properties of ranks of sets

A set $A$ is said to be grounded, if $A\subseteq V_{\alpha }$ in the cumulative hierarchy for some ordinal $\alpha$. The smallest such $\alpha$ such that $A\subseteq V_{\alpha }$ is called the rank of $A$, and is denoted by $\rho (A)$.

In this entry, we list derive some basic properties of groundedness and ranks of sets. Proofs of these properties require an understanding of some of the basic properties of ordinals.

1. $\mathrm{\varnothing }$ is grounded, whose rank is itself. This is obvious.

2. If $A$ is grounded, so is every $x\in A$, and .

   Proof.

   $A\subseteq V_{\rho (A)}$, so $x\in V_{\rho (A)}$, which means $x\subseteq V_{\beta }$ for some . This shows that $x$ is grounded. Then $\rho (x)\le \beta$, and hence .∎

3. If every $x\in A$ is grounded, so is $A$, and $\rho (A)=sup\{\rho {(x)}^{+}\mid x\in A\}$.

   Proof.

   Let $B=\{\rho {(x)}^{+}\mid x\in A\}$. Then $B$ is a set of ordinals, so that $\beta :=\bigcup B=supB$ is an ordinal. Since each $x\in V_{\rho {(x)}^{+}}$, we have $x\in V_{\beta }$. So $A\subseteq V_{\beta }$, showing that $A$ is grounded. If , then for some $x\in A$, , which means $x\notin V_{\alpha }$, and therefore $A⊈V_{\alpha }$. This shows that $\rho (A)=\beta$.∎

4. If $A$ is grounded, so is $\{A\}$, and $\rho (\{A\})=\rho {(A)}^{+}$. This is a direct consequence of the previous result.

5. If $A,B$ are grounded, so is $A\cup B$, and $\rho (A\cup B)=\max (\rho (A),\rho (B))$.

   Proof.

   Since $A,B$ are grounded, every element of $A\cup B$ is grounded by property 2, so that $A\cup B$ is also grounded by property 3. Then $\rho (A\cup B)=sup\{\rho {(x)}^{+}\mid x\in A\cup B\}=\max (sup\{\rho {(x)}^{+}\mid x\in A\},sup\{\rho {(x)}^{+}\mid x\in B\})=\max (\rho (A),\rho (B))$.∎

6. If $A$ is grounded, so is $B\subseteq A$, and $\rho (B)\le \rho (A)$.

   Proof.

   Every element of $B$, as an element of the grounded set $A$, is grounded, and therefore $B$ is grounded. So $\rho (B)=sup\{\rho {(x)}^{+}\mid x\in B\}\le sup\{\rho {(x)}^{+}\mid x\in A\}=\rho (A)$. Since $\rho (B)$ and $\rho (A)$ are both ordinals, $\rho (B)\le \rho (A)$.∎

7. If $A$ is grounded, so is $P(A)$, and $\rho (P(A))=\rho {(A)}^{+}$.

   Proof.

   Every subset of $A$ is grounded, since $A$ is by property 6. So $P(A)$ is grounded. Furthermore, $P(A)=sup\{\rho {(x)}^{+}\mid x\in P(A)\}$. Since $\rho (B)\le \rho (A)$ for any $B\in P(A)$, and $A\in P(A)$, we have $P(A)=\rho {(A)}^{+}$ as a result.∎

8. If $A$ is grounded, so is $\bigcup A$, and $\rho (\bigcup A)=sup\{\rho (x)\mid x\in A\}$.

   Proof.

   Since $A$ is grounded, every $x\in A$ is grounded. Let $B=\{\rho (x)\mid x\in A\}$. Then $\beta :=\bigcup B=supB$ is an ordinal. Since $\rho (x)\le \beta$, $V_{\rho (x)}=V_{\beta }$ or $V_{\rho (x)}\in V_{\beta }$. In either case, $V_{\rho (x)}\subseteq V_{\beta }$, since $V_{\alpha }$ is a transitive set for any ordinal $\alpha$. Since $x\subseteq V_{\rho (x)}$, $x\subseteq V_{\beta }$ for every $x\in A$. This means $\bigcup A\subseteq V_{\beta }$, showing that $\bigcup A$ is grounded. If , then for some $\rho (x)\le \beta$, which means $x⊈V_{\alpha }$, or $\bigcup A⊈V_{\alpha }$ as a result. Therefore $\rho (\bigcup A)=\beta$.∎

9. Every ordinal is grounded, whose rank is itself.

   Proof.

   If $\alpha =0$, then apply property 1. If $\alpha$ is a successor ordinal, apply properties 4 and 5, so that $\rho (\alpha )=\rho ({\beta }^{+})=\rho (\beta \cup \{\beta \})=\max (\rho (\beta ),\rho (\{\beta \}))=\max (\rho (\beta ),\rho {(\beta )}^{+})=\rho {(\beta )}^{+}$. If $\alpha$ is a limit ordinal, then apply property 8 and transfinite induction, so that .∎