# Sets of Cardinals without choice

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.

• The question of whether (without choice) "any set is in bijection with a set of distinct cardinals" is very interesting, and as far as I know, it is open. I strongly suspect the answer is no, even if we relax the requirement to "For every $X$ there is a set $Y$ of distinct cardinalities, and a surjection from $Y$ onto $X$". – Andrés E. Caicedo May 9 '14 at 2:08
• Isn't it true that every set $A$ of cardinals is well-ordered ? (the order of the ordinals restricted to $A$). If so, the statement "any set is in bijection with a set of distinct cardinals" should imply the axiom of choice, right ? – Archimondain May 9 '14 at 3:06
• @Archimondain: Every set of cardinals is wellordered if you assume choice; without choice they needn't even be linearly ordered. – Malice Vidrine May 9 '14 at 3:20
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.