> I don't see how you can be _skeptical_ of those ideas.

Well you can be skeptical of anything and everything, and I would argue should be.

Addressing your issue directly, the Axiom of Choice is actively debated: https://en.wikipedia.org/wiki/Axiom_of_choice#Criticism_and_...

I understand the construction and the argument, but personally I find the argument of diagonalization should be criticized for using finities to prove statements about infinities.

You must first accept that an infinity can have any enumeration before proving its enumerations lack the specified enumeration you have constructed.

https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument

> Math is math, if you start with ZFC axioms

This always bothers me. "Math is math" speaks little to the "truth" of a statement. Math is less objective as much as it rigorously defines its subjectivities.

https://news.ycombinator.com/item?id=44739315

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SabrinaJewson 7 days ago | parent | next [-]

> Addressing your issue directly, the Axiom of Choice is actively debated:

The axiom of choice is not required to prove Cantor’s theorem, that any set has strictly smaller cardinality than its powerset.

Actually, I can recount the proof here: Suppose there is an injection f: Powerset(A) ↪ A from the powerset of a set A to the set A. Now consider the set S = {x ∈ A | ∃ s ⊆ A, f(s) = x and x ∉ s}, i.e. the subset of A that is both mapped to by f and not included in the set that maps to it. We know that f(S) ∉ S: suppose f(S) ∈ S, then we would have existence of an s ⊆ A such that f(s) = f(S) and f(S) ∉ s; by injectivity, of course s = S and therefore f(S) ∉ S, which contradicts our premise. However, we can now easily prove that there exists an s ⊆ A satisfying f(s) = f(S) and f(S) ∉ s (of course, by setting s = S), thereby showing that f(S) ∈ S, a contradiction.

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codebje 7 days ago | parent [-]

Perhaps this is an ignorant question, but wouldn't you need AC to select the s ⊆ A whose existence the contradiction depends on? A constructive proof, at least the ones I'm trying to build in my head, stumbles when needing to produce that s to use in the following arguments.

	▲	SabrinaJewson 7 days ago \| parent [-]
		No, because you only have to choose _one_ s for the proof to work, and a finite number of choices is valid in intuitionistic and constructive mathematics.

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blueplanet200 7 days ago | parent | prev [-]

The axiom of choice is debated as a matter of if its inclusion into our mathematics produces useful math.

I don't think it's debated on the ground of if it's true or not.

And I was imprecise with language, but by saying "math is math" I meant that there are things that logically follow from the ZFC axioms. That is hard to debate or be skeptical of. The point I was driving was that it's strange to be skeptical of an axiom. You either accept it or not. Same as the parallel postulate in geometry, where you get flat geometry if you take it, and you get other geometries if you don't, like spherical or hyperbolic ones...

To give what I would consider to be a good counterargument, if one could produce an actual inconsistency with ZFC set theory that would be strong evidence that it is "wrong" to accept it.

	▲	egorelik 6 days ago \| parent [-]
		Skepticism of a ZFC axiom in particular could just be in terms of its standard status. I don't think anyone debates that ZFC in a particular logic doesn't imply this or that, but people can get into philosophical questions about whether it is the right foundation. There are also purely mathematical reasons to care - an extra axiom may allow you to produce more useful math, but it also potentially blocks you from other interesting math by keeping you out of models where, e.g., Choice is false.