Does the assertion that every two cardinalities are comparable imply the axiom of choice?

If, for any two sets $A$ and $B$, Either $|A|<|B|, |B|<|A|$ or $|A|=|B|$ holds, does the axiom of choice holds? Why?

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This is Hartogs theorem.

Suppose that $A$ is a set, let $\aleph(A)$ be the minimal ordinal $\alpha$ such that $|\alpha|\nleq|A|$. We cannot have that $\aleph(A)\leq|A|$, so if we assume that all cardinalities are comparable we have to have that $|A|<\aleph(A)$. This means that $A$ can be well ordered, as it can be injected into an ordinal.

It is also known that if every set can be well ordered then the axiom of choice holds.

(The ordinal $\aleph(A)$ is known as the Hartogs number of $A$ and it plays an important role in many of these constructions)

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The details can be found online in this paper by Leonard Gillman in the Monthly; see Section 7. – Brian M. Scott May 31 '12 at 6:47
@Brian: I believe that I wrote many of these things on this site before. Some even more than once. – Asaf Karagila May 31 '12 at 6:48
I shouldn’t be surprised. But the Gillman reference came to hand immediately, so I used it. It also has the virtue of being a largely self-contained full discussion in one place. – Brian M. Scott May 31 '12 at 6:52

This is the famous Trichotomy. It implies the Axiom of Choice. You can find a partial list of equivalents of AC in Wikipedia.

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