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Careful, now. Ultrafinitists exist within the mathematical community and sometimes we overstate our opinions for funsies. Doron is much more reasonable in per

Careful, now. Ultrafinitists exist within the mathematical community and sometimes we overstate our opinions for funsies. Doron is much more reasonable in person than one may conclude from this website. To me, ultrafinitism is an intellectual curiosity. Proofs which depend upon the axiom of choice, or a notion of infinity, do not always correspond to algorithms which terminate in finite time. Such results are "useless" to a computational mathematician. The goal of determining how much of mathematics can translate to an finite universe is, in my opinion, a noble one. A hill I would die on, for sport: "the logarithm of the biggest number I see reason to believe in is the combined size of my RAM and HDD." The key word being "believe." Mathematicians do not tend to operate on faith. Proofs invoking unrealistic axioms are perfectly acceptable, valid proofs. One needn't "believe" anything.

cubefox 3y ago
There are three levels: - Infinitism, people who believe in the existence of completed infinite sets, and transfinite numbers more generally. - Finitism, people who believe that "infinite" means "arbitrarily large" or "unbounded", but nothing more. - Ultrafinitism, people who think every quantity has an upper bound due to mathematics not being separate from the physical world. Zeilberger identifies as an ultrafinitist. I personally think finitism is the most plausible one. It accepts an intuitive notion of potential infinity, but it avoids basically all the mathematical paradoxes of infinity and the philosophically extravagant mathematics of transfinite set theory.
- A_D_E_P_T 3y ago
  > philosophically extravagant mathematics of transfinite set theory And poorly-founded, I would add. Cantor's diagonal proof is, quite literally, question begging. Sure, you can construct a number not in the infinite set of natural numbers via that diagonalization method, but only if you're at infinity, or if you make very liberal use of "..." -- otherwise, you cannot. And Skolem was quick to discredit Cantor's extravagances, on grounds that tend to support finitism generally.
  - epgui 3y ago
    I don't think I agree, but this is difficult to communicate in words. Cantor's diagonalization argument doesn't rely on the contents of an infinite list of numbers, or on numbers themselves. You don't need to know anything about what "..." represents and you don't need to be "at" anywhere. There is no need to compute anything with concrete numbers. The only thing the argument relies upon is whether the fundamental structure of numbers allows for defining a bijection. The proof/computation is an abstract relation, not a concrete one with actual number objects.
    - cubefox 3y ago
      > The only thing the argument relies upon is whether the fundamental structure of numbers allows for defining a bijection. That there is no bijection must be the conclusion, not an assumption, otherwise the argument would simply be question begging. If you can't define a new diagonal number in the first place, you don't have an argument why you can't have a bijection. With finitism it doesn't seem like you can define a diagonal number, because then Cantor's infinite table is merely potentially infinite in width and height. For the diagonal to be finished, the table has to be at least as wide as it is tall. But if you list numbers by their lexicographic order, the list grows much faster in height than in width. For base 10, for n columns (digits), you get 10^n rows (numbers). Then you can't ever finish a diagonal that isn't on the list. The number of digits doesn't grow fast enough to cover the entire list. You don't get the diagonal in the limit. Now if you assume infinitism, you can simply declare that the height and width of the list is the same (or even that the width is larger), and therefore the diagonal can be constructed. But that's not a convincing assumption for a finitist (or ultrafinitist).
- flir 3y ago
