##
**Axioms of multiple choice.**
*(English)*
Zbl 0134.24805

In this paper the author proves certain theorems concerning the relative strength and weakness of various axioms of multiple choice. Among other things the author gives an affirmative answer to a question posed by the reviewer [Fundam. Math. 49, 11–14 (1960; Zbl 0113.245)], who himself was originally motivated by an abstract of R. L. Vaught [On the equivalence of the axiom of choice and a maximal principle. Bull. Am. Math. Soc. 58, 66 (1952; doi:10.1090/S0002-9904-1952-09549-5)].

Let the axiom of choice be abbreviated as AC. Let \(Z(n)\) express the fact that on every set \(x\) of non-void sets there exists a function \(f\) such that, for every \(y\in x\), \(0\ne f (y)\subset y\) and there exists a one-to-one mapping of \(f(y)\) into \(n\). Let \(Z(\infty)\) differ from \(Z(n)\) only in that the sets \(f(y)\) are assumed to be finite. Let \(C(\infty)\) express the fact that on every set \(x\) of non-void finite sets there exists a function \(g\) such that \(g(y)\in y\) for every \(y\in x\). Let \(C(n)\) differ from \(C(\infty)\) only in that every \(y\in x\) is assumed to have exactly \(n\) elements.

The author shows that each \(Z(n)\) is equivalent with AC, and so is \((\exists n) Z(n)\). However, \(Z(\infty)\) does not imply AC in a suitable set theory with urelements. Also, while \(Z(\infty)\wedge C(\infty)\) implies AC, \(Z(\infty)\wedge (\forall n) C(n)\) does not imply AC. A corollary of this fact is that \((\forall n) C(n)\) does not imply \(C(\infty)\). The methods used in the metamathematical part of this paper originated with Fraenkel and were developed by Mostowski.

Let the axiom of choice be abbreviated as AC. Let \(Z(n)\) express the fact that on every set \(x\) of non-void sets there exists a function \(f\) such that, for every \(y\in x\), \(0\ne f (y)\subset y\) and there exists a one-to-one mapping of \(f(y)\) into \(n\). Let \(Z(\infty)\) differ from \(Z(n)\) only in that the sets \(f(y)\) are assumed to be finite. Let \(C(\infty)\) express the fact that on every set \(x\) of non-void finite sets there exists a function \(g\) such that \(g(y)\in y\) for every \(y\in x\). Let \(C(n)\) differ from \(C(\infty)\) only in that every \(y\in x\) is assumed to have exactly \(n\) elements.

The author shows that each \(Z(n)\) is equivalent with AC, and so is \((\exists n) Z(n)\). However, \(Z(\infty)\) does not imply AC in a suitable set theory with urelements. Also, while \(Z(\infty)\wedge C(\infty)\) implies AC, \(Z(\infty)\wedge (\forall n) C(n)\) does not imply AC. A corollary of this fact is that \((\forall n) C(n)\) does not imply \(C(\infty)\). The methods used in the metamathematical part of this paper originated with Fraenkel and were developed by Mostowski.

Reviewer: C.-C. Chang (M. R. 25 # 2960)

### MSC:

03E25 | Axiom of choice and related propositions |