Proof of Basic Facts about Intervals of the Real Line and Their Interior Points

Throughout we use the elementary order arithmetic of Elementary Order Arithmetic in an Ordered Field, whose clause numbers are cited explicitly.

1. (The real line.) $\mathbb{R}$ is an interval: the defining condition asks that $y\in\mathbb{R}$ whenever $x,z\in\mathbb{R}$, $y\in\mathbb{R}$ and $x\le y\le z$, which holds trivially.

By clause 6, $0<1$, so $0\ne 1$; hence $\mathbb{R}$ contains the two distinct points $0$ and $1$.

Let $x\in\mathbb{R}$. From $0<1$ and clause 1 (strict compatibility with addition, adding $x$) we get $x+0<x+1$, that is, $x<x+1$. From $0<1$ and clause 4 (sign reversal) we get $-1<0$, and adding $x$ gives $x+(-1)<x+0$, that is, $x-1<x$. Since $x-1,x+1\in\mathbb{R}$ and $x-1<x<x+1$, the point $x$ is an interior point of $\mathbb{R}$.

2. (Closed subintervals.) Let $u,v\in I$ with $u\le v$, and let $y\in[u,v]$, so that $u\le y\le v$. Since $u,v\in I$ and $I$ is an interval, $y\in I$. Hence $[u,v]\subseteq I$.

3. (Points strictly between.) Let $p,q\in I$ with $p<x<q$. Then $p\le x\le q$, so $x\in I$ because $I$ is an interval. Since $p,q\in I$ and $p<x<q$, the point $x$ is an interior point of $I$ by the definition of an interior point.

4. (Open intervals.) First, $(p,q)$ is an interval. Let $x,z\in(p,q)$ and let $y\in\mathbb{R}$ with $x\le y\le z$. From $p<x$ and $x\le y$, clause 2 (mixed transitivity) gives $p<y$; from $y\le z$ and $z<q$, the same clause gives $y<q$. Hence $y\in(p,q)$.

Now let $x\in(p,q)$, so $p<x<q$. By clause 1, $p<x$ gives $0<x-p$, and $x<q$ gives $0<q-x$. By clause 8 (halving), $0<(x-p)\cdot 2^{-1}$ and $(x-p)\cdot 2^{-1}<x-p$, and likewise $0<(q-x)\cdot 2^{-1}<q-x$. Put

Adding $p$ to $0<(x-p)\cdot 2^{-1}<x-p$ and using clause 1 gives $p<u<p+(x-p)=x$. Adding $x$ to $0<(q-x)\cdot 2^{-1}<q-x$ gives $x<v<x+(q-x)=q$. Hence $p<u<x<v<q$, so $u,v\in(p,q)$ and $u<x<v$; therefore $x$ is an interior point of $(p,q)$.

5. (Closed intervals.) Let $a\le b$. Then $[a,b]$ is an interval: if $x,z\in[a,b]$ and $x\le y\le z$, then $a\le x\le y$ and $y\le z\le b$, so $a\le y\le b$ and hence $y\in[a,b]$.

Suppose moreover $a<b$. Then $a,b\in[a,b]$ and $a\ne b$, so $[a,b]$ contains at least two points. Finally, let $x\in(a,b)$, so $a<x<b$; since $a,b\in[a,b]$, claim 3 applied to the interval $[a,b]$ shows that $x$ is an interior point of $[a,b]$.