Proof of Discreteness of the Integers

Each result cited below is universally quantified over the data in its own statement and is applied to the data named where it is cited. The argument takes place in $\mathbb{Z}$ and $\mathbb{N}_{0}$ only. By The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §integers and The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ordered-ring, $\le$ is a total order on $\mathbb{Z}$ with strict relation $<$, and $\mathbb{Z}$ is an ordered ring, so $u\le v$ implies $u+w\le v+w$ for all $u,v,w\in\mathbb{Z}$ by Ordered Rings §ordered-ring. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring and Commutative Rings §ring, addition in $\mathbb{Z}$ is associative and commutative, $u+0_{\mathbb{Z}}=u$, and $u+(-u)=0_{\mathbb{Z}}$ for $u\in\mathbb{Z}$; and $u-v=u+(-v)$ by The Integers §operations. In $m+1$, the $1$ is $1_{\mathbb{Z}}$: as an element of $\mathbb{Z}$ it is $1_{\mathbb{Z}}$ by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §integers, and read as the natural number $1$ it denotes $\iota_{0}(1)$ by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §numerals and The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §identification, which is $1_{\mathbb{Z}}$ because $\iota_{0}$ preserves $1$, by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §embeddings.

Let $m<n$. By Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization, $m\le n$ and $m\neq n$. Adding $-m$ to both sides of $m\le n$ gives $0_{\mathbb{Z}}=m+(-m)\le n+(-m)=n-m$. Moreover $n-m\neq0_{\mathbb{Z}}$: otherwise

$$n=n+0_{\mathbb{Z}}=n+((-m)+m)=(n+(-m))+m=0_{\mathbb{Z}}+m=m,$$

contradicting $m\neq n$. Hence $0_{\mathbb{Z}}<n-m$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization.

By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §positive, there is $d\in\mathbb{N}$ with $n-m=\iota(d)$, and $\iota(d)=\iota_{0}(d)$ by The Natural Numbers with Zero and Their Embedding into the Integers §embedding; this is the description of the image of $\mathbb{N}$ under $\iota_{0}$ as the set of positive integers recorded in The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §embeddings. Since $1$ is the least natural number, by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §order, $1\le d$ in $\mathbb{N}_{0}$. By The Natural Numbers with Zero and Their Embedding into the Integers §embedding, $\iota_{0}$ preserves the order, so $\iota_{0}(1)\le\iota_{0}(d)$, that is, $1_{\mathbb{Z}}\le n-m$.

Adding $m$ to both sides gives $1_{\mathbb{Z}}+m\le(n-m)+m$. Here $1_{\mathbb{Z}}+m=m+1_{\mathbb{Z}}$ by commutativity, and $(n-m)+m=n+((-m)+m)=n+0_{\mathbb{Z}}=n$ by associativity, commutativity and $n+0_{\mathbb{Z}}=n$. Hence $m+1\le n$.