proof of Zermelo’s postulate

The following is a proof that the axiom of choice implies Zermelo’s postulate.

Proof.

Let $ℱ$ be a disjoint family of nonempty sets. Let $f:ℱ\to {\displaystyle \bigcup ℱ}$ be a choice function. Let $A,B\in ℱ$ with $A\ne B$. Since $ℱ$ is a disjoint family of sets, $A\cap B=\mathrm{\varnothing }$. Since $f$ is a choice function, $f(A)\in A$ and $f(B)\in B$. Thus, $f(A)\notin B$. Hence, $f(A)\ne f(B)$. It follows that $f$ is injective.

Let $C=\{f(B)\in {\displaystyle \bigcup ℱ}:B\in ℱ\}$. Then $C$ is a set.

Let $A\in ℱ$. Since $f$ is injective, $A\cap C=\{f(A)\}$. ∎

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