# Existence of a rank function over a well founded set

I need help in understanding a step from Jech Set Theory, theorem 2.27

In the proof, he defines a series of subsets of a well founded set $P$ under relation $E$:

$P_0=\emptyset, \hspace{2mm} P_{\alpha+1}=\{ x\in P : \forall y \hspace{1mm} (yEx \rightarrow y\in P_{\alpha} \}, \hspace{2mm} P_{\alpha}=\bigcup_\limits{\xi<\alpha}{P_{\xi}} \hspace{2mm}$ for limit $\alpha$

He claims that there exists some ordinal $\theta$ such that $P_{\theta} = P_{\theta+1}$ (by replacment).

My question is why it exists and why it follows from replacement.

Thanks!

• Otherwise, $\mathsf{ORD}$ would be a set. Commented Jul 21, 2015 at 12:19
• One does not need replacement. Anyway, Jech is probably arguing along the following lines: If the sequence does not stabilize, one obtains an injection from $\mathsf{ORD}$ into $\mathcal P(P)$, and therefore a surjection of the latter onto the former. But then replacement gives us that $\mathsf{ORD}$ is a set, and this is a contradiction. Commented Jul 21, 2015 at 12:22
• yea, replacement for $P^{-1}$ , my bad, ty Commented Jul 21, 2015 at 12:24

1. $P_\alpha\subseteq P_{\alpha+1}$ for all $\alpha$ (induction)
2. $P_\alpha\subseteq P_{\beta}$ whenever $\alpha\leq \beta$ (induction on $\beta$ and item 1.)
3. $P=\bigcup_\alpha P_\alpha$ (if not, take an $E$-minimal element in $P\setminus \bigcup_\alpha P_\alpha$ and work from there. You will need to use Replacement.)
4. Use Replacement to conclude that $P=P_\theta$ for some $\theta$ (use 3 and 2: for each $y$, consider the least $\alpha(y)$ such that $y\in P_{\alpha(y)}$ and let $\theta=\sup_y\alpha(y)$.)
The $\theta$ in item 4. will satisfy $P_{\theta}=P_{\theta+1}$, but in fact you can just work with the stronger property $P=P_{\theta}$ throughout the rest of the proof of 2.27.