proof of Zermelo’s postulate

The following is a proof that the axiom of choiceMathworldPlanetmath implies Zermelo’s postulateMathworldPlanetmath.


Let be a disjoint family of nonempty sets. Let f: be a choice function. Let A,B with AB. Since is a disjoint family of sets, AB=. Since f is a choice function, f(A)A and f(B)B. Thus, f(A)B. Hence, f(A)f(B). It follows that f is injectivePlanetmathPlanetmath.

Let C={f(B):B}. Then C is a set.

Let A. Since f is injective, AC={f(A)}. ∎

Title proof of Zermelo’s postulate
Canonical name ProofOfZermelosPostulate
Date of creation 2013-03-22 16:14:25
Last modified on 2013-03-22 16:14:25
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 9
Author Wkbj79 (1863)
Entry type Proof
Classification msc 03E25