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In topology, if you have a continuous surjective map X --> Y, then it might have a continuous splitting (a map the other way which is a "partial" inverse in the

In topology, if you have a continuous surjective map X --> Y, then it might have a continuous splitting (a map the other way which is a "partial" inverse in the sense that Y ---> X ---> Y is the identity) e.g. there are lots of splittings of the projection R^2 ---> R, you could include the line back as the x-axis but also the graph of any continuous function is a splitting. On the other hand, there's no continuous splitting of the map from the unit interval to the circle that glues together the two endpoints. So the category of topological spaces does not have the property "every epimorphism splits." As the article mentions, the axiom of choice says that the category of sets has this property. So we can think of the various independence results of the 20th century as saying, hey, (assuming ZFC is consistent) there's this category, Set, with this rule, and there's this other category called idk Snet, that satisfies the ZF axioms but where there are some surjections that don't split, and that's ok too. Then whatever, if you want to study something like rings but you don't like the axiom of choice, define a rning to be a snet with two binary operations such that blah blah blah, and you've got a nice category Rning and your various theorems about rnings and maybe they don't all have maximal ideals, even though rings do. You're not arguing about ontology or the nature of truth, you're just picking which category to work in.

karmakurtisaani 1y ago
Yeah, it's important to think of these axioms as choosing the rules of the game, rather than what intuitively makes sense. The real question is if playing the game produces useful results.
- btilly 1y ago
  Spoken like a true formalist. It doesn't really have to mean anything when we say that the reals are a larger set than the natural numbers - that's just the conclusion of the game that we are playing. What fraction of people who "know" that there are more reals than natural numbers, do you think really understand that this is not an eternal verity of mathematics, but only a conclusion that follows from a particular set of rules that we're playing the mathematics game with?
  - skissane 1y ago
    > What fraction of people who "know" that there are more reals than natural numbers, do you think really understand that this is not an eternal verity of mathematics, but only a conclusion that follows from a particular set of rules that we're playing the mathematics game with? The claim that there are more reals than naturals holds given classical ZF(C) set theory. But there are alternative set theories in which the reals are countable, e.g. NFU+AxCount. These alternative set theories ensure all reals are countable by rendering Cantor’s diagonalisation argument invalid, since their axioms are too weak to validate it. But, they contain all the same reals as the high school mathematics concept of “reals”. So, there are many reals, and that some of them are countable and others are not are indeed “eternal truths” (it is an eternal truth that whatever axioms have the consequences they do), but the everyday (non-expert) concept of reals isn’t any of them in particular - and it is unclear if the dominance of classical notions in mainstream professional mathematics was historically inevitable or a historical accident - maybe, on the other side of the galaxy, there exists some alien civilisation, in which different foundations of mathematics are mainstream, because their mathematics took a different evolutionary course from ours - maybe for them, reals are classically countable, and uncountability is an exotic notion belonging to alternative foundations of mathematics
    - btilly 1y ago
      As I pointed out at https://news.ycombinator.com/item?id=44271589 https://news.ycombinator.com/item?id=44271589, there are systems that can accept Cantor's argument, without concluding that there are more reals than rational numbers. As you point out, there are many mathematical systems that contain all of the numbers in the high school mathematics concept of "reals". Since those with a high school understanding of reals do not know which of those systems they would agree with, they should not be asked to accept as true, any results that hold in only some of those systems. And that is why I don't like mathematicians telling lay audiences that there are more reals than rationals.
