Here is the beginning of a standard proof that the Nested Set Theorem implies the Axiom of Completeness (e.g. see here and here):

Suppose $E$ is a non-empty set bounded above by $b$ and let $a$ be an element of $E$. Define the following sequence of sets: \begin{gather*} [a_1, b_1] = [a,b] \\[8pt] [a_{n+1}, b_{n+1}] = \begin{cases} \left[a_n, \dfrac{a_n + b_n}{2}\right] & \text{if $\, \dfrac{a_n + b_n}{2}\, $ is an upper bound of $E$} \\[8pt] \left[\dfrac{a_n + b_n}{2}, b_n\right]& \text{otherwise} \end{cases} \end{gather*} Then $[a_1, b_1]$ is bounded and $\{[a_n, b_n]\}_{n=1}^\infty$ is a descending countable collection of nonempty closed sets of real numbers, so the Nested Set Theorem implies that the intersection $\bigcap_{n=1}^\infty [a_n,b_n]$ is non-empty. Let $x$ be an element of this intersection.

Correct me if I'm wrong, but to show that $x$ is the least upper bound of $E$ you need to know that $b_n \to x$ and $a_n \to x$. Is it possible to know this without invoking the Archemedean property of the reals?

  • $\begingroup$ I think that we need the archimedean property to define a sequence, i.e. a sequence is a function with denumerable domain. $\endgroup$ – Masacroso Jul 10 '16 at 20:54
  • $\begingroup$ @Masacroso Can I define a sequence as a real-valued function whose domain is the set of natural numbers, define the set of natural numbers to be the intersection of all inductive subsets of R, and define a set $E$ to be inductive if $1 \in E$ and $x \in E \, \Rightarrow \, x+1 \in E$? $\endgroup$ – David Jul 10 '16 at 21:08
  • $\begingroup$ The natural numbers are defined directly from the axiom of infinity of ZF. But Im not sure that the axiom of infinity implies archimedean property. I hope someone more qualified can answer your question. $\endgroup$ – Masacroso Jul 10 '16 at 21:33
  • $\begingroup$ @Masocroso: I think you are rather confused about the archimedean property. There is no connection between the archimedean property on ordered monoid and the ability to define sequences of elements of that monoid. $\endgroup$ – Rob Arthan Jul 10 '16 at 21:56
  • $\begingroup$ @David - could I think of the question as asking if the implication in the title also holds in the hyperreals? $\endgroup$ – Ben Blum-Smith Jul 10 '16 at 21:56

You did not say what exactly you mean by nested set theorem, but I will assume that you mean the nested interval property:

If $\{F_n\}$ is a sequence of closed bounded intervals such that $F_{n+1} \subseteq {F_n}$ for every $n \in \mathbb{N}$, then $\bigcap_{n=1}^\infty \ne \emptyset$.

This, combined with the ordered field axioms, does not imply order-completeness. A useful reference for this type of question is James Propp's Real analysis in reverse. On page 13 (of version 4) it says

The Nested Interval Property (18) does not imply completeness

and gives a number of references for counterexamples.

So you do need some extra hypothesis, such as the Archimedean property.


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