Proof of Pointwise Operations on Real-Valued Maps Exist and Are Unique

Each result cited is universally quantified over the data in its own statement.

Throughout, $\mathbb{R}^{I}$ is a set whose elements are exactly the maps from $I$ to $\mathbb{R}$, by Sets and Maps: Ordinary Notation §maps, The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set §set and The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set §members; and $\mathbb{R}^{I}\times\mathbb{R}^{I}$ and $\mathbb{R}\times\mathbb{R}^{I}$ are sets by Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set §set. Two elements of $\mathbb{R}^{I}$ both have domain $I$ by Functions, Values of a Function, and Functions from One Class to Another §map, so they are equal as soon as their values agree at every $p\in I$, by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §equality; we refer to this as (E).

Step 1: pointwise maps. Let $e(p)$ be one of the expressions

$$f(p)+g(p),\qquad f(p)\cdot g(p),\qquad t\,f(p),\qquad -f(p),\qquad c,$$

whose parameters, among $f,g\in\mathbb{R}^{I}$ and $t,c\in\mathbb{R}$, are those that occur in it. Let $p\in I$. Then $f(p),g(p)\in\mathbb{R}$, since $f$ and $g$ are maps from $I$ to $\mathbb{R}$; the values of the binary operations $+$ and $\cdot$ of $\mathbb{R}$ lie in $\mathbb{R}$ by Binary Operations on a Set §notation, and $-f(p)\in\mathbb{R}$ by Negatives, Differences, Reciprocals and Quotients §negative and Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima §negative; here $\mathbb{R}$, with $+$, $\cdot$, $0$ and $1$, is a commutative ring, since it is an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, hence a field and so a commutative ring by Ordered Fields §ordered-field and Fields §field. Hence $e(p)\in\mathbb{R}$ for every $p\in I$. The expression $e(p)$ is built from defined set symbols used properly (values of maps by Functions, Values of a Function, and Functions from One Class to Another §value, values of binary operations by Binary Operations on a Set §notation, and the negative, the unique $w\in\mathbb{R}$ with $f(p)+w=0$, by Negatives, Differences, Reciprocals and Quotients §negative), and no quantifier in it is applied to a class variable; so the expression $e(p)$ is predicative in the sense of Formulas of the Language of Class Theory, Free Variables, Predicative Formulas and Abbreviations §defined-symbols, by Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols §closure (for the substitution of values of maps and of binary operations, and for the negative, see also Predicative Substitution of Values of Functions and Binary Operations, and Predicative Formulas for Negatives, Differences, Absolute Values, Distances, Reciprocals and Quotients in an Ordered Field §value-substitution, Predicative Substitution of Values of Functions and Binary Operations, and Predicative Formulas for Negatives, Differences, Absolute Values, Distances, Reciprocals and Quotients in an Ordered Field §operation-substitution and Predicative Substitution of Values of Functions and Binary Operations, and Predicative Formulas for Negatives, Differences, Absolute Values, Distances, Reciprocals and Quotients in an Ordered Field §negative-formula). Therefore Maps and Relations Given by Formulas §map, applied with the sets $I$ and $\mathbb{R}$, as in Sets and Maps: Ordinary Notation §maps, gives exactly one map $h:I\to\mathbb{R}$ with $h(p)=e(p)$ for every $p\in I$; it lies in $\mathbb{R}^{I}$.

Step 2: the pointwise map as a defined set symbol. Let $\chi(h,I,\mathbb{R})$ be the formula written $\mu(F,A,B)$ in Predicativity under Renaming of Set Variables by Variables, of Formulas with Ordered Pairs, and of Formulas for Relations, Functions, Maps, Injectivity, Surjectivity and Bijectivity §map, with $h$, $I$ and $\mathbb{R}$ for $F$, $A$ and $B$; it is predicative and holds if and only if $h$ is a map $h:I\to\mathbb{R}$, by that clause. Let $\theta_{e}$ be the formula

$$\chi(h,I,\mathbb{R})\wedge\forall p\,\bigl(p\in I\Rightarrow h(p)=e(p)\bigr),$$

