Godel's pairing function and proving c = c*c for aleph cardinals

Solution 1:

I think that you miss a certain point there. The proof is not that the ordinal $P(\alpha,\beta)\leq\max\{|\alpha|,|\beta|\}^2$, but rather that $|P(\alpha,\beta)|$ is less than that.

The inequality is of cardinalities, not order types. And indeed, note that $P(\alpha,\beta)$ when both are countable, is a countable ordinal.

Now we can go about our induction at peace. Suppose that for all $\lambda<\kappa$ we prove that $\lambda\times\lambda=\lambda$, and again -- that equality is of cardinals!

Now we will show that the order type of $\kappa\times\kappa$ (as a set of pairs of ordinals!) is $\kappa$, but that much is easy. Given any point $(\alpha,\beta)\in\kappa\times\kappa$ then by the induction hypothesis we have that $P(\alpha,\beta)\leq\lambda$, where $\lambda=\max\{|\alpha|,|\beta|\}$. Therefore the order type of $P(\alpha,\beta)$ is smaller than the order type of $\kappa$.

It is not hard to show that $\kappa$ is the unique well-order that every proper initial segment has size $<\kappa$, but the order itself has cardinality of at least $\kappa$. Since clearly $\kappa\times\kappa$ has at least $\kappa$ many elements, but every initial segment must have less, $P$, restricted to $\kappa\times\kappa$ is in fact a bijection between $\kappa$ and $\kappa\times\kappa$ as wanted.

Note that you have to keep close attention on when the discussion is about cardinals and when it is about ordinals, and sometimes it is just about sets of ordinals. That's a delicate point, and one of the less fortunate typecasts and overloads in set theory; but on the other hands that same less fortunate feature sometimes allows us to save on notation and simplify things.

Also related: For all infinite cardinals $\kappa, \ (\kappa \times \kappa, <_{cw}) \cong (\kappa, \in).$

If you don't want that, you can use a Zorn-like argument, mixing both transfinite induction and "fun". The point is that we don't actually appeal to Zorn's lemma (which would have made the whole proof much shorter), but rather prove the existence of a maximal element directly, or at least a "maximal enough" element.

Suppose that for all $\lambda<\kappa$ we prove that $\lambda\times\lambda=\lambda$. Let $P$ be the partial order whose elements are $(A,f)$ where $A\subseteq\kappa$ and $f\colon A\to A\times A$ is a bijection. We say that $(A,f)\leq(B,g)$ if $A\subseteq B$ and $f\subseteq g$.

By the induction hypothesis this set is not empty. Also note that every chain has a least upper bound (the union of each coordinate in the elements of the chain, of course). So it suffices to find a maximal chain.

Let $I_0$ be any chain, and assume that it includes its least upper bound $(A_0,f_0)$. We construct by induction chains, if $A_0$ has cardinality $\kappa$ we are done, since this means that there is a bijection from $\kappa$ to $\kappa\times\kappa$. Suppose not, let $\alpha_0$ be the least not in $A_0$. consider $A'_0=A_0\cup\{\alpha\}$, then we have (one can, and should verify these equalities for sets of ordinals in $\sf ZF$): $$|A'_0|\cdot|A'_0|=|A_0+1|\cdot|A_0+1|=|A_0^2+2\cdot A_0+1|=|A_0|.$$

Therefore there exists $f'_0\colon A'_0\to A'_0\times A'_0$. Moreover we can ensure that such $f'_0$ extends $f_0$, by combinatorial juggling of elements and basic cardinal arithmetics. Then we extend $I_1$ to be $I_0\cup\{(A'_0,f'_0)\}$.

Now continue with the induction, at limit stages take the union of previous chains and add the least upper bound.

After $\kappa$ steps we must have exhausted our possible domains, and have constructed a maximal element whose domain is a set of size $\kappa$, and therefore we have the wanted conclusion: $\kappa$ and $\kappa\times\kappa$ are equipollent.