Equivalence of Formulations of Axiom of Choice/Formulation 1 iff Formulation 4

From ProofWiki
Jump to navigation Jump to search

Theorem

The following formulation of the Axiom of Choice:

Formulation 1

For every set of non-empty sets, it is possible to provide a mechanism for choosing one element of each element of the set.

$\ds \forall s: \paren {\O \notin s \implies \exists \paren {f: s \to \bigcup s}: \forall t \in s: \map f t \in t}$

That is, one can always create a choice function for selecting one element from each element of the set.


implies the following formulation of the Axiom of Choice:

Formulation 4

Let $A$ be a non-empty set.

Then there exists a mapping $f: \powerset A \to A$ such that:

for every non-empty proper subset $x$ of $A$: $\map f x \in x$

where $\powerset A$ denotes the power set of $A$.


Proof

We note from Set equals Union of Power Set that:

$x = \ds \map \bigcup {\powerset x}$

Setting $\powerset A =: s$, we see that from Formulation $1$:

$\ds \paren {\O \notin \powerset A \implies \exists \paren {f: \powerset A \to \bigcup \powerset A}: \forall x \in \powerset A: \map f x \in x}$


That is:

for every non-empty proper subset of $A$

there exists a mapping $f: \powerset A \to A$ such that:

for every non-empty proper subset $x$ of $A$: $\map f x \in x$.

That is Formulation $4$ of the Axiom of Choice

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Formulations_of_Axiom_of_Choice/Formulation_1_iff_Formulation_4&oldid=630468"