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Is the class of cardinals totally ordered?

Intuitively, it seems like for any sets $A,B$ either $\lvert A\rvert\leq \lvert B\rvert$ or $\lvert B \rvert \leq \lvert A\rvert$. How can I prove this?

Using the definition of cardinality, the problem reduces to proving that for all sets $A,B$, there is either an injection from $A$ to $B$ or from $B$ to $A$. However, I don't see how to proceed from there. Is AC necessary?

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tba
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    Yes, AC is necessary. – Brian M. Scott Aug 10 '12 at 07:17
  • The proposition that cardinality of any two sets is comparable is equivalent to the axiom of choice. If every set can be well-ordered, then cardinality of any two sets is comparable. And if there exists some set which can't be well-ordered, then by Hartogs theorem there exists a well-ordered set whose cardinality is incomparable to it. (The proposition that every set can be well-ordered is equivalent to the axiom of choice.) – Mike Rosoft Dec 08 '19 at 14:25

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Yes, the axiom of choice is necessary. Without it you can have an infinite, Dedekind-finite set. If $A$ is such a set, there is no injection of $\omega$ into $A$ and no injection of $A$ into $\omega$.

Brian M. Scott
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