The Cumulative Hierarchy of Sets

Transitive Set

Define $z$ to be a transitive set iff for all $x$ and $y$, if $x\in y$ and $y\in z$, then $x\in z$. This is equivalent to saying that, for every $y$, if $y\in z$, then $y\subseteq z$.

Transitive Closure

Let $A$ be a set. By recursion on $n\in \omega$, define $B_0=A$ and $B_{n+1}=\bigcup B_n.$ Let $TC(A)=\bigcup \{B_n\mid n<\omega \}.$ We call $TC(A)$ the transitive closure of $A$.

We need the Replacement Axiom to guarantee that $TC(A)$ as defined above is a set, and we also discussed that it actually is a transitive set (as the term transitive closure suggests).

Lemma

Every nonempty class $C$ has an $\in$-minimal element.

Let $S\in C$. If $S\cap C=\varnothing$ then $S$ is $\in$-minimal. If $S\cap C\ne \varnothing$ take $X=T\cap C$, with $T=TC(S)$, and by separation axiom, $X$ is a set. $X$ is nonempty so by regularity, there is $x\in X$ such that $x\cap X=\varnothing$. It follows that $x\cap C=\varnothing$. If $y\in x,y\in C$, $x\in X\subset T$ so $x\in T$ since $T$ transitive closure, thus $y\in T$ as $y\in x\in T$ and $T$ transitive closure, so $y\in x\cap C\cap T$ contradiction.

Cumulative Hierarchy of Sets

We define, by transfinite induction, $V_0=\varnothing ,V_{\alpha +1}=\mathcal{P}(V_{\alpha }),$ $V_{\alpha }={\bigcup }_{\beta <\alpha }{V}_{\beta }\text{if }\alpha \text{ is a limit ordinal.}$ The sets $V_{\alpha }$ have the following properties (by induction):

• (i) Each $V_{\alpha }$ is transitive.
• (ii) If $\alpha <\beta$, then $V_{\alpha }\subseteq V_{\beta }$.
• (iii) $\alpha \subseteq V_{\alpha }$.

The Axiom of Regularity implies that every set is in some $V_{\alpha }$:

Lemma

For every $x$ there is $\alpha$ such that $x\in V_{\alpha }$. As a statement about classes: ${\bigcup }_{\alpha \in \mathrm{Ord}}{V}_{\alpha }=V.$

Proof. Let $C$ be the class of all $x$ that are not in any $V_{\alpha }$. If $C$ is nonempty, then $C$ has an $\in$-minimal element $x$. That is, $x\in C$, and $z\in {\bigcup }_{\alpha }{V}_{\alpha }$ for every $z\in x$. Hence $x\subseteq {\bigcup }_{\alpha \in \mathrm{Ord}}{V}_{\alpha }$. Define $F:x\to \text{Ord}$ with $F(z)=\text{the least element}\alpha \text{that}z\in V_{\alpha }$, this function is well-defined since $\text{Ord}$ is a well-ordering. By Replacement, $F(x)$ is a set, so there exists an ordinal $\gamma$ such that $x\subseteq {\bigcup }_{\alpha <\gamma }{V}_{\alpha }$. Hence $x\subseteq V_{\gamma }$ and so $x\in V_{\gamma +1}$. Thus $C$ is empty and we have the required result. $\square$

Since every $x$ is in some $V_{\alpha }$, we may define the rank of $x$: $\mathrm{rank}(x)=\text{the least }\alpha \text{ such that }x\in V_{\alpha +1}.$

Transitive

$z$ is transitive iff for all x and y, if $x\in y$ and $y\in z$ then $x\in z$. Equivalently $\forall yy\in z⟹y\subseteq z$

A = {{a}, {a,b},{{ b}}} B1 = { {a}, {a,b}, {b} }