# Question on the proof of the existence of rank function

As defined on Page 64, Set Theory, Jech(2006), by transfinite induction:

• $V_0=\emptyset$,
• $V_{\alpha+1}=P(V_{\alpha})$,
• $V_{\alpha}=\bigcup_{\beta<\alpha}V_\beta$, if $\alpha$ is a limit ordinal.

In the proof of Lemma 6.3(For every $x$ there is $\alpha$, such that $x \in V_\alpha$), which is a Reductio ad absurdum.

$x$ is $\in$-minimal element of the class whose element dosn't belong to any $V_\alpha$. The existence of $z$ such that $z \in x$and $z \in \bigcup_{\alpha \in Ord}V_\alpha$ implies $x \subset \bigcup_{\alpha \in Ord}V_\alpha$.

But I don't know what does the following mean, especially why replacement is in need?

Hence $x \subset \bigcup_{\alpha \in Ord}V_\alpha.$ By Replacement, there exists an ordinal $\lambda$ such that $x \subset \bigcup_{\alpha \in \lambda}V_\alpha$.

Assuming AC, we have $f$ as a choice function with the domain $\{V_\alpha:\alpha \in Ord\}$. Let $x=\{f(V_\alpha) \in V_\alpha: \alpha \in Ord\}$. Could it suffice to be counterexample?

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Suppose that there are sets that don’t belong to any $V_\alpha$, and let $x$ be an $\in$-minimal set of that kind. Then the $\in$-minimality of $x$ implies that $\forall y\in x\,\exists \alpha\in\mathbf{On}(y\in V_\alpha)$. This further implies

$$\forall y\in x\,\exists!\alpha\in\mathbf{On}\big(y\in V_\alpha\land\forall\beta\in\mathbf{On}(y\in V_\beta\to\beta\ge\alpha)\big)\;.\tag{1}$$

$(1)$ shows that the predicate $\varphi(x,\alpha)$ given by

$$\varphi(y,\alpha)\leftrightarrow\Big(y\in V_\alpha\land\forall\beta\in\mathbf{On}(y\in V_\beta\to\beta\ge\alpha)\Big)\lor(y\notin x\land\alpha=0)$$

is a class function, and the axiom schema of replacement then says that the image of any set under this class function is a set. In particular, the image of $x$ under this class function is a set, and the image of $x$ under this class function is a set $A$ of ordinals with the property that for each $y\in x$ there is an $\alpha\in A$ such that $y\in V_\alpha$. Thus, $$x\subseteq\bigcup_{\alpha\in A}V_\alpha\;.$$

The last step is to use the fact that since $A$ is a set of ordinals, the ordinal $\lambda=\sup A=\bigcup A$ exists. Then since $$\bigcup_{\alpha\in A}V_\alpha=\bigcup_{\alpha\in\lambda}V_\alpha$$ by the definition of the sets $V_\xi$, we have $$x\subseteq\bigcup_{\alpha\in\lambda}V_\alpha\;.$$

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Read back about the Replacement Schema, and you will see that this situation is exactly where it is useful. If to each element $u$ of some set $z$ we associate an object $\varphi ( u )$ in a uniform manner (i.e., by some specific and fixed formula of set theory), then the class $\{ \varphi (u) : u \in z \}$ is itself a set.

As for your counterexample, there are a few words to mention.

• The Axiom of Choice does not imply that any class of nonempty sets has a choice function. This principle is called the Axiom of Global Choice, and is quite a bit stronger, though not exactly expressible in the language of ZFC.
• Even so, there are some simple choice functions you can define for $\{ V_\alpha : 0 < \alpha \in \ORD \}$.
• One that springs to mind is $f : V_\alpha \mapsto \emptyset$. With this, note that $\{ f ( V_\alpha ) : 0 < \alpha \in \ORD \} = \{ \emptyset \}$ is clearly a set.
• Another one is $g : V_{\alpha+1} \mapsto \alpha$, and $g : V_\lambda \mapsto \emptyset$ for limit $\lambda$. In this case $\{ g ( V_\alpha ) : 0 < \alpha \in \ORD \} = \ORD$, is a proper class.
• In the case where $\{ f(V_\alpha) : 0 < \alpha \in \ORD \}$ is a proper class, the theorem does not say anything: in ZFC only sets have existence, and so proper classes a priori cannot be elements of any $V_\alpha$.
• If $x = \{ f(V_\alpha) : 0 < \alpha \in \ORD \}$ is a set, then you can apply the Axiom of Choice to choose for each $u \in x$ some $\alpha$ such that $u = f ( V_\alpha )$. By Replacement the collection of such $\alpha$ will be bounded, and the remainder of the proof goes through.
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Is $\mathbf{Orf}$ another thing you learned about in Prague? :-) –  Asaf Karagila Dec 4 '12 at 18:31
@Asaf: Fanks!!! –  Arthur Fischer Dec 4 '12 at 18:33
@Arfur: Fure! No froblemf! Ffff ff ff fff? Ffff! Fffff? –  Asaf Karagila Dec 4 '12 at 18:36