A Question about the Cumulative Hierarchy.

Question

I am self-studying set theory, following the book "The Joy of Sets" by Keith Devlin.

In section 2.2, page 37, the author introduced two equivalent definitions of the cumulative hierarchy.

First, for each ordinal number $\alpha$, there is a set $V_{\alpha}$. It is defined as follows:

1. $V_{0}=\emptyset$
2. For each successor ordinal $\alpha+1$, the set $V_{\alpha+1}$ is defined as the power set of $V_{\alpha}$, i.e $$V_{\alpha+1}=\mathcal{P}(V_{\alpha})$$
3. For each limit ordinal $\alpha$, one might consider $$V_{\alpha}=\mathcal{P}\left(\underset{\beta<\alpha}{\bigcup}V_{\beta}\right).\tag{1}$$

This definition seems totally fine for me, but then the author claims that when we come to investigate the set-theoretic hierarchy more thoroughly we shall see that

$$V_{\alpha+1}=\mathcal{P}\left(\underset{\beta\leq\alpha}{\bigcup}V_{\beta}\right).\tag{2}$$

There's a very subtle difference between the presumed naive definition (1) and the set-theoretical defnition (2). How much is the correct definition (2) different from (1)?

After that, the author gives an equivalent definition for the limit ordinal number case:

$$V_{\alpha}=\underset{\beta<\alpha}{\bigcup}V_{\beta} \tag{3}.$$

How to show that definition (3) is indeed equivalent to definition (2)?

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So far, my guess is that the author might have made a mistake. According to Wikipedia, (1) should be $V_{\alpha}=\underset{\beta<\alpha}{\bigcup}\mathcal{P}(V_{\beta})$, and (3) should be equivalent to (1) for some reason which I don't yet.

Answer 1

My favorite definition of $V_\alpha$ (which I think ought to be standard) is the recursion that you quote from Wikipedia: $V_\alpha=\bigcup_{\beta<\alpha}\mathcal P(V_\beta)$. Notice that this immediately implies that $V_0=\varnothing$ (because there is no $\beta<0$) and that if $\alpha\leq\gamma$ then $V_\alpha \subseteq V_\gamma$ and, as a consequence, $\mathcal P(V_\alpha)\subseteq\mathcal P(V_\gamma)$.

In view of these facts, the definition also gives that $V_{\alpha+1}=\bigcup_{\beta\leq\alpha}\mathcal P(V_\beta)=\mathcal P(V_\alpha)$. So we have the $0$ and successor clauses of the definition you quoted from Devlin.

Finally, for a limit ordinal $\alpha$, we have (using the fact that the successor ordinals $<\alpha$ are cofinal among all the ordinals $<\alpha$) that $V_\alpha=\bigcup_{\beta<\alpha}\mathcal P(V_\beta) =\bigcup_{\beta<\alpha}V_{\beta+1}=\bigcup_{\beta<\alpha}V_\beta$, which is formula (3) in the question.

Answer 2

The author must have made a mistake. According to Wikipedia, the cumulative hierarchy is a collection of sets $V_{\alpha}$ defined in the following way:

1. Let $V_{0}=\emptyset$.
2. For any ordinal number $\beta$, $V_{\beta+1}$ is defined as the power set of $V_{\beta}$. i.e $$V_{\beta+1}=\mathcal{P}(V_{\beta}).$$
3. For any limit ordinal $\lambda$, $V_{\lambda}$ is given by the union $$V_{\lambda}=\underset{\beta<\lambda}{\bigcup}V_{\beta}.$$

Steps $1$ and $2$ are very intuitive. The reason behind step $3$ is that when we form a limit ordinal, we are really just collecting together all the previous ordinals without introducing anything new. An equivalent definition is $$V_{\alpha}=\underset{\beta<\alpha}{\bigcup}\mathcal{P}(V_{\beta}).$$

In general, $\mathcal{P}(A)\cup\mathcal{P}(B)\subseteq\mathcal{P}(A\cup B)$, and the converse inclusion does not necessarily hold.

Now the problem is to prove that the above two definitions are equivalent. This can be done by applying transfinite induction. To begin with, the first definition of the cumulative hierarchy implies the following theorem:

Theorem $\star$: For any two ordinals $\alpha$ and $\gamma$,
(a) $V_{\alpha}$ is transitive, and
(b) $\gamma<\alpha\Rightarrow V_{\gamma}\subset V_{\alpha},$
where we call a set $V_{\alpha}$ transitive iff $\forall A\in V_{\alpha}$, and $\forall a\in A$, we have $a\in V_{\alpha}$.

Before giving the proof of the above theorem, it is easy to show that the definition of transitive sets is equivalent to the following:

The set $V_{\alpha}$ is transitive iff: $$\forall a\in V_{\alpha}\Rightarrow a\subseteq V_{\alpha}.$$

It is easy to see that the above definition implies:

I: Every subset of a transitive set is transitive.
II: Every union of transitive sets is transitive.
III: The power set of a transitive set is transitive.