  > Zeilberger identifies as an ultrafinitist. So, um, does that mean there's a "biggest number"? One you can't +1?
  - epgui 3y ago
    Within certain parameters, that's right. And that probably makes as much sense (if not more) as the alternative for all I know. There is a way to think of mathematics as corresponding to computation in a way where anything that cannot be computed cannot be proven. If you cannot "construct" or "compute" infinity, then you cannot say that it exists. I think what I am describing corresponds to the "constructivist" view of mathematics, but tbh I don't know enough about this to say whether the framework allows for infinities.
    - flir 3y ago
      <sotto voce>These people are crazy
      - epgui 3y ago
        What do you mean?
  - Animats 3y ago
    Yes. It's like writing code that's aware of MAX_INT. The code doesn't need to know the value of MAX_INT. It just has to handle special cases near that number.
  - boothby 3y ago
    Yeah! Programmers are quite familiar with this universe. In many programming languages, you can't represent 2^64 with a built-in integral datatype. While it's certainly possible to implement data structures and algorithms to support arbitrary-size integer arithmetic, most programmers don't! They just pretend that numbers larger than 2^64-1 do not exist! And, that's a fairly safe assumption for many real-world applications. But even with that arbitrary-size integer arithmetic, your mathematical universe is still bounded by your computational environment's resources.
mensetmanusman 3y ago
Axioms are faiths.
- boothby 3y ago
  No, axioms are hypotheses. In general, a mathematical proof has the form "If [conditions] are met, then [result] is guaranteed." This requires no faith whatsoever that [conditions] will ever be met. But those numbers that require the totality of my computer's N bits of storage to represent? I can, without reservation, believe that every integer between zero and 2^N-1 exists. But incrementing that biggest number will be challenging. I hear you scoff; surely there are bigger numbers, you can just buy more hard drive. And I can! But there are only finitely many hard drives available for purchase, many more than I can afford, so I'd rather save my money and believe in less.
  - epgui 3y ago
    > No, axioms are hypotheses IMO they're neither "faith" nor "hypotheses", although "whether a set of axioms is self-consistent" is probably a hypothesis. Axioms are just building blocks. They don't even need to be "true" in any real sense. They're just arbitrary determinations or rules that we make up, and then we figure out what the consequences of these choices are. In practice we often try to come up with axioms that have some practical value in the real world, but this is not necessarily or always how we go about doing maths.
  - mensetmanusman 3y ago
    Whether you believe the axioms are true is an act of faith, because by definition they are the lowest bits.
IngoBlechschmid 3y ago
> To me, ultrafinitism is an intellectual curiosity. Proofs which depend upon the axiom of choice, or a notion of infinity, do not always correspond to algorithms which terminate in finite time. Such results are "useless" to a computational mathematician. Fun fact: For certain kinds of results, there is an explicit procedure for transforming any given ZFC-proof into an IZF-proof. By ZFC, I mean classical "uncomputable" mathematics with the axiom of choice and with the law of excluded middle. By IZF, I mean (a specific flavor of) constructive mathematics without those two contested axioms. In particular, this metatheorem applies to termination results. So feel free to use the axiom of choice and the law of excluded middle to your heart's content when proving that a program terminates; you can trust that at the end of the day, your proof can be transformed to a constructive proof. Slides which I hope are also accessible to non-specialists in proof theory: https://www.speicherleck.de/iblech/stuff/37c3-axiom-of-choice.pdf https://www.speicherleck.de/iblech/stuff/37c3-axiom-of-choic... More pointers to the literature: https://math.stanford.edu/~feferman/papers/reductive.pdf https://math.stanford.edu/~feferman/papers/reductive.pdf

