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Mathematicians who are finitists.
#41
RE: Mathematicians who are finitists.
It's also interesting that it is possible to adopt a 'countable only' philosophy that allows *countably* infinite sets, but not uncountable ones.

This system has the advantage of allowing construction of specific real numbers even though it does not allow the construction of the set of *all* real numbers. Also, it is possible to do a lot of the work with Turing machines and proof theory without the circumlocutions required in a purely finitist system.

The biggest problem with the 'countable only' system is that it does NOT allow the construction of the power set of a set: the set of all subsets of a set. All the rest of the standard axioms are satisfied, however, and this allows a great deal of math to be done.

In contrast, the finitist system uses all of the standard axioms except the axiom of infinity (which assumes the existence of an infinite set).
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#42
RE: Mathematicians who are finitists.
@polymath257 what is the english name of this axiom: "Consider two convergent sequences A(n), B(n) with n belonging to the natural numbers and a sequence of intervals where [A(n), B(n)] is a superset of [A(n+1),B(n+1)] for all n. Then, there exists one and only one real number c that belongs to all the intervals, same as saying the intersection of these intervals is not an empty set"

I really dunno the english name for this.
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#43
RE: Mathematicians who are finitists.
(April 15, 2019 at 12:32 pm)LastPoet Wrote: @polymath257 what is the english name of this axiom: "Consider two convergent sequences A(n), B(n) with n belonging to the natural numbers and a sequence of intervals where [A(n), B(n)] is a superset of [A(n+1),B(n+1)] for all n. Then, there exists one and only one real number c that belongs to all the intervals, same as saying the intersection of these intervals is not an empty set"

I really dunno the english name for this.

It is one of the many results due to Cantor and often carries his name.

Some care is required, however. As stated, the hupotheses show the intersection is non-empty, but they do NOT show that there is only one point in common to all. For that, you need, in addition, that B(n)-A(n)-->0.

For example, if A(n)=-1/n and B(n)=1+(1/n), then the intervals [A(n),B(n)] satisfy your hypotheses, but the intersection is the whole interval [0,1].

The result you stated is usually associated these days with compactness and it is in the subject of topology that it gets its full expression.
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#44
RE: Mathematicians who are finitists.
(April 15, 2019 at 5:06 pm)polymath257 Wrote:
(April 15, 2019 at 12:32 pm)LastPoet Wrote: @polymath257 what is the english name of this axiom: "Consider two convergent sequences A(n), B(n) with n belonging to the natural numbers and a sequence of intervals where [A(n), B(n)] is a superset of [A(n+1),B(n+1)] for all n. Then, there exists one and only one real number c that belongs to all the intervals, same as saying the intersection of these intervals is not an empty set"

I really dunno the english name for this.

It is one of the many results due to Cantor and often carries his name.

Some care is required, however. As stated, the hupotheses show the intersection is non-empty, but they do NOT show that there is only one point in common to all. For that, you need, in addition, that B(n)-A(n)-->0.

For example, if A(n)=-1/n and B(n)=1+(1/n), then the intervals [A(n),B(n)] satisfy your hypotheses, but the intersection is the whole interval [0,1].

The result you stated is usually associated these days with compactness and it is in the subject of topology that it gets its full expression.

I apologise for my broken English (I am Portuguese) and I wrote that on mobile, so I forgot to add in the hypothesis that both A(n) and B(n) converge to c (it is a pain to use brackets and other symbols on the devices) .

Its rougly translated as the principle of "fitting" but that doesn't sound good O.o;

Perhaps I should read some books in english to better know the math vocabulary in the language.
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#45
RE: Mathematicians who are finitists.
(April 16, 2019 at 12:58 pm)LastPoet Wrote:
(April 15, 2019 at 5:06 pm)polymath257 Wrote: It is one of the many results due to Cantor and often carries his name.

Some care is required, however. As stated, the hupotheses show the intersection is non-empty, but they do NOT show that there is only one point in common to all. For that, you need, in addition, that B(n)-A(n)-->0.

For example, if A(n)=-1/n and B(n)=1+(1/n), then the intervals [A(n),B(n)] satisfy your hypotheses, but the intersection is the whole interval [0,1].

