# Sum of cardinals without AC

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?

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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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I think the following is easier: Let $A$ be an amorphous set. Clearly $|A \times \{0\}| \leq |A \times \{1\}|$. However $|A \times \{0\} \cup A \times \{1\}| \neq |A \times \{1\}| = |A|$ because if there was a bijection $f$, then $A \times \{1\} = f(A \times \{0\}) \cup f(A \times \{1\})$. Hence $A$ is the union of two infinite disjoint subset. –  William Aug 10 '12 at 9:44
@William: Yes, that’s easier; I just didn’t think of it right away, and I did think of the other example. –  Brian M. Scott Aug 10 '12 at 9:48
@William and Brian: Doesn't William's example show that without AC $x+x=x$ need not hold for an infinite cardinal $x$? Brian's example answers my question. –  Joel Adler Aug 10 '12 at 10:24
@Joel: Now that I look more closely, I see that William’s example doesn’t quite answer your question, because it has $|A|\le|B|$ rather than $|A|<|B|$. –  Brian M. Scott Aug 10 '12 at 10:31
@Brian: I don't see, why $E$ has to be finite. But independent of the finiteness of $E\cap (A\times\{0\})$ the partition of the amorphous set $A\times\{0\}$ into the disjoint infinite sets $C_0$ and $D_0\cup F$, where $F$ can also be finite, which is impossible. Thanks a lot for your answer! –  Joel Adler Aug 10 '12 at 14:07

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.

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