# 3 - Analysis in reverse.
# Are the completeness axioms all equivalent?
A natural question we can ask is the reverse, if an ordered field $R$ is monotone complete, is it necessarily Dedekind complete? The answer is yes!
First we show if $R$ is monotone complete, then it must be Archimedean. $\blacktriangleright$ Suppose not, that there exists some $x \in R$ such that $x$ is not bounded by any positive integers. Then $x$ is an upper bound of the increasing sequence $a_{n}=n$. Since $R$ is monotone complete, $a_{n} \to a$ for some $a\in R$. But consider $\epsilon = \frac{1}{4}$, this means there exists infinitely many integers $n$ that lies in the interval $(a-\frac{1}{4},a+\frac{1}{4})$, forcing any two such integer with a difference of $|n-m| \le |n - a| + |m-a| < \frac{1}{4}+\frac{1}{4} = \frac{1}{2}$. But any two distinct integers have distance of at least $1$, a contradiction. Hence $R$ is Archimedean. $\blacksquare$
Next, with Archimedean, we show $R$ is monotone complete implies it is Dedekind complete. $\blacktriangleright$ Take a nonempty subset $A \subset R$ that is bounded above by some $b_{0} \in R$. Now take $a_{0} \in A$, and we can consider the interval $I_{0}=[a_{0},b_{0}]$. Note this closed interval contains an element of $A$, with the left end point $b_{0}$ an upper bound of $A$. Now cut this interval $I_{0}$ into two halves at the midpoint $\frac{a_{0}+b_{0}}{2}$, and denote $I_{1} = [a_{1},b_{1}]$ to be the right closed interval $[\frac{a_{0}+b_{0}}{2},b_{0}]$ if it contains an element of $A$, otherwise let $I_{1}$ be the left closed interval $[a_{0},\frac{a_{0}+b_{0}}{2}]$. Now we repeat the process, cut the interval $I_{n}$ into two halves at the midpoint $\frac{a_{n}+b_{n}}{2}$, and set $I_{n+1}=[a_{n+1},b_{n+1}]$ to be the right closed interval $[\frac{a_{n}+b_{n}}{2},b_{n}]$ if it contains an element of $A$, otherwise set $I_{n}$ to be the left closed interval $[a_{n}, \frac{a_{n}+b_{n}}{2}]$. Repeating this gives us a nested set of closed intervals $I_{0}\supset I_{1}\supset I_{2}\supset\cdots$, where the right end points $(b_{n})$ is a bounded below monotone decreasing sequence of upper bounds of $A$, and that each $I_{n}$ contains an element of $A$. Note also the length of the intervals $I_{n}$ is $b_{n}-a_{n} = \frac{1}{2^{n}}(b_{0}-a_{0})$, which converges to zero as $R$ is Archimedean. Note $b_{n} \to b$ by monotone completeness, and we claim $b = \sup A$. Indeed, $b$ is an upper bound of $A$: If to the contrary that some $a \in A$ is such that $a > b$, take $\epsilon = a-b$, and take $b_{n}$ such that $|b-b_{n}| < \epsilon$, so $b_{n} < b+\epsilon = a$. This means $a$ is outside of the interval $I_{n}= [a_{n},b_{n}]$, a contradiction. So $b$ is an upper bound of $A$. Next, $b$ is the least upper bound of $A$. Suppose to the contrary that $b'$ is an upper bound of $A$ such that $b' < b$. Then by Archimedean property, there exists integer $N$ such that $\frac{1}{2^{N}}(b_{0}-a_{0}) < b-b'$. Take this interval $I_{N}=[a_{N},b_{N}]$, and note $b \le b_{N}$. Note also $a_{N} > b'$ because $b_{N} = a_{N} + \frac{1}{2^{N}}(b_{0}-a_{0}) < a_{N} +b-b'$, which implies $a_{N} > b_{N}-b+b'\ge b'$. So $I_{N}=[a_{N},b_{N}]$ is disjoint from $b'$. But there exists an element $a\in A$ in the interval $I_{N}$, which $a \ge b_{N} > b'$, contradicting $b'$ is an upper bound of $A$. Hence $b = \sup A$, and hence $R$ is Dedekind complete! $\blacksquare$
Hence we showed, if $R$ is an ordered field, $$
R \text{ is monotone complete} \implies \begin{array}{l}
R \text{ is Archimedean } \\ \text{and Dedekind complete.}
\end{array}
$$In other words, we can use monotone completeness as an alternate characterization of the reals!
