Proof of The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences

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. Throughout, elements of $\mathbb{N}$, of $\mathbb{N}_{0}$ and of $\mathbb{Z}$ standing where a real number is required are read as in The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §identification, and by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement, sums, negatives, differences and the order agree whether formed in $\mathbb{N}_{0}$, in $\mathbb{Z}$ or in $\mathbb{R}$; so a fact proved in $\mathbb{N}_{0}$ or in $\mathbb{Z}$ about sums, differences and the order holds for the corresponding real numbers, and conversely an inequality between such real numbers holds between the elements of $\mathbb{N}_{0}$ or the integers they stand for. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ordered-ring, $\le$ is a total order on $\mathbb{Z}$ and $\mathbb{Z}$ is an ordered ring, so $u\le v$ implies $u+w\le v+w$ for $u,v,w\in\mathbb{Z}$ by Ordered Rings §ordered-ring. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring, $\mathbb{Z}$ is a commutative ring with $u+(-u)=0_{\mathbb{Z}}$, and $u-v=u+(-v)$ by The Integers §operations. By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, $\mathbb{R}$ is an ordered field, hence a field and so a commutative ring, with $u+(-u)=0$ and $u-v=u+(-v)$ by Negatives, Differences, Reciprocals and Quotients §negative. By the ring laws we mean the identities of Commutative Rings §ring (associativity and commutativity of $+$, and $u+0=u$) together with $u+(-u)=0$, in $\mathbb{Z}$ or in $\mathbb{R}$. They give, for all $u,v$ in $\mathbb{Z}$ or in $\mathbb{R}$,

$$(u+v)+(-v)=u,\qquad(v+u)+(-v)=u,\qquad v+(u-v)=u,$$

since $(u+v)+(-v)=u+(v+(-v))=u+0=u$, $(v+u)+(-v)=(u+v)+(-v)$, and $v+(u-v)=v+(u+(-v))=(v+u)+(-v)$. Moreover, by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §archimedean, $\mathbb{R}$ has the Archimedean property: for every $z\in\mathbb{R}$ there is $N\in\mathbb{N}$ with $z<N$. The integers are discrete: for $m,n\in\mathbb{Z}$ with $m<n$, $m+1\le n$, by Discreteness of the Integers §discrete.

Clause greatest. Let $S\subseteq\mathbb{Z}$ be nonempty and $b\in\mathbb{R}$ with $s\le b$ for every $s\in S$. Choose $s_{0}\in S$, and by the Archimedean property choose $N\in\mathbb{N}$ with $b-s_{0}<N$. Let

$$T=\{t\in\mathbb{N}_{0}:s_{0}+t\in S\},$$

where $t$ stands for the integer $\iota_{0}(t)$. If $t\in T$, then $s_{0}+t\le b$ in $\mathbb{R}$, so $(s_{0}+t)+(-s_{0})\le b+(-s_{0})$ by adding $-s_{0}$ to both sides (Rules of Arithmetic and Order in an Ordered Field §order-sum, with $z=w=-s_{0}$), that is $t\le b-s_{0}$, since $(s_{0}+t)+(-s_{0})=t$ by the ring laws of $\mathbb{R}$ and $b+(-s_{0})=b-s_{0}$; as $b-s_{0}<N$, we get $t<N$ in $\mathbb{R}$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed; moving this inequality back to $\mathbb{N}_{0}$ by the readings above (The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement) gives $t<N$ in $\mathbb{N}_{0}$, so $t\le N$ there by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization. Hence $T$ is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §naturals. Let $f:T\to\mathbb{Z}$ be the map $t\mapsto s_{0}+t$; its image $f(T)$ is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §image.

We claim $f(T)=S'$, where $S'=\{s\in S:s_{0}\le s\}$. If $t\in T$, then $f(t)=s_{0}+t\in S$, and $s_{0}\le s_{0}+t$: indeed $0_{\mathbb{Z}}\le\iota_{0}(t)$, the image of $\iota_{0}$ being $\{x\in\mathbb{Z}:0_{\mathbb{Z}}\le x\}$ by The Natural Numbers with Zero and Their Embedding into the Integers §embedding, so adding $s_{0}$ gives $0_{\mathbb{Z}}+s_{0}\le\iota_{0}(t)+s_{0}$, where $0_{\mathbb{Z}}+s_{0}=s_{0}$ and $\iota_{0}(t)+s_{0}=s_{0}+t$ by the ring laws; so $f(t)\in S'$. Conversely, let $s\in S'$. Adding $-s_{0}$ to $s_{0}\le s$ gives $s_{0}+(-s_{0})\le s+(-s_{0})$, that is $0_{\mathbb{Z}}\le s-s_{0}$ by the ring laws, so by the same clause $s-s_{0}=\iota_{0}(t)$ for some $t\in\mathbb{N}_{0}$; then $s_{0}+t=s_{0}+(s-s_{0})=s$ by the ring laws, so $s_{0}+t\in S$, $t\in T$ and $s=f(t)$.

