# Does Rolle's Theorem imply Dedekind completeness?

I think the answer to the title question is "yes", but Gerald Edgar, in his comment on Does antidifferentiability of continuous functions imply Dedekind completeness? , points out an article (actually a solution to a Monthly problem) in which M. J. Pelling claims to construct a non-Archimedean ordered field in which Rolle's Theorem holds; see http://www.jstor.org/pss/2321145 .

Since I'm working on an article for the Monthly ( http://jamespropp.org/reverse.pdf ) that claims to prove that the Mean Value Theorem (which follows from Rolle's Theorem) implies Dedekind completeness, I'd like to know who's right!

First, I'll show that the Mean Value Theorem implies the Constant Value Theorem. Then I'll show that the latter implies the Cut Property. Then I'll show that the Cut Property implies Dedekind completeness. And then one of you will tell me where my mistake is, or where Pelling's mistake is.

Step 1: Suppose $f$ is a continuous function on $[0,1]$ that is differentiable on $(0,1)$, with $f'(x)=0$ for all $x$ in $(0,1)$. We'd like to show that $f$ is constant. Suppose not; consider any $a > 0$ with $f(a) \neq f(0)$. Then (by the Mean Value Theorem) there must exist $c$ in $(0,a)$ with $f'(c) = (f(a)-f(0))/(a-0) \neq 0$, which is a contradiction. Hence the Constant Value Theorem holds.

Step 2: Suppose $A$ and $B$ are disjoint sets whose union is the whole ordered field, such that $a < b$ for all $a \in A$, $b \in B$. We'd like to show that there exists $c$ in the ordered field such that $a \leq c$ for all $a \in A$ and $b \geq c$ for all $b \in B$. Suppose not. Then for all $a \in A$ there exists $a' \in A$ with $a' > a$, and for all $b \in B$ there exists $b' \in B$ with $b' < b$, from which it follows that the indicator function of the set $A$ is continuous and differentiable everywhere, with derivative constantly 0. This contradicts the Constant Value Theorem. Hence the Cut Property holds.

Step 3: Let $S$ be a bounded non-empty subset of the ordered field. We'd like to show that $S$ has a least upper bound. Let $B$ be the set of upper bounds of $S$, and let $A$ be the complement of $B$. Then $A$ and $B$ satisfy the hypotheses of the Cut Property, so there exists $c$ such that $a \leq c$ for all $a \in A$ and $b \geq c$ for all $b \in B$. It is easy to check that $c$ is a least upper bound for $S$.

This argument seems fairly straightforward to me, so I'm inclined to think that the error must be in Pelling's more complicated argument (which I won't pretend to understand). Though perhaps it will turn out that Pelling and I mean different things when we say than an ordered field satisfies Rolle's theorem.

Can anyone shed light on this?

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This is just a random thought--I have not read through your proof--but the exactly what is meant by a "continuous" function might be an important point here. In particular, a non-Archimidean field often comes with its own topology, which I can imagine might not agree with the order topology if the field happens to be ordered. – Charles Staats Feb 7 '12 at 23:48

It turns out that Pelling and I mean different things by "Rolle's Theorem": his form of Rolle's Theorem treats only polynomials. I figured that out when I came across the following passage: "... Then $F$ is not real-closed since, e.g., $z^{1/2} \in L - F$, and it remains to show that Rolle's theorem is valid in $F$. By linear changes of variable if necessary it suffices to prove that if $p(x)$ is a polynomial [emphasis mine] over $F$ such that $p(0)=p(1)=0$ then $p'(x)$ has a root in the open interval $(0,1)$ in $F$." Case closed (I think).