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.

  • 1
    $\begingroup$ $n$ 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 $a_n$. $\endgroup$ – Andrés E. Caicedo Feb 20 '11 at 3:03

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.