Thus $S'$ is a finite subset of $\mathbb{Z}$, nonempty since $s_{0}\in S'$, as $s_{0}\in S$ and $s_{0}\le s_{0}$ by reflexivity of the partial order $\le$ of $\mathbb{Z}$. As $\le$ is a total order on the set $\mathbb{Z}$, Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §extremes gives a greatest element $m$ of $S'$. Then $m\in S$, and $m$ is a greatest element of $S$: let $s\in S$; if $s_{0}\le s$, then $s\in S'$ and $s\le m$; otherwise $s<s_{0}$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, and $s_{0}\le m$ because $s_{0}\in S'$, so $s\le m$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization and transitivity.

Clause floor. Let $S=\{m\in\mathbb{Z}:m\le x\}$, the inequality being read in $\mathbb{R}$. By the Archimedean property there is $N\in\mathbb{N}$ with $-x<N$, so $-N<-(-x)$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-negative, and $-(-x)=x$ by Rules of Arithmetic and Order in an Ordered Field §signs, so $-N<x$; here $-N$ is the real number standing for the integer $-N$, that is $-\iota_{0}(N)$, since negatives agree in $\mathbb{Z}$ and in $\mathbb{R}$ by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement. Then $-N\le x$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization. Hence the integer $-N$ lies in $S$ and $S\neq\emptyset$. Every element of $S$ is at most $x$, so by the clause greatest, proved above, $S$ has a greatest element $n$. Then $n\le x$. Moreover $0_{\mathbb{Z}}<1_{\mathbb{Z}}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §positive, since $1_{\mathbb{Z}}=\iota(1)$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding; so $0_{\mathbb{Z}}\le1_{\mathbb{Z}}$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization, and adding $n$ gives $0_{\mathbb{Z}}+n\le1_{\mathbb{Z}}+n$, that is $n\le n+1$, as $0_{\mathbb{Z}}+n=n+0_{\mathbb{Z}}=n$ and $1_{\mathbb{Z}}+n=n+1$ by the ring laws. Also $n\neq n+1$, since otherwise $0_{\mathbb{Z}}=1_{\mathbb{Z}}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring; so $n<n+1$ in $\mathbb{Z}$. Then $n+1\notin S$: otherwise $n+1\le n$, as $n$ is greatest in $S$, and this together with $n<n+1$ is impossible: $n+1\le n$ and $n\le n+1$ give $n=n+1$ by antisymmetry of the partial order $\le$ of $\mathbb{Z}$, contradicting $n\neq n+1$. That is, $n+1\le x$ fails, and $x<n+1$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, the order of $\mathbb{R}$ being total. This proves existence.

For uniqueness, let $n,n'\in\mathbb{Z}$ with $n\le x<n+1$ and $n'\le x<n'+1$. If $n<n'$, then $n+1\le n'$ by Discreteness of the Integers §discrete, and in $\mathbb{R}$ this gives $n'\le x<n+1\le n'$, so $n'<n'$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed, which is impossible by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive, applied to the order of $\mathbb{R}$. By symmetry $n'<n$ is impossible too, so $n=n'$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §trichotomy, the order of $\mathbb{Z}$ being total.

Clause floor-lower. Let $n\in\mathbb{Z}$ with $n\le x<n+1$. From $x<n+1$, adding $-1$ gives $x+(-1)<(n+1)+(-1)$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. Here $x+(-1)=x-1$ by Negatives, Differences, Reciprocals and Quotients §negative, and $(n+1)+(-1)=n$ by the ring laws of $\mathbb{R}$; so $x-1<n$.

Clause floor-nonnegative. Let $n\in\mathbb{Z}$ with $n\le x<n+1$, and let $x\ge0$. If $n<0_{\mathbb{Z}}$, Discreteness of the Integers §discrete gives $n+1\le0_{\mathbb{Z}}$, so $x<n+1\le0$ in $\mathbb{R}$ and $x<0$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed, contradicting $x\ge0$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation. Hence $n<0_{\mathbb{Z}}$ fails, and $n\ge0$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, the order of $\mathbb{Z}$ being total.

Clause reciprocal. Let $\varepsilon\in\mathbb{R}$ be positive. Then $0<\varepsilon^{-1}$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal. By the Archimedean property there is $n\in\mathbb{N}$ with $\varepsilon^{-1}<n$. Read in $\mathbb{R}$, $0<n$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §naturals, so $n\neq0$ and $n^{-1}$ is defined. Since $0<\varepsilon^{-1}<n$, Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal gives $n^{-1}<(\varepsilon^{-1})^{-1}$, and $(\varepsilon^{-1})^{-1}=\varepsilon$ by Rules of Arithmetic and Order in an Ordered Field §reciprocals, applied with $x=y=\varepsilon$, which is nonzero since $0<\varepsilon$. Finally $1/n=1\cdot n^{-1}=n^{-1}$ by Negatives, Differences, Reciprocals and Quotients §reciprocal and Commutative Rings §ring. Thus $1/n<\varepsilon$.

Clause eventually. By the Archimedean property there is $N\in\mathbb{N}$ with $x<N$. Let $n\in\mathbb{N}$ with $n\ge N$. Read in $\mathbb{R}$, this gives $N\le n$ by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement, so $x<n$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed.