[SabrinaJewson]

> 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.

[codebje]

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]

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.