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

Question

I don't understand the proof to the a/m claim. How we know that $\eta < \kappa$ and $(\alpha,\beta)<_{cw} (0, \eta) $ and hence $h: \mu \to \eta \times \eta$ is injective? Appreciate if anyone could advise.

Here is an excerpt of the proof:

For an infinite ordinal $\kappa,$ the canonical well-ordering of $\kappa \times \kappa,$ denoted by $<_{cw}$ is defined as follows: $(\alpha_1, \beta_1)<_{cw} < (\alpha_2, \beta_2)$ iff either one of the following holds.

$(1): \max\{\alpha_1,\beta_1\} < \max\{\alpha_2,\beta_2\}$

$(2): \max\{\alpha_1,\beta_1\} =\max\{\alpha_2,\beta_2\}$ and $\alpha_1<\alpha_2$

$(3): \max\{\alpha_1,\beta_1\} =\max\{\alpha_2,\beta_2\}$ and $\alpha_1 = \alpha_2$ and $\beta_1 < \beta_2.$

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

Proof: (It has been show that it is true for $\kappa = \omega$) Assume there is a counterexample $\kappa.$ Let $A= \{\lambda \in \kappa + 1| \lambda\geq \omega \ is \ a \ cardinal \ and \ (\lambda \times \lambda, <_{cw}) \not\cong (\lambda, \in)\}.$ Then, $\kappa \in A.$ Let $\mu$ be least element of $A.$ Thus, $\mu > \omega$ and for all $\lambda,$ if $\omega \leq \lambda < \mu$ and $\lambda$ is a cardinal, then $(\lambda \times \lambda, <_{cw}) \cong (\lambda, \in).$

By Comparability theorem, $(\mu, \in)$ is isomorphic to initial segment of $(\mu \times \mu, <_{cw})$ since $|\mu \times \mu| \geq \mu$ and $\mu$ is cardinal. Let $(\alpha, \beta) \in \mu \times \mu$ such that $(\mu, \in)$ is isomorphic to initial segment of $(\mu \times \mu, <_{cw})$ given by $(\alpha, \beta).$ Let $h$ be the isomorphism. Let $\eta= \max\{\alpha, \beta\}+ \omega.$ Then $\eta < \kappa$ and $(\alpha,\beta)<_{cw} (0, \eta).$ HENCE, $h: \mu \to \eta \times \eta$ is injective.

Answer 1

The proof is really by induction. It goes through the cardinals, and if the claim is false then there is a least one which is a counterexample. Moreover this cardinal is not $\omega$ since for $\omega$ we can check this explicitly. We write $\mu$ for that least counterexample.

Because every two well orders are comparable in the relation "isomorphic to an initial segment", $(\mu,\in)$ and $(\mu\times\mu,<_{cw})$ are comparable. So one is an initial segment of the other, perhaps a proper one. By the fact that $\mu$ is a cardinal and $|\mu|\leq|\mu\times\mu|$ it is impossible that the latter is a proper initial segment of the former, and it is necessary that $\mu$ is an initial segment of $\mu\times\mu$.

But by our assumption that the claim fails at $\mu$, $(\mu,\in)$ is only isomorphic to some proper initial segment. So there is some $\eta<\mu$ such that there is an embedding from $(\mu,\in)$ into $(\eta\times\eta,<_{cw})$ -- this is true because you can always find $\eta$ large enough for that: if $\mu$ is isomorphic to the initial segment below $(\alpha,\beta)$ then taking $\eta=\max\{\alpha,\beta\}+\omega$ we guarantee that $\eta\times\eta$ contains all the information we want. Therefore there is some injection from $\mu$ into $\eta$, which is a contradiction to the fact that $\mu$ is a cardinal and $\eta<\mu$.

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Another way of looking at it is to see that an ordinal $\alpha$ is an initial ordinal, or cardinal, if and only if it satisfies the following property:

Every proper initial segment of the order $(\alpha,\in)$ has a strictly smaller cardinality than $\alpha$.

The proof can be given in a positive way instead.

Suppose that the claim is true for all infinite $\lambda<\mu$, it is clear that $(\mu,\in)$ is isomorphic to an initial segment of $(\mu\times\mu,<_{cw})$, by the fact that $\mu$ is an initial ordinal.

It suffices to show that every proper initial segment of $\mu\times\mu$ has cardinality $<\mu$. But if $(\alpha,\beta)$ is any point in $\mu\times\mu$, and $\eta=\max\{\alpha,\beta\}$ then $(\alpha,\beta)\leq_{cw}(\eta,\eta)=\max(\eta+1)\times(\eta+1)$, where the last $\max$ is in the $<_{cw}$ order.

Since $|\eta|=|\eta+1|=\lambda<\mu$, for some $\lambda$, by the induction hypothesis $|\eta\times\eta|=|\lambda\times\lambda|=|\lambda|=\lambda$.

It follows, if so, that every proper initial segment of $\mu\times\mu$ must have cardinality strictly less than $\mu$, and therefore $\mu\times\mu$ is order isomorphic to $\mu$.

Answer 2

We have $\eta < \kappa$ because $\alpha, \beta < \kappa$ (hence $\max\{\alpha,\beta\} < \kappa$) and $\kappa > \omega$ is a cardinal. (Consider the cardinality of $\eta$, for instance.)

We have $(\alpha, \beta) <_{cw} (0, \eta)$ because $\max\{\alpha, \beta\} < \eta = \max\{0,\eta\}$ by definition of $\eta$.

Now the portion of $\mu\times\mu$ which is $<_{cw} (0,\eta)$ is exactly $\eta\times\eta$ by definition of $<_{cw}$. Since $(\alpha,\beta) <_{cw} (0,\eta)$ this means that $h$, which injects $\mu$ to the portion of $\mu\times\mu$ which is $<_{cw} (\alpha,\beta)$ is an injection into the portion of $\mu\times\mu$ which is $<_{cw} (0, \eta)$, which is $\eta\times\eta$.