Idempotent law of cardinal arithmetic

Question

I'm reading a (surprisingly short) proof of $\pi \times \pi = \pi$ if $\pi$ is an infinite cardinal from nLab (Theorem 3.1). (I call it surprising because I know that a traditional proof uses Zorn's lemma and is very long.)

However, I can't figure out where the proof uses the condition of infinite cardinal. And I can't see why the last line ``regarding $\alpha$ as an ordinal less than $\pi$'' is true (clearly $\alpha \leq \pi$ but why $\alpha \neq \pi?$).

I would be extremely appreciative of any assistance!

Answer 1

When the proof says

Thus $|\alpha|^2=|\alpha|$ for all ordinals $\alpha<\pi$.

it should state this only for infinite ordinals $\alpha<\pi$. This is because the statement being proved only applies to infinite cardinals, so assuming $\pi$ is a minimal counterexample only tells you that $\kappa^2=\kappa$ for all infinite cardinals $\kappa<\pi$.

So then, in the final step, in order to say that $|\alpha|^2=|\alpha|<|\pi|$, you need to know that $\alpha$ is infinite. Also, in order to say $|S(\alpha,\alpha,\alpha)|=|\alpha|^2$, you need to know that $\alpha$ is infinite. Indeed, $S(\alpha,\alpha,\alpha)$ is in bijection with $(\alpha+1)\times(\alpha+1)\setminus\{(\alpha,\alpha)\}$ by mapping $(x,y,z)\in S(\alpha,\alpha,\alpha)$ to $(y,z)$. When $\alpha$ is infinite, $|\alpha+1|=|\alpha|$ and so $S(\alpha,\alpha,\alpha)$ is in bijection with $\alpha^2$ with one element removed (which again does not change the cardinality since it is infinite), but this does not work if $\alpha$ is finite.

So, if $\alpha$ is finite, an alternate argument is needed. Assuming $\pi$ is infinite you can instead just say that if $\alpha$ is finite then $|\alpha|^2$ is finite and thus less than $|\pi|$. It is this alternate argument which fails if $\pi$ is finite.

(As for why $\alpha<\pi$, this is just because $(\alpha,\beta,\gamma)$ is some element of $\pi^3$ which means $\alpha\in\pi$, i.e. $\alpha<\pi$.)

Answer 2

FYI. Here is a different proof, by transfinite induction. Let $p$ be an infinite cardinal ordinal. Suppose that $|q\times q|<p$ for every ordinal $q\in p.$

We use the fact that if $<_W$ is a well-order on a set $A$ and if $|B|<p$ for every $<_W$-initial segment $B$ of $A$ then $|A|\le p.$

For $(x,y), (x',y')\in p\times p$ let $(x,y)<_W (x',y')$ iff

$(i).$ $\max (x,y)<\max (x',y')$ OR

$(ii).$ $\max (x,y)=\max (x',y')$ and $x<x'$ OR

$(iii)$. $\max (x,y)=\max (x',y')$ and $x=x'$ and $y<y'.$

That is, in $(ii)$ and $(iii)$, if $\max (x,y)=\max (x,y')$ then $(x,y), (x',y')$ are $<_W$-ordered lexicographically.

Confirm that $<_W$ is a well-order.

Now if $(x',y')\in p\times p$ and if $B=\{z:z<_W (x',y')\}$ is a $<_W$-initial segment, let $q=\max (x',y')+1$. Then $q\in p$ and $B\subseteq q\times q ,$ so $|B|\le |q\times q|< p.$

Therefore $[\,\forall q\in p\,(|q\times q|<p) \,] \implies |p\times p|=p.$

Now if there existed an infinite cardinal ordinal $p$ such that $|p\times p|>p$ then consider the $least$ such $p.$ The reasoning above shows there must exist some $q\in p$ with |$q\times q|\ge p,$ so $|q|$ cannot be a finite cardinal. But then $|(|q|\times |q|)|=|q\times q|\ge p >|q|$ contradicts the minimality of $p.$