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> The Cartesian product of nonempty sets is nonempty. > > This is obvious!, you might say — obviously we can just pick one element from each set and be done wit
by SabrinaJewson 1y ago
> The Cartesian product of nonempty sets is nonempty.
>
> This is obvious!, you might say — obviously we can just pick one element from each set and be done with it. But the statement that we can pick an element from each set is the axiom of choice.
No? I don’t see how this relates to AC at all. AC is about making an infinite number of choices at once – if you’re just making two choices (or, more generally any finite number of choices), as is needed here to prove that this Cartesian product is nonempty, then that’s completely fine without extra axioms. See for example https://mathoverflow.net/q/32538 https://mathoverflow.net/q/32538
E.g. in type theory, one term of type `Nonempty(A) → Nonempty(B) → Nonempty(A × B)` (supposing that `Nonempty` is defined as the [bracket type](https://ncatlab.org/nlab/show/bracket+type https://ncatlab.org/nlab/show/bracket+type)) would just be `λ [a] ↦ λ [b] ↦ [(a, b)]`.
- deleted 1y ago[deleted]
- gjm11 1y agoWhat AC's equivalent to is "the Cartesian product of any set of nonempty sets is nonempty". Not just of two nonempty sets, for which indeed you don't need AC.
- Kranar 1y agoThe two statements imply each other, they are logically equivalent: https://en.wikipedia.org/wiki/Product_topology#Axiom_of_choice https://en.wikipedia.org/wiki/Product_topology#Axiom_of_choi... >One of many ways to express the axiom of choice is to say that it is equivalent to the statement that the Cartesian product of a collection of non-empty sets is non-empty.