Proof of The Archimedean Property of the Real Numbers

Throughout we use the properties of $\iota$ recorded in Properties of the Canonical Map from the Natural Numbers to an Ordered Field and the elementary order arithmetic of Elementary Order Arithmetic in an Ordered Field. Rearrangements of sums and products below use the commutativity and associativity of the addition and multiplication of the field $\mathbb{R}$, the identity $a\cdot 1=a$, and the identity $a\,a^{-1}=1$ valid for $a\ne 0$, all part of Field.

1. Unboundedness. Suppose, for contradiction, that there is $x\in\mathbb{R}$ with $\iota(n)\le x$ for every $n\in\mathbb{N}$. Let

$$E=\{\iota(n):n\in\mathbb{N}\}$$

be the image of $\iota$. Then $E$ is nonempty, since $\iota(1)\in E$, and $x$ is an upper bound for $E$. By the least upper bound property stated in The Real Numbers the set $E$ has a least upper bound, unique by Uniqueness of the Supremum and of the Infimum; write $c=\sup E$.

Statement 6 of Elementary Order Arithmetic in an Ordered Field gives $0<1$, so statement 4 of the same result gives $-1<0$ and then statement 1 gives $c-1<c+0$; since $c+0=c$ by Additive Cancellation and Elementary Additive Identities in a Field we obtain $c-1<c$. Statement 1 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ therefore produces an element of $E$ strictly above $c-1$, that is, a natural number $n$ with $c-1<\iota(n)$.

Adding $1$ to both sides, which preserves strict inequality by statement 1 of Elementary Order Arithmetic in an Ordered Field, and using $(c-1)+1=c$ from Additive Cancellation and Elementary Additive Identities in a Field, we get $c<\iota(n)+1$. By statement 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field we have $\iota(n)+1=\iota(n+1)$, so $c<\iota(n+1)$. But $\iota(n+1)\in E$ and $c$ is an upper bound for $E$, so $\iota(n+1)\le c$; together with $c\le\iota(n+1)$ antisymmetry gives $c=\iota(n+1)$, contradicting $c\ne\iota(n+1)$. This contradiction proves claim 1.

2. Archimedean property. Let $x,\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. In particular $\varepsilon\ne 0$, so $\varepsilon^{-1}$ exists and $0<\varepsilon^{-1}$ by statement 7 of Elementary Order Arithmetic in an Ordered Field. Applying claim 1 to the real number $x\,\varepsilon^{-1}$ gives $n\in\mathbb{N}$ with $x\,\varepsilon^{-1}<\iota(n)$. Multiplying by the positive element $\varepsilon$, which preserves strict inequality by statement 10 of Elementary Order Arithmetic in an Ordered Field, gives

$$\varepsilon\,(x\,\varepsilon^{-1})<\varepsilon\,\iota(n).$$

Now $\varepsilon\,(x\,\varepsilon^{-1})=x\,(\varepsilon\,\varepsilon^{-1})=x\cdot 1=x$ and $\varepsilon\,\iota(n)=\iota(n)\,\varepsilon$, so $x<\iota(n)\,\varepsilon$.

3. Small reciprocals. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$, so that $\varepsilon^{-1}$ exists and $0<\varepsilon^{-1}$ as in claim 2. Applying claim 1 to $\varepsilon^{-1}$ gives $n\in\mathbb{N}$ with $\varepsilon^{-1}<\iota(n)$. By statement 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field the inverse $\iota(n)^{-1}$ exists and satisfies $0<\iota(n)^{-1}$.

Put $c=\varepsilon\,\iota(n)^{-1}$. Since $0<\varepsilon$ and $0<\iota(n)^{-1}$, statement 5 of Elementary Order Arithmetic in an Ordered Field gives $0<c$. Multiplying the inequality $\varepsilon^{-1}<\iota(n)$ by $c$, which preserves strict inequality by statement 10 of Elementary Order Arithmetic in an Ordered Field, gives $c\,\varepsilon^{-1}<c\,\iota(n)$. Now

$$c\,\varepsilon^{-1}=(\varepsilon\,\varepsilon^{-1})\,\iota(n)^{-1}=\iota(n)^{-1},\qquad c\,\iota(n)=\varepsilon\,(\iota(n)^{-1}\iota(n))=\varepsilon ,$$

so $\iota(n)^{-1}<\varepsilon$. Together with $0<\iota(n)^{-1}$ this is the assertion.