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Without the axiom of infinity you can't show that there exists an infinite set in ZF. i.e. you can't use anything infinite, including natural numbers. It's act

by qsort 4y ago

Without the axiom of infinity you can't show that there exists an infinite set in ZF. i.e. you can't use anything infinite, including natural numbers. It's actually extremely restrictive to get rid of it.

amelius 4y ago
Not a mathematician, but can't you parameterize e.g. the natural numbers by its largest element, then write whatever you want to do with the set and put lim n->inf around it? That way, you keep the objects finite, but you still allow them to become arbitrarily large.
- Ferrotin 4y ago
  That parameterization, as you describe it, would be a an infinite set.
  - naasking 4y ago
    I don't see why. A parameter provides an upper bound but there's no need to also assume that that upper bound must go to infinity.
    - rfergie 4y ago
      One construction of the natural numbers that I've seen is 1. There exists an element 0 that is a natural number 2. For every natural number there is a "sucessor" that is also a natural number. (i.e. if n is a natural number then n+1 is a natural number) This construction means there can't be an upper bound N because then step 2 couldn't be applied to N. Maybe there are other constructions that could workaround this? I'm guessing not because you'd still struggle to define the usual rules of addition for all numbers in a bounded set
      - naasking 4y ago
        > This construction means there can't be an upper bound N because then step 2 couldn't be applied to N. Bendegem discusses this problem at length in his paper [1]. As programming-heavy site, I assume we're all aware that computers have finite resources. The universe too has finite resources so no matter how big a computer you build, it too will be finite. Therefore the infinity that is so pervasive in math is unphysical in a very real sense. So what would math look like and how would theorems change if this finiteness were formalized? That's what various flavours of finitism aim to achieve. So to get back to your question as to the nature of the naturals, it seems evident that yes, at some point, you literally can fail construct the natural number N+1 if you are given N, because you will run out of particles in the universe. What implications this will have for various theorems will be interesting for sure, but it isn't clear yet because finitism isn't given much funding. Edit: however, it's clear that some very unintuitive results follow from the infinities embedded in mathematics, and that a finitist approach resolves some of them. For instance, the argument that "0.9999... = 1" is true in classical mathematics while this equality is arguably not true under strict finitism because "0.999..." does not exist, because infinite objects do not exist, and so it will never equal 1. [1] See the section on continuous counting, https://www.researchgate.net/publication/288354797_A_Defense_of_Strict_Finitism https://www.researchgate.net/publication/288354797_A_Defense...
        Ferrotin 4y ago
        Calling “0.999… = 1” very unintuitive is a very strange thing to say, because that makes perfect sense to most children. I’d like to see a result that truly is unintuitive, like what we get with the axiom of choice.
        naasking 4y ago
        > Calling “0.999… = 1” very unintuitive is a very strange thing to say, because that makes perfect sense to most children. Gotta call bullshit on that. First I don't think you have any robust empirical data on that question. Second what's convincing to children that don't have enough knowledge of math to have formed any intuitions about it is not a compelling argument.
    - Ferrotin 4y ago
      Define lim without infinite sets. Define how you quantify over all values of n without infinite sets. The parameterization you describe would be a function mapping an infinite set of sets of natural numbers to their maximum elements, that function itself being (or containing) an infinite set of ordered pairs. You might be able to pull it off without creating something equivalent to the axiom of infinity — I wouldn’t know if it’s possible. But the naïve implementation of your ideas involves the construction of infinite sets. I think you can do this rigorously outside the theory, talking about it, but not inside.
- qsort 4y ago
  That "lim n->inf" is doing a lot of work there. How would you even define "limit"? You can certainly work with bounded integers, but now suddenly your arithmetic has undefined behavior.
andrewla 4y ago
I need more elucidation on that. Why can't you use the natural numbers? You can construct them even within the constraints of ZF. Is it just that you can't use it the same way as a set in all cases? It seems like you could define a new class, call them seits, that include both the set as axiomatically defined as well as these sorts of unbounded objects, and then prove that you can define similar operations as exist for sets on seits, and then go on and live your life. When you say you can't "use" anything infinite, what's an example of a theorem that cannot be proven without the Axiom of Infinity?
