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Mathematicians who are finitists.
#91
RE: Mathematicians who are finitists.
Why not accept Rosen's definition (page 127), "A set is said to be infinite if it is not finite." Seems reasonable to me, especially in light of Cantor's diagonalization proof (page 183) that the cardinality of some infinite sets is greater than others.
And without delay Peter went quickly out of the synagogue (assembly) and went unto the house of Marcellus, where Simon lodged: and much people followed him...And Peter turned unto the people that followed him and said: Ye shall now see a great and marvellous wonder. And Peter seeing a great dog bound with a strong chain, went to him and loosed him, and when he was loosed the dog received a man's voice and said unto Peter: What dost thou bid me to do, thou servant of the unspeakable and living God? Peter said unto him: Go in and say unto Simon in the midst of his company: Peter saith unto thee, Come forth abroad, for thy sake am I come to Rome, thou wicked one and deceiver of simple souls. And immediately the dog ran and entered in, and rushed into the midst of them that were with Simon, and lifted up his forefeet and in a loud voice said: Thou Simon, Peter the servant of Christ who standeth at the door saith unto thee: Come forth abroad, for thy sake am I come to Rome, thou most wicked one and deceiver of simple souls. And when Simon heard it, and beheld the incredible sight, he lost the words wherewith he was deceiving them that stood by, and all of them were amazed. (The Acts of Peter, 9)
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#92
RE: Mathematicians who are finitists.
(July 9, 2019 at 1:34 pm)A Toy Windmill Wrote:
(July 9, 2019 at 9:56 am)polymath257 Wrote: After thinking a bit, this does not quite prove the convergence since you need to show that
|x_n^2 -2|<eps for ALL n>=p where p is that smallest exponent.

Is there a finitistic definition that gets around this issue?
Yes, you're right, but that use of a quantifier is still legal.

n >= p implies that |x_n^2 - 2| <= 1/2^n <= 1/2^p < ε.

OK, so this particular unbounded quantifier is allowed? Why is that? Are quantifiers of the form
for all n>=p, P(n)
allowed?
If so, take p=1 and we have quantification over all of N.

(July 9, 2019 at 2:29 pm)Jehanne Wrote: Why not accept Rosen's definition (page 127), "A set is said to be infinite if it is not finite."  Seems reasonable to me, especially in light of Cantor's diagonalization proof (page 183) that the cardinality of some infinite sets is greater than others.

A good definition, but in a purely finitistic world, there are no infinite sets. We cannot even talk about the set of natural numbers, let alone the set of real numbers, so the diagonalization argument can't even be started.

Once you accept that there *is* a set of real numbers, then you can do the Cantor trick. But it is an *axiom* that there even exists an infinite set. Without that axiom (which the finitists reject), there may only be finite sets.
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#93
RE: Mathematicians who are finitists.
(July 10, 2019 at 4:55 am)polymath257 Wrote: OK, so this particular unbounded quantifier is allowed? Why is that? Are quantifiers of the form
for all n>=p, P(n)
allowed?
Quantifiers are allowed if they're all outermost. Such quantifiers are then dropped, and you're left with a statement that just has free variables.

When every statement in every proof has this form, you can treat them all as schema or templates. That means that, for my proof above, if someone has concrete numerals for the numerator and denominator of ε, I can plug them in and run that proof down to applied axiom schemes for primitive recursive arithmetic.

I can't do this if quantifiers ever appear left of an implication, as in, say, the principle of mathematical induction:

for any property P, if P holds of 0, and holding of n implies holding of (n+1), then P holds for all natural numbers.

Primitive recursive arithmetic cannot have this as an axiom, because the quantifier position is illegal. Instead, it takes induction as an inference rule.
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#94
RE: Mathematicians who are finitists.
(July 10, 2019 at 4:55 am)polymath257 Wrote: Once you accept that there *is* a set of real numbers, then you can do the Cantor trick. But it is an *axiom* that there even exists an infinite set. Without that axiom (which the finitists reject), there may only be finite sets.

