# Division of cardinals

Given two distinct infinite cardinals, $\mu<\pi$, Wikipedia states that $\kappa=\pi$ is the only possible solution of the equation $\mu\cdot\kappa=\pi$, so that one could say that $\pi/\mu=\pi$. It also states that this relies on the axiom of choice.

My question is, how can one see this?

If I have $\lvert X\rvert=\pi$ and $\lvert Y\rvert=\mu$, how can I describe and prove the existence of an injection from $X\times Y$ to $X$, using the axiom of choice?

(This question arose out of some considerations related to this post).

Note that the axiom of choice implies that $|X|=|X\times X|$.

Since there is an injection from $X\times Y$ into $X\times X$ (assuming that $|Y|\leq|X|$, as you did), this is an easy consequence of the Cantor-Bernstein theorem.

One can also try and do it a bit more explicitly if $X$ and $Y$ are not just any sets, but ordinals. Then the axiom of choice is used to find the bijection between $X$ and $Y$ and the ordinals.

(You may want to read How to divide aleph numbers as well.)

• Can you elaborate or suggest pointers for that first sentence? I found that this statement relates to Tarski's theorem but there the implication is reversed (AC is the consequence), and I could not (yet) find an intuitive explanation as to why AC should imply $\lvert X\rvert=\lvert X\times X\rvert$. The rest is very clear, thank you very much. – MvG Sep 5 '14 at 13:56
• math.stackexchange.com/questions/608538/… (you can look at the linked questions too) has information on how to do it for ordinals. Then with the axiom of choice you can do it for every infinite set by well-ordering; or you can do it directly using Zorn's lemma (you just need the fact that infinite sets have countably infinite subsets which you can prove separately). – Asaf Karagila Sep 5 '14 at 14:07
• I think that you should point not that choice is not required for $\kappa^2=\kappa$. For sets $X$ choice well orders the set and reduced it to the last case. – Rene Schipperus Sep 5 '14 at 14:36
• @Rene: If you assume that $\kappa$ is an ordinal, sure. If you just use it to denote some arbitrary cardinal, then no. I did mention that for ordinals you get it for free, it's in the third sentence of the answer. – Asaf Karagila Sep 5 '14 at 14:39
• My comment was not directed to you so much as wikapedia. – Rene Schipperus Sep 5 '14 at 14:51