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How to show that Rolle's theorem, the Mean Value Theorem are equivalent to the least upper bound property?

I'm thinking of starting like this: Let F be an ordered field that does not satisfy the least upper bound property, and then deduce that F does not satisfy either Rolle's or MVT. But I'm not sure how to continue, any help please? thanks!

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Sorry what is the least upper bound property? – Jon Beardsley Mar 9 '11 at 2:34
The ordered field F satisfies the least upper bound property if every non-empty subset of F that is bounded above has a least upper bound. For example, the real numbers have the least upper bound property. – Lindsay Duran Mar 9 '11 at 2:38
It's not clear to me that these things should be equivalent. For example, the integers satisfy the least upper bound property (right?), but I don't get what Rolle's theorem or the MVT would even mean in this case. – Daenerys Naharis Mar 9 '11 at 2:44
Have you seen the wikipedia page? – PEV Mar 9 '11 at 2:54
@Joseph: presumably we're supposed to show these three properties are equivalent for an ordered field $F$. – Chris Eagle Mar 9 '11 at 3:08
up vote 8 down vote accepted

Suppose $F$ is an ordered field, and $A \subset F$ is bounded above but has no least upper bound. Define a function $f \colon F \to F$ by $f(x)=1$ if $x$ is an upper bound of $A$, $f(x)=0$ otherwise.

I claim that for every $x$, there is an open interval around $x$ on which $f$ is constant. If $x$ is an upper bound of $A$, then since $x$ is not a least-upper-bound there exists an upper bound $y$ of $A$ with $y < x$, and so $(y,x+1)$ works. If $x$ is not an upper bound of $A$, then there exists an element $a$ of $A$ such that $a>x$, and so $(x-1,a)$ works.

The existence of these intervals shows that $f$ is everywhere differentiable with derivative $0$. Since $f$ is not constant, this contradicts the mean value theorem. Thus the MVT implies the least-upper-bound property.

The fact that the least-upper-bound property implies Rolle's theorem, and that Rolle's theorem implies the MVT, is proved in every first course of analysis.

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I'd be grateful if you could tell me the source of that proof (I'm interested in the history of these results). – Bill Dubuque Mar 9 '11 at 3:35
@Bill: I just made it up now. I'm pretty sure I've seen the same function used to prove that the intermediate value theorem implies the least-upper-bound property, but I don't recall where. – Chris Eagle Mar 9 '11 at 3:40

The purpose of this answer is to provide some supplementary insight into Chris Eagle's answer and to show in particular that his proof of MVT $\implies$ LUB is a very natural one.

In $\S 4$ of this note, I discuss the notions of completeness and Dedekind completeness in linearly ordered sets in terms of the induced order topology. (Recall an ordered set is Dedekind complete if it satisfies LUB -- every nonempty subset which is bounded above has a least upper bound -- and is complete if it is Dedekind complete and has a minimal element and a maximal element.) In particular, I show that a linear order is dense and Dedekind complete iff it is connected in the order topology. Every ordering on a field (compatible with the field structure, of course) is dense, so we see that an ordered field satisfies LUB iff it is connected in the order topology.

(In fact Chris Eagle's answer gives a snappy proof of "connected $\implies$ LUB": if LUB does not hold, take a nonempty set $A$ which is bounded above but has no least upper bound. Then the set of upper bounds of $A$ is nonempty, proper, open and closed. The other implication is the one which is more familiar from real analysis, and is not quite so easy. In the above note, I take the perspective that a very natural proof is given using the technique of real induction.)

So if we want to show that MVT $\implies$ LUB (as Chris Eagle points out, LUB $\implies$ Rolle's Theorem $\implies$ MVT is part of the standard honors calculus / elementary real analysis curriculum), it is natural to go by contrapositive: if an ordered field $(F,+,\cdot,<)$ does not satisfy LUB, then it is not connected, i.e., there exists a nonconstant continuous function $f: F \rightarrow \{0,1\}$. It follows immediately from the definition of the order topology that for each $x \in F$, there exists an open interval $I$ containing $x$ such that $f|_I$ is constant. Thus $f$ is a nonconstant function which is everywhere differentiable with derivative identically zero, contradicting MVT.

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Thank you for the insight from topology point of view! It is really helpful – Lindsay Duran Mar 9 '11 at 18:07

First lets prove equivalence of the Mean Value Theorem (MVT) and Rolles theorem.

$\text{MVT}\iff \text{Rolles}$:

Clearly MVT implies Rolles. For the converse, consider a differentiable function $f$ on the interval $[a,b]$ with $$\frac{f(b)-f(a)}{b-a}=c.$$ Then set $g(x)=f(x)-c(x-a)$. Then $g(a)=f(a)$ and $g(b)=f(b)-(f(b)-f(a))=f(a)$ so we can apply rolles theorem to find a point such that $g'(x)=0$. Since $$g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}$$ we have proven the MVT.

Chris Eagle showed why the MVT implies the least upper bound property. Lets do the other direction:

$\text{Least Upper Bound}\Rightarrow \text{Rolles}$:

Suppose $f$ is differentiable on the interval $(a,b)$ with $f(a)=f(b)$, and continuous on $[a,b]$. Since it will be uniformly continuous, it attains a maximum and minimum in $[a,b]$. These cannot both be endpoints (otherwise it is constant), so assume WLOG the maximum occurs in $(a,b)$ at the point $\gamma$. Then look at the difference quotients.

In some neighborhood of $\gamma$, if $x\lt \gamma$ then $$ \frac{f(x)-f(\gamma)}{x-\gamma }\geq 0$$ and if $x\gt\gamma$ then $$\frac{f(x)-f(\gamma)}{x-\gamma}\leq 0$$ since $f(\gamma)$ is a maximum. Since $f$ is differentiable, the right and left hand limits must be equal. Hence $f'(\gamma)=0$ as desired.

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