Proof of Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting

Throughout, $k$ ranges over $\mathbb{N}_{0}$. We use the following facts about $\mathbb{N}_{0}$. Its order $\le$ is a total order by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §well-order, and $k<n$ means $k\le n$ and $k\neq n$ by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §strict. Since $n+1$ is the successor of $n$ 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, The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §successor gives $k<n+1$ if and only if $k\le n$, in particular $n<n+1$, so that $n+1\le n$ fails; and The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §successor-below gives $n<k$ if and only if $n+1\le k$.

Segment. An element $k$ of $\mathbb{N}_{0}$ lies in $\mathbb{N}$ if and only if $1\le k$: if $k\in\mathbb{N}$, then $1\le k$ by Arithmetic and Order of the Natural Numbers §least; if $k\notin\mathbb{N}$, then $k=0$ by The Natural Numbers with Zero and Their Embedding into the Integers §naturals, and $1\le0$ fails because $1$ is the successor of $0$ by The Set of Natural Numbers and the Number One §one, that is, $1=0+1$, and $0+1\le0$ fails. Hence $[n]=\{k\in\mathbb{N}_{0}:1\le k\le n\}=\{k\in\mathbb{N}:k\le n\}$. If $k\in\mathbb{N}$ and $k\le0$, then $1\le k\le0$, which is impossible; so $[0]=\emptyset$. If $k\in\mathbb{N}$ and $k\le1$, then $1\le k$ gives $k=1$ by antisymmetry; and $1\in\mathbb{N}$ with $1\le1$; so $[1]=\{1\}$.

Successor. For $k\in\mathbb{N}$, $k\le n+1$ holds if and only if $k<n+1$ or $k=n+1$, that is, if and only if $k\le n$ or $k=n+1$. Since $n+1\in\mathbb{N}$ 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 §successors, the description of segments above gives $[n+1]=[n]\cup\{n+1\}$. As $n+1\le n$ fails, $n+1\notin[n]$.

Split. Let $m\le n+1$. The intervals $\{m,\dots,n\}$ and $\{n+1,\dots,n+p\}$ are disjoint, since $k\le n$ and $n+1\le k$ would give $n+1\le n$. Both lie in $\{m,\dots,n+p\}$: we have $n\le n+p$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §difference, so $m\le k\le n$ implies $m\le k\le n+p$; and $n+1\le k\le n+p$ implies $m\le n+1\le k\le n+p$. Conversely, let $m\le k\le n+p$. By totality, $k\le n$ or $n<k$; in the first case $k\in\{m,\dots,n\}$, and in the second $n+1\le k$, so $k\in\{n+1,\dots,n+p\}$.

Shift. By Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §order and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §commutative, $m\le k\le n$ holds if and only if $m+p\le k+p\le n+p$. Hence $\varphi:k\mapsto k+p$ is a map from $\{m,\dots,n\}$ to $\{m+p,\dots,n+p\}$. It is injective, since $k+p=k'+p$ implies $k=k'$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §cancellation and commutativity. It is surjective: let $m+p\le j\le n+p$. Since $p\le p+m=m+p$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §difference, we get $p\le j$, so by the same clause $j=p+k=k+p$ for some $k\in\mathbb{N}_{0}$; then $m+p\le k+p\le n+p$ gives $m\le k\le n$, and $j=\varphi(k)$. So $\varphi$ is a bijection.

Inclusion. If $m\le n$ and $k\in[m]$, then $k\le m\le n$, so $k\in[n]$. Conversely, let $[m]\subseteq[n]$. If $m=0$, then $m\le n$ by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §zero-least. Otherwise $m\in\mathbb{N}$ by The Natural Numbers with Zero and Their Embedding into the Integers §naturals, and $m\le m$ gives $m\in[m]\subseteq[n]$, so $m\le n$.