# For $\alpha$ a limit ordinal, show $V_{\alpha}=\bigcup_{\beta\lt\alpha}P(V_{\beta})$

For $\alpha$ a limit ordinal, I would like to show

$V_{\alpha}=\bigcup_{\beta\lt\alpha}P(V_{\beta})$

where the $V$'s are members of the cumulative hierarchy and $P$ is the power set.

(This is a continuation of Showing equivalence of definitions for Zermelo Hierarchy, where Max ostensibly provided a solution in for a successor ordinal)

In the cumulative hierarchy, for a limit ordinal $\alpha$, $V_{\alpha}:=\bigcup_{\beta\lt\alpha}V_{\beta}$, so I would like to show

$\bigcup_{\beta\lt\alpha}V_{\beta}=\bigcup_{\beta\lt\alpha}P(V_{\beta})$

Can I say that for any $\beta\lt\alpha$, there is a $\beta$' with $\beta\lt\beta'\lt\alpha$ such that $V_{\beta'}=P(V_{\beta})$.

And conversely, for any such $\beta'$, there is a $\beta$ again with $V_{\beta'}=P(V_{\beta})$, establishing inclusion in both dorections?

Thanks.

In the following $\alpha$ is a limit ordinal (and $\alpha > 0$):

Your approach almost works. First of all, for all $\beta < \alpha$ $$V_{\beta+1} = \mathcal P (V_\beta),$$ so that $$\{ \mathcal P(V_\beta) \mid \beta < \alpha\} \subseteq \{ V_{\beta} \mid \beta < \alpha\}$$

The inverse inclusion does not hold, since for limit ordinals $\beta < \alpha$ (including $\beta = 0$) there is no $\gamma$ such that $V_{\beta} = \mathcal P(V_{\gamma})$.

But if $\beta < \alpha$ is a limit ordinal, then $V_{\beta} = \bigcup_{\gamma < \beta} V_{\gamma}$ which inductively yields that for all $x \in V_{\beta}$ there is some $\gamma < \beta$ such that $x \in V_{\gamma+1} = \mathcal P(V_{\gamma})$, so that $$\bigcup_{\beta < \alpha} V_\beta = \bigcup_{\beta < \alpha} \mathcal P(V_{\beta})$$ after all.

• Thanks, Stefan. Do you have to modify anything if you are dealing with, for example, $\omega+n$ where $n$ is a natural number. I.e., a limit ordinal plus a successor ordinal. With regards, – user12802 Dec 21 '17 at 17:53
• For successor ordinals, the formula becomes rather useless: $V_{\lambda + n} = \mathcal P^n (V_{\lambda}) = \bigcup_{\alpha \le \lambda + n} V_{\alpha} = \bigcup_{\alpha < \lambda + n} P(V_{\alpha})$. – Stefan Mesken Dec 21 '17 at 17:57
• I'm reviewing, so maybe you would not mind a follow-up. In the first part, is $\alpha$ a limit ordinal? If not, how is it that the LHS which would equal $\{V_{\beta+1}\}$ be contained in $\{V_{\beta}\}$. I.e., would there not be a $\beta+1=\alpha$? Whereas if $\alpha$ is a limit ordinal then you could always still have $\beta+1\lt\alpha$. But then, if $\alpha$ is a limit ordinal, what happens to $\alpha$ in the second part where I presume you are indicating $\beta$ is a limit ordinal? I hope I expressed this clearly and sorry for the imposition. Thanks, – user12802 Jan 11 '18 at 15:10
• @Andrew I'm afraid I don't understand what you are asking. – Stefan Mesken Jan 11 '18 at 15:38
• @Andrew Ralf's book can be read with little background in set theory (like Kunen's Set Theory) but it's fast paced and therefore I wouldn't necessarily recommend it as a first textbook in advanced set theory. Instead, I think it's best to pick up after having read Kunen or Jech (Kunen is easier to read than Jech) since it offers advanced concepts (and often rather different proofs) that you won't find in either of those. Also: It contains some really tough exercises. Don't be discouraged if you don't solve all of them. – Stefan Mesken Apr 26 '18 at 22:17

The main idea for the reverse inclusion, I think, is that $V_\beta \subset P(V_\beta)$ because $V_\beta$ is transitive.

For the first inclusion, what you said was enough, also perhaps you should say more clearly that (with your notations) $\beta' = \beta +1$, so that $\beta < \alpha \implies \beta' < \alpha$ (as $\alpha$ is limit)

$b<a\implies b+1<a\implies P(V_b)=V_{b+1}\subset \cup_{c<a}V_c=V_a.$

So $\cup_{b<a}V_{b+1}\subset V_a.$

By transfinite induction every $V_b$ is a transitive set. If $x$ is transitive then $x\subset P(x).$ So $V_b\subset V_{b+1}.$ So $$V_a=\cup_{b<a}V_b\subset\cup_{b<a}V_{b+1}\subset V_a.$$

More generally if $S\subset a=\cup a$ then $\cup S=a\iff \cup_{b\in S}V_b=V_a.$