Does the above non-Archimedean but ordered field satisfy Nested interval property?

Question

Consider the ordered non-Archimedean field $ \mathbb{R}(t)$, the field of rational function. My question is:

$ \text{Does the above non-Archimedean but ordered field satify Nested interval property?} $

Answer: The field of rational function $\mathbb{R}(t)$ is expressed as

$ \mathbb{R}(t)=\left\{\frac{p(t)}{q(t)}: p(t), q(t) \in \mathbb{R}[t], \ q(t) \neq 0 \right\}$

This field is totally ordered field but non-Archimedean. How to conclude about Nested interval property?

A totally ordered field $\Bbb F$ is said to have the Nested interval property if every decreasing sequence of closed and bounded intervals in $\Bbb F$ has a nonempty intersection.

Answer 1

$\mathbb{R}(t)$ does not satisfy the Nested interval property.

Proof 1: Let's consider the order in $\mathbb{R}(t)$ such that $0<t<a$ for every $a\in\mathbb{R}^+$. Then the closed intervals $[-x^n,x^n]$ ($n\in\mathbb{N}$) form a nest with intersection $\{0\}$. Notice that the formal power series $f(t)=\sum_{i=1}^\infty t^{i^2}$ is not in $\mathbb{R}(t)$ but the partial sums $s_n(t)=\sum_{i=1}^n t^{i^2}\in \mathbb{R}(t)$ converges to $f(t)$ in $\mathbb{R}((t))$ (proofs of these statements here).

Now consider the nest formed by intervals of the form $I_n=s_n(t)+[-x^{n^2},x^{n^2}]$ for $n\in\mathbb{N}$. Notice that $I_{n+1}\subset I_n$. If these intervals are considered in $\mathbb{R}((t))$ then its intersection will be $\{f(t)\}$. Therefore, if this nest is considered in $\mathbb{R}(t)$, then its intersection is empty.

Proof 2: When an ordered field satisfies the Nested interval property, then it is Cauchy-complete i.e. every Cauchy sequence is convergent in the order topology.

In fact, an ordered field is Cauchy complete whenever every nest of intervals with diameters tending to $0$ has nonempty intersection.

Notice that the field $\mathbb{R}((t))$ is the completion of $\mathbb{R}(t)$. Since $\mathbb{R}(t)$ is not Cauchy complete, it cannot satisfy the Nested interval property.

Answer 2

The nested interval property does not hold in $\mathbb{R}(t)$. Here is a proof, although it is probably not the simplest.

$\mathbb{R}(t)$ can be ordered in more than one way. I will assume that the ordering is the one where the polynomial $a_dt^d + a_{d+1}t^{d+1} + a_{d+2}t^{d+2} + \cdots$ is positive iff $a_d > 0$.

(It seems more usual to take the ordering in which a polynomial is positive iff its leading coefficient is positive; in that case, use the infinitesimal element $1/t$ in place of $t$ in what follows.)

The element $u = 1 + t$ has no square root in $\mathbb{R}(t)$, because if $(p(t)/q(t))^2 = 1 + t$, then $p(t)^2 = (1 + t)q(t)^2$, which is impossible, because the highest power of $t$ on the left is even and the highest power on the right is odd.

(In the case of the alternative ordering of $\mathbb{R}(t)$: if $(p(t)/q(t))^2 = 1 + 1/t$, then $tp(t)^2 = (1 + t)q(t)^2$, which is impossible, because the lowest power of $t$ on the left is odd and the lowest power of $t$ on the right is even.)

Consider the fate of an iterative process which attempts to compute a square root of $u$ in $\mathbb{R}(t)$.

