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

Question

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

Answer 1

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)

Answer 2

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