0,\infty]$, and $(a_{k})$ is also a sequence in $\mathbb{R}$. Let $r_{n}=\sum_{k=1}^{n}a_{k}$ be the partial sums of the real series, which are iterated sums along $[n]$ of the sum of $\mathbb{R}$ by Natural Numbers as Indices: ℕ ⊆ ℕ₀, 1 ∈ ℕ₀, and the Two Readings of an Operation from 1 to n (v), and real. We show by induction on $n$ that the partial sums $s_{n}$ in $[0,\infty]$ equal the real ones: let $K=\{n\in\mathbb{N}:s_{n}=r_{n}\}$. By Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (i), applied to the restriction of $(a_{k})$ to $[1]$, once for the sum $+$ on $[0,\infty]$ and once for the sum of $\mathbb{R}$, $s_{1}=a_{1}=r_{1}$; so $1\in K$. Let $n\in K$. By Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (i), applied to the restriction of $(a_{k})$ to $[n+1]$ for both operations, $s_{n+1}=s_{n}+a_{n+1}$, the sum taken in $[0,\infty]$, and $r_{n+1}=r_{n}+a_{n+1}$, the sum taken in $\mathbb{R}$. As $s_{n}=r_{n}$ and $a_{n+1}$ are real elements of $[0,\infty]$, their sum in $[0,\infty]$ is their real sum, by Addition and Multiplication on [0,∞] (i); so $s_{n+1}=r_{n}+a_{n+1}=r_{n+1}$, and $n+1\in K$. By Arithmetic and Order of the Natural Numbers (xiii), $K=\mathbb{N}$. So $P=\{s_{n}:n\in\mathbb{N}\}$ is a nonempty subset of $\mathbb{R}$, it is the set of terms of $(r_{n})$, and $\Sigma a=\sup P$. If $P$ is bounded above in $\mathbb{R}$, then the series $\sum_{k=1}^{\infty}a_{k}$ converges with sum $\sup_{\mathbb{R}}P$ by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (i) (with Bounded Sequences of Real Numbers (i)), and $\Sigma a=\sup_{\mathbb{R}}P$ by Order and Suprema in the Extended Real Line (iv); so $\Sigma a\in\mathbb{R}$ and the two sums agree. If $P$ is not bounded above in $\mathbb{R}$, then the series does not converge, by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (i), and $\Sigma a=\infty\notin\mathbb{R}$ by Order and Suprema in the Extended Real Line (v). These two cases give the first assertion. If $a_{k}=\infty$ for some $k$, then $\infty\le\Sigma a$ by (iii), so $\Sigma a=\infty$ by Order and Suprema in the Extended Real Line (ii).

Proof of (vii). Let $a_{k}=0$ for every $k\in\mathbb{N}$ with $n<k$, and let $F\subseteq\mathbb{N}$ be finite. If $k\in F\setminus[n]$, then $1\le k$ by Arithmetic and Order of the Natural Numbers (x), so $k\le n$ fails by Intervals of Natural Numbers (ii), and $n<k$ by Arithmetic and Order of the Natural Numbers (vii) and Arithmetic and Order of the Natural Numbers (vi); hence $a_{k}=0$. The sets $F\cap[n]$ and $F\setminus[n]$ are finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iv), disjoint, with union $F$. By Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (ii), Suprema and Finite Sums in [0,∞] (vii) (the terms over $F\setminus[n]$ being $0$), Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i), Suprema and Finite Sums in [0,∞] (v) and (S1),

$$\sum_{k\in F}a_{k}=\sum_{k\in F\cap[n]}a_{k}+\sum_{k\in F\setminus[n]}0=\sum_{k\in F\cap[n]}a_{k}\le\sum_{k\in[n]}a_{k}=s_{n}.$$

So $\Sigma a\le s_{n}$ by (v), and $s_{n}\le\Sigma a$ by (ii); hence $\Sigma a=s_{n}$.