- qsort 4y ago
  > Why can't you use the natural numbers? Because the rest of the axioms of ZF aren't sufficient to construct them. > what's an example of a theorem that cannot be proven without the Axiom of Infinity? That there exists a set equal to the natural numbers up to isomorphism, that there exist well-defined operations "+", "-", "*", "/" with usual semantics, that there exist the usual sets of rational, real, complex numbers.
- klodolph 4y ago
  You can construct any natural number without the axiom of infinity, but you cannot construct the set of natural numbers itself. > It seems like you could define a new class, call them seits, that include both the set as axiomatically defined as well as these sorts of unbounded objects, and then prove that you can define similar operations as exist for sets on seits, and then go on and live your life. Imagine you’re at a vegan restaurant. Your food arrives, you pull a fistful of ground beef out of your pocket and stick it on top. You’re “still eating vegan food”, you’ve just added beef to it. Finitism is a lot like a dietary restriction. It’s asking “What can you do WITHOUT these axioms?” The question is whether you still have a balanced diet afterwards, whether you’re satisfied, whether you prefer a vegan or omnivorous diet, etc. If you add infinities to your finite sets but call them classes instead you’re just cheating on your diet. (Or you’re talking about models of finitism.) I suspect that there are a ton of results that can’t be proven without the axiom of infinity, but I don’t keep track of them.
Chinjut 4y ago
Well, you can't create the infinite set of all natural numbers without the Axiom of Infinity. You can still reason about natural numbers themselves without the Axiom of Infinity (and if you are inclined to speak in such terms, the natural numbers will still comprise a "class", just a proper class rather than a set, when the Axiom of Infinity is negated). ZF(C) with the Axiom of Infinity replaced by its negation is essentially equivalent to Peano Arithmetic, and you can do an awful lot of math just fine in Peano Arithmetic.
- Kranar 4y ago
  You can't create the set of natural numbers even with the axiom of infinity either, at least not in any first order logic. No first order logic is sufficient to strictly define the set of all natural numbers. You can technically construct natural numbers in second order logic, at least syntactically, but then the semantics of natural numbers in second order logic does not end up corresponding to what is actually meant by natural numbers. A good article on this is linked here: https://www.lesswrong.com/posts/MLqhJ8eDy5smbtGrf/completeness-incompleteness-and-what-it-all-means-first https://www.lesswrong.com/posts/MLqhJ8eDy5smbtGrf/completene...
  - dwohnitmok 4y ago
    > You can't create the set of natural numbers even with the axiom of infinity either, at least not in any first order logic. No first order logic is sufficient to strictly define the set of all natural numbers. Eh that's a fundamentally different problem though, although that is part of the reason why finitism and ultra-finitism are attractive. You can define a set that is the natural numbers for all intents and purposes from the viewpoint of your theory. If you were willing accept that something like PA is enough to do large parts of real analysis (and also believed in the usefulness of real analysis), then philosophically you should be totally fine with using theories that have models with nonstandard natural numbers, since "you can't observe any of the consequences from your theory" is the same bar in both cases (in the case of finitistic approaches to analysis it's even trickier because often you can only deal with encodings of objects rather than objects themselves and must rely on your metatheory to do the association of the encoding to the object).
skulk 4y ago
> you can't use anything infinite, including natural numbers. This is false, you don't need the axiom of infinity to prove theorems about all natural numbers. https://math.stackexchange.com/questions/2850064/does-countable-induction-over-mathbb-n-require-the-axiom-of-infinity https://math.stackexchange.com/questions/2850064/does-counta... Particularly: > Notice here, we are just treating the natural numbers as a class, not necessarily a set. In fact we can prove from ZF minus infinity that any of the (translations of the) axioms of PA, including all the induction axioms, hold in this class.

by qsort 4y ago

amelius 4y ago

Not a mathematician, but can't you parameterize e.g. the natural numbers by its largest element, then write whatever you want to do with the set and put lim n->inf around it? That way, you keep the objects finite, but you still allow them to become arbitrarily large.

