Multiplication on Omega

In the setting of Class Theory NBG: the Axioms, Standing Conventions and Basic Notation, let $\omega$ and $0$ be as in The Class Omega of Natural Numbers with Zero §omega and The Class Omega of Natural Numbers with Zero §zero, let $S(x)$ denote the successor of a set $x$, and let $+$ be the addition on $\omega$. By Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §set, $\omega$ is a set; by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §inductive, $0\in\omega$ and $S(k)\in\omega$ for every $k\in\omega$; and by Addition on Omega §addition, $+$ is a map $\omega\times\omega\to\omega$, so $u+m\in\omega$ for all $u,m\in\omega$. Below, $m,n,k,u,z,f,p$ are set variables, and each class is formed by class abstraction from a formula quantifying over set variables only, in which the ordered pair, $S(k)$, sums, the values of functions and the products $m\cdot n$ introduced below are defined set symbols; so it is predicative as Class Theory NBG: the Axioms, Standing Conventions and Basic Notation §comprehension requires.

¶ For $m\in\omega$, let $\tau_{m}=\{p:\exists n\,\exists u\,(n\in\omega\wedge u\in\omega\wedge p=((n,u),u+m))\}$. Exactly as for the map $\sigma$ of Addition on Omega §addition, with $u+m\in\omega$ in place of $S(u)\in\omega$, $\tau_{m}$ is a map $\tau_{m}:\omega\times\omega\to\omega$ with $\tau_{m}((n,u))=u+m$. The Recursion Theorem on Omega §recursion, applied with $a=\omega$, $c=0$ and $g=\tau_{m}$, gives exactly one map $f:\omega\to\omega$ with $f(0)=0$ and $f(S(k))=\tau_{m}((k,f(k)))=f(k)+m$ for every $k\in\omega$; it is a set by Images of Sets under Functions Are Sets, and a Function with a Set Domain Is a Set §function-set, its domain $\omega$ being a set. For $m,n\in\omega$, the product $m\cdot n$ is the unique set $z$ satisfying

$$\exists f\,\bigl(f:\omega\to\omega\wedge f(0)=0\wedge\forall k\,(k\in\omega\Rightarrow f(S(k))=f(k)+m)\wedge z=f(n)\bigr):$$

the map just described is a set witnessing this formula, and every set $f$ witnessing it satisfies $f(S(k))=f(k)+m=\tau_{m}((k,f(k)))$ for every $k\in\omega$, hence equals that map by the uniqueness in The Recursion Theorem on Omega §recursion; so $m\cdot n$ is a defined set symbol, used only for $m,n\in\omega$, and $m\cdot n=f(n)\in\omega$, since $f(n)$ lies in the range of $f$, which is included in $\omega$ by Functions, Values of a Function, and Functions from One Class to Another §map. The multiplication on $\omega$ is the class

$${\cdot}=\{p:\exists m\,\exists n\,(m\in\omega\wedge n\in\omega\wedge p=((m,n),m\cdot n))\};$$

exactly as for $\sigma$, with $m\cdot n\in\omega$ in place of $S(u)\in\omega$, it is a map ${\cdot}:\omega\times\omega\to\omega$, that is, a binary operation on $\omega$, whose value at $(m,n)$ is $m\cdot n$ by Functions, Values of a Function, and Functions from One Class to Another §value, since $((m,n),m\cdot n)\in{\cdot}$.

¶ In expressions combining $+$ and $\cdot$, the product is formed first: $k\cdot m+n$ stands for $(k\cdot m)+n$, and $n+k\cdot m$ for $n+(k\cdot m)$.