# How do you prove the trichotomy law for cardinal numbers?

Law of trichotomy is that for any two cardinals $a$ and $b$, exactly one of the conditions $a<b$, $a=b$, or $a>b$ holds.

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Depends on your starting axioms, and your definition of "cardinal numbers". – Cameron Buie Aug 14 '12 at 9:15
That any two cardinals (as opposed to $\aleph$s) are $\leq$ comparable is equivalent to the Axiom of Choice. The anti-symmetry of the relation is the Cantor–Bernstein–Schroeder Theorem. – inactive... for now Aug 14 '12 at 9:16
The book that I'm reading states it as a theorem and the author only comments that the proofs uses tools of transfinite induction and that it's one of the direct consequences of the well-ordering theorem. – Mark Aug 14 '12 at 9:18

First let us prove this only for $\aleph$ numbers.

Suppose that $\aleph_\alpha$ and $\aleph_\beta$ are two cardinals. We take $\omega_\alpha$ and $\omega_\beta$ to be the corresponding initial ordinals (i.e. transitive sets which are well-ordered by $\in$, and are not in bijection with any of their members).

Since every two ordinals are comparable in $\subseteq$, either $\omega_\alpha\subseteq\omega_\beta$, and then $\aleph_\alpha\leq\aleph_\beta$; either $\omega_\beta\subseteq\omega_\alpha$ and then $\aleph_\beta\leq\aleph_\alpha$.

If both things occur then $\omega_\alpha=\omega_\beta$ by extensionality and therefore $\aleph_\alpha=\aleph_\beta$.

Now comes the point where we use the axiom of choice (or rather the well-ordering theorem), given two infinite sets they can be put in bijection with an ordinal and therefore with an initial ordinal. The above shows that the trichotomy holds.

See that as a direct and nice corollary we have the Cantor-Bernstein theorem. Take two [infinite] sets, use bijections to replace them momentarily with the corresponding initial ordinals, and you will have that the initial ordinals are equal and the bijections you used can form the bijection between your two original sets.

(This, to my knowledge, was Cantor's original proof of the theorem)

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If $A$, $B$ are any well-ordered sets then one of them is isomorphic to initial segment of the other one. Some authors call this fundamental theorem for well-ordered sets, e.g. Suppes: Axiomatic Set Theory, p.233, Theorem 80.

If you know from well-ordering theorem there for any two sets $A$, $B$ there is a well-order than the above implies that there is injective map either from $A$ to $B$ or the other way round. (Simply take arbitrary well-orderings of $A$ and $B$ and then there exists an isomorphism between one of the sets and a subset -- an initial segment -- of the other one. This isomorphism is necessarily a bijection.)

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