cubefox 3y ago

There are three levels: - Infinitism, people who believe in the existence of completed infinite sets, and transfinite numbers more generally. - Finitism, people who believe that "infinite" means "arbitrarily large" or "unbounded", but nothing more. - Ultrafinitism, people who think every quantity has an upper bound due to mathematics not being separate from the physical world. Zeilberger identifies as an ultrafinitist. I personally think finitism is the most plausible one. It accepts an intuitive notion of potential infinity, but it avoids basically all the mathematical paradoxes of infinity and the philosophically extravagant mathematics of transfinite set theory.

A_D_E_P_T 3y ago
> philosophically extravagant mathematics of transfinite set theory And poorly-founded, I would add. Cantor's diagonal proof is, quite literally, question begging. Sure, you can construct a number not in the infinite set of natural numbers via that diagonalization method, but only if you're at infinity, or if you make very liberal use of "..." -- otherwise, you cannot. And Skolem was quick to discredit Cantor's extravagances, on grounds that tend to support finitism generally.
- epgui 3y ago
  I don't think I agree, but this is difficult to communicate in words. Cantor's diagonalization argument doesn't rely on the contents of an infinite list of numbers, or on numbers themselves. You don't need to know anything about what "..." represents and you don't need to be "at" anywhere. There is no need to compute anything with concrete numbers. The only thing the argument relies upon is whether the fundamental structure of numbers allows for defining a bijection. The proof/computation is an abstract relation, not a concrete one with actual number objects.
  - cubefox 3y ago
    > The only thing the argument relies upon is whether the fundamental structure of numbers allows for defining a bijection. That there is no bijection must be the conclusion, not an assumption, otherwise the argument would simply be question begging. If you can't define a new diagonal number in the first place, you don't have an argument why you can't have a bijection. With finitism it doesn't seem like you can define a diagonal number, because then Cantor's infinite table is merely potentially infinite in width and height. For the diagonal to be finished, the table has to be at least as wide as it is tall. But if you list numbers by their lexicographic order, the list grows much faster in height than in width. For base 10, for n columns (digits), you get 10^n rows (numbers). Then you can't ever finish a diagonal that isn't on the list. The number of digits doesn't grow fast enough to cover the entire list. You don't get the diagonal in the limit. Now if you assume infinitism, you can simply declare that the height and width of the list is the same (or even that the width is larger), and therefore the diagonal can be constructed. But that's not a convincing assumption for a finitist (or ultrafinitist).
flir 3y ago
> Zeilberger identifies as an ultrafinitist. So, um, does that mean there's a "biggest number"? One you can't +1?
- epgui 3y ago
  Within certain parameters, that's right. And that probably makes as much sense (if not more) as the alternative for all I know. There is a way to think of mathematics as corresponding to computation in a way where anything that cannot be computed cannot be proven. If you cannot "construct" or "compute" infinity, then you cannot say that it exists. I think what I am describing corresponds to the "constructivist" view of mathematics, but tbh I don't know enough about this to say whether the framework allows for infinities.
  - flir 3y ago
    <sotto voce>These people are crazy
    - epgui 3y ago
      What do you mean?
- Animats 3y ago
  Yes. It's like writing code that's aware of MAX_INT. The code doesn't need to know the value of MAX_INT. It just has to handle special cases near that number.
- boothby 3y ago
  Yeah! Programmers are quite familiar with this universe. In many programming languages, you can't represent 2^64 with a built-in integral datatype. While it's certainly possible to implement data structures and algorithms to support arbitrary-size integer arithmetic, most programmers don't! They just pretend that numbers larger than 2^64-1 do not exist! And, that's a fairly safe assumption for many real-world applications. But even with that arbitrary-size integer arithmetic, your mathematical universe is still bounded by your computational environment's resources.

mensetmanusman 3y ago

Axioms are faiths.

boothby 3y ago
No, axioms are hypotheses. In general, a mathematical proof has the form "If [conditions] are met, then [result] is guaranteed." This requires no faith whatsoever that [conditions] will ever be met. But those numbers that require the totality of my computer's N bits of storage to represent? I can, without reservation, believe that every integer between zero and 2^N-1 exists. But incrementing that biggest number will be challenging. I hear you scoff; surely there are bigger numbers, you can just buy more hard drive. And I can! But there are only finitely many hard drives available for purchase, many more than I can afford, so I'd rather save my money and believe in less.
- epgui 3y ago
  > No, axioms are hypotheses IMO they're neither "faith" nor "hypotheses", although "whether a set of axioms is self-consistent" is probably a hypothesis. Axioms are just building blocks. They don't even need to be "true" in any real sense. They're just arbitrary determinations or rules that we make up, and then we figure out what the consequences of these choices are. In practice we often try to come up with axioms that have some practical value in the real world, but this is not necessarily or always how we go about doing maths.
- mensetmanusman 3y ago
  Whether you believe the axioms are true is an act of faith, because by definition they are the lowest bits.

IngoBlechschmid 3y ago

> To me, ultrafinitism is an intellectual curiosity. Proofs which depend upon the axiom of choice, or a notion of infinity, do not always correspond to algorithms which terminate in finite time. Such results are "useless" to a computational mathematician. Fun fact: For certain kinds of results, there is an explicit procedure for transforming any given ZFC-proof into an IZF-proof. By ZFC, I mean classical "uncomputable" mathematics with the axiom of choice and with the law of excluded middle. By IZF, I mean (a specific flavor of) constructive mathematics without those two contested axioms. In particular, this metatheorem applies to termination results. So feel free to use the axiom of choice and the law of excluded middle to your heart's content when proving that a program terminates; you can trust that at the end of the day, your proof can be transformed to a constructive proof. Slides which I hope are also accessible to non-specialists in proof theory: https://www.speicherleck.de/iblech/stuff/37c3-axiom-of-choice.pdf https://www.speicherleck.de/iblech/stuff/37c3-axiom-of-choic... More pointers to the literature: https://math.stanford.edu/~feferman/papers/reductive.pdf https://math.stanford.edu/~feferman/papers/reductive.pdf