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I am interested in learning a little more about finitism, currently about which I only know a few encyclopedic paragraphs.

I know that during some time, some mathematicians like Kronecker thought that finitism is the right choice, so I guess that an important theory such as calculus would somehow be obtained in such an axiom system.

So I have two questions along these lines:

1) Is there a construction of calculus within a finitist axiom system? If so, does it include the important theorems that are taught to a first year student, (like the extreme value theorem, and fundamental theorem of calculus, with an appropriate definition of function)? Are the proofs much more complicated than the standard calculus?

2) Could you give some fundemantal axioms, and define what a function means in such a system? I am especially curious about constructing some real numbers with a definition like this Wikipedia example: http://en.wikipedia.org/wiki/Constructivism_%28mathematics%29#Example_from_real_analysis , but I don't know what a function would mean.

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    $\begingroup$ Well, if you are a sufficiently finite finitist, then you reject calculus itself: math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/real.html $\endgroup$ Jul 25, 2010 at 15:05
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    $\begingroup$ @Qiaochu Yuan: That is not what is usually meant by "finitism", a finitist accepts existence of any natural number. The position that does not accept this is usually called "ultra-finitism". $\endgroup$
    – Kaveh
    Jul 25, 2010 at 16:53
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    $\begingroup$ Calculus can be interpreted as a scaling limit of discrete mathematics. From an ultrafinitistic point of view, you would start with functions defined on a finite grid, after scaling and averaging over cells that become a new grid point, you obtain smooth functions in the scaling limit. This scaling limit is taken at the end of calculations. So, this is analogous to how in ordinary math one avoids working with infinitesimals or infinite quantities by using limits. $\endgroup$ Nov 4, 2016 at 0:45

3 Answers 3

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The book Simpson, Stephen G. Subsystems of second order arithmetic. Perspectives in Logic. Cambridge University Press, Cambridge; Association for Symbolic Logic, ISBN: 978-0-521-88439-6 MR2517689 will tell you far more than you want to know about this topic. It explains exactly what assumptions have to be added to a basic finitisitic system to prove various common theorems of calculus. The idea is to start with a basic form of second order arithmetic equivalent in strength to primitive recursive arithmetic (which is what is sometimes meant by finitisitic mathematics) and show that theorems of calculus are equivalent over this weak system to various axioms (such as weak Konig's lemma). You can also check http://en.wikipedia.org/wiki/Reverse_mathematics for some details.

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It is not completely clear for me what is the intended meaning of "a finitist axiom system". AFAIK, Kronecker was not a finitist, but rather a semi-intuitionist. Do you mean something similar to Primitive Recursive Arithmetic (PRA) (which is considered by some experts to correspond to Hilbert's finitism?). Do you consider first-order Peano Arithmetic (PA) as a finitist axiom system?

If you mean a system that does not accept existence of infinite objects but only finite numbers/strings/..., then there are various approaches toward mathematical analysis, which would satisfy this condition. For example there is Markov/Russian School of constructivism, there are computability schools, ... . One important school which is completely compatible with classical mathematics is Bishop school, see books by Errett Bishop and Douglas Bridges.

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  • $\begingroup$ Thanks for your answer! I am not an expert so I don't know about PRA, but what I mean by a finitist system is exactly as you described in your second paragraph. So I think first order PA is a finitist system. I'm sure your pointers will be very helpful to my question. $\endgroup$
    – AgCl
    Jul 25, 2010 at 19:08
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    $\begingroup$ The problem is many finitist do not accept quantification over infinite domains, like natural numbers. Since PA includes induction over formulas with arbitrary first-order quantification, they do not accept it as finitist. If you are considering PA as acceptable, then Simpson's book mentioned by Richard is a good reference (though it is a logic book) which tries to find the weakest system containing PRA that one can prove a theorem. PRA is a very similar to PA. Basically, we have function symbols and definitions for all primitive recursive functions, and induction for quantifier-free formulas. $\endgroup$
    – Kaveh
    Jul 26, 2010 at 11:06
  • $\begingroup$ By the way, there is also a book by M. Beeson, "Foundations of Constructive Mathematics" which you may also like. It considers different constructive approaches to mathematics. $\endgroup$
    – Kaveh
    Jul 26, 2010 at 11:15
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Digital computers "do" calculus.

Digital computers are always used to find numerical solutions to problems in analytic mathematics. Unless the solution is by rare chance a rational number (0 or 1 perhaps) that is the only way. “The demand for continuous description was encouraged by the fact that the mathematician claims to be able to indicate simple continuous descriptions of some of his simple mental constructions… Physical dependences can always be approximated by this simple kind of functions (the mathematician calls them ‘analytical’, which means something like ‘they can be analysed’). But to assume that physical dependence is of this simple type, is a bold epistemological step, and ptobably sn inadmissible step.” (Erwin Schrödinger, 1951, “Science and Humanism.)

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    $\begingroup$ You should expand your answer or request that it be made a comment. $\endgroup$ Jan 31, 2016 at 11:55
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    $\begingroup$ There is software for 'doing' lots of different types of mathematics, but that doesn't entail a correspondingly finitist axiom system. $\endgroup$
    – Todd Trimble
    Jan 31, 2016 at 13:12
  • $\begingroup$ @ToddTrimble The fact that a finite state machine can do calculus implies that calculus can be re-interpreted in terms of only finitistic concepts. $\endgroup$ Nov 4, 2016 at 0:25
  • $\begingroup$ @CountIblis I don't consider "doing" calculus, whatever that means, as remotely the same as giving a committedly finitist axiom system on which to found the calculus. Why don't you spell out what you mean in an actual answer? $\endgroup$
    – Todd Trimble
    Nov 4, 2016 at 1:08
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    $\begingroup$ Indeed, this interpretation would imply all mathematics is finitist, since it is given by finite strings of symbols produced by finitely many inference rules. $\endgroup$
    – Jim Conant
    Nov 4, 2016 at 1:14

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