Without axioms, there is no mathematics.
And without delay Peter went quickly out of the synagogue (assembly) and went unto the house of Marcellus, where Simon lodged: and much people followed him...And Peter turned unto the people that followed him and said: Ye shall now see a great and marvellous wonder. And Peter seeing a great dog bound with a strong chain, went to him and loosed him, and when he was loosed the dog received a man's voice and said unto Peter: What dost thou bid me to do, thou servant of the unspeakable and living God? Peter said unto him: Go in and say unto Simon in the midst of his company: Peter saith unto thee, Come forth abroad, for thy sake am I come to Rome, thou wicked one and deceiver of simple souls. And immediately the dog ran and entered in, and rushed into the midst of them that were with Simon, and lifted up his forefeet and in a loud voice said: Thou Simon, Peter the servant of Christ who standeth at the door saith unto thee: Come forth abroad, for thy sake am I come to Rome, thou most wicked one and deceiver of simple souls. And when Simon heard it, and beheld the incredible sight, he lost the words wherewith he was deceiving them that stood by, and all of them were amazed. (The Acts of Peter, 9)
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#95
RE: Mathematicians who are finitists.
(July 10, 2019 at 7:55 am)Jehanne Wrote:
(July 10, 2019 at 4:55 am)polymath257 Wrote: Once you accept that there *is* a set of real numbers, then you can do the Cantor trick. But it is an *axiom* that there even exists an infinite set. Without that axiom (which the finitists reject), there may only be finite sets.

Without axioms, there is no mathematics.

Yes, precisely. So the question is *which* axioms. How do we choose the axioms? And it seems different people have different opinions on how to choose.
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#96
RE: Mathematicians who are finitists.
(July 10, 2019 at 8:34 am)polymath257 Wrote:
(July 10, 2019 at 7:55 am)Jehanne Wrote: Without axioms, there is no mathematics.

Yes, precisely. So the question is *which* axioms. How do we choose the axioms? And it seems different people have different opinions on how to choose.

Ultimately, consensus among experts. I doubt that finitism even has a chapter in any number theory, abstract alebgra or complex analysis textbook, but, I admit that I could be wrong on this one.
And without delay Peter went quickly out of the synagogue (assembly) and went unto the house of Marcellus, where Simon lodged: and much people followed him...And Peter turned unto the people that followed him and said: Ye shall now see a great and marvellous wonder. And Peter seeing a great dog bound with a strong chain, went to him and loosed him, and when he was loosed the dog received a man's voice and said unto Peter: What dost thou bid me to do, thou servant of the unspeakable and living God? Peter said unto him: Go in and say unto Simon in the midst of his company: Peter saith unto thee, Come forth abroad, for thy sake am I come to Rome, thou wicked one and deceiver of simple souls. And immediately the dog ran and entered in, and rushed into the midst of them that were with Simon, and lifted up his forefeet and in a loud voice said: Thou Simon, Peter the servant of Christ who standeth at the door saith unto thee: Come forth abroad, for thy sake am I come to Rome, thou most wicked one and deceiver of simple souls. And when Simon heard it, and beheld the incredible sight, he lost the words wherewith he was deceiving them that stood by, and all of them were amazed. (The Acts of Peter, 9)
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#97
RE: Mathematicians who are finitists.
(July 10, 2019 at 8:19 pm)Jehanne Wrote:
(July 10, 2019 at 8:34 am)polymath257 Wrote: Yes, precisely. So the question is *which* axioms. How do we choose the axioms? And it seems different people have different opinions on how to choose.

Ultimately, consensus among experts.  I doubt that finitism even has a chapter in any number theory, abstract alebgra or complex analysis textbook, but, I admit that I could be wrong on this one.

