# Trouble understanding that the rank of an ordinal is equal to itself

In the course I'm taking, we defined the von Neumann hierarchy by $V_0 = \emptyset$, $V_{\alpha + 1} = \mathcal{P}(V_\alpha)$, and $V_\lambda = \bigcup_{\alpha < \lambda} V_\alpha$ for limit ordinals $\lambda$.

We defined the class of well-founded sets $WF$ as the collection of all $x$ such that there exists an ordinal $\alpha$ such that $x \in V_\alpha$. Finally, we defined the rank of a set $x$ by the least $\alpha$ such that $x \in V_\alpha$.

Here is what I fail to understand: if each ordinal has rank equal to itself, then for each ordinal $\alpha$, $\alpha \in V_\alpha$. However, obviously $0 = \emptyset \notin \emptyset = V_0$. So the rank of $0$ is not $0$, but rather $1$.

The same problem occurs for $1$, and I imagine for all the finite ordinals. What's the matter with this setup? It it that one of the definitions I have is incorrect?

• No, the rank of a set $x$ is the least $\alpha$ such that $x\in V_{\alpha+1}$. Note that if $\alpha$ is limit, any $x\in V_\alpha$ belongs to some $V_\beta$ with $\beta<\alpha$, so under the definition you suggest, many ordinals (0 and all limit ordinals) are not the rank of any set. This is not the only reason to use instead the definition I indicate, but it suggests why we would go for that version instead of the variant you have. – Andrés E. Caicedo Feb 5 '17 at 16:47
• I see. I figured it was simply a mistake in one of my definitions, but I wasn't sure which. Furthermore, the proof I have that the rank of an ordinal is equal to itself doesn't seem to make sense using the definition I have, yet with yours it does. I'll accept your comment as an answer if you post it as one. – Jacob Errington Feb 5 '17 at 16:49

## 1 Answer

No, the rank of a set $x$ is the least $\alpha$ such that $x\in V_{\alpha+1}$. Note that if $\alpha$ is limit, any $x\in V_\alpha$ belongs to some $V_\beta$ with $\beta<\alpha$ so, under the definition you suggest, many ordinals (0 and all limit ordinals) are not the rank of any set. This is not the only reason to use instead the definition I indicate, but it suggests why we would go for that version instead of the variant you have.

With the standard definition of rank, that $\operatorname{rk}(\alpha)=\alpha$ for all ordinals $\alpha$ can be proved in a straightforward fashion by transfinite induction.