Proof of Basic Facts about Intervals of the Real Line

$\mathbb{R}$ is an ordered field, so Rules of Arithmetic and Order in an Ordered Field and Absolute Value in an Ordered Field (i) apply to it. For the sets of Intervals of the Real Line (ii) and Intervals of the Real Line (iii), $y\in[a,b]$ if and only if $y\in\mathbb{R}$ and $a\le y\le b$, and likewise for the others. Throughout, $\le$ is a total order on $\mathbb{R}$, and its rules, including $w<w'\le w''\Rightarrow w<w''$ and $w\le w'<w''\Rightarrow w<w''$, are used without citation.

(i). Every $y\in\mathbb{R}$ belongs to $\mathbb{R}$, so $\mathbb{R}$ is an interval by Intervals of the Real Line (i). Since $0<1$, $0$ and $1$ are two distinct points of $\mathbb{R}$, and $s=x-1$ and $t=x+1$ satisfy $s<x<t$.

(ii). Let $y\in[u,v]$, so that $u\le y\le v$. Since $u,v\in I$ and $I$ is an interval, $y\in I$ by Intervals of the Real Line (i). Hence $[u,v]\subseteq I$.

(iii). From $p<x<q$ we get $p\le x\le q$, and since $p,q\in I$, $x\in I$ by Intervals of the Real Line (i).

(vi). Let $J$ be one of the eight sets, let $x',z'\in J$ and let $y\in\mathbb{R}$ with $x'\le y\le z'$. By Intervals of the Real Line (ii) and Intervals of the Real Line (iii), membership of $w\in\mathbb{R}$ in $J$ is given by at most one lower condition, $a\le w$ or $a<w$, together with at most one upper condition, $w\le b$ or $w<b$. If $J$ has a lower condition, $x'$ satisfies it, and so does $y$: $a\le x'\le y$ gives $a\le y$, and $a<x'\le y$ gives $a<y$. If $J$ has an upper condition, $z'$ satisfies it, and so does $y$: $y\le z'\le b$ gives $y\le b$, and $y\le z'<b$ gives $y<b$. Hence $y\in J$, and $J$ is an interval by Intervals of the Real Line (i).

(iv). $(p,q)$ is an interval by (vi). Let $x\in(p,q)$, so $p<x<q$, and put $s=(p+x)/(1+1)$ and $t=(x+q)/(1+1)$. By Rules of Arithmetic and Order in an Ordered Field (ix), $p<s<x$ and $x<t<q$. Then $s<x<q$ gives $s<q$, and $p<x<t$ gives $p<t$; so $s,t\in(p,q)$ and $s<x<t$.

(v). Let $a\le b$. $[a,b]$ is an interval by (vi). Since $a\le a\le b$ and $a\le b\le b$, $a,b\in[a,b]$. If $y\in(a,b)$, then $a<y<b$ gives $a\le y\le b$, so $y\in[a,b]$; hence $(a,b)\subseteq[a,b]$. If moreover $a<b$, then $a\ne b$, and $a$ and $b$ are two distinct points of $[a,b]$.

(vii). Let $c,c'\in I$ with $c\ne c'$. By totality $c\le c'$ or $c'\le c$, and the inequality is strict since $c\ne c'$; let $c_{1}$ be the smaller and $c_{2}$ the larger of the two, so that $c_{1},c_{2}\in I$ and $c_{1}<c_{2}$. Let $x\in I$ and $\delta>0$. In each case below $t$ lies between $x$ and $c_{1}$ or $c_{2}$, which belong to $I$, so $t\in I$ by Intervals of the Real Line (i). All additions of a number to both sides of an inequality are by Rules of Arithmetic and Order in an Ordered Field (iv).

Case 1: $x<c_{2}$. Adding $-x$ gives $0<c_{2}-x$. If $c_{2}-x<\delta$, take $t=c_{2}$; then $t\in I$ and $0<t-x<\delta$. Otherwise $\delta\le c_{2}-x$ by totality, and adding $x$ gives $x+\delta\le c_{2}$; adding $x$ to $0<\delta$ gives $x<x+\delta$. Take $t=(x+(x+\delta))/(1+1)$. By Rules of Arithmetic and Order in an Ordered Field (ix), $x<t<x+\delta$; hence $x\le t\le c_{2}$, so $t\in I$, and adding $-x$ gives $0<t-x<\delta$. In both subcases $0\le t-x$, so $|t-x|=t-x$ by Absolute Value in an Ordered Field (i), and $0<|t-x|<\delta$.

Case 2: $c_{2}\le x$. Then $c_{1}<c_{2}\le x$ gives $c_{1}<x$, and adding $-c_{1}$ gives $0<x-c_{1}$. If $x-c_{1}<\delta$, take $t=c_{1}$; then $t\in I$ and $0<x-t<\delta$. Otherwise $\delta\le x-c_{1}$, and adding $c_{1}-\delta$ gives $c_{1}\le x-\delta$; adding $x-\delta$ to $0<\delta$ gives $x-\delta<x$. Take $t=((x-\delta)+x)/(1+1)$. By Rules of Arithmetic and Order in an Ordered Field (ix), $x-\delta<t<x$; hence $c_{1}\le t\le x$, so $t\in I$; adding $\delta-t$ to $x-\delta<t$ gives $x-t<\delta$, and adding $-t$ to $t<x$ gives $0<x-t$. In both subcases $t<x$, so $0\le t-x$ fails, since by Rules of Arithmetic and Order in an Ordered Field (iv) it would give $x\le t$. Hence $|t-x|=-(t-x)=x-t$ by Absolute Value in an Ordered Field (i) and Rules of Arithmetic and Order in an Ordered Field (ii), and $0<|t-x|<\delta$.