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In cumulative hierarchy, rank function $R(x)$ is defined as the least ordinal $\alpha$ that $x\in V_{\alpha +1}$. For any $x\in y$, $R(x)<R(y)$. I need an exact proof of it.

Eugene Zhang
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2 Answers2

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Let $\alpha = R(y)$ and $x \in y$. Then $y \in V_{\alpha+1}$ and since $V_{\alpha+1} = \mathcal P(V_{\alpha})$, we know that $y \subseteq V_{\alpha}$ and it hence follows that $x \in V_{\alpha}$. If $\alpha = \beta +1$ is a successor ordinal, we have that $R(x) \le \beta < \alpha$.

Otherwise $\alpha$ is a limit ordinal and since, in this case, $V_{\alpha} = \bigcup_{\beta < \alpha} V_{\beta}$, there is some $\beta < \alpha$ such that $x \in V_{\beta+1}$ -- again witnessing that $R(x) \le \beta < \alpha$.

Stefan Mesken
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  • It's often also useful to know that, unless $y = \emptyset$, $R(y) = \sup {R(x) + 1 \mid x \in y }$, which trivially implies the claim. It's an easy exercise to verify this formula. – Stefan Mesken Dec 16 '17 at 23:37
  • $V_{\alpha+1}$ is - per definition - the power set of $V_{\alpha}$. – Stefan Mesken Dec 16 '17 at 23:52
  • The formula is valid for all $y.$.... If $y=\emptyset $ then $\cup {R(x)+1| x\in y}=$ $\cup {R(x)+1:x\in \emptyset}=$ $\cup \emptyset=$ $\emptyset .$ And $\emptyset \in V_1$ but $\emptyset \not \in V_0=\emptyset.$..... If we use the formula as the def'n of $R(y)$ then we should prove that $R(y)=\min {a:y\in V_{a+1}}.$ – DanielWainfleet Dec 17 '17 at 01:50
  • @DanielWainfleet Oh, yeah. Thanks. I was off by one... – Stefan Mesken Dec 17 '17 at 07:52
  • I couldn't resist spelling it out in detail. There's so much you can say about the empty set. – DanielWainfleet Dec 18 '17 at 13:33
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Def'n: If $a=\cup a\in On$ then $V_a=\cup_{b<a}V_b.$ If $a\in On$ then $V_{a+1}=P(V_a).$

By transfinite induction on $a,$ every $V_a$ is a transitive set. That is $\forall x\in V_a\;(x\subset V_a).$

if $S$ is a transitive set then $(\forall x \; (x\in S\implies x\subset S\implies x\in P(S)\;)\;)\implies S\subset P(S).$ So $V_a\subset V_{a+1}$ for every $a$.

Let $x\in y$ with $R(y)=a.$ Then $x\in y\in V_{a+1}\implies x\in y\in P(V_a)\implies x\in y\subset V_a \implies x\in V_a.$

Now if $a=\cup a$ then $V_a=\cup_{b<a}V_{b+1}$ (see Footnote). So $x\in V_{b+1}$ for some $b+1<a.$ So $R(x)\leq b<a=R(y).$

Or if $a=a'+1$ then $x\in V_{a'+1}.$ So $ R(x)\leq a'<a =R(y).$

Footnote: For the case $a=\cup a$ we have $V_b\subset V_{b+1}$ for all $b<a$. So $V_a=\cup_{b<a}V_b\subset \cup_{b<a}V_{b+1}.$ But also $a=\cup a \implies \{b+1:b<a\}\subset a=\{b:b<a\},$ so $\cup_{b<a}V_{b+1}\subset \cup_{b<a}V_b=V_a.$ Therefore if $a=\cup a$ then $V_a=\cup_{b<a}V_{b+1}.$

Remark: We can also define $R(y)$ recursively as $R(y)=\cup \{R(x)+1:x\in y\}.$ Then it is immediate that $x\in y\implies R(x)<R(y),$ but we should then have to prove that $R(x)=\min \{b:x\in V_{b+1}\}.$

DanielWainfleet
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