Proof of (viii). Let $s'_{n}=\sum_{k=1}^{n}b_{k}$ (maps $n\mapsto\sum_{k=1}^{n}u_{k}$ of partial iterates, here and below, exist by Maps Given by Iterated Operations: Partial Iterates, Images of Natural Numbers, Powers and Operations over a Finite Set (i)). By Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (v), the $n$-th partial sum of $(a_{k}+b_{k})$ is $s_{n}+s'_{n}$. The sequences $(s_{n})$ and $(s'_{n})$ satisfy $s_{n}\le s_{n+1}$ by (i), and $s'_{n}\le s'_{n+1}$ by (i) (applied to $(b_{k})$), so by Suprema and Finite Sums in [0,∞] (iii), $\sum_{k\in\mathbb{N}}(a_{k}+b_{k})=\sup_{n}(s_{n}+s'_{n})=\sup_{n}s_{n}+\sup_{n}s'_{n}=\Sigma a+\Sigma b$.

Proof of (ix). Let $c\in[0,\infty]$ and $\varphi:[0,\infty]\to[0,\infty]$, $x\mapsto cx$; then $\varphi(x+y)=\varphi(x)+\varphi(y)$ by Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (iii), so the $n$-th partial sum of $(ca_{k})$ is $cs_{n}$ by Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (ii). By Suprema and Finite Sums in [0,∞] (ii), applied to the nonempty set of partial sums, $\sum_{k\in\mathbb{N}}ca_{k}=\sup\{cs_{n}:n\in\mathbb{N}\}=c\,\Sigma a$.

Proof of (x). Let $a_{k}\le b_{k}$ for every $k$. For $n\in\mathbb{N}$, by (S1), Suprema and Finite Sums in [0,∞] (iv) and (ii) (applied to $(b_{k})$), $s_{n}=\sum_{k\in[n]}a_{k}\le\sum_{k\in[n]}b_{k}\le\Sigma b$. So $\Sigma b$ is an upper bound of the partial sums of $(a_{k})$, and $\Sigma a\le\Sigma b$.

Proof of (xi). Let $c_{k}=a_{\sigma(k)}$, so $(c_{k})$ is a sequence in $[0,\infty]$. For a finite $F\subseteq\mathbb{N}$, $\sigma(F)$ is a finite subset of $\mathbb{N}$, and by (S3) and (iv), $\sum_{k\in F}c_{k}=\sum_{j\in\sigma(F)}a_{j}\le\Sigma a$. By (v), applied to $(c_{k})$, $\sum_{k\in\mathbb{N}}c_{k}\le\Sigma a$. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vi), $\sigma^{-1}:\mathbb{N}\to\mathbb{N}$ is a bijection with $\sigma(\sigma^{-1}(j))=j$, so $a_{j}=c_{\sigma^{-1}(j)}$ for every $j\in\mathbb{N}$; the same argument with $(c_{k})$ and $\sigma^{-1}$ in place of $(a_{k})$ and $\sigma$ gives $\Sigma a\le\sum_{k\in\mathbb{N}}c_{k}$. By antisymmetry the two sums are equal.

Proof of (xii). For $m,N\in\mathbb{N}$ let $R_{m}=\sum_{n\in\mathbb{N}}w(m,n)$ and let $R_{m}(N)=\sum_{n=1}^{N}w(m,n)=\sum_{n\in[N]}w(m,n)$ (by (S1)) be the $N$-th partial sum of $n\mapsto w(m,n)$; so $R_{m}=\sup_{N}R_{m}(N)$ by Sum of a Sequence in [0,∞] (ii), and $R_{m}(N)\le R_{m}(N+1)$ and $R_{m}(N)\le R_{m}$ by (i) and (ii) (applied to $n\mapsto w(m,n)$). The map $m\mapsto R_{m}$ is a sequence in $[0,\infty]$ (maps $u\mapsto\sup t(u)$, here and below, exist by Predicative Formulas for Suprema and Infima, and Maps Defined by Suprema and Infima (iii), $\le$ being a partial order on $[-\infty,\infty]$).

