# proof of Zermelo’s postulate

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

###### Proof.

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

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

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

Title proof of Zermelo’s postulate ProofOfZermelosPostulate 2013-03-22 16:14:25 2013-03-22 16:14:25 Wkbj79 (1863) Wkbj79 (1863) 9 Wkbj79 (1863) Proof msc 03E25