      - zozbot234 1y ago
        "Cantor's diagonalization argument" is best understood as a mere special case of Lawvere's fixed-point theorem. Lawvere's theorem is really the meat of the argument, and it's also the part that is very easy for exotic systems to "accept", since it's close to a purely logical argument. Whether these systems truly accept "Cantor's argument" is perhaps only a matter of perception and intuition, that people may perhaps disagree about.
- woopsn 1y ago
  Axioms are also introduced in practical terms just to make proofs and results "better". Usually we talk in terms of what propositions are provable, saying that indicates the strength/power of these assumptions, but beyond this there are issues of proof length and complexity. For example in arithmetic without induction, roughly, theorems remain the same (those which can still be expressed) but may have exponentially longer proofs because of the loss of those `∀n P(n)`-type propositions. In this sense it does sometimes come back to intuition. If for all n we can prove P(n), then `∀n P(n)` should be an acceptable proposition and doesn't really change "the game" we are trying to play. It just makes it more intuitive and playable.
  - karmakurtisaani 1y ago
    Good point. I would argue, however, that having nicer proofs is a "useful" result of the game.
  - SabrinaJewson 1y ago
    I’m not sure what you mean by “theorems remain the same”. If you take away induction from Peano arithmetic, you get Robinson arithmetic, which has many more models, including (from https://math.stackexchange.com/a/4076545 https://math.stackexchange.com/a/4076545): - ℕ ∪ {∞} - Cardinal arithmetic - ℤ[x]⁺ Obviously, not all theorems that are true for the natural numbers are true for cardinals, so it seems misleading to say that theorems remain the same. I also believe that the addition of induction increases the consistency strength of the theory, so it’s not “just” a matter of expressing the theorems in a different way. I would agree more for axioms that don’t affect consistency strength, like foundation or choice (over the rest of the ZF axioms).
    - deleted 1y ago
      [deleted]
    - woopsn 1y ago
      If I had to write again I might say "same theorems about natural numbers" and capitalize ROUGHLY. It is a conversation, what exactly I am weaseling around (not just nonstandard model theoretic issues), and I take your caveat about consistency strength - with that said would you still call it misleading? Why is it that eg x+y=y+x for x y given takes exponential length proof in Robinson compared to PA? For the reason stated, which is true in a very broad sense.
alexey-salmin 1y ago
Doesn't "continuous" make all the difference here? AC doesn't contain a comparable limitation, so the analogy doesn't work that week.
- gsf_emergency 1y ago
  Careful here, you might wake up the diabol (of pure algebra)
- pfortuny 1y ago
  Yes, but the parent comment is trying to say "imagine the world would only be made up of topological spaces and continuous maps". Then the retraction principle would not hold.
LudwigNagasena 1y ago
How is it different from using ZF as a meta-theory to study ZF(C)? Is there anything special about category theory vis-à-vis ZF as a meta-theory? You're not arguing about ontology or the nature of truth, because you've picked category theory as your ontology just like you could pick ZF or ZFC.
- gylterud 1y ago
  Category theory gives a structural framework for discussing these things. The various categories live side by side and can be related with functors. This allows a broader view and makes it easier perhaps, to understand that there isn’t a right answer to “what is true” about sets in the absolute.
  - LudwigNagasena 1y ago
    But then you would think there is a right answer to “what is true” about categories, and you would face AC again.
    - gylterud 1y ago
      For sure there would have to be a meta theory of some sort. But, I think that many classical mathematicians would be happy with that meta theory to be weaker, rather than stronger. I don't think there is a need for that theory to have AC. After all, thanks to its independence of from the other axioms, AC does not add logical strength to a classical theory. (For die-hard constructivists, such as myself, the story is of course different. But that story is for a another time. I am presenting the classical view here.)

