There are many equivalent statements of the axiom of choice. One of them is that every set can be well-ordered.
I would like to state the axiom of choice as
Trichotomy holds for cardinality of sets.
What I mean by this is that if A and B are sets, (and the vertical bars denote cardinality,) then one and only one of the following three statements are true:
- $|A|<|B|$;
- $|A|=|B|$;
- $|A|>|B|$.
Definitions
- $|A|\le|B|$ if and only if there exists an injective function $$F:A\rightarrowtail B.$$
- $|A|<|B| := |A|\le|B|\land\lnot|B|\le|A|$;
- $|A|=|B| := |A|\le|B|\land|B|\le|A|$, (note Cantor-Bernstein theorem);
- $|A|>|B| := \lnot|A|\le|B|\land|B|\le|A|$.
Choice implies trichotomy
If the axiom of choice is accepted, then every set S can be well-ordered, and in fact brought into one-to-one correspondence with an ordinal number, (of the class ON.) Of any two ordinal numbers, one is included as a set in the other. Therefore if $\alpha$ and $\beta$ are ordinal numbers such that $\alpha\subseteq\beta$, there exists an injective function, namely a restriction of the identity function: $$I|_\alpha:\alpha\rightarrowtail\beta.$$ If A and B are sets, then there are ordinal numbers $\alpha$ and $\beta$ such that bijections $$G:\alpha\rightarrow A$$ $$H:\beta\rightarrow B$$ exist.
If $\alpha\subseteq\beta$, then $$H\circ I|_\alpha\circ G^{-1}:A\rightarrowtail B$$ is an injective function and $|A|\le|B|$.
Otherwise $\beta\subseteq\alpha$ and $$G\circ I|_\beta\circ H^{-1}:B\rightarrowtail A$$ is an injective function and $|B|\le|A|$.
Negation of choice implies that trichotomy fails
When we reject the axiom of choice we have to accept the existence of a set S that cannot be well-ordered. In fact there is no injective function from such a set S to any ordinal number, because even without the axiom of choice, every set of ordinal numbers can be well-ordered.
Suppose there is some ordinal number $\sigma\in{\bf ON}$ such that no injective function $F:\sigma\rightarrowtail S$ exists. Then neither $|\sigma|\le|S|$ nor $|S|\le|\sigma|$, and so trichotomy fails.
Otherwise for all $\sigma\in{\bf ON}$, there exists an injective function $$F_\sigma:\sigma\rightarrowtail S,$$ and $$\bigcup\limits_{\sigma\in{\bf ON}}F_\sigma`(\sigma)$$ (where the tick mark $`$ denotes the image under the function) is at the same time a proper class and a subset of the set S, which would violate the axiom schema of comprehension.
Question Is this proof valid? Is it well known? Is there a simpler or better way to express this? Are there statements I've made that need to be better qualified?