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Let $R$ be an arbitrary Ordered Field.
Order Completeness Property: If $S \subset R$ is bounded above, then $\exists \ c \in R$ that is an upperbound of S and for every upperbound $b$ of S, we have $c \leq b.$

Nested Interval Property: If $I_1,I_2,...,I_n,...$ be a collection of nested closed intervals, i.e $I_1 \supseteq I_2 \supseteq ... \supseteq I_n \supseteq ...$ then $\bigcap_{1}^{\infty} I_i \neq \phi$.

Order Completeness $\implies$ Nested Interval Property.

This part is easy to prove.

Nested Interval Property $\implies$ Order Completeness.

Consider $A \underset{bdd.}\subseteq R$,
$b_0$: an upperbound of $A$,
$a_0$: not an upperbound of $A$.
$I_0:=[a_0,b_0] \implies I_0 \cap A \neq \phi$.
Let $c=(a_0+b_0)/2_R$, If $\exists \ a \in A$ with $c<a$ then choose $a_1=c$, $b_1=b_0$.
Otherwise choose $a_1=a_0$, $b_1=c$.
Proceeding this way one can construct a sequence of closed nested intervals $I_0 \supseteq I_1 \supseteq ... \supseteq I_n \supseteq ...$, in which $I_n=[a_n,b_n]$ where
$a_n$: not an upperbound of A,
$b_n$: upperbound of A.

Now $\operatorname{diam}(I_n)=\frac{b_0-a_0}{2^n}$. Assuming R enjoys Archemedian Property I can conclude $\operatorname{diam}(I_n) \to 0$
$\implies$ $\bigcap_{1}^{\infty} I_i=\{s\}$.

Remains to show that $s=\operatorname{lub}A$.

If there exists $a \in A$ such that $s<a$ since $\operatorname{diam}(I_n) \to 0$ there exists $p \in \mathbb{N}$ such that $\operatorname{diam}(I_p)<a-s$. Now $s \in I_p$ and $a \notin I_p \implies b_p<a$, contradiction to the fact that $b_p$ is an upperbound of $A$.

Hence $s$ is an upperbound of A.

Again let $l<s$ then there exists $q \in \mathbb{N}$ such that $\operatorname{diam}(I_q)<s-l$.
We have $s \in I_q$ then $l \notin I_q$ $\implies l<a_q$. Now there exists $b \in A$ with $l<a_q<b \implies l$ is not an upperbound of A.

Hence there exists $s \in R$ such that $s=\operatorname{lub}A$.

Now my question is If I drop the assumption of Archimedian Property can we still conclude this result?

Saikat
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    If you drop the Archimedean property, do you keep the nested interval property verbatim, or do you change it to the intersection of all non-empty chains of non-empty closed intervals being non-empty? Or to something else? –  Jan 05 '21 at 09:55
  • I keep it the same as I stated above. Will, there be any problem in keeping the definition the same after dropping Archimedean Property? – Saikat Jan 05 '21 at 10:17
  • note that the intervals in the nested interval property must have endpoints in $R$ - I am not sure if this is implied by the expression "closed intervals" – nombre Jan 05 '21 at 13:12

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Certain ordered fields $R$ are $\mathbf{\aleph_1}$-saturated as linear orders. This means that given countable subsets $A,B\subset R$ with $A<B$, there is $x \in R$ with $A<x<B$. Such an ordered field is not archimedean, since there is $x \in R$ with $\mathbb{N}<x<\varnothing$.

But such a field satisfies the nested interval property because writing $I_n=[a_n,b_n]$ for each $n \in \mathbb{N}$ and considering $x \in R$ with $\{a_n: n \in \mathbb{N}\}<x<\{b_n: n \in \mathbb{N}\}$, one has $x \in \bigcap \limits_{n \in \mathbb{N}} I_n$ (in the non-degenerate case where $(a_n)_{n \in \mathbb{N}}$ or $(b_n)_{n \in \mathbb{N}}$ is not stationnary).

So on its own the nested interval property does not imply completeness.

An example of an $\aleph_1$-saturated ordered field is the field $\mathbf{No}(\omega_1)$ of surreal numbers of countable length.

nombre
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  • Does $A < B$ means for any $x \in A$, $x<y$ for all $y \in B$? – Saikat Jan 05 '21 at 14:41
  • Does the set of hyperreal numbers have this property? I mean it having Nested Interval Property and not having Order Completeness? – Saikat Jan 05 '21 at 15:43
  • @Saikat Yes and yes. And $A<x<B$ means that $A<{x}$ and ${x}<B$. Fields of hyperreal numbers constructed using free ultrafilters on $\mathbb{N}$ are $\aleph_1$-saturated, so they have the nested interval property and are not complete. The notion of $\aleph_1$-saturation appears in a more general context in model theory. – nombre Jan 05 '21 at 19:56
  • Can you tell me what is a free ultrafilter? – Saikat Jan 06 '21 at 06:47
  • I read about ultrafilters on wikipedia. You wrote Hyperreals constructed using ultrafilter on $\mathbb{N}$ has this property. My question: In whatever way I construct Hyperreals I think it must have the above mentioned property. Am I correct on this? By "this" I mean it having NIP and not having order completeness. – Saikat Jan 06 '21 at 06:54
  • For order completeness yes, as long as $\mathbb{R}$ itself is not considered a hyperreal field. I wrote this because there are different notions of what a "hyperreal field" is. If hyperreal fields are those of Dale and Woodin, then they do satisfy the NIP, in fact they are $\aleph_1$-saturated. According to other authors, for instance S. Shelah and V. Benci, hyperreal fields are characterized by the fact that they satisfy a transfer principle, in which case I am not sure they need be $\aleph_1$-saturated. – nombre Jan 06 '21 at 08:45
  • I am a beginning to study this. Can you tell me where can I learn about N_1 saturation? I am completely new to this topic. – Saikat Jan 06 '21 at 08:53
  • Saturation is a basic notion in model theory, which you can learn about in classical textbooks. For instance David Marker's Model Theory: An Introduction should not be too difficult to find. In the case of dense linear orders without endpoints, or similarly real-closed fields, the notion is much simplified and coincides with that which I gave. I don't have very nice references for this more specific context, but the notion can be traced back to Hausdorff, and you can look around for it by searching for the term $\eta$-set (eta set). – nombre Jan 06 '21 at 09:02