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.