(i) We show $\sum_{m\in\mathbb{N}}R_{m}=\sup_{M}\sup_{N}\sum_{m\in[M]}\sum_{n\in[N]}w(m,n)$, the maps $N\mapsto\sum_{m\in[M]}R_{m}(N)$ and $m\mapsto\sum_{n\in[N]}w(m,n)$ existing by Maps Given by Iterated Operations: Partial Iterates, Images of Natural Numbers, Powers and Operations over a Finite Set (v) (as do the maps $x\mapsto\sum_{y\in B}f(x,y)$ over a fixed finite set $B$ below). Let $M\in\mathbb{N}$. The map $\mathbb{N}\times[M]\to[0,\infty]$, $(N,m)\mapsto R_{m}(N)$, is the restriction to $\mathbb{N}\times[M]$ of the map $g:\mathbb{N}\times\mathbb{N}\to[0,\infty]$ with $g(N,m)=\sum_{n=1}^{N}w(m,n)=R_{m}(N)$ given by Partial Iterates of a Two-Variable Map, Row by Row and Column by Column (i), applied with $A=\mathbb{N}$, $X=[0,\infty]$, $\ast$ the sum $+$ of Addition and Multiplication on [0,∞] (i), and $f=w$. It is nondecreasing in $N$, so by Suprema and Finite Sums in [0,∞] (viii),

$$\sum_{m\in[M]}R_{m}=\sum_{m\in[M]}\sup_{N}R_{m}(N)=\sup_{N}\sum_{m\in[M]}R_{m}(N)=\sup_{N}\sum_{m\in[M]}\sum_{n\in[N]}w(m,n).$$

By (S1), $\sum_{m\in[M]}R_{m}$ is the $M$-th partial sum of $(R_{m})$, so the claim follows from Sum of a Sequence in [0,∞] (ii).

(ii) Let $w':\mathbb{N}\times\mathbb{N}\to[0,\infty]$ be the map with $w'(n,m)=w(m,n)$ for all $m,n\in\mathbb{N}$, given by Partial Iterates of a Two-Variable Map, Row by Row and Column by Column (iii) with $A=B=\mathbb{N}$, $X=[0,\infty]$ and $f=w$. Then (i), applied to $w'$ with the names of the two variables exchanged, gives

$$\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}w(m,n)=\sup_{N}\sup_{M}\sum_{n\in[N]}\sum_{m\in[M]}w(m,n).$$

(iii) For $M,N\in\mathbb{N}$ let $T(M,N)=\sum_{(m,n)\in[M]\times[N]}w(m,n)$, the set $[M]\times[N]$ being finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (v). By Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (iv),

$$T(M,N)=\sum_{m\in[M]}\sum_{n\in[N]}w(m,n)=\sum_{n\in[N]}\sum_{m\in[M]}w(m,n).$$

So by (i) and (ii) it remains to show that $J_{1}=\sup_{M}\sup_{N}T(M,N)$ equals $J_{2}=\sup_{N}\sup_{M}T(M,N)$. For all $M,N\in\mathbb{N}$, $T(M,N)\le\sup_{M'}T(M',N)\le J_{2}$. Hence, for each $M$, $J_{2}$ is an upper bound of $\{T(M,N):N\in\mathbb{N}\}$, so $\sup_{N}T(M,N)\le J_{2}$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv); so $J_{2}$ is an upper bound of $\{\sup_{N}T(M,N):M\in\mathbb{N}\}$, and $J_{1}\le J_{2}$. Exchanging the roles of $M$ and $N$ gives $J_{2}\le J_{1}$, and $J_{1}=J_{2}$ by antisymmetry (Order and Suprema in the Extended Real Line (i)).

Proof of (xiii). The first equality is (xii). Let $R_{m}$ and $R_{m}(N)$ be as in the proof of (xii), and $t_{k}=w(\tau(k))$ for $k\in\mathbb{N}$. It remains to show $\sum_{m\in\mathbb{N}}R_{m}=\sum_{k\in\mathbb{N}}t_{k}$.

