# Calculus in finitistic systems

I was just curious if there were some approaches to prove major theorems of calculus in finitistic systems like PRA?

Some related questions are, e.g.,

https://mathoverflow.net/questions/551/does-finite-math-need-the-axiom-of-infinity

Math without infinity

If all sets were finite, how could the real numbers be defined?

• You can do a surprising amount of calculus in the theory of a real-closed field, such as the field of algebraic real numbers. If you are interested see, van Den Dries's book Tame Topology and O-minimal structures or Bochnak, Coste and Roy's Real Algebraic Geometry. – Rob Arthan Nov 3 '13 at 16:14
• Thank you for the comment. Short question: which axiomatic system lies in basics of this approach? Can it be classified as finitistic? – Rubi Shnol Nov 4 '13 at 9:09
• The usual axiomatisation of a real-closed field says it is an ordered field such that every positive element is a square and every polynomial of odd degree has a root. The minimal model of this system is the field of algebraic reals and the members of this model all have finite representations ($x$ is represented by giving $f$ $a$ and $b$ where $x$ is the unique root of the polynomial $f$ with $a < x < b$ and where $a$, $b$ and the coefficients of $f$ are all rational. The field operations can all be carried out effectively. Does that help? – Rob Arthan Nov 4 '13 at 13:29
• I should have said that the theory of real-closed fields is actually much weaker than PA (in fact is is decidable). All the infinite sets that you can define are describable by finite data, e.g., the subsets of the field that you can define are finite unions of points and open intervals. – Rob Arthan Nov 4 '13 at 14:33

One issue here is the meaning of "finitistic system". $\mathsf{PRA}$ is usually considered a finitistic system, but every model of $\mathsf{PRA}$ is infinite. So the mere presence of an axiom of infinity is not an obstacle for a system to be finitistic.
Also, it is well known that if we let $\mathsf{ZFC}^i$ be $\mathsf{ZFC}$ with the axiom of infinity replaced by its negation, then $\mathsf{ZFC}^i$ is mutually interpretable with Peano arithmetic. Peano arithmetic is not usually considered a finitistic system, and so neither is $\mathsf{ZFC}^i$. Thus just removing the axiom of infinity does not make $\mathsf{ZFC}$ become finitistic.
In Simpson's book, quite a bit of calculus is formalized in a system $\mathsf{WKL}_0$. This system is not itself finitisitic, but it is conservative over $\mathsf{PRA}$ for $\Pi^0_2$ sentences. So, even though $\mathsf{WKL}_0$ is not finitistic, if it proves a sentence of the form that a finitist would recognize, that sentence is finitistically provable (in $\mathsf{PRA}$). This is the sense in which Simpson's book does some calculus in a way compatible with finitism.
To actually do calculus in $\mathsf{PRA}$ or another first-order arithmetic, it would be necessary to represent all the objects in question with natural numbers. The result would be something like a weak form of "Russian-style" computable analysis. That is the school of computable analysis in which all objects are coded by natural numbers.
• But ZF with no infinite sets interprets Peano arithmetic. Are you saying that Peano arithmetic is finitistic? In that case, you could move to the second-order system $\mathsf{ACA}_0$, which is conservative over Peano arithmetic for sentences of arithmetic, and where we can formalize the entire standard calculus curriculum. – Carl Mummert Nov 5 '13 at 13:08
• $\mathsf{ACA}_0$ is mutually interpretable with Peano arithmetic - that is how the conservation result is proved, actually, by giving an iterpretation of $\mathsf{ACA}_0$ into $\mathsf{PA}$. So, if you wish to consider Peano arithmetic finitistic, there is no harm in working in $\mathsf{ACA}_0$ instead. – Carl Mummert Nov 5 '13 at 21:30