If $A$ and $B$ are infinite sets, is it true that the cardinality of its Cartesian product ($A \times B$) is equal to max $(A,B)$? If it is true, why? (We assume the axiom of choice)
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1With axiom of choice it is true in case of $A=B$. – AlvinL May 06 '20 at 20:07

1It is true if we assume the axiom of choice; see **MikeMathMan**’s answer to [this question](https://math.stackexchange.com/questions/1383755/cardinalityofthecartesianproductoftwoequinumerousinfinitesets?rq=1). It also implies the axiom of choice; see [this question](https://math.stackexchange.com/questions/56466/foreveryinfinitessstimessimpliestheaxiomofchoice?noredirect=1&lq=1). – Brian M. Scott May 06 '20 at 20:08
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Correct me please if I'm not right.
$A \times B \geq \max (A,B)$
$ A \cup B = \max (A, B)$
$ (A \cup B) \times (A \cup B) = A \cup B = \max (A, B)$
$A \times B \leq  (A \cup B) \times (A \cup B) = \max (A, B)$
$\max (A, B) \leq A \times B \leq \max (A, B)$
then $A \times B = \max (A, B)$
Chickenmancer
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The claim that $(A\cup B)\times(A\cup B)=A\cup B$ (and more generally, that for an infinite set $C$, $C\times C$ is bijectable with $C$, is equivalent to the Axiom of Choice (theorem of Tarski) – Arturo Magidin May 06 '20 at 21:47