Ferrotin 4y ago
That parameterization, as you describe it, would be a an infinite set.
- naasking 4y ago
  I don't see why. A parameter provides an upper bound but there's no need to also assume that that upper bound must go to infinity.
  - rfergie 4y ago
    One construction of the natural numbers that I've seen is 1. There exists an element 0 that is a natural number 2. For every natural number there is a "sucessor" that is also a natural number. (i.e. if n is a natural number then n+1 is a natural number) This construction means there can't be an upper bound N because then step 2 couldn't be applied to N. Maybe there are other constructions that could workaround this? I'm guessing not because you'd still struggle to define the usual rules of addition for all numbers in a bounded set
    - naasking 4y ago
      > This construction means there can't be an upper bound N because then step 2 couldn't be applied to N. Bendegem discusses this problem at length in his paper [1]. As programming-heavy site, I assume we're all aware that computers have finite resources. The universe too has finite resources so no matter how big a computer you build, it too will be finite. Therefore the infinity that is so pervasive in math is unphysical in a very real sense. So what would math look like and how would theorems change if this finiteness were formalized? That's what various flavours of finitism aim to achieve. So to get back to your question as to the nature of the naturals, it seems evident that yes, at some point, you literally can fail construct the natural number N+1 if you are given N, because you will run out of particles in the universe. What implications this will have for various theorems will be interesting for sure, but it isn't clear yet because finitism isn't given much funding. Edit: however, it's clear that some very unintuitive results follow from the infinities embedded in mathematics, and that a finitist approach resolves some of them. For instance, the argument that "0.9999... = 1" is true in classical mathematics while this equality is arguably not true under strict finitism because "0.999..." does not exist, because infinite objects do not exist, and so it will never equal 1. [1] See the section on continuous counting, https://www.researchgate.net/publication/288354797_A_Defense_of_Strict_Finitism https://www.researchgate.net/publication/288354797_A_Defense...
      - Ferrotin 4y ago
        Calling “0.999… = 1” very unintuitive is a very strange thing to say, because that makes perfect sense to most children. I’d like to see a result that truly is unintuitive, like what we get with the axiom of choice.
        naasking 4y ago
        > Calling “0.999… = 1” very unintuitive is a very strange thing to say, because that makes perfect sense to most children. Gotta call bullshit on that. First I don't think you have any robust empirical data on that question. Second what's convincing to children that don't have enough knowledge of math to have formed any intuitions about it is not a compelling argument.
  - Ferrotin 4y ago
    Define lim without infinite sets. Define how you quantify over all values of n without infinite sets. The parameterization you describe would be a function mapping an infinite set of sets of natural numbers to their maximum elements, that function itself being (or containing) an infinite set of ordered pairs. You might be able to pull it off without creating something equivalent to the axiom of infinity — I wouldn’t know if it’s possible. But the naïve implementation of your ideas involves the construction of infinite sets. I think you can do this rigorously outside the theory, talking about it, but not inside.
qsort 4y ago
That "lim n->inf" is doing a lot of work there. How would you even define "limit"? You can certainly work with bounded integers, but now suddenly your arithmetic has undefined behavior.

andrewla 4y ago

I need more elucidation on that. Why can't you use the natural numbers? You can construct them even within the constraints of ZF. Is it just that you can't use it the same way as a set in all cases? It seems like you could define a new class, call them seits, that include both the set as axiomatically defined as well as these sorts of unbounded objects, and then prove that you can define similar operations as exist for sets on seits, and then go on and live your life. When you say you can't "use" anything infinite, what's an example of a theorem that cannot be proven without the Axiom of Infinity?

