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I'm having trouble proving that if $A$ is an infinite set, then $A\approx A\times A$ using Zorn's Lemma directly.

My problem is I don't know how to choose the correct set to which apply Zorn; I have tried $$X=\{f_{B}\colon B\times B\longrightarrow B\mid B\subseteq A,\ f_{B}\ {\rm injective}\}$$ since it's obvious that $A\preceq A\times A$, and with $(X,\leq)$ (the order $f_{B}\leq f_{B'}$ if $B\subseteq B'$ and $f_{B'}$ restricted to $B\times B$ is $f_{B}$) I manage to get a maximal function of that kind, but then I can't "expand it" as usual to prove it must have domain $A\times A$. Do you know which set $X$ should I work with?

valkyriebel
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So, you're ordering such injective functions $f_B: B \times B \rightarrow B$ by the relation that $f_B \leq f_{B'}$ if and only if $B \subseteq B'$ and $f_{B'}$ restricts to $f_B$ on $B \times B$, right?

By Zorn's lemma, there is a maximal such injective function $f_B: B \times B \rightarrow B$. If $B$ is properly contained in $A$, there are two cases (i): $A -B$ is finite or (ii): $A-B$ is infinite. In the first case, you're already done, because $B$ and $A$ would have the same cardinality. In the second case, you have plenty of leeway to extend $f_B$ to an injective function on $C \times C \rightarrow C$, where $C - B$ is infinite.

D_S
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    Yes, I'm using that order. As for the second case, I don't quite get how to extend $f_{B}$ since I would have to inject at least $(C-B)\times(C-B)$ (infinite) into $C-B$... isn't that in a way what I'm trying to prove possible? – valkyriebel Jun 07 '21 at 21:15
  • @maryvwxz But this time you have control over the cardinality of $C-B$, by picking a nice $C$. So you can let it be $\aleph_0$. Which is a case you need to prove directly anyway, as you need to show your partially ordered set is non-empty in order to use Zorn's lemma. – Arthur Jun 07 '21 at 21:56
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    Still, I get stuck at considering that $B$ might not be countable and so $(C-B)\times B$ would be an uncountable subset of $(C\times C)-(B\times B)$ that I need to inject in $C-B$ which is in this case countable – valkyriebel Jun 07 '21 at 22:36