100 years of Zermelo's axiom of choice: What was the problem with it? (2006)
research.mietek.io119 points by Bogdanp 3 days ago
119 points by Bogdanp 3 days ago
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.
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.
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.
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):
- 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).
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.
> If for all n we can prove P(n), then `∀n P(n)` should be an acceptable proposition
But how can you prove that P(n) for all n without induction? Maybe I misinterpret what you're saying, or I'm naive about something in formal languages, but if we can prove P(n) for all n. then `∀n P(n)` just looks like a trivial transcription of the conclusion into different symbols.
I think the crux of the matter is that we accept inductive arguments as valid, and so we formalize it via the inductive axiom (of Peano arithmetic). i.e., we accept induction as a principle of mathematical reasoning, but we can't derive it from something else so we postualte it when we come around to doing formalizations. Maybe that's what you mean by it coming down to intuition, now that I reread it...
Poincaré has a nice discussion of induction in "On the nature of mathematical reasoning", reprinted in Benacerraf & Putnam Philosophy of Mathematics, where he explicates it as an infinite sequence of modus ponens steps, but irreducible to any more basic logical rule like the principle of (non-)contradiction
>> If for all n we can prove P(n), then `∀n P(n)` should be an acceptable proposition
> But how can you prove that P(n) for all n without induction?
Just to be clear to all readers, the axiom of COUNTABLE choice is uncontroversial. Nobody is disturbed by induction.
The issue it that when you allow UNCOUNTABLE choice - choices being made for all real numbers (in a non-algorithmic way, I believe - so not a simple formula) - there are some unpleasant consequences.
> But how can you prove that P(n) for all n without induction? Maybe I misinterpret what you're saying, or I'm naive about something in formal languages, but if we can prove P(n) for all n. then `∀n P(n)` just looks like a trivial transcription of the conclusion into different symbols.
Of course it is likely that an interesting result about all positive integers, that is "really" about positive integers, is proved by induction, but you certainly don't need induction to prove P(n): n = 1.n, or, more boringly, P(n): 0 = 0. (These statements are obviously un-interesting, both in the human sense of the word and in the sense that they are just statements about semi-rings, of which the non-negative integers are an example.)
My understanding is that the difference between "For every positive integer n, I can prove P(n)" and "I can prove ∀n.P(n)" is that the former only guarantees that we can come up with some terrible ad hoc proof for P(1), some different terrible ad hoc proof for P(2), and so on. How could I be sure I have all these infinitely many different terrible ad hoc proofs without induction? I dunno, but that's all that the first statement guarantees. Whereas the second statement, in the context of computability, guarantees that there is some computable function that takes a positive integer n and produces a proof of P(n); that is, there is some sort of guaranteed uniformity to the proofs.
I think it may be easier to picture if connected with math_comment_21's analogy in https://news.ycombinator.com/item?id=44269153: the analogous statements in the category of topological spaces (I think one actually has to work about topos, but I don't know enough about topos theory to speak in that language) about a map f : X \to Y are "every element of Y has a pre-image under f in X" versus "I can continuously select, for each element of Y, a pre-image of it under f in X", i.e., "there is a continuous pre-inverse g : Y \to X of f."
Rejecting induction could be quite useful if you want to be very precise about the implications of your constructions wrt. computational complexity. This is of course only a mildly strengthened variant of the usual arguments for constructivism.
Good point. I would argue, however, that having nicer proofs is a "useful" result of the game.
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?
> 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
As I pointed out at 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.
"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.
It does not matter what your best understanding of Cantor's diagonalization argument is. In some mathematical systems it means, "there are more reals than natural numbers", and in others it means, "the reals encode self-reference in a more direct way than the natural numbers do".
The result is that it is possible for the acceptance of the argument to lead to very different consequences about what we then conclude.
> "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.
Okay, but can you prove Lawvere’s theorem in NFU+AxCount?
And even if you can, since NFU+AxCount proves that the reals are countable, if NFU+AxCount proves Lawvere, then (to echo what btilly says in a sibling comment) NFU+AxCount+Lawvere couldn’t entail the countability of the reals, since that would render NFU+AxCount trivially inconsistent, and we know it is isn’t trivially inconsistent (as with any formal system, consistency is ultimately unprovable, but if a system is taken seriously as an object of mathematical research, then any inconsistency must be highly non-trivial.)
I agree, but I also want to clarify that cantors argument was about subsets of the naturals (N), or more precisely functions from N to Bool (the decidable subsets). This is where the diagonal argument makes sense.
So to conclude that there are more reals than naturals, the classical mathematical argument is:
a) There are more functions N to Bool than naturals.
b) There are as many reals as functions from N to Bool.
Now, we of course agree the mistake is in b) not in a).
> So to conclude that there are more reals than naturals, the classical mathematical argument is:
> a) There are more functions N to Bool than naturals.
> b) There are as many reals as functions from N to Bool.
> Now, we of course agree the mistake is in b) not in a).
Given certain foundations, (a) is false. For example, in the Russian constructivist school (as in Andrey Markov Jr), functions only exist if they are computable, and there are only countably many computable functions from N to Bool. More generally, viewing functions as sets, if you sufficiently restrict the axiom schema of separation/specification, then only countably many sets encoding functions N-to-Bool exist, rendering (a) false
Indeed, what you write is true from an external point of view; just note that within this flavor of constructive mathematics, the set of functions from N to Bool is uncountable again.
There is no paradox: Externally, there is an enumeration of all computable functions N -> Bool, but no such enumeration is computable.
Is it internally uncountable in the strong sense that the system can actually prove the theorem “this set is uncountable”, or only in the weaker sense that it can’t prove the theorem “this set is countable”, but can’t prove its negation either?
If the latter, what happens if you add to it the (admittedly non-constructive) axiom that the set in question is countable?