# Does antidifferentiability of continuous functions imply Dedekind completeness?

Let $R$ be an ordered field, and let $I$ be {$x \in R: a < x < b$} for some $a < b$ in $R$. Define notions of $R$-continuity and $R$-differentiability for functions $f : I \rightarrow R$ by replacing every reference to the real numbers in the standard real-analysis definitions by a reference to $R$. (E.g., the quantification "For all (real) $\epsilon > 0$" is replaced by quantification over all $\epsilon \in R$ satisfying $\epsilon > 0_R$, where $0_R$ is the additive identity in $R$.)

Say that $R$ has the all-continuous-functions-on-closed-bounded-intervals-have-antiderivatives property (let's call it "The Property" for short) iff for every $R$-continuous function $f : I \rightarrow R$ there exists an $R$-differentiable function $F : I \rightarrow R$ with $F'=f$.

Can a non-archimedean ordered field have The Property? In my article-in-progress "Real Analysis in Reverse" ( http://jamespropp.org/reverse.pdf ), I made the mistake (on page 14) of saying that an article of Schikhof gives an affirmative answer, but as Matt Baker pointed out to me, my understanding of Schikhof's article was incorrect.

It's not too hard to show that every archimedean ordered field with The Property is Dedekind complete (see pages 15 and 16), so if the answer to the question in the preceding paragraph is "no", then The Property is equivalent to Dedekind completeness. Hence the title of this posting.

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This reminds me vaguely of an old Monthly problem ... 5861, See the solution by M. Pelling, where he constructs an unusual nonarchimedean ordered field. jstor.org/pss/2321145 ... In that example, Rolle's Theorem holds but not every positive element has a square-root. – Gerald Edgar Feb 7 '12 at 5:11
Thanks, Gerald! But I am not sure that Pelling's argument can be correct; doesn't Rolle's Theorem imply completeness? I can't explain the implication in just 600 characters (that's the length-limit for comments), but I'll try to say more about this below. If there's no mistake in my argument, then there must be something wrong with Pelling's. I'd like to know which! – James Propp Feb 7 '12 at 22:24
It seemed simplest to start a new MathOverflow thread on the subject of Pelling's argument and mine, so I did: mathoverflow.net/questions/87848/… . But I am troubled by the fact that only people with access to JSTOR (or to a copy of the February 1981 issue of the Monthly) will be able to read Pelling's argument. I don't know of a good solution to this problem. What constitutes fair use? Are there precedents for this? – James Propp Feb 7 '12 at 23:00
Subsequent paper ... full text, open access ... projecteuclid.org/… – Gerald Edgar Feb 8 '12 at 15:10