The Completeness Theorem



Let $\QTR{cal}{S}$ be a fine and consistent family of real intervals.

Let $\QTR{bf}{X}$ be the family of all rational intervals MATH such that $I$ overspreads at least one interval in $\QTR{cal}{S}$. This means that for $[r,s]$ to be in $\QTR{bf}{X}$ there must be some $[A,B]$ in $\QTR{cal}{S}$ --- depending of $I$ --- such that MATH.




The Completeness Theorem says that $\QTR{bf}{X}$ is a real number (i.e. is fine and consistent) and, as a real number, MATH for every $[A,B]$ in $\QTR{cal}{S}$. Furthermore, $\QTR{bf}{X}$ is the only real which is in all the intervals of $\QTR{cal}{S}$.



Fineness: Suppose we are given some $\epsilon >0$. Choose $[A,B]$ in $\QTR{cal}{S}$ with $B-A<\epsilon /3$ (we can do this because $\QTR{cal}{S}$ is given to be fine) and choose rational intervals $[r,s]$ in $A$ with $s-r<\epsilon /3$ and $[u,v]$ in $B$ with $v-u<\epsilon /3 $. (We can do this because $A$ and $B$ are real numbers, hence fine families of rational intervals.). Note that we have $r\leq A\leq s$ and $u\leq B\leq v $ (by Corollary 1.3.13). Then $[r,v]$ lies in $\QTR{bf}{X}$ because MATH, so $[r,v]$ overspreads the interval $[A,B]$ which is in $\QTR{cal}{S}$. We have: MATH

Since we have produced an interval in $\QTR{bf}{X}$ of length less than an arbitrary $\epsilon >0$, $\QTR{bf}{X}$ is fine.

Consistency: Let $[r,s]$ and $[u,v]$ be in $\QTR{bf}{X}$; we must show that they intersect. We can find intervals $[A,B]$ and $[C,D]$ in $\QTR{cal}{S}$ such that MATH and MATH. Then $A\leq D$ and $C\leq B$ since $\QTR{cal}{S}$ is given to be consistent. We then have MATH and MATH. Thus, $[r,s]$ and $[u,v]$ intersect since $r\leq v$ and $u\leq s$.

$\QTR{bf}{X}$ lies in each $[A,B]$ of $S$: To prove MATH we must show MATH. To prove that $A\leq \QTR{bf}{X}$ we show first that $A$ is $\leq $ every upper bound of $\QTR{bf}{X}$. So suppose $[r,s]$ is any interval of $\QTR{bf}{X}.$We must show $A\leq s$. Since $[r,s]$ is in $\QTR{bf}{X}$, there is some interval $[C,D]$ in $\QTR{cal}{S}$ overspread by $[r,s]$; i.e. MATHSince $\QTR{cal}{S}$ is consistent, $[C,D]$ meets $[A,B]$, so there is a $W$ with MATHNow we do the usual tracing through of inequalities: MATH. Thus, any lower bound of $A$ is $\leq $ any upper bound of $\QTR{bf}{X}$, which proves $A\leq \QTR{bf}{X}$. Now we must show $\QTR{bf}{X}\leq B$ We have MATH. Thus $r$, an arbitrary lower bound of $\QTR{bf}{X}$, is $\leq $ any upper bound of $B$, we have $\QTR{bf}{X}\leq B$: done.




Uniqueness: Suppose $Y$ is also in every interval of $\QTR{cal}{S}$. Let $\epsilon >0$ be arbitrary, and choose $[A,B]$ an interval in $\QTR{cal}{S}$ of length $<\epsilon $. Since $Y\in \lbrack A,B]$, $Y\leq B$; since MATH, $A\leq \QTR{bf}{X}$ so MATH. Adding these inequalities we get: MATH so MATH. We conclude that $Y\leq \QTR{bf}{X}$ by the Wiggle Lemma for reals. But the same argument, starting with $\QTR{bf}{X}-Y$ proves $\QTR{bf}{X}\leq Y$; thus, $\QTR{bf}{X}=Y$.