# Two well orderings of an infinite cardinal agree on a large set

I've seen this question but I'm having trouble following the proof given.

This is an exercise from Kunen: If $$\kappa$$ is an infinite cardinal and $$\triangleleft$$ is a well ordering on $$\kappa$$, then $$\exists X \subseteq \kappa$$ with $$|X|= \kappa$$ such that $$\lt$$ and $$\triangleleft$$ agree on $$X$$.

The proof is easy for $$\omega$$ and it's easy for successor cardinals. I understand that the argument for successor cardinals works for any regular cardinal. The problem is singular cardinals.

It seems to me that the first proof given works by taking a cofinal sequence of regular cardinals and iteratively building the set $$X$$ at each stage, taking care that $$\lt$$ and $$\triangleleft$$ continue to agree at each step. But I don't see why that's possible.

Here is the proof reproduced:

As you have already seen the case for $$\kappa$$ regular is not very difficult.

Now suppose $$\operatorname{cf}(\kappa)<\kappa$$. Let $$\langle \kappa_\alpha:\alpha<\operatorname{cf}(\kappa)\rangle$$ be an increasing sequence of regular cardinals less than $$\kappa$$ cofinal in $$\kappa$$ such that for limit $$\alpha$$ we have $$\kappa_\alpha>\sup\{\kappa_\eta:\eta<\alpha\}$$

First, let $$A_0\subseteq \kappa_0$$ have size $$\kappa_0$$ be such that $$\triangleleft$$ and $$<$$ coincide. For $$1\leq\alpha<\operatorname{cf}(\kappa)$$, let $$A_\alpha$$ be a subset of $$\kappa_\alpha\setminus\lambda$$ of size $$\kappa_\alpha$$ where $$\triangleleft$$ and $$<$$ coincide, and for all $$\eta<\alpha$$ and all $$x\in A_\eta$$ we have $$x\triangleleft\min A_\alpha$$, where $$\lambda=\sup\{\kappa_\eta:\eta<\alpha\}$$; this can be done as the order type of $$(\kappa_\alpha\setminus\lambda,\in)$$ is $$\kappa_\alpha$$ and $$\kappa_\alpha$$ is regular.

Finally $$X=\bigcup_{\alpha<\operatorname{cf}(\kappa)}A_\alpha$$ is a subset of $$\kappa$$ of size $$\kappa$$ where $$\triangleleft$$ and $$<$$ agree.

But suppose $$\kappa_1 = \omega_1, \kappa_2 = \omega_2$$. Suppose $$\triangleleft$$ restricted to $$\omega_2$$ has order type $$\omega_2+\omega_1$$ with every element of $$\omega_1 \triangleleft$$-greater than every element of $$\omega_2 \setminus \omega_1$$. Then there's no way to build the set $$X$$ iteratively in this way, and I don't see how the answer presented rules this possibility out. Specifically, I don't see why the last phrase of the third paragraph is true.

Help understanding this would be appreciated. I've spent enough time on this problem and it's time to move on. Thanks in advance.

• I posted on MathOverflow the question of finding a combinatorial solution to this problem that does not use the axiom of choice. There is a very nice proof found by Clinton Conley, which I think is better than the usual approach (using choice) of splitting into cases according to whether the cardinal in question is regular or not. – Andrés E. Caicedo Mar 13 at 11:44
• @AndrésE.Caicedo Thanks. May I make two requests? First, if someone would explicitly confirm I'm right that the proof above is not correct for the reason I've stated, it would set my mind at ease. (What is the protocol when a well received and accepted answer to a question turns out to have been wrong?) Second, may I ask that you provide the answer here, with perhaps more details spelled out to assist my understanding in light of my absence from the field for nearly 30 years? – Robert Shore Mar 13 at 16:24
• The two orders agree on each $A_{\alpha}$ but I don't see that they must agree even on $A_0\cup A_1$, just as you showed. I suspect some additional constraint is missing from the construction. – DanielWainfleet Mar 13 at 18:55