**Remark.** Naturally now we ask, does nested completeness imply Dedekind completeness? The answer is no but involves a tricky example of using Conway's surreal numbers. However, if one further assumes Archimedean, then yes, an Archimedean nested complete ordered field is indeed Dedekind complete.
**Remark.** Also, being Cauchy complete does not imply Dedekind complete either. What is missing is Archimedean property. In fact, if we assume an Archimedean ordered field, then the following are indeed equivalent: Dedekind complete, monotone complete, nested complete, Weierstrass complete, Bolzano complete, and Cauchy complete.
Interested reader should check out the article by James Propp: [Real analsis in reverse](https://faculty.uml.edu/jpropp/reverse.pdf).
One should also try to find or look up examples of ordered fields that fail these completeness axioms.
## Calculus theorems as axioms?
One can in fact take it further. Since we derive all the calculus results from the axiomatic definition of the reals, like intermediate value theorem, extreme value theorem, mean value theorem, intervals are topologically connected, etc. One can ask the following reverse question: If we desire an ordered field $R$ to do calculus on, such that your favorite calculus theorem $T$ must always hold, must $R$ be the reals?
Certainly there are ordered fields where calculus theorem fails, for example on $\mathbf{Q}$, we do not have intermediate value theorem.
Again, one can look at James Propp's paper.
But for a taste, let us say our favorite calculus theorem $T$ is intermediate value theorem. We shall show that if $R$ is an ordered field such that intermediate value theorem is observed always, then $R$ must be the reals!
We define the following.
Let $R$ be an ordered field. Let $f:R\to R$ be a function. We say $f$ is **continuous** at $p \in R$ if for every $\epsilon > 0$, there exists $\delta > 0$ such that $|x - p| < \delta$ implies $|f(x) - f(p)| < \epsilon$. And we say $f$ is continuous on $R$ if it is continuous at each $p \in R$.
Now suppose $R$ is an ordered field such that every continuous function $f$ on $R$ satisfies the **intermediate value property**: On any closed interval $[a,b]$, $a < b$, for any value $y$ strictly between $f(a)$ and $f(b)$, there exists some point $c \in (a,b)$ such that $f(c) = y$.
We claim such a field $R$ must be Dedekind complete.
Take $A$ any nonempty subset of $R$ that is bounded above. Suppose to the contrary that $\sup A$ does not exist. In this scenario $\max A$ does not exist.
Consider the function $$
f(x) = \begin{cases}
1 & \text{if \(x\) is an upper bound of \(A\)} \\
0 & \text{otherwise}
\end{cases}
$$
This function is then continuous on $R$!
Indeed, say we are at a point $p$ where $p$ is an upper bound of $A$, so $f(p)=1$. Take any $\epsilon > 0$. Since $A$ has no least upper bound, there exists another upper bound $p' < p$. Then for $\delta = p - p'$, if $|x - p| < \delta$ then $x$ is an upper bound of $A$, so $f(x) = 1$, and hence $|f(x)-f(p)| = 0 < \epsilon$.
And if we are at a point $p$ that is not an upper bound of $A$, then there is some $a \in A$ such that $a > p$. Now take $\delta = a - p$. Then for any $x \in R$ such that $|x - p| < \delta$, we have $|f(x)-f(p)| = |0-0| = 0 < \epsilon$. Hence $f$ is continuous on $R$.
But every continuous function in $R$ satisfies the intermediate value property, so on the interval $[a,u]$, where $a \in A$, $u$ an upper bound of $A$, take the number $\frac{1}{2}$ that is between $f(a)= 0$ and $f(u)=1$. But $\frac{1}{2}$ is not in the range of $f$, a contradiction!
Hence if $R$ is an ordered field where every continuous function satisfies the intermediate value property, $R$ must be Dedekind complete, and hence the reals!
**Exercise.** Adapted above to show: If $R$ is an ordered field such that every differentiable function satisfies the mean value theorem, then $R$ must be the reals. You could define what differentiable means on an ordered field, but the argument at the end would be clear why certain function is "differentiable", as it would be locally constant.
B / 7 2024