Ordinals and the Von Neumann universe

Question

Let ON be the proper class of all ordinals, and $R(\alpha)$ for each $\alpha\in$ON defined by transfinite recursion as:

1. $R(\emptyset)=\emptyset$

2. If $\alpha$ is such that $\alpha=\beta+1$ then $R(\alpha)=\mathcal{P}(R(\beta))$.

3. If $\alpha$ is a limit ordinal then $R(\alpha)=\bigcup_{\beta<\alpha}R(\beta)$

There is a step on a proof (in the book "Set theory" by Kunen, last edition, lemma II.6.2) that assumes that for each $\alpha\in$ON $$R(\alpha)\cap\text{ON}=\alpha$$

I am sure it is very easy but I haven't managed to get that proved yet, by transfinite induction I get that if it is true for $\alpha$ then $$\alpha+1\subset R(\alpha+1)\cap \text{ON}$$ since $R(\alpha+1)\cap \text{ON}$ is an ordinal (it is transitive and contained in ON), that means that it is $\geq\alpha+1$, intuitively I can see why it can't be greater, since there aren't "enough parenthesis" after doing $\mathcal{P}(\cdot)$ just once, but how to show that properly?

Answer 1

If $R(\alpha+1)\cap ON>\alpha+1$, then $\alpha+1\in R(\alpha+1)$, so $\alpha+1\subseteq R(\alpha)$. In particular, since $\alpha\in \alpha+1$, this means $\alpha\in R(\alpha)$. This violates the induction hypothesis that $R(\alpha)\cap ON=\alpha$.

Answer 2

It is true that the cardinality of each $R(\alpha)$ goes exponentially at each step. However, as far as "longer" sets go, they only grow by "one pair of braces".

Look at the power set of $\omega$, for example, while uncountable, all the elements there are subsets of $\omega$. So the largest ordinal you could muster is $\omega+1$, since that is the last ordinal whose elements are all subsets of $\omega$.

Here is a simple claim that you should be able to prove: If $\alpha$ is an ordinal, then $\alpha+1\subseteq\mathcal P(\alpha)$, and $\alpha+2\nsubseteq\mathcal P(\alpha)$.

From this simple claim, the proof you seek shall follow.

Answer 3

By trasnfinite induction you only need to prove that $\alpha\in{R(\alpha+1)\setminus{R(\alpha)}}$. For the base and limit cases the conclution is trivial. Now if $\alpha\in{R(\alpha+1)\setminus{R(\alpha)}}$ then $\alpha\subset{R(\alpha+1)}$ by transitivity, so $\alpha+1\subset{R(\alpha+1)}$ and this implies tha $\alpha+1\in{R(\alpha+2)}$. Now, if $\alpha+1\in{R(\alpha+1)=\mathcal{P(R(\alpha))}}$ then $\alpha\in\alpha+1\subset{R(\alpha)}$. Absurd!