# Whether an ordered field $K$ equipped with the Nested Interval Property or The Cauchy Complete Property is an archimedean ordered field [closed]

The Nested Interval Property

For all sequences $$\{a_n\},\{b_n\}\in K^\mathbb N$$ satisfied that for every element $$\varepsilon\in K_{>0}$$ , there exists an $$N\in\mathbb N$$ such shat $$|a_n-b_n|<\varepsilon$$ for all $$n>N$$ and$$a_0\le a_1<\cdots\le a_n<\cdots\le b_n\le \cdots\le b_1\le b_0$$then there is $$c\in K$$ such that $$\bigcap_n[a_n,b_n]=\{c\}$$

The Cauchy Complete Property

For all $$\{a_n\}\in K^\mathbb N$$ satisfied forall $$\varepsilon\in K_{>0}$$, there is a $$N\in \mathbb N$$ such that $$|a_n-a_m|<\varepsilon$$ for all $$n,m>N$$ , then there is a $$c\in K$$ such that for all $$\varepsilon\in K_{>0}$$ there is a $$N\in \mathbb N$$ such that $$|a_n-c|<\varepsilon$$ for all $$n>N$$

The Archimedean Property

For all $$x\in K$$ and $$y\in K_{>0}$$ , there is $$n\in\mathbb N$$ such that $$n\cdot y>x$$ .

$$n\cdot y$$ is $$\underbrace{y+y+\cdots+y}_{n \ \text{times}}$$

I know an archimedean ordered field equipped with the Nested Interval Property or the Cauchy Complete Property is isomorphic to $$\mathbb R$$ and I'm wandering whether it can be proved except the Archimedean Property.

No, the Archimedian property is necessary. The ease of constructing a counterexample depends on your mathematical background.

Before I begin: (Cauchy) completeness is easily shown to be equivalent to the Nested Interval Property. I will refer to the two interchangeably as completeness.

At the level of intuition, the problem is that the decimal expansion of a real number is "only" as long as $$\mathbb{N}$$. There are analogues with longer "decimal sequences"Fn. 1, called the surreal numbers. Wikipedia's (linked) construction is quite good, and no more complicated than Rudin's construction of $$\mathbb{R}$$ in Principles of Mathematical Analysis.

The Archimedian property encodes the length of the decimals when it restricts $$n\in\mathbb{N}$$. Your definition of completenessFn. 2 does not; it depends on $$\epsilon\in K^+$$, which may be too small for decimals. It admits more examples than the Archimedian property: the simplest is the Levi-Civita field; the most complex (IMHO) is the aforementioned surreals.

If you've seen some mathematical logic, then the nonstandard reals $${}^*\mathbb{R}$$ are a slick counterexample. That $${}^*\mathbb{R}$$ is non-Archimedean is well-known: for any nonprincipal ultrafilter $$\mathcal{U}$$ and any $$r\in\mathbb{R}$$, $$(0,1,2,3,\dots)/\mathcal{U}>(r,r,r,\dots)/\mathcal{U}$$ But Cauchy Completeness is easily formalizable in first-order logic with quantification over $$\mathbb{N}$$, and so holds in $${}^*\mathbb{R}$$ by the transfer principle. (The transfer principle shows completeness for sequences indexed by $${}^*\mathbb{N}$$, but this implies standard completeness a fortiori.)

Fn. 1 Technically, Dedekind cuts.

Fn. 2 There are other definitions of completeness that "bake in" the reals; these are equivalent to Archimediacy.