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. $\varnothing$ is grounded, whose rank is itself. This is obvious.
2. If $A$ is grounded, so is every $x\in A$ and $\rho(x)< \rho(A)$
   Proof. $A\subseteq V_{\rho(A)}$ so $x\in V_{\rho(A)}$ which means $x\subseteq V_{\beta}$ for some $\beta < \rho(A)$ This shows that $x$ is grounded. Then $\rho(x)\le \beta$ and hence $\rho(x)< \rho(A)$
3. If every $x\in A$ is grounded, so is $A$ and $\rho(A)=\sup \lbrace \rho(x)^+ \mid x\in A\rbrace$
   Proof. Let $B=\lbrace \rho(x)^+ \mid x\in A\rbrace$ Then $B$ is a set of ordinals, so that $\beta:=\bigcup B = \sup B$ 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 $\alpha <\beta$ then for some $x\in A$ $\alpha < \rho(x)^+$ which means $x\notin V_{\alpha}$ and therefore $A\nsubseteq V_{\alpha}$ This shows that $\rho(A)=\beta$
4. If $A$ is grounded, so is $\lbrace A\rbrace$ and $\rho(\lbrace A\rbrace)=\rho(A)^+$ 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 $\rho(A\cup B) = \sup \lbrace \rho(x)^+ \mid x\in A\cup B\rbrace = \max(\sup \lbrace \rho(x)^+ \mid x\in A \rbrace, \sup \lbrace \rho(x)^+ \mid x\in B\rbrace) = \max(\rho(A),\rho(B))$
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 $\rho(B)= \sup \lbrace \rho(x)^+ \mid x\in B\rbrace \le \sup \lbrace \rho(x)^+ \mid x\in A\rbrace = \rho(A)$ 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, $P(A) = \sup \lbrace \rho(x)^+ \mid x\in P(A)\rbrace$ 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 $\rho(\bigcup A)=\sup \lbrace \rho(x)\mid x\in A\rbrace$
   Proof. Since $A$ is grounded, every $x\in A$ is grounded. Let $B=\lbrace \rho(x)\mid x\in A\rbrace$ Then $\beta:=\bigcup B =\sup B$ 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 $\alpha < \beta$ then $\alpha < \rho(x)$ for some $\rho(x) \le \beta$ which means $x \nsubseteq V_{\alpha}$ or $\bigcup A\nsubseteq V_{\alpha}$ 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 $\rho(\alpha)=\rho(\beta^+) = \rho(\beta\cup \lbrace \beta\rbrace) = \max(\rho(\beta), \rho(\lbrace \beta \rbrace)) = \max(\rho(\beta), \rho(\beta)^+) = \rho(\beta)^+$ If $\alpha$ is a limit ordinal, then apply property 8 and transfinite induction, so that $\rho(\alpha)=\rho(\bigcup \alpha)=\sup \lbrace \rho(\beta) \mid \beta < \alpha\rbrace = \sup \lbrace \beta \mid \beta< \alpha\rbrace =\alpha$

Bibliography

1

H. Enderton, Elements of Set Theory, Academic Press, Orlando, FL (1977).

2

A. Levy, Basic Set Theory, Dover Publications Inc., (2002).