This is second attempt of me to prove:
The Nested Interval Property implies the Axiom of Completeness of the real numbers.
Nested interval property: If $I_1 \supseteq I_2 \supseteq I_3 \dots$ are closed intervals then $\bigcap_n I_n$ is not empty.
Axiom of completeness: If $S$ is a non-empty set in $\mathbb R$ that has an upper bound then $S$ has a least upper bound.
A first attempt is here.
Please can you check my proof again?
Proof: Let $K$ be an upper bound of $S$. Pick $s \in S$. Let $I_1 = [s,K]$. If $K$ is not the least upper bound there is a smaller upper bound $K_2$. Let $I_2 = [s,K_2]$. And so on. If no $K_n$ is a least upper bound for $S$ then because of nested interval property the intersection $I=\bigcap_n I_n$ is non-empty. Also, it is closed. Then the maximum $M$ of $I$ is a least upper bound of $S$: For all $K_n$ it holds that all $s \in S$ are $\le K_n$. The $M$ is the limit of the sequence $K_n$ therefore also $s \le M$ for every $s$. Also $M$ is the least upper bound because if it is not the least upper bound then by the construction $K_n = M$ for some $n$ and there is a smaller upper bound $K_{n+1}$. Then $M \notin \bigcap_n I_n$ which contradicts that $M$ is the maximum in the closed set $\bigcap_n I_n$.
Best Answer
You need to make the gap between the upper bounds and $S$ shrink to $0$.
HINT: Given $s_n\in S$ and an upper bound $K_n$ of $S$, let $x_n=\frac12(s_n+K_n)$. If $x_n$ is an upper bound for $S$, let $s_{n+1}=s_n$ and $K_{n+1}=x_n$. If not, choose $s_{n+1}\in[x_n,K_n]$, and let $K_{n+1}=K_n$. Now consider the intervals $I_n=[s_n,K_n]$.