The Sum of $n$ Ones is Strictly Increasing in $n$

Let $\mathbb{N}$ be the set of \reftext{def:natural-numbers-2026a}{natural numbers}, with successor map $S$ and \reftext{def:order-natural-numbers-2026a}{order relations} $<$ and $\le$, and for $n\in\mathbb{N}$ let $[n]$ be the \reftext{def:initial-segment-natural-numbers-2026a}{initial segment} it determines. Let $\mathbb{R}$ be the \reftext{def:ordered-field-c54-2026b}{ordered field} of \reftext{def:real-numbers-c54-2026c}{real numbers}, with additive identity $0$ and multiplicative identity $1$.

For $n\in\mathbb{N}$ let $\mathbf{1}^{(n)}\in\mathbb{R}^{n}$ be the \reftext{def:finite-tuple-power-2026a}{$n$-tuple} with every component equal to $1$, and let $\sigma:\mathbb{N}\to\mathbb{R}$ be given by the \reftext{def:finite-sum-field-2026b}{finite sum}

Then the following hold for all $m,n\in\mathbb{N}$, inequalities between natural numbers being those of $\mathbb{N}$ and inequalities between values of $\sigma$ those of $\mathbb{R}$.

\textbf{1. (Recursion)} $\sigma(1)=1$ and $\sigma(S(n))=\sigma(n)+1$.

\textbf{2. (Strict monotonicity)} If $m<n$, then $\sigma(m)+1\le\sigma(n)$; in particular $\sigma(m)\le\sigma(n)$ and $\sigma(m)\ne\sigma(n)$.

\textbf{3. (Injectivity)} If $\sigma(m)=\sigma(n)$, then $m=n$.