# 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 $f:\mathcal{F}\to {\displaystyle \bigcup \mathcal{F}}$ be a choice function. Let $A,B\in \mathcal{F}$ with $A\ne B$. Since $\mathcal{F}$ 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 $C=\{f(B)\in {\displaystyle \bigcup \mathcal{F}}:B\in \mathcal{F}\}$. Then $C$ is a set.

Let $A\in \mathcal{F}$. 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 |