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blueplanet200 | 6 months ago
Math is math, if you start with ZFC axioms you get uncountable infinites.
Maybe you don't start with those axioms. But that has nothing to do with truth, it's just a different mathematical setting.
blueplanet200 | 6 months ago
Math is math, if you start with ZFC axioms you get uncountable infinites.
Maybe you don't start with those axioms. But that has nothing to do with truth, it's just a different mathematical setting.
andrewla|6 months ago
So yes, generally not starting with ZFC.
I can't speak to "truth" in that sense. The skepticism here is skepticism of the utility of the ideas stemming from Cantor's Paradise. It ends up in a very naval-gazing place where you prove obviously false things (like Banach-Tarski) from the axioms but have no way to map these wildly non-constructive ideas back into the real world. Or where you construct a version of the reals where the reals that we can produce via any computation is a set of measure 0 in the reals.
CyLith|6 months ago
dullcrisp|6 months ago
axolotliom|6 months ago
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
SabrinaJewson|6 months ago
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.
blueplanet200|6 months ago
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.
orangecat|6 months ago