The proof in my text is as such:
Let $a_1,a_2,a_3,\ldots$ and $b_1, b_2, b_3,\ldots$ be the labels of the left- and right- hand endpoints respectively.
Consider the set A of the left-hand endpoints of the intervals, and let x = sup A. Since x is an upper bound for A, we have $a_n \leq x$. Since each $b_n$ is an upper bound for A, we have $x\leq b_n$. Then since $a_n\leq x \leq b_n$, we can conclude that $x\in I_n$ for every choice of $n\in \mathbb{N}$. Hence x is in the infinite intersection of nested intervals.
My question is.. could the proof work with $x = a_n$ instead? It seems that all the key properties would still hold — $a_n \leq a_n \leq b_n$ for all n. This does not seem to require the AoC to be true.
Best Answer
Take some irrational number, e.g. $\sqrt{2}$. Consider its decimal expansion $$ 1.4142\ldots $$ Now construct a sequence of nested intervals: $$ [1,2] \supset [1.4,1.5] \supset [1.41,1.42] \supset [1.414,1.415] \supset [1.4142,1.4143] \supset \cdots $$ These are nested intervals with rational endpoints, but their intersection contains no rational number. So without completeness the Nested Interval Property is false.