Exercise 2.5.4

Let $A$ be a bounded set, we’re basically going to binary search for $\sup <!--\; FUNCTION\; APPLICATION\; -->A$ and then use NIP to prove the limit exists.

Let $M$ be an upper bound on $A$, and pick any $L\in A$ as our starting lower bound for $\sup <!--\; FUNCTION\; APPLICATION\; -->A$ and define $I_1=[L,M]$. Doing binary search gives $I_{n+1}\subseteq I_n$ with length proportional to ${(1∕2)}^n$. Applying the Nested Interval Property gives

As the length ${(1∕2)}^n$ goes to zero, there is a single $s\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{I}_n$ which must be the least upper bound since $I_n=[L_n,M_n]$ gives $L_n\le x\le M_n$ for all $n$ meaning $s=\sup <!--\; FUNCTION\; APPLICATION\; -->A$ since

(i)

$s\ge L_n$ implies $s$ is an upper bound

(ii)

$s\le M_n$ implies $s$ is the least upper bound

The assumption that $\left(1∕2^n\right)\to 0$ is necessary because otherwise, we would need to invoke the Archimedian Property (Theorem 1.4.2), which is proved using the Axiom of Completeness.