Nested interval property does not solely imply the completeness

Motivated by a problematic exercise in an analysis textbook, I decided to prove the following proposition.

1. The statement

Theorem. There exists an ordered field $F$ such that it satisfies the nested interval property but does not satisfy the completeness axiom.

Before the proof, we should explain what a non-principal ultrafilter means. Given a set $X$, a subset $\mathcal{U}$ of $\mathcal{P}(X)$ is called an ultrafilter on $X$ if it satisfies the following four properties:^[1]

1. $\varnothing \notin \mathcal{U}$.
2. If $A \subset B \subset X$ and $A \in \mathcal{U}$, then $B \in \mathcal{U}$.
3. If $A, B \in \mathcal{U}$, then $A \cap B \in \mathcal{U}$.
4. If $A \subset X$, then either $A \in \mathcal{U}$ or $X \setminus A \in \mathcal{U}$. (Note: axioms 1 and 3 imply that $A$ and $X \setminus A$ cannot both be elements of $\mathcal{U}$.)

It is plain to observe that, for a given element $x \in X$, the family $$ \mathcal{U}_{x} = \{ A \subset X : x \in A \}$$ defines an ultrafilter. This type of ultrafilter is called principal. An ultrafilter which is not principal is called non-principal. The existence of a non-principal ultrafilter on an infinite set is guaranteed if we assume ZFC.

2. The constuction

Let $U$ be a non-principal ultrafilter on $\omega = \Bbb{N}$ and $\Bbb{R}^{*} = \Bbb{R}^{\omega} / U$ be the corresponding ultraproduct of countably many copies of $\Bbb{R}$. Clearly $\Bbb{R}^{*}$ is an ordered field, and in fact it is a model of the hyperreal number system. Let $$ I_n = [ [\mathrm{a}_{n}], [\mathrm{b}_{n}] ] $$ be a sequence of nested intervals in $\Bbb{R}^{*}$, where $[\mathrm{a}_{n}], [\mathrm{b}_{n}] \in \Bbb{R}^{*}$ are hyperreal numbers written in the equivalence relation notation and the real-valued sequences $\mathrm{a}_{n} = (a_{n,k})_{k=1}^{\infty}, \mathrm{b}_{n} = (b_{n,k})_{k=1}^{\infty}$ are the corresponding representatives. Now we modify these sequences as follows:

• Let $A_0 = \{ k : a_{1,k} \leq b_{1,k} \}$. Since $A_0 \in U$, the definitions $$ \tilde{\mathrm{a}}_{1} : \tilde{a}_{1,k} = \begin{cases} a_{1,k} & k \in A_0 \\ 0 & k \notin A_0 \end{cases} \quad \text{and} \quad \tilde{\mathrm{b}}_{1} : \tilde{b}_{1,k} = \begin{cases} b_{1,k} & k \in A_0 \\ 0 & k \notin A_0 \end{cases}, $$ yield another representatives of $\mathrm{a}_{1}$ and $\mathrm{b}_{1}$, respectively, as follows: $$ [\tilde{\mathrm{a}}_{1}] = [\mathrm{a}_{1}] \quad \text{and} \quad [\tilde{\mathrm{b}}_{1}] = [\mathrm{b}_{1}]. $$
• Let $n \geq 1$ and assume that we have chosen $\tilde{\mathrm{a}}_{1}, \cdots, \tilde{\mathrm{a}}_{n}, \tilde{\mathrm{b}}_{1}, \cdots, \tilde{\mathrm{b}}_{n}$ as follows:
  1. $ [\tilde{\mathrm{a}}_{j}] = [\mathrm{a}_{j}] $ and $ [\tilde{\mathrm{b}}_{j}] = [\mathrm{b}_{j}] $ for $j = 1, \cdots, n$.
  2. $ \tilde{a}_{i,k} \leq \tilde{a}_{j,k} \leq \tilde{b}_{j,k} \leq \tilde{b}_{i,k} $ for any $1 \leq i \leq j \leq n$ and any $k \geq 1$.
  Let $A_n = \{ k : \tilde{a}_{n,k} \leq a_{n+1,k} \leq b_{n+1,k} \leq \tilde{b}_{n,k} \}$. Since $$ [\tilde{\mathrm{a}}_{n}] = [\mathrm{a}_{n}] \leq [\mathrm{a}_{n+1}] \leq [\mathrm{b}_{n+1}] \leq [\mathrm{b}_{n}] = [\tilde{\mathrm{b}}_{n}], $$ we have $A_n \in U$. Then by defining $$ \tilde{\mathrm{a}}_{n+1} : \tilde{a}_{n+1,k} = \begin{cases} a_{n+1,k} & k \in A_n \\ \tilde{a}_{n,k} & k \notin A_n \end{cases} \quad \text{and} \quad \tilde{\mathrm{b}}_{n+1} : \tilde{b}_{n+1,k} = \begin{cases} b_{n+1,k} & k \in A_n \\ \tilde{b}_{n,k} & k \notin A_n \end{cases}, $$ we obtain the following two properties.
  1. $ [\tilde{\mathrm{a}}_{n+1}] = [\mathrm{a}_{n+1}] $ and $ [\tilde{\mathrm{b}}_{n+1}] = [\mathrm{b}_{n+1}] $.
  2. $ \tilde{a}_{i,k} \leq \tilde{a}_{n+1,k} \leq \tilde{b}_{n+1,k} \leq \tilde{b}_{i,k} $ for any $1 \leq i \leq n$ and any $k \geq 1$.

Finally, let $$ \mathrm{a} : a_{k} = \sup_{n} \tilde{a}_{n,k} = \lim_{n} a_{n,k} \quad \text{and} \quad \mathrm{b} : b_{k} = \inf_{n} \tilde{b}_{n,k} = \lim_{n} b_{n,k}. $$ Then clearly $$ [\mathrm{a}_{n}] = [\tilde{\mathrm{a}}_{n}] \leq [\mathrm{a}] \leq [\mathrm{b}] \leq [\tilde{\mathrm{b}}_{n}] = [\mathrm{b}_{n}] $$ and hence $ [[\mathrm{a}], [\mathrm{b}]] =: I \subset I_{n} $ for all $n$. This proves the nested interval property.

Since $\Bbb{R}^{*}$ is not order-isomorphic to $\Bbb{R}$, it follows that $\Bbb{R}^{*}$ is an ordered field which has the nested interval property yet lacks the completeness.

3. References

1. Ultrafilter/Formal definition - Wikipedia, the free encyclopedia.