Equivalence of Versions of Axiom of Choice/Formulation 1 implies Formulation 3

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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.

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

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 3

Let $\SS$ be a set of non-empty pairwise disjoint sets.

Then there is a set $C$ such that for all $S \in \SS$, $C \cap S$ has exactly one element.

Symbolically:

$\forall s: \paren {\paren {\O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O} \implies \exists c: \forall t \in s: \exists x: t \cap c = \set x}$


Proof

Let $\SS$ be the set:

$\SS = \set {s: \O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O}$

Let $c$ be a choice function on $\SS$ and consider the image set $c \sqbrk \SS$:

$c \sqbrk \SS = \set {\map c s: \O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O}$

By the definition of choice function:

$\map c s \in s$

By construction of $\SS$, for any $s \in \SS$:

$s \cap c \sqbrk \SS = \set {\map c s}$

$\blacksquare$


Sources