So, as the title says I'd like to give a proof of the fact that if $C$ is an infinite set then it is equipotent or equivalent to its cartesian product $C\times C$ using Zorn's Lemma (and of course some of its implications like the fact that $C$ has an infinite countable subset which I think might be very useful).
The main problem that I have is that I'm not supposed to use any theorem or result involving cardinal numbers since I'm still taking an elementary set theory course which hasn't covered that topic yet and all the proofs that I've read so far use cardinals arithmetic at some point.
Another thing that I think might be useful is a lemma which was proved via Zorn's Lemma in an answer to the question Prove that if $A$ is an infinite set then $A \times 2$ is equipotent to $A$ which states that given the infinite set $C$ there exists a non-empty set $B$ such that $B\times \mathbb{N}$ is equipotent to $C$. Then it suffices to give a bijection from $(B\times \mathbb{N})$ $\times$ $(B\times \mathbb{N})$ to $B\times \mathbb{N}$
So, any suggestion on the direct proof (without cardinals, unfortuntely) via Zorn's Lemma or an actual bijection from $(B\times \mathbb{N})$ $\times$ $(B\times \mathbb{N})$ to $B\times \mathbb{N}$ would be highly appreciated. Thanks in advance.
Best Answer
You can't give an actual bijection, since the theorem you are trying to prove is in fact equivalent to the axiom of choice and Zorn's lemma. So for some cases, like $\Bbb N$ or so we can write it down, but in general we can't.
You need to appeal to Zorn's lemma three times here:
Every infinite set has a countably infinite subset, which is quite straightforward.
If $X$ is an infinite set, then $|X|+|X|=|X|$. This can be done by considering $(A,f)$ such that $A\subseteq X$ and $f\colon A\times\{0,1\}\to A$ is a bijection, ordered by extension on both coordinates.
From this you can prove directly that $|X\times\mathbb N|=|X|$ for every infinite set $X$ (you can also try and prove that directly with Zorn's lemma, but I don't see an easy way to do so).
Now we can prove $|X|^2=|X|$ for all infinite $X$, by considering the partial order whose elements are $(A,f)$ such that $A\subseteq X$ and $f\colon A\times A\to A$ is a bijection. Again, order this by extension on both coordinates.
Now if $(A,f)$ is a maximal element, then $|A|=|X|$ or else there is a countably infinite set $B\subseteq X\setminus A$. Then you can prove that: $$X\times X=A\times A\cup A\times B\cup B\times A\cup B\times B$$ and conclude via cardinal arithmetic and the previous steps that $|X|=|A|$ after all. Therefore $|X|^2=|X|$.
While this path to a proof does rely on some cardinal arithmetic, it is relatively painless. The only difficult part is showing that $|X\times\Bbb N|=|X|$, which is an application of induction on the existing bijection $f\colon X\times\{0,1\}\to X$. Zorn's lemma won't be as helpful there.