PlanetMath: proof of Zermelo's postulate

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (200)
Orphanage
Unclass'd
Unproven (490)
Corrections (6)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

proof of Zermelo's postulate

(Proof)

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

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

is injective.

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

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

"proof of Zermelo's postulate" is owned by Wkbj79.

(view preamble)

This object's parent.

Log in to rate this entry.
(view current ratings)

Cross-references: injective, choice function, disjoint, Zermelo's postulate, implies, axiom of choice, proof
There is 1 reference to this entry.

This is version 6 of proof of Zermelo's postulate, born on 2006-09-13, modified 2007-05-30.
Object id is 8342, canonical name is ProofOfZermelosPostulate.
Accessed 797 times total.

Classification:

AMS MSC:

03E25 (Mathematical logic and foundations :: Set theory :: Axiom of choice and related propositions)

Pending Errata and Addenda

None.

Discussion

forum policy

No messages.

Interact

post | correct | update request | add example | add (any)