Proof of One Modulus and a Sum Bound for a Finite Family

Conventions. Of the setting adopted by the statement only the real numbers with their order, the natural numbers and the initial segments $[n]$ are used; no Euclidean or matrix notation enters. Elementary order facts are those of Elementary Order Arithmetic in an Ordered Field, whose claim 1 is the compatibility of the strict order with addition, the non-strict law being an axiom of the ordered field $\mathbb{R}$; order facts about $\mathbb{N}$ are those of Properties of the Order on the Natural Numbers. Recall that $[n]=\{k\in\mathbb{N}:k\le n\}$.

Proof of claim 2. Every real $c$ satisfies $c\cdot 1=c$, by the multiplicative identity axiom of the ordered field $\mathbb{R}$. Hence, applying claim 3 of Properties of Finite Sums to the $n$-tuple all of whose components are $1$ and to the scalar $c$,

the last equality by commutativity of multiplication. For the first assertion, $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, so $0\le 1$ and the $n$-tuple of ones has nonnegative components; claim 6 of Properties of Finite Sums then gives $1\le\sigma_{n}$, applied with the index $1$, which lies in $[n]$ because $1\le n$ by claim 4 of Properties of the Order on the Natural Numbers.

Proof of claim 3. Let $b\in\mathbb{R}^{n}$ be the $n$-tuple all of whose components equal $t$. Then $a_{k}\le b_{k}$ for every $k\in[n]$, so claim 1 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers gives

the last equality by claim 2.

Proof of claim 1. Let $B$ be the set of those $n\in\mathbb{N}$ with the following property: for all metric spaces $(X',d')$ and $(Y',d'_{Y})$, every $A'\subseteq X'$, every $x'\in A'$, every assignment to each $k\in[n]$ of a function $f_{k}:A'\to Y'$ that is continuous at $x'$ relative to $A'$, and every positive $\varepsilon\in\mathbb{R}$, there is a positive $\delta\in\mathbb{R}$ such that every $y\in A'$ with $d'(x',y)<\delta$ satisfies $d'_{Y}(f_{k}(y),f_{k}(x'))<\varepsilon$ for every $k\in[n]$. Claim 1 asserts that $B=\mathbb{N}$.

The number $1$ lies in $B$. Let data as above be given with $n=1$. Continuity of $f_{1}$ at $x'$ relative to $A'$ provides a positive $\delta$ such that every $y\in A'$ with $d'(x',y)<\delta$ satisfies $d'_{Y}(f_{1}(y),f_{1}(x'))<\varepsilon$. If $k\in[1]$ then $k\le 1$, while $1\le k$ by claim 4 of Properties of the Order on the Natural Numbers, so $k=1$ by claim 2 of that lemma; the displayed bound therefore holds for every $k\in[1]$.

The step. Let $n\in B$ and let data as above be given for $S(n)$ in place of $n$. Every $k\in[n]$ satisfies $k\le n$ and $n<S(n)$, hence $k\le S(n)$ by claim 1 of Properties of the Order on the Natural Numbers, so $k\in[S(n)]$; the given assignment therefore restricts to an assignment on $[n]$ of functions continuous at $x'$ relative to $A'$, and since $n\in B$ there is a positive $\delta'$ such that every $y\in A'$ with $d'(x',y)<\delta'$ satisfies $d'_{Y}(f_{k}(y),f_{k}(x'))<\varepsilon$ for every $k\in[n]$. Continuity of $f_{S(n)}$ at $x'$ relative to $A'$ provides a positive $\delta''$ such that every $y\in A'$ with $d'(x',y)<\delta''$ satisfies $d'_{Y}(f_{S(n)}(y),f_{S(n)}(x'))<\varepsilon$. By claim 9 of Elementary Order Arithmetic in an Ordered Field there is $\delta\in\mathbb{R}$ with $\delta\le\delta'$, $\delta\le\delta''$ and $\delta=\delta'$ or $\delta=\delta''$; in either case $\delta$ is positive.

Let $y\in A'$ satisfy $d'(x',y)<\delta$ and let $k\in[S(n)]$. By the mixed transitivity of claim 2 of Elementary Order Arithmetic in an Ordered Field, $d'(x',y)<\delta'$ and $d'(x',y)<\delta''$. If $k=S(n)$, the required bound holds by the choice of $\delta''$. Otherwise $k\ne S(n)$ and $k\le S(n)$, so $k\le n$ by claim 5 of Properties of the Order on the Natural Numbers, that is $k\in[n]$, and the required bound holds by the choice of $\delta'$. Hence $S(n)\in B$.

By Principle of Induction for the Natural Numbers, $B=\mathbb{N}$, which is claim 1. $\blacksquare$