TITLE: Sum of cardinals without AC
QUESTION [7 upvotes]: Let $A$ and $B$ be infinite sets. To show $|A\cup B|=\max\{|A|,|B|\}$ we need AC.
Now let us assume $|A|<|B|$. Can we show $|A\cup B|=|B|$ without AC?

REPLY [5 votes]: The axiom of choice is in fact equivalent to the assertion that $A+B=\max\{A,B\}$, for every two infinite cardinals.
To see the non-trivial implication, consider $A$ to be any set and $B=\aleph(A)$, the Hartogs number of $A$.
On the other hand, if we already assume that $A<B$, things may be a bit trickier, it requires some choice but not everything.

If there exists an infintie Dedekind-finite set (a set that is larger than any proper subset), then clearly $1<A$, but $A+1$ is strictly larger than $A$.
On the other hand, consider Sageev's model in which every infinite set has the property that $A+A=A$. In this model, however, there is a countable family without a choice function so not even countable choice holds.
Suppose $A<B$, then there is a subset $B'\subseteq B$ such that $A\sim B'$. In particular there is a bijective map from $A\cup B$ into $B$ which maps $A\cup B'$ onto $B'$. Therefore $A+B=B$. 

So without the axiom of choice we may have models in which this is true, and others in which this is false.