A question about finitism

[Kapantais Doukas]

Dear all,

In Skolem’s formal Arithmetic (and also in PRA and Hilbert’s finitary metamathematics – with variations) quantification is bounded. No quantifier runs through any infinite domain. Instead, we have schemata. For what used to be the claim “for all x, so and so”, we now have the schema “… so and so”. The idea being that the schema comes out true for every substitution instance for ‘…’. The schema itself is allowed into the formalism and its interpretation is the above.

QUESTION: Does the question “How many substitution instances are there?” makes any finitistic sense with regard to the metatheoretical natural language quantifier in the underlined phrase above?

With kind regards,
Doukas Kapantais
Research Centre for Greek Philosophy
Academy of Athens


From: FOM <fom-bounces at cs.nyu.edu> on behalf of Arnon Avron <aa at tauex.tau.ac.il>
Date: Sunday, 5 February 2023 - 5:05 PM
To: Haim Gaifman <hg17 at columbia.edu>, Vaughan Pratt <pratt at cs.stanford.edu>
Cc: fom at cs.nyu.edu <fom at cs.nyu.edu>, Arnon Avron <aa at tauex.tau.ac.il>
Subject: A question about finitism
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> on behalf of Haim Gaifman <hg17 at columbia.edu>
Sent: Sunday, January 29, 2023 5:03 AM
To: Vaughan Pratt <pratt at cs.stanford.edu>
Cc: fom at cs.nyu.edu <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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