karmakurtisaani 1y ago

Yeah, it's important to think of these axioms as choosing the rules of the game, rather than what intuitively makes sense. The real question is if playing the game produces useful results.

btilly 1y ago
Spoken like a true formalist. It doesn't really have to mean anything when we say that the reals are a larger set than the natural numbers - that's just the conclusion of the game that we are playing. What fraction of people who "know" that there are more reals than natural numbers, do you think really understand that this is not an eternal verity of mathematics, but only a conclusion that follows from a particular set of rules that we're playing the mathematics game with?
- skissane 1y ago
  > What fraction of people who "know" that there are more reals than natural numbers, do you think really understand that this is not an eternal verity of mathematics, but only a conclusion that follows from a particular set of rules that we're playing the mathematics game with? The claim that there are more reals than naturals holds given classical ZF(C) set theory. But there are alternative set theories in which the reals are countable, e.g. NFU+AxCount. These alternative set theories ensure all reals are countable by rendering Cantor’s diagonalisation argument invalid, since their axioms are too weak to validate it. But, they contain all the same reals as the high school mathematics concept of “reals”. So, there are many reals, and that some of them are countable and others are not are indeed “eternal truths” (it is an eternal truth that whatever axioms have the consequences they do), but the everyday (non-expert) concept of reals isn’t any of them in particular - and it is unclear if the dominance of classical notions in mainstream professional mathematics was historically inevitable or a historical accident - maybe, on the other side of the galaxy, there exists some alien civilisation, in which different foundations of mathematics are mainstream, because their mathematics took a different evolutionary course from ours - maybe for them, reals are classically countable, and uncountability is an exotic notion belonging to alternative foundations of mathematics
  - btilly 1y ago
    As I pointed out at https://news.ycombinator.com/item?id=44271589 https://news.ycombinator.com/item?id=44271589, there are systems that can accept Cantor's argument, without concluding that there are more reals than rational numbers. As you point out, there are many mathematical systems that contain all of the numbers in the high school mathematics concept of "reals". Since those with a high school understanding of reals do not know which of those systems they would agree with, they should not be asked to accept as true, any results that hold in only some of those systems. And that is why I don't like mathematicians telling lay audiences that there are more reals than rationals.
    - zozbot234 1y ago
      "Cantor's diagonalization argument" is best understood as a mere special case of Lawvere's fixed-point theorem. Lawvere's theorem is really the meat of the argument, and it's also the part that is very easy for exotic systems to "accept", since it's close to a purely logical argument. Whether these systems truly accept "Cantor's argument" is perhaps only a matter of perception and intuition, that people may perhaps disagree about.
woopsn 1y ago
Axioms are also introduced in practical terms just to make proofs and results "better". Usually we talk in terms of what propositions are provable, saying that indicates the strength/power of these assumptions, but beyond this there are issues of proof length and complexity. For example in arithmetic without induction, roughly, theorems remain the same (those which can still be expressed) but may have exponentially longer proofs because of the loss of those `∀n P(n)`-type propositions. In this sense it does sometimes come back to intuition. If for all n we can prove P(n), then `∀n P(n)` should be an acceptable proposition and doesn't really change "the game" we are trying to play. It just makes it more intuitive and playable.
- karmakurtisaani 1y ago
  Good point. I would argue, however, that having nicer proofs is a "useful" result of the game.
- SabrinaJewson 1y ago
  I’m not sure what you mean by “theorems remain the same”. If you take away induction from Peano arithmetic, you get Robinson arithmetic, which has many more models, including (from https://math.stackexchange.com/a/4076545 https://math.stackexchange.com/a/4076545): - ℕ ∪ {∞} - Cardinal arithmetic - ℤ[x]⁺ Obviously, not all theorems that are true for the natural numbers are true for cardinals, so it seems misleading to say that theorems remain the same. I also believe that the addition of induction increases the consistency strength of the theory, so it’s not “just” a matter of expressing the theorems in a different way. I would agree more for axioms that don’t affect consistency strength, like foundation or choice (over the rest of the ZF axioms).
  - deleted 1y ago
    [deleted]
  - woopsn 1y ago
    If I had to write again I might say "same theorems about natural numbers" and capitalize ROUGHLY. It is a conversation, what exactly I am weaseling around (not just nonstandard model theoretic issues), and I take your caveat about consistency strength - with that said would you still call it misleading? Why is it that eg x+y=y+x for x y given takes exponential length proof in Robinson compared to PA? For the reason stated, which is true in a very broad sense.

alexey-salmin 1y ago

Doesn't "continuous" make all the difference here? AC doesn't contain a comparable limitation, so the analogy doesn't work that week.

gsf_emergency 1y ago
Careful here, you might wake up the diabol (of pure algebra)
pfortuny 1y ago
Yes, but the parent comment is trying to say "imagine the world would only be made up of topological spaces and continuous maps". Then the retraction principle would not hold.

LudwigNagasena 1y ago

How is it different from using ZF as a meta-theory to study ZF(C)? Is there anything special about category theory vis-à-vis ZF as a meta-theory? You're not arguing about ontology or the nature of truth, because you've picked category theory as your ontology just like you could pick ZF or ZFC.

gylterud 1y ago
Category theory gives a structural framework for discussing these things. The various categories live side by side and can be related with functors. This allows a broader view and makes it easier perhaps, to understand that there isn’t a right answer to “what is true” about sets in the absolute.
- LudwigNagasena 1y ago
  But then you would think there is a right answer to “what is true” about categories, and you would face AC again.
  - gylterud 1y ago
    For sure there would have to be a meta theory of some sort. But, I think that many classical mathematicians would be happy with that meta theory to be weaker, rather than stronger. I don't think there is a need for that theory to have AC. After all, thanks to its independence of from the other axioms, AC does not add logical strength to a classical theory. (For die-hard constructivists, such as myself, the story is of course different. But that story is for a another time. I am presenting the classical view here.)