Well, I wouldn't expect to see much about finitism in such textbooks. I have, however, seen good discussions in introductory books on axiomatic set theory and the issues come up extensively in proof theory. Kunen's book on the Foundations of Mathemtics talks quite a bit about finitism, formalism, and Platonism in the context of introductory set theory. The point is that the meta-theory used to discuss proofs tends to be finitistic by nature.
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#98
RE: Mathematicians who are finitists.
(July 11, 2019 at 2:52 am)polymath257 Wrote:
(July 10, 2019 at 8:19 pm)Jehanne Wrote: Ultimately, consensus among experts.  I doubt that finitism even has a chapter in any number theory, abstract alebgra or complex analysis textbook, but, I admit that I could be wrong on this one.

Well, I wouldn't expect to see much about finitism in such textbooks. I have, however, seen good discussions in introductory books on axiomatic set theory and the issues come up extensively in proof theory. Kunen's book on the Foundations of Mathemtics talks quite a bit about finitism, formalism, and Platonism in the context of introductory set theory. The point is that the meta-theory used to discuss proofs tends to be finitistic by nature.

Fact is that most professional mathematicians accept ZFC and the proofs of Cantor, which was the whole point of my OP. I don't think that it is an either/or proposition -- one can accept ZFC while having intelligent conversations about finitism; without having to reject the former while working within the confines of the latter.
And without delay Peter went quickly out of the synagogue (assembly) and went unto the house of Marcellus, where Simon lodged: and much people followed him...And Peter turned unto the people that followed him and said: Ye shall now see a great and marvellous wonder. And Peter seeing a great dog bound with a strong chain, went to him and loosed him, and when he was loosed the dog received a man's voice and said unto Peter: What dost thou bid me to do, thou servant of the unspeakable and living God? Peter said unto him: Go in and say unto Simon in the midst of his company: Peter saith unto thee, Come forth abroad, for thy sake am I come to Rome, thou wicked one and deceiver of simple souls. And immediately the dog ran and entered in, and rushed into the midst of them that were with Simon, and lifted up his forefeet and in a loud voice said: Thou Simon, Peter the servant of Christ who standeth at the door saith unto thee: Come forth abroad, for thy sake am I come to Rome, thou most wicked one and deceiver of simple souls. And when Simon heard it, and beheld the incredible sight, he lost the words wherewith he was deceiving them that stood by, and all of them were amazed. (The Acts of Peter, 9)
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#99
RE: Mathematicians who are finitists.
(July 11, 2019 at 9:30 am)Jehanne Wrote:
(July 11, 2019 at 2:52 am)polymath257 Wrote: Well, I wouldn't expect to see much about finitism in such textbooks. I have, however, seen good discussions in introductory books on axiomatic set theory and the issues come up extensively in proof theory. Kunen's book on the Foundations of Mathemtics talks quite a bit about finitism, formalism, and Platonism in the context of introductory set theory. The point is that the meta-theory used to discuss proofs tends to be finitistic by nature.

Fact is that most professional mathematicians accept ZFC and the proofs of Cantor, which was the whole point of my OP.  I don't think that it is an either/or proposition -- one can accept ZFC while having intelligent conversations about finitism; without having to reject the former while working within the confines of the latter.

Philosophically, I am a formalist. You tell me which axioms to work with, and I will work with them.

I *prefer* a set of axioms that has classes and a single choice function for all sets, but that's just me. I can live with ZFC. Smile

That said, working in the hereditarily finite sets has advantages in some contexts.
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RE: Mathematicians who are finitists.
(July 11, 2019 at 9:30 am)Jehanne Wrote: Fact is that most professional mathematicians accept ZFC and the proofs of Cantor, which was the whole point of my OP. I don't think that it is an either/or proposition -- one can accept ZFC while having intelligent conversations about finitism; without having to reject the former while working within the confines of the latter.
Yes, we can. Please try to do so next time, without making silly comparisons to creationism.
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