Sum of cardinals without AC

Question

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?

Answer 1

In the absence of choice we may take $X$ be an amorphous set, let $A=X\times\{0\}$, and let $B=\big(X\times\{1\}\big)\cup\omega$. Suppose that $f:A\cup B\to B$ is a bijection. Let $C=f[A]$, $D=f[X\times\{1\}]$, and $E=f[\omega]$; then $\{C,D,E\}$ is a partition of $B=\big(X\times\{1\}\big)\cup\omega$.

Since $X$ is amorphous, so are $A$ and $X\times\{1\}$, so $E\cap\big(X\times\{1\}\big)$, $C\cap\omega$, and $D\cap\omega$ must be finite. Let $C_0=C\setminus\omega$, $D_0=D\setminus\omega$, and $F=\big(X\times\{1\}\big)\setminus E$; then $C_0,D_0$, and $F$ are amorphous, and $F$ is the disjoint union of $C_0$ and $D_0$, which is impossible. Thus, $|A\cup B|\ne|B|$.

Answer 2

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.