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# 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. $\varnothing$ 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 $\rho(x)\leq\beta$, and hence $\rho(x)<\rho(A)$. ∎

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 $\beta:=\bigcup B=\sup B$ 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 $\{A\}$, 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 $\rho(B)\leq\rho(A)$.

###### Proof.

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

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 $\rho(B)\leq\rho(A)$ 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 $\beta:=\bigcup B=\sup B$ is an ordinal. Since $\rho(x)\leq\beta$, $V_{{\rho(x)}}=V_{{\beta}}$ 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 $\rho(x)\leq\beta$, 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 $\rho(\alpha)=\rho(\beta^{+})=\rho(\beta\cup\{\beta\})=\max(\rho(\beta),\rho(\{% \beta\}))=\max(\rho(\beta),\rho(\beta)^{+})=\rho(\beta)^{+}$. If $\alpha$ is a limit ordinal, then apply property 8 and transfinite induction, so that $\rho(\alpha)=\rho(\bigcup\alpha)=\sup\{\rho(\beta)\mid\beta<\alpha\}=\sup\{% \beta\mid\beta<\alpha\}=\alpha$. ∎

# References

- 1 H. Enderton, Elements of Set Theory, Academic Press, Orlando, FL (1977).
- 2 A. Levy, Basic Set Theory, Dover Publications Inc., (2002).

## Mathematics Subject Classification

03E99*no label found*

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