Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality

In the setting of Commutative Rings, Fields and Ordered Fields: Standard Notation, let $A$ and $B$ be finite sets; $A\times B$ and the subsets of $A$ 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 §union 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 §subset.

Let $R$, with $+$, $\cdot$, $0$ and $1$, be a commutative ring, let $\#$ be as in The Number of Elements of a Finite Set §cardinality, and let $f,g:A\to R$, $h:B\to R$ and $c\in R$.

¶ $c\sum_{x\in A}f(x)=\sum_{x\in A}c\,f(x)$.

¶ $\sum_{x\in A}\big(f(x)-g(x)\big)=\sum_{x\in A}f(x)-\sum_{x\in A}g(x)$.

¶ $\Big(\sum_{x\in A}f(x)\Big)\Big(\sum_{y\in B}h(y)\Big)=\sum_{(x,y)\in A\times B}f(x)\,h(y)$.

¶ If $E\subseteq A$ and $f(x)=0$ for every $x\in A\setminus E$, then $\sum_{x\in A}f(x)=\sum_{x\in E}f(x)$; in particular, if $y\in A$ and $f(x)=0$ for every $x\in A$ with $x\neq y$, then $\sum_{x\in A}f(x)=f(y)$.

¶ $\sum_{x\in A}c=\#A\cdot c$; in particular $\sum_{x\in A}1=\#A$.

Let $(B_{x})_{x\in A}$ be a family of finite sets, $U=\bigcup_{z\in A}B_{z}$, which 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 §family-union, and $T=\{(x,y)\in A\times U:y\in B_{x}\}$, which 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 §union 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 §subset; sums over $T$ are written $\sum_{(x,y)\in T}u(x,y)$. Let $u:T\to R$, and let $P=\prod_{x\in A}B_{x}$ be the product of the family, that is the set of maps $\varphi:A\to U$ with $\varphi(x)\in B_{x}$ for every $x\in A$, which 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 §maps 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 §subset.

¶ $\sum_{x\in A}\sum_{y\in B_{x}}u(x,y)=\sum_{(x,y)\in T}u(x,y)$.

¶ $\prod_{x\in A}\sum_{y\in B_{x}}u(x,y)=\sum_{\varphi\in P}\prod_{x\in A}u(x,\varphi(x))$.

Let now $F$ be an ordered field, and let $f,g:A\to F$.

¶ If $f(x)\le g(x)$ for every $x\in A$, then $\sum_{x\in A}f(x)\le\sum_{x\in A}g(x)$.

¶ If $f(x)\ge0$ for every $x\in A$ and $S\subseteq A$, then $0\le\sum_{x\in S}f(x)\le\sum_{x\in A}f(x)$; in particular $f(y)\le\sum_{x\in A}f(x)$ for every $y\in A$.

¶ $\Big|\sum_{x\in A}f(x)\Big|\le\sum_{x\in A}|f(x)|$.

¶ If $f(x)\ge0$ for every $x\in A$ and $\sum_{x\in A}f(x)=0$, then $f(x)=0$ for every $x\in A$.