Proof of Products over Finite Sets: a Single Factor, Vanishing in a Field, and Signs and Monotonicity in an Ordered Ring

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named where it is cited.

Products over finite sets are as in Commutative Rings, Fields and Ordered Fields: Standard Notation §rings; the empty product is $1$ there. The ring laws of Commutative Rings §ring are used without further mention. For a finite set $B$, $\#B\in\mathbb{N}_{0}$ is as in The Number of Elements of a Finite Set §cardinality.

(I) Removing one factor. Let $B$ be a finite set, $y\in B$, $B'=B\setminus\{y\}$, and $h:B\to R$. Then $B'$ 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 §subset, $B'\cup\{y\}=B$ as $y\in B$, and $B'\cap\{y\}=\emptyset$. The products over $B'$ and $\{y\}$ are those of the restrictions of $h$, by Sums and Products over a Finite Set and over an Interval §subsets, with values those of $h$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §restriction. Hence, by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §disjoint-union and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §singleton, for the operation $\cdot$ of $R$,

$$\prod_{x\in B}h(x)=\Big(\prod_{x\in B'}h(x)\Big)\cdot h(y).$$

Moreover $\#B=\#B'+\#\{y\}=\#B'+1$ by Counting: Intervals, Empty Sets and Singletons, Injective Images, Subsets, Unions and Products §union and Counting: Intervals, Empty Sets and Singletons, Injective Images, Subsets, Unions and Products §empty.

(II) Induction on the number of elements. Let $\Phi(B)$ be a property of finite sets $B$ expressed by a formula quantifying over sets only (possibly with parameters), such that $\Phi(\emptyset)$ holds and, for every nonempty finite set $B$ and every $y\in B$, $\Phi(B\setminus\{y\})$ implies $\Phi(B)$. Then $\Phi(A)$ holds for every finite set $A$. Indeed, let $K$ be the set of $n\in\mathbb{N}_{0}$ such that $\Phi(B)$ holds for every finite set $B$ with $\#B=n$. If $\#B=0$, then $B=\emptyset$ by Counting: Intervals, Empty Sets and Singletons, Injective Images, Subsets, Unions and Products §empty; so $0\in K$. Let $n\in K$ and $\#B=n+1$. Then $B\neq\emptyset$, since $\#\emptyset=0$ by Counting: Intervals, Empty Sets and Singletons, Injective Images, Subsets, Unions and Products §empty and $n+1\neq0$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §successor-nonzero, the successor of $n$ being $n+1$ by Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication §plus-one. Choose $y\in B$; by (I), $\#(B\setminus\{y\})+1=n+1$, so $\#(B\setminus\{y\})=n$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §cancellation and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §commutative, hence $\Phi(B\setminus\{y\})$ and so $\Phi(B)$. Thus $n+1\in K$, and $K=\mathbb{N}_{0}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction; as $\#A\in\mathbb{N}_{0}$, $\Phi(A)$ holds.

Clause single. Let $\Phi(B)$ say: every map $h:B\to R$ with $h(x)=1$ for all $x\in B$ satisfies $\prod_{x\in B}h(x)=1$. $\Phi(\emptyset)$ holds as the empty product is $1$. If $B\neq\emptyset$, $y\in B$ and $\Phi(B\setminus\{y\})$ holds, then for such $h$, by (I), $\prod_{x\in B}h(x)=1\cdot h(y)=1$, as the restriction of $h$ to $B\setminus\{y\}$ is again constantly $1$. By (II), $\Phi(B)$ holds for every finite $B$. Now let $x_{0}\in A$ and $f$ be as in the clause, and $A'=A\setminus\{x_{0}\}$. The restriction of $f$ to $A'$ is constantly $1$, so by (I) and $\Phi(A')$, $\prod_{x\in A}f(x)=1\cdot f(x_{0})=f(x_{0})$.

