Sets of Cardinals without choice

Question

I have a theory that the axiom of choice is equivalent to the statement that sets of distinct cardinals are well ordered by cardinality. I can prove that the axiom of choice implies this. However I am having trouble proving the other way. I want to prove, without choice, that any set is in bijection with a set of distinct cardinals. I don't know if this is true, but if it is could someone provide a proof for me? Thank you.

Answer 1

If every set of cardinal numbers is well-ordered by cardinality, then in particular any two cardinals are comparable by cardinality (i.e., trichotomy holds).

To see that trichotomy implies the well-ordering theorem: Let $X$ be any set that we want to well-order, and let $\alpha$ be its Hartogs number, which is the first ordinal that cannot be injected into $X$. (The existence of such an ordinal can be proved without Choice; it is the union of the ordinals that represent the order types of relations that well-order subsets of $X$). Since by assumption $|\alpha|\le|X|$ is not true, trichotomy tells us that $|X|<|\alpha|$, so there exists an injection $X\to\alpha$, and therefore $X$ can be well-ordered.