Define a sequence $(x_n)_{n\geqslant0}$ by $x_0 = 1$ and $x_{n+1} = f(x_n)$, where: $$ f(x) = 1 + \frac{t}{1 + x} = \frac{u + x}{1 + x}. $$ By induction on $n$, we have $x_n \geqslant 1$ for all $n$. A simple calculation gives: $$ f(x)^2 - u = \frac{u^2 + 2ux + x^2 - u - 2ux - ux^2}{(1 + x)^2} = -t\frac{x^2 - u}{(1 + x)^2}, $$ whence, by another induction on $n$: $$ \left\lvert x_n^2 - u \right\rvert \leqslant \frac{t^{n+1}}{2^{2n}} \quad (n = 0, 1, 2, \ldots). $$ By another simple calculation: $$ f(x) - x = \frac{u + x - x - x^2}{1 + x} = -\frac{x^2 - u}{1 + x}, $$ whence, by induction on $n$ again, $$ \left\lvert x_{n+1} - x_n \right\rvert \leqslant \frac{t^{n+1}}{2^{2n+1}} \quad (n = 0, 1, 2, \ldots). $$ As for the order properties of the sequence $(x_n)_{n\geqslant0}$, first note that, by induction on $m$, using the results that have already been proved, $$ x_{2m}^2 < u < x_{2m+1}^2 \quad (m = 0, 1, 2, \ldots). $$ Then, a third simple calculation: $$ f^2(x) - x = \frac{u + \frac{u + x}{1 + x}}{1 + \frac{u + x}{1 + x}} - x = \frac{(u + 1)x + 2u}{2x + u + 1} - x = -2\frac{x^2 - u}{2x + u + 1}, $$ followed by yet another inductive argument, gives: $$ x_0 < x_2 < x_4 < \cdots < x_5 < x_3 < x_1. $$ Suppose, if possible, that there exists $x$ such that: $$ x \in \bigcap_{m=0}^\infty [x_{2m}, x_{2m+1}]. $$ Then, because $f$ is a decreasing function, we have: $$ x_{2m+2} = f(x_{2m+1}) \leqslant f(x) \leqslant f(x_{2m}) = x_{2m+1} \quad (m = 0, 1, 2, \ldots). $$ Therefore: $$ \left\lvert f(x) - x \right\rvert \leqslant \left\lvert x_{2m+1} - x_{2m} \right\rvert \leqslant \frac{t^{2m+1}}{2^{4m+1}} \quad (m = 0, 1, 2, \ldots). $$ Let $\left\lvert f(x) - x \right\rvert = p(t)/q(t)$, where $p(t), q(t) \in \mathbb{R}[t]$ and $q(t) \ne 0$. Then: $$ p(t) \leqslant \frac{t^{2m+1}q(t)}{2^{4m+1}} \quad (m = 0, 1, 2, \ldots). $$ This is impossible for strictly positive $p(t)$, if $2m + 1$ exceeds the lowest power of $t$ in $p(t)$. Therefore $p(t) = 0$; therefore $f(x) = x$; therefore $x^2 = u$. But we saw that such a value of $x$ does not exist in $\mathbb{R}(t)$, so we have a contradiction, proving that the intersection of the nested sequence of closed and bounded intervals $[x_{2m}, x_{2m+1}]$ is empty. $\square$

(In the case of the alternative ordering of $\mathbb{R}(t)$, we have, instead: $$ 2^{4m+1}t^{2m+1}p(t) \leqslant q(t), $$ which leads to a contradiction whenever $2m + 1$ exceeds the degree of $q(t)$. $\square$)

Answer 3

Assume the ordering with $t>a$ for all $a \in \mathbb R$. The ordering is: $$ \frac{p(t)}{q(t)} > 0 \quad\Longleftrightarrow\quad \text { there exists } x_0 \in \mathbb R \text { with } \frac{p(x)}{q(x)}>0 \text { for all } x > x_0 . $$ So my $t$ is $1/t$ in Chilote's answer.

Use closed bounded intervals $$ \left[n,\frac{1}{n}\;t\right]\qquad n=1,2,3,\dots $$ We claim $Q=\varnothing$, where $$ Q := \bigcap_{n=1}^\infty \left[n,\frac{1}{n}\;t\right] . $$

First, if $\frac{p(t)}{q(t)} \le 0$ then $\frac{p(t)}{q(t)} \notin Q$.
Let $\frac{p(t)}{q(t)} > 0$.
Then $\frac{p(t)}{q(t)} \ge n$ for all $n$ implies $\deg p > \deg q$.
Also $\frac{p(t)}{q(t)} \le \frac{1}{n}\,t$ for all $n$ implies $\deg p \le \deg q$.
Thus $\frac{p(t)}{q(t)} \notin Q$.