The result you stated is usually associated these days with compactness and it is in the subject of topology that it gets its full expression.

I apologise for my broken English (I am Portuguese) and I wrote that on mobile, so I forgot to add in the hypothesis that both A(n) and B(n) converge to c (it is a pain to use brackets and other symbols on the devices) .

Its rougly translated as the principle of "fitting" but that doesn't sound good O.o;

Perhaps I should read some books in english to better know the math vocabulary in the language.

Need to learn Mathanese! Big Grin
If you get to thinking you’re a person of some influence, try ordering somebody else’s dog around.
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#46
RE: Mathematicians who are finitists.
(April 16, 2019 at 12:58 pm)LastPoet Wrote:
(April 15, 2019 at 5:06 pm)polymath257 Wrote: It is one of the many results due to Cantor and often carries his name.

Some care is required, however. As stated, the hupotheses show the intersection is non-empty, but they do NOT show that there is only one point in common to all. For that, you need, in addition, that B(n)-A(n)-->0.

For example, if A(n)=-1/n and B(n)=1+(1/n), then the intervals [A(n),B(n)] satisfy your hypotheses, but the intersection is the whole interval [0,1].

The result you stated is usually associated these days with compactness and it is in the subject of topology that it gets its full expression.

I apologise for my broken English (I am Portuguese) and I wrote that on mobile, so I forgot to add in the hypothesis that both A(n) and B(n) converge to c (it is a pain to use brackets and other symbols on the devices) .

Its rougly translated as the principle of "fitting" but that doesn't sound good O.o;

Perhaps I should read some books in english to better know the math vocabulary in the language.

It's closest to Cantor's Intersection Theorem.

In this, if X is a complete metric space and C_n are closed subspaces with C_{n+1} a subset of C_n and if diam(C_n )-->0, then the intersection of the C_n has exactly one point.

In your case, it is better to assume B(n)-A(n)-->0 than to assume they both converge to the same point (that is part of what follows).
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#47
RE: Mathematicians who are finitists.
Finitism is stricter than simply assuming there are no infinite sets. The finitist goes so far as to say that one cannot even quantify over the natural numbers! The most cited formal system which tries to capture what limited reasoning is left is "Primitive Recursive Arithmetic."

The finitist will accept that "n + 1 > n", but will regard this as a schema. to be filled in with particular numbers that they can muster.
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#48
RE: Mathematicians who are finitists.
(June 29, 2019 at 5:55 pm)A Toy Windmill Wrote: Finitism is stricter than simply assuming there are no infinite sets. The finitist goes so far as to say that one cannot even quantify over the natural numbers! The most cited formal system which tries to capture what limited reasoning is left is "Primitive Recursive Arithmetic."

The finitist will accept that "n + 1 > n", but will regard this as a schema. to be filled in with particular numbers that they can muster.

Well, that does depend on the brand of finitism you subscribe to. Yours is one of the more extreme versions.
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#49
RE: Mathematicians who are finitists.
Math Duel! Blackboard erasers at 30 Yards!

Big Grin

Seriously though, I like seeing this difference. It's like Cosmology. People approach it from different directions, and all kinds of new information comes from the competing theories.
If you get to thinking you’re a person of some influence, try ordering somebody else’s dog around.
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#50
RE: Mathematicians who are finitists.
(June 29, 2019 at 6:44 pm)polymath257 Wrote:
(June 29, 2019 at 5:55 pm)A Toy Windmill Wrote: Finitism is stricter than simply assuming there are no infinite sets. The finitist goes so far as to say that one cannot even quantify over the natural numbers! The most cited formal system which tries to capture what limited reasoning is left is "Primitive Recursive Arithmetic."

The finitist will accept that "n + 1 > n", but will regard this as a schema. to be filled in with particular numbers that they can muster.

Well, that does depend on the brand of finitism you subscribe to. Yours is one of the more extreme versions.
It's the standard one, advanced by Hilbert and maintained by Goodstein (who was mentioned in the OP). For a nice introductory paper, see Hilbert's On the Infinite. That paper doesn't give a formal system, but primitive recursive arithmetic is usually taken to scope finitist reasoning.

Finitism is still being developed, in the sense that there is still interest in just how much of contemporary mathematics could, in principle, go through in things like PRA.
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