Proof of Theorem $\star$:
The claims are trivial when $\alpha=0$. Now assume (a) and (b) are true for all $\alpha$ and $\gamma$ such that $\gamma<\alpha$. If $\alpha$ is a successor ordinal, say $\alpha=\beta+1$, then by definition, $V_{\alpha}=\mathcal{P}(V_{\beta})$. Following the assumption and claim III, one finds $V_{\alpha}$ must be transitive. From the assumption, $V_{\gamma}\subset V_{\beta}$, if $\forall a\in V_{\gamma}$, then $a\in V_{\beta}$. Since $V_{\beta}$ is transitive, $a\subseteq V_{\beta}$. In other words, $a\in\mathcal{P}(V_{\beta})=V_{\beta+1}=V_{\alpha}$. If $\alpha$ is a limit ordinal, then $V_{\alpha}=\underset{\lambda<\alpha}{\bigcup}V_{\lambda}$. Using claim II, $V_{\alpha}$ is transitive. Also, $V_{\gamma}\subset\underset{\lambda<\alpha}{\bigcup}V_{\lambda}$ is clear.

Finally, the proof the the equivalence:

For $\alpha=0$, the equivalence is completely trivial.
If $\alpha$ is a successor ordinal, one has to prove $$V_{\alpha+1}=\underset{\beta<\alpha+1}{\bigcup}\mathcal{P}(V_{\beta})=\mathcal{P}(V_{\alpha}).$$ But from Theorem $\star$, for all $\beta<\alpha$, $V_{\beta}\subset V_{\alpha}$, and hence $\mathcal{P}(V_{\beta})\subset\mathcal{P}(V_{\alpha})$. Therefore, $$\underset{\beta<\alpha+1}{\bigcup}\mathcal{P}(V_{\beta})\subseteq\mathcal{P}(V_{\alpha}).$$ On the other hand, it is clear that $$\mathcal{P}(V_{\alpha})\subseteq\underset{\beta<\alpha+1}{\bigcup}\mathcal{P}(V_{\beta}).$$ Thus, the two definitions are equivalent when $\alpha$ is a successor ordinal.
The case when $\alpha$ is a limit ordinal can be found here. First, for $\forall\beta$ such that $\beta<\alpha$, one has $$\beta+1<\alpha.$$ Consider the following class $$\left\{\mathcal{P}(V_{\beta})|\beta<\alpha\right\}.$$ Since for any $\beta<\alpha$, one has $$\mathcal{P}(V_{\beta})=V_{\beta+1}\in\left\{V_{\beta}|\beta<\alpha\right\},$$ one has $$\left\{\mathcal{P}(V_{\beta})|\beta<\alpha\right\}\subseteq\left\{V_{\beta}|\beta<\alpha\right\}.$$ Thus, one has $$\underset{\beta<\alpha}{\bigcup}\mathcal{P}(V_{\beta})\subseteq\underset{\beta<\alpha}{\bigcup}V_{\beta}.$$ If $\beta$ is a successor ordinal, then it is clear that for any $V_{\beta}\in\left\{V_{\beta}|\beta<\alpha\right\}$, there exists $\mathcal{P}(V_{\beta-1})\in\left\{\mathcal{P}(V_{\beta})|\beta<\alpha\right\}$.
As is shown by Stefan Mesken, if $\beta$ is a limit ordinal, the converse inclusion does not hold, but in such a case, $V_{\beta}=\underset{\gamma<\beta}{\bigcup}V_{\gamma}$, and hence, for any $a\in V_{\beta}$, there is some $\gamma+1<\beta$ such that $x\in V_{\gamma+1}=\mathcal{P}(V_{\gamma})$. Thus, for any limit ordinal $\beta$ such that $\beta<\alpha$, there exists some $\gamma<\beta$ such that $V_{\beta}\subseteq\mathcal{P}(V_{\gamma})$, i.e one must have $$\underset{\beta<\alpha}{\bigcup}\mathcal{P}(V_{\beta})=\underset{\beta<\alpha}{\bigcup}V_{\beta}.$$

I am still not entirely sure about my proof. If anyone finds a mistake, please correct me.

Answer 3

I don't have access to the book but, from your account, it does not appear that the author has made a mistake. He puts forward (1) as a possible definition but then settles on (3), which is the standard. He does not say that they are equivalent; they are not. (2) is not a definition, it does not tell you what $V_\alpha$ is for a limit $\alpha$. It is a consequence of either definition.

Both definitions give the same universe $V$, but the stage at which particular sets appear can differ. For any infinite ordinal $\alpha$, the $V_\alpha$ given by definition (1) is the $V_{\alpha+1}$ of definition (3). If $\alpha$ is a limit ordinal, the standard $V_\alpha$ does not appear in the hierarchy given by definition (1). Presumably one reason for rejecting that definition.