Let (A,R) be a graph. Define by transfinite recursion:
$ W_{0}=\emptyset \\ W_{\alpha+1}=\{a \in A : ext_{R}(a) \subseteq W_{\alpha +1}\} \\ W_{\alpha}=\cup_{\beta < \alpha} W_{\beta} \text{if alpha is limit} $
Show that there exists $\lambda$ such that $W_{\lambda}=W_{\lambda +1}$.
So far, I have been trying to approach this by contradiction.. If no such $\lambda$ exists, then for any ordinal $\alpha$, there is some $x \in A$ such that $ext_{R}(x)$ is contained in $W_{\alpha}$ but $x \not\in W_{\alpha}$.
If I find such a $W_{\lambda}$, it can be shown to be the well-founded part of A. Does anyone have an idea of how to continue from here? Any direction would be really appreciated!
Show that if $\alpha < \beta$ then $W_\alpha\subseteq W_\beta$.
Therefore, if you let $\delta$ be an ordinal that is larger (cardinality-wise) than $A$ -- such an ordinal exists by Hartogs' theorem, then you can't keep adding new elements to $W$ for each step up to $W_\delta$, because there are too many steps for each of them to have an element they add.