(a) Let $F\subseteq\mathbb{N}$ be finite. Then $\tau(F)$ is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vi), and so is the set $G$ of the components $p_{1}$ and $p_{2}$ of the elements $p\in\tau(F)$, a subset of $\mathbb{N}$ that is the union of the images of $\tau(F)$ under the two component maps $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ (Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vi) and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (v)). By (S2), $G\subseteq[N]$ for some $N\in\mathbb{N}$, so $\tau(F)\subseteq Q=[N]\times[N]$, a finite set by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (v). By (S3), Suprema and Finite Sums in [0,∞] (v), Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (iv), Suprema and Finite Sums in [0,∞] (iv) (with $R_{m}(N)\le R_{m}$), and (iv),

$$\sum_{k\in F}t_{k}=\sum_{p\in\tau(F)}w(p)\le\sum_{p\in Q}w(p)=\sum_{m\in[N]}R_{m}(N)\le\sum_{m\in[N]}R_{m}\le\sum_{m\in\mathbb{N}}R_{m}.$$

By (v), $\sum_{k\in\mathbb{N}}t_{k}\le\sum_{m\in\mathbb{N}}R_{m}$.

(b) Let $M\in\mathbb{N}$. The map $\mathbb{N}\times[M]\to[0,\infty]$, $(N,m)\mapsto R_{m}(N)$, the restriction to $\mathbb{N}\times[M]$ of the map given by Partial Iterates of a Two-Variable Map, Row by Row and Column by Column (i) as in (i) of the proof of (xii), is nondecreasing in $N$, so by Suprema and Finite Sums in [0,∞] (viii), $\sum_{m\in[M]}R_{m}=\sup_{N}\sum_{m\in[M]}R_{m}(N)$. Let $N\in\mathbb{N}$, $Q=[M]\times[N]$, finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (v), and $H=\{k\in\mathbb{N}:\tau(k)\in Q\}$. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vi), $\tau^{-1}:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ is a bijection with $\tau(\tau^{-1}(q))=q$ and $\tau^{-1}(\tau(k))=k$; hence $H=\tau^{-1}(Q)$ is the image of $Q$ under $\tau^{-1}$, finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vi), and $\tau(H)=Q$. By Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (iv), (S3) and (iv),

$$\sum_{m\in[M]}R_{m}(N)=\sum_{p\in Q}w(p)=\sum_{k\in H}t_{k}\le\sum_{k\in\mathbb{N}}t_{k}.$$

Hence $\sum_{m\in[M]}R_{m}\le\sum_{k\in\mathbb{N}}t_{k}$, which by (S1) bounds the $M$-th partial sum of $(R_{m})$; as $M$ is arbitrary, $\sum_{m\in\mathbb{N}}R_{m}\le\sum_{k\in\mathbb{N}}t_{k}$. With (a), $\sum_{m\in\mathbb{N}}R_{m}=\sum_{k\in\mathbb{N}}t_{k}$.

Proof of (xiv). Let $g_{k}=\sup_{n}w(n,k)$ for $k\in\mathbb{N}$, $S_{n}=\sum_{k\in\mathbb{N}}w(n,k)$ for $n\in\mathbb{N}$, and $L=\sup_{n}S_{n}$, all in $[0,\infty]$ by Order and Suprema in the Extended Real Line (ix). As $w(n,k)\le g_{k}$ for all $n,k$, (x) gives $S_{n}\le\sum_{k\in\mathbb{N}}g_{k}$ for every $n$, so $L\le\sum_{k\in\mathbb{N}}g_{k}$. Conversely let $N\in\mathbb{N}$. By (S1) and Suprema and Finite Sums in [0,∞] (viii), applied to the restriction of $w$ to $\mathbb{N}\times[N]$,

$$\sum_{k=1}^{N}g_{k}=\sum_{k\in[N]}\sup_{n}w(n,k)=\sup_{n}\sum_{k\in[N]}w(n,k),$$

and $\sum_{k\in[N]}w(n,k)\le S_{n}\le L$ for every $n$, by (S1) and (ii) (applied to $k\mapsto w(n,k)$). So $\sum_{k=1}^{N}g_{k}\le L$ for every $N$, and $\sum_{k\in\mathbb{N}}g_{k}\le L$. By antisymmetry, $\sum_{k\in\mathbb{N}}g_{k}=L$.