For some reason, I think this is true. I suppose it is clear why $|A| \leq |A_1 \cup A_2|$. I wonder if there's a way to show the existence of an injection $\phi: A_1 \cup A_2 \to A$.
I would appreciate any help. Thanks in advance.
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@lulu Sorry, I forgot to mention that these sets have to be infinite obviously. My bad. – Hoji Aug 20 '22 at 22:00
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no worries. $\quad$ – lulu Aug 20 '22 at 22:00
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2Tarski showed that "$S$ is always in bijection with $S\times S$" (for infinite $S$) is equivalent to the Axiom of Choice. Of course that implies your result, though perhaps one can get there without AOC. – lulu Aug 20 '22 at 22:03
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3Alas, no: Amorphous sets can be constructed if you drop choice. – lulu Aug 20 '22 at 22:06
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Does this answer your question? About a paper of Zermelo Even $A\times A$ is as big as $A$, for $2A$ it then follows from Schröder–Bernstein theorem. – Aug 20 '22 at 22:26
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Thank you guys. Your comments have been very helpful. I understand the answer now. – Hoji Aug 21 '22 at 20:16