I want to prove that if $A$ in an infinite set, then the cartesian product of $A$ with 2 (the set whose only elements are 0 and 1) is equipotent to $A$.
I'm allowed to use Zorn's Lemma, but I can't use anything about cardinal numbers or cardinal arithmetic (since we haven't sotten to that topic in the course).
I read a proof of the fact that if $a$ is an infinite cardinal number, then $a+a=a$, which is something similar to what I want to prove.
Any suggestions will be appreciated 🙂
Best Answer
Consider the collection $S$ of pairs $(X,f)$ where $X\subseteq A$ is infinite and $f:X\times 2\hookrightarrow X$. Then $S$ is nonempty, since $A$ being infinite contains a copy of $\Bbb N$, and it is known $\Bbb N\times 2\simeq \Bbb N$. It should be evident that under the ordering of extension, this set always has $(X,f)\in S$ if $X=\bigcup X_i$ and $f=\bigcup f_i$ with each $(X_i,f_i)\in S, i\in I$ and $I$ a total order. It follows by Zorn's lemma that there is a maximal element $(A',g)$ in $S$. If we show that $A'$ has the same cardinality as $A$, we're done. Now it is clear $\# A'\leqslant \# A$ since $A'$ is a subset of $A$. So assume $\#A'<\#A$. Then $A\setminus A'$ must be infinite, since else we would have by $A=A'\sqcup (A\setminus A')$ an equality (recall that if $\mathfrak a$ is infinite and $\mathfrak b$ finite, $\mathfrak a+\mathfrak b=\mathfrak a$) , so there is some $A''\subset A$ disjoint from $A'$ with cardinality $\aleph_0$. But then by patching things together, we would be able to extend the injection $A'\times 2\to A'$ to a larger $(A'\sqcup A'')\times 2\to (A'\sqcup A'')$.