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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.

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$n$ is not a fixed number here, it is a variable for 1, 2, 3, ... When you say "$x=a_n$", what do you mean? $x=a_1$? Then maybe $x<a_2$. If you mean $x=a_2$, then maybe $x<a_3$, and so on. You need an $x$ that for sure is larger than or equal to each $a_n$. –  Andres Caicedo Feb 20 '11 at 3:03

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up vote 4 down vote accepted

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.

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Hmm... So the NIP is essentially the AoC but from 'two directions' (the left and right endpoints) instead of one? –  int3 Feb 20 '11 at 3:13
    
I believe they're equivalent for ordered (Archimedean?) rings - you can trying proving this. –  Yuval Filmus Feb 20 '11 at 3:41

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