qsort 4y ago
> Why can't you use the natural numbers? Because the rest of the axioms of ZF aren't sufficient to construct them. > what's an example of a theorem that cannot be proven without the Axiom of Infinity? That there exists a set equal to the natural numbers up to isomorphism, that there exist well-defined operations "+", "-", "*", "/" with usual semantics, that there exist the usual sets of rational, real, complex numbers.
klodolph 4y ago
You can construct any natural number without the axiom of infinity, but you cannot construct the set of natural numbers itself. > It seems like you could define a new class, call them seits, that include both the set as axiomatically defined as well as these sorts of unbounded objects, and then prove that you can define similar operations as exist for sets on seits, and then go on and live your life. Imagine you’re at a vegan restaurant. Your food arrives, you pull a fistful of ground beef out of your pocket and stick it on top. You’re “still eating vegan food”, you’ve just added beef to it. Finitism is a lot like a dietary restriction. It’s asking “What can you do WITHOUT these axioms?” The question is whether you still have a balanced diet afterwards, whether you’re satisfied, whether you prefer a vegan or omnivorous diet, etc. If you add infinities to your finite sets but call them classes instead you’re just cheating on your diet. (Or you’re talking about models of finitism.) I suspect that there are a ton of results that can’t be proven without the axiom of infinity, but I don’t keep track of them.

Chinjut 4y ago

Well, you can't create the infinite set of all natural numbers without the Axiom of Infinity. You can still reason about natural numbers themselves without the Axiom of Infinity (and if you are inclined to speak in such terms, the natural numbers will still comprise a "class", just a proper class rather than a set, when the Axiom of Infinity is negated). ZF(C) with the Axiom of Infinity replaced by its negation is essentially equivalent to Peano Arithmetic, and you can do an awful lot of math just fine in Peano Arithmetic.

Kranar 4y ago
You can't create the set of natural numbers even with the axiom of infinity either, at least not in any first order logic. No first order logic is sufficient to strictly define the set of all natural numbers. You can technically construct natural numbers in second order logic, at least syntactically, but then the semantics of natural numbers in second order logic does not end up corresponding to what is actually meant by natural numbers. A good article on this is linked here: https://www.lesswrong.com/posts/MLqhJ8eDy5smbtGrf/completeness-incompleteness-and-what-it-all-means-first https://www.lesswrong.com/posts/MLqhJ8eDy5smbtGrf/completene...
- dwohnitmok 4y ago
  > You can't create the set of natural numbers even with the axiom of infinity either, at least not in any first order logic. No first order logic is sufficient to strictly define the set of all natural numbers. Eh that's a fundamentally different problem though, although that is part of the reason why finitism and ultra-finitism are attractive. You can define a set that is the natural numbers for all intents and purposes from the viewpoint of your theory. If you were willing accept that something like PA is enough to do large parts of real analysis (and also believed in the usefulness of real analysis), then philosophically you should be totally fine with using theories that have models with nonstandard natural numbers, since "you can't observe any of the consequences from your theory" is the same bar in both cases (in the case of finitistic approaches to analysis it's even trickier because often you can only deal with encodings of objects rather than objects themselves and must rely on your metatheory to do the association of the encoding to the object).

skulk 4y ago

> you can't use anything infinite, including natural numbers. This is false, you don't need the axiom of infinity to prove theorems about all natural numbers. https://math.stackexchange.com/questions/2850064/does-countable-induction-over-mathbb-n-require-the-axiom-of-infinity https://math.stackexchange.com/questions/2850064/does-counta... Particularly: > Notice here, we are just treating the natural numbers as a class, not necessarily a set. In fact we can prove from ZF minus infinity that any of the (translations of the) axioms of PA, including all the induction axioms, hold in this class.