22 февр. 2018 г. · Once you have f, use the assumption that A is nonempty to choose x∈A. Now, you can define a sequence recursively by a0=x, an+1=f(an).

23 февр. 2018 г. · The axiom of dependent choice states that this [Theorem 1] is always possible. But I can only infer Theorem 1 from Axiom of Choice, not from Axiom of Dependent ...

12 мар. 2018 г. · I have formalized Noah Schweber's idea to prove that Axiom of Dependent Choice implies Axiom of Countable Choice, but I'm not sure if my ...

5 июн. 2017 г. · DC implies CC (Countable Choice). CC is: If f is a set-valued function with dom(f)=ω, and f(n)≠ϕ for all n, then ∏n∈ωf(n)≠ϕ.

10 мар. 2018 г. · Countable choice is exactly the special case of dependent choice in which the later choices don't depend on the previous ones.

11 янв. 2022 г. · Axiom of dependent choice (DC): Let A be a non-empty set and R⊆A×A satisfy ∀a∈A∃b∈A:aRb. Then there exists a sequence (an)n such that anRan+1∀n.

9 авг. 2015 г. · the generalized axiom of dependent choice implies the axiom of choice on the ordinals (although it is not clear if they are also equivalent).

10 апр. 2015 г. · The failure of the principle of dependent choice means not just that equivalent statements fail, but also that anything stronger than it will fail.

24 апр. 2018 г. · R satisfies the requirement of Axiom of Dependent Choice. Hence there exists a sequence (xi∣i∈N) such that xiRxi+1 where for some n, xi∈Ai+n for ...

31 мар. 2019 г. · There are weaker versions of choice axioms, and among them, DC(Axiom of Dependent Choice) is heavily used in, for example, recursive defining ...