in the variable $h$ and the parameters of $e$. Its second conjunct is predicative by Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols §closure, so $\theta_{e}$ is predicative. By Step 1, whenever the parameters $f,g$ of $e$ lie in $\mathbb{R}^{I}$ and $t,c$ lie in $\mathbb{R}$, exactly one set $h$ satisfies $\theta_{e}$. Write $T_{e}$ for it, with the parameters of $e$ as arguments. Thus $T_{e}$ is a defined set symbol in the sense of Formulas of the Language of Class Theory, Free Variables, Predicative Formulas and Abbreviations §defined-set-symbols, used properly whenever its arguments lie in $\mathbb{R}^{I}$, respectively $\mathbb{R}$, that is, whenever they satisfy the condition $\gamma$ of Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols §elimination which is the conjunction of $\chi(f,I,\mathbb{R})$, $\chi(g,I,\mathbb{R})$, $t\in\mathbb{R}$ and $c\in\mathbb{R}$, each conjunct taken only for those of $f,g,t,c$ that are parameters of $e$ (by Predicativity under Renaming of Set Variables by Variables, of Formulas with Ordered Pairs, and of Formulas for Relations, Functions, Maps, Injectivity, Surjectivity and Bijectivity §map and The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set §members, $\chi(f,I,\mathbb{R})$ holds exactly when $f\in\mathbb{R}^{I}$, and likewise for $g$); $\gamma$ is predicative, as a conjunction of predicative formulas; so formulas containing $T_{e}$ are predicative by Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols §elimination; and $T_{e}\in\mathbb{R}^{I}$ with $T_{e}(p)=e(p)$ for every $p\in I$.

Clause binary. Let $\ast$ be $+$ or $\cdot$ of $\mathbb{R}$, and let $T(f,g)$ be $T_{e}$ for $e(p)=f(p)\ast g(p)$. Since $T(f,g)\in\mathbb{R}^{I}$ for all $f,g\in\mathbb{R}^{I}$, Maps and Relations Given by Formulas §binary, applied with $b=c=d=\mathbb{R}^{I}$, gives exactly one map $\circledast:\mathbb{R}^{I}\times\mathbb{R}^{I}\to\mathbb{R}^{I}$ with $\circledast((f,g))=T(f,g)$ for all $f,g\in\mathbb{R}^{I}$. It is a binary operation on $\mathbb{R}^{I}$, and by Binary Operations on a Set §notation and Step 2, $(f\circledast g)(p)=T(f,g)(p)=f(p)\ast g(p)$ for all $f,g\in\mathbb{R}^{I}$ and $p\in I$. For uniqueness, let $\circledast'$ be a binary operation on $\mathbb{R}^{I}$ with $(f\circledast'g)(p)=f(p)\ast g(p)$ for all $f,g$ and $p$. For $f,g\in\mathbb{R}^{I}$, the elements $f\circledast'g$ and $T(f,g)$ of $\mathbb{R}^{I}$ have the same value at every $p\in I$, so they are equal by (E). Thus $\circledast'$, a map from $\mathbb{R}^{I}\times\mathbb{R}^{I}$ to $\mathbb{R}^{I}$ by Binary Operations on a Set §operation, satisfies $\circledast'((f,g))=T(f,g)$ for all $f,g\in\mathbb{R}^{I}$, and $\circledast'=\circledast$ by the uniqueness in Maps and Relations Given by Formulas §binary.

Clause scalar-multiple. Let $S(t,f)$ be $T_{e}$ for $e(p)=t\,f(p)$. Since $S(t,f)\in\mathbb{R}^{I}$ for all $t\in\mathbb{R}$ and $f\in\mathbb{R}^{I}$, Maps and Relations Given by Formulas §binary, applied with $c=\mathbb{R}$ and $b=d=\mathbb{R}^{I}$, gives exactly one map $\mu:\mathbb{R}\times\mathbb{R}^{I}\to\mathbb{R}^{I}$ with $\mu(t,f)=S(t,f)$, so that $\mu(t,f)(p)=t\,f(p)$ for all $t$, $f$ and $p\in I$. If $\mu'$ is a second such map, then $\mu'(t,f)$ and $S(t,f)$ have the same values, so $\mu'(t,f)=S(t,f)$ by (E) for all $t$ and $f$, and $\mu'=\mu$ by the uniqueness in Maps and Relations Given by Formulas §binary.

Clause negative. Let $N(f)$ be $T_{e}$ for $e(p)=-f(p)$. Since $N(f)\in\mathbb{R}^{I}$ for every $f\in\mathbb{R}^{I}$, Maps and Relations Given by Formulas §map, applied with $a=b=\mathbb{R}^{I}$, gives exactly one map $\nu:\mathbb{R}^{I}\to\mathbb{R}^{I}$ with $\nu(f)=N(f)$, so that $\nu(f)(p)=-f(p)$ for all $f$ and $p\in I$. If $\nu'$ is a second such map, then $\nu'(f)=N(f)$ for every $f$ by (E), and $\nu'=\nu$ by the uniqueness in Maps and Relations Given by Formulas §map.

Clause constant. For $c\in\mathbb{R}$, Step 1 with $e(p)=c$ gives $h\in\mathbb{R}^{I}$ with $h(p)=c$ for every $p\in I$; and any two elements of $\mathbb{R}^{I}$ with value $c$ at every $p\in I$ are equal by (E).