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?
Sum of cardinals without AC
2 Answers
In the absence of choice we may suppose that $A$ is an amorphous set and $B=\big(A\times\{0\}\big)\cup\omega$. Suppose that $f:A\cup B\to B$ is a bijection. Let $C=f[A]$, $D=f[A\times\{0\}]$, and $E=f[\omega]$; then $\{C,D,E\}$ is a partition of $\big(A\times\{0\}\big)\cup\omega$. Since $A$ is amorphous, $E\cap\big(A\times\{0\}\big)$, $C\cap\omega$, and $D\cap\omega$ must be finite. Let $C_0=C\setminus\omega$, $D_0=D\setminus\omega$, and $F=\big(A\times\{0\}\big)\setminus E$; 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, $|A\cup B|\ne|B|$.
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0@Brian: Yes, of course. Thank you! – 2012-08-11
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, 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, 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, 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.