A question about finitism
Haim Gaifman
hg17 at columbia.edu
Sun Feb 12 19:42:30 EST 2023
Dear Arnon,
You appeal to understanding—a vague, difficult concept.
Suppose I define a natural number as follows:
A natural number is either 0 or any number obtained from 0 by adding 1 a finite
number of times: 0, 0+1, 0+1+1,….
This is not a full definition since “a finite number of times”
has not been defined. Nonetheless it provides one with a certain understanding:
Consider the process of adding ones. At each finite stage of this process one gets a finite number, n, of the form 0+1+1+…+1.
We might however go further and imagine what comes after all these stages, an infinitary stage, at
which we get the finished totality consisting of all the natural numbers.
You seem to imply that without taking this last step we do not understand what the
natural numbers are. But this, I claim, is not true. It is merely a matter of certain habits. One can understand very well what
the natural numbers are, without assuming the completed entity of all the natural numbers. It is sufficient
that we understand the rule that generates the sequence. You can also see this by considering the following theory: Let ZF be Zermelo Fraenkel set theory;
and let ZF|* be obtained by replacing in it the axiom of infinity by its negation (which says that there are no infinite sets).
This theory will constitute a rather weak version of finitism. The strongest (and much more interesting) version is obtained by adopting
a suggestion of Bill Tait and using Skolem’s system PRA (Primitive Recursive Arithmetic) as the theory of potential, non-actual infinity.
Best, Haim
> On Feb 4, 2023, at 5:15 PM, Arnon Avron <aa at tauex.tau.ac.il> wrote:
>
> Dear Haim,
>
> I was never able to understand the coherence of `finitism'.
>
> For example, in your first reply to Vaughan you wrote:
>
> "Only structures based on proper initial segments of the natural numbers: {0, 1, 2,…, m} are accepted as legitimate, but for every m, if n > m, one accepts also the extension based on {0, 1, …., m, m+1,…., n}."
>
> First question: at least to me it seems that if one understands that X is an initial segment of the natural numbers, it means that somehow
> he understands that there is something that X s an initial segment of, so he understands that there is the collection of the natural numbers.
> So why pretending not to understand that collection?
>
> Second question: you explicitly wrote that "for every m, if n>m ...". But if I understood you correctly (almost certainly I did not) a finitist is not allowed
> to make claims about every m!
>
> And the final question: is there any way for a finitist to explain his principles (even to himself!)
> without violating these principles? I doubt it...
>
> Best regards,
>
> Arnon
>
>
>
>
>
>
>
> From: FOM <fom-bounces at cs.nyu.edu <mailto:fom-bounces at cs.nyu.edu>> on behalf of Haim Gaifman <hg17 at columbia.edu <mailto:hg17 at columbia.edu>>
> Sent: Sunday, January 29, 2023 5:03 AM
> To: Vaughan Pratt <pratt at cs.stanford.edu <mailto:pratt at cs.stanford.edu>>
> Cc: fom at cs.nyu.edu <mailto:fom at cs.nyu.edu> <fom at cs.nyu.edu <mailto:fom at cs.nyu.edu>>
> Subject: Re: Re: Mathematics with the potential infinite
>
> Dear Vaughan,
> Long time no hear no see, and it is very nice to hear from you.
> The restriction of subscribing only to potential infinities (which can be traced back to Aristoteles) is Hilbert’s so called finitist position; Abraham Robinson agrees with him. Only structures based on proper initial segments of the natural numbers: {0, 1, 2,…, m} are accepted as legitimate, but for every m, if n > m, one accepts also the extension based on {0, 1, …., m, m+1,…., n}. The functions and/or relations that come with these structures are the usual functions and/or relations of PA (Peano Arithmetic). Of course, the functions are partial functions , given the restrictions on the domain.
>
> PA, which is based on the standard model N of natural numbers, is much much… stronger than the theories
> that arise within the framework of potential infinity.
> One such interesting theory has been proposed by Skolem
> and is known as PRA for Primitive Recursive Arithmetic.
>
> Now your question, if I understand you correctly, asks for a way of describing an uncountable structure using only potential infinities.This would be impossible, unless you allow countable non-standard model for the theory linearly ordered groups.
>
> Best, Haim Gaifman
>
>> On Jan 28, 2023, at 3:13 AM, Vaughan Pratt <pratt at cs.stanford.edu <mailto:pratt at cs.stanford.edu>> wrote:
>>
>> My apologies for not having previously followed threads on this topic. However after seeing Stephen Simpson's message just now (Friday) it occurred to me to ask whether an uncountable set could be described using only potential infinities, for example the real numbers (R, *, 0, <=) as a linearly ordered group under addition, compatibly ordered in the sense that each of the group multiplication's two arguments is monotone: if x <= y then x*z <= y*z, and likewise for the right argument. (* = +.)
>>
>> Define a *geodesic* to be a nondegenerate linearly ordered group (G, *, 0, <=). (Although G is not assumed abelian, the linear order makes it abelian.) Examples include the integers, the dyadic rationals, every field between the rationals and the reals, and many non-Archimedean extensions thereof.
>>
>> Call a geodesic G *gapless* when (i) it is dense, and (ii) for every nonempty suborder (U, <=) of (G, <=) having no least element, and every nonempty suborder (L, <=) of (G, <=) with L < U and having no greatest element, such that there is at most one element of G between L and U; then there exists an element of G between L and U.
>>
>> I claim that every gapless geodesic is isomorphic to R with the above structure.
>>
>> (Proof outline: Take any element x of G with 0 < x and pair 0 and x with 0 and 1 in R. Pair the integers in R with the subgroup of G generated by x, cyclic and therefore abelian. Repeatedly divide the intervals in (n, n+1) in G into two equal parts and pair the results with the dyadic rationals in (0,1), a dense set. Pair each dyadic irrational q in R with the unique x given by the gaplessness condition for any L and U in G whose counterpart in R converges to q from each side. Lastly, G must be Archimedean or there would be an empty gap between the finite and infinite elements of G.)
>>
>> 1. Do these definitions, claims, and constructions meet the criteria for only potential infinities?
>>
>> 2. Can R be shown to be uncountable using only potential infinities?
>>
>> (Those familiar with Otto Hoelder's 1901 paper showing that every Archimedean linearly ordered group is isomorphic to some subgroup of R under addition, which may be anywhere between Z and R, may see some similarity of ideas in the above.)
>>
>> Vaughan Pratt
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