Clause zero. Let $R$ be a field. If $f(x_{0})=0$ for some $x_{0}\in A$, then by (I) and Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §zero, $\prod_{x\in A}f(x)=\big(\prod_{x\in A\setminus\{x_{0}\}}f(x)\big)\cdot0=0$. Conversely, let $\Phi(B)$ say: every map $h:B\to R$ with $\prod_{x\in B}h(x)=0$ has $h(x)=0$ for some $x\in B$. $\Phi(\emptyset)$ holds vacuously, because the empty product is $1$ and $1\neq0$ by Fields §field. Let $B\neq\emptyset$, $y\in B$, $B'=B\setminus\{y\}$, and assume $\Phi(B')$. If $\prod_{x\in B}h(x)=0$, then by (I) $\big(\prod_{x\in B'}h(x)\big)\cdot h(y)=0$, so by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §field either $h(y)=0$, or $\prod_{x\in B'}h(x)=0$ and then, by $\Phi(B')$ applied to the restriction of $h$ to $B'$, $h(x)=0$ for some $x\in B'\subseteq B$. So $\Phi(B)$ holds, and by (II) $\Phi(A)$ holds; applied to $f$, this is the remaining implication.

Rules in an ordered ring. Let now $R$, with $\le$, be an ordered ring; it is a commutative ring by Ordered Rings §ordered-ring, so (I), (II) and Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field apply to it. The order is total, reflexive and transitive by Commutative Rings, Fields and Ordered Fields: Standard Notation §ordered-rings and Partial and Total Orders on a Set and the Associated Strict Relation §partial. Let $a,b,c,d,z\in R$. By Ordered Rings §ordered-ring: (A) if $a\le b$, then $a+z\le b+z$; (M) if $0\le a$ and $0\le b$, then $0\le ab$. Negatives and differences are as in Negatives, Differences, Reciprocals and Quotients §negative, so $a+(-a)=0$ and $b-a=b+(-a)$.

(R1) If $a\le b$, then $0\le b-a$: by (A) with $z=-a$, $0=a+(-a)\le b+(-a)=b-a$.

(R2) If $0\le b-a$, then $a\le b$: by (A) with $z=a$, $a=0+a\le(b+(-a))+a=b+(a+(-a))=b$.

(R3) $0\le1$: as $\le$ is total, $0\le1$ or $1\le0$. If $1\le0$, then by (A) with $z=-1$, $0=1+(-1)\le0+(-1)=-1$, so by (M) $0\le(-1)(-1)=1\cdot1=1$, using Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs. In either case $0\le1$.

(R4) If $0\le a\le b$ and $0\le c\le d$, then $ac\le bd$. By (R1), $0\le b-a$ and $0\le d-c$. By (M) and Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs, $0\le(b-a)c=bc-ac$, so $ac\le bc$ by (R2). As $0\le a\le b$, $0\le b$ by transitivity, so by (M) and Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs, $0\le(d-c)b=db-cb=bd-bc$, so $bc\le bd$ by (R2). By transitivity, $ac\le bd$.

Clause nonnegative. Let $\Phi(B)$ say: every map $h:B\to R$ with $h(x)\ge0$ for all $x\in B$ satisfies $\prod_{x\in B}h(x)\ge0$. $\Phi(\emptyset)$ holds by (R3), the empty product being $1$. Let $B\neq\emptyset$, $y\in B$, and assume $\Phi(B\setminus\{y\})$. For such $h$, $\prod_{x\in B\setminus\{y\}}h(x)\ge0$ by $\Phi(B\setminus\{y\})$ applied to the restriction of $h$, and $h(y)\ge0$; so by (I) and (M), $\prod_{x\in B}h(x)\ge0$. By (II), $\Phi(A)$ holds; applied to $u$, this is the clause.

Clause monotone. Let $\Phi(B)$ say: all maps $h,k:B\to R$ with $0\le h(x)\le k(x)$ for all $x\in B$ satisfy $\prod_{x\in B}h(x)\le\prod_{x\in B}k(x)$. $\Phi(\emptyset)$ holds as $1\le1$ by reflexivity. Let $B\neq\emptyset$, $y\in B$, $B'=B\setminus\{y\}$, and assume $\Phi(B')$. For such $h$ and $k$ put $P=\prod_{x\in B'}h(x)$ and $Q=\prod_{x\in B'}k(x)$. Then $0\le P$ by the clause nonnegative, proved above, applied to the finite set $B'$ and the restriction of $h$ to it; $P\le Q$ by $\Phi(B')$ applied to the restrictions of $h$ and $k$; and $0\le h(y)\le k(y)$. By (I) and (R4), $\prod_{x\in B}h(x)=P\,h(y)\le Q\,k(y)=\prod_{x\in B}k(x)$. By (II), $\Phi(A)$ holds; applied to $u$ and $v$, this is the clause.