I am having problems with calculating ranks of ordinals. In particular, I do not understand why $\text{rank}~{\mathbb Q}=\omega+3$.

I know that the rank is defined as such: $\text{rank} (A)=\text{least $\alpha\in\mathbf{ORD}$ such that A$\subseteq V_\alpha$}$. I have also read other posts here that describe formulaic means of computing ranks of sets: $\text{rank}(A)=\sup\{\text{rank}(B)+1:B\in A\}$.

It seems that since $\mathbb{Q}=\{<a,b>:a,b\in\omega, b\neq 0\}$, it follows that $\mathbb{Q}\subset \omega\times\omega \subset \mathcal P(\mathcal P(\omega))$. That is, we apply the power set function twice on $\omega$. Since $\omega$ has rank $\omega$, shouldn't we get $\mathbb Q$ to have rank at most $\omega+2$ (because we applied the power set function twice)?

Specifically, I am arguing that since $\mathcal P(\omega)$ has rank $\omega+1$, and $\mathcal P(\mathcal P(\omega))$ has rank $\omega+1+1$, so that $\mathbb{Q}$ has rank at most $\omega+2$. Could someone explain to me where I am mistaken? Also, if this is not too much of a hassle, could someone EXPLICITLY demonstrate that $\mathbb{Q}$ indeed has rank $\omega+3$?

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    $\begingroup$ How do you represent $-1$ in your definition of $\mathbb{Q}$. And $2/3$? Typically $\mathbb{Q}$ is defined as some quotient set of $\mathbb{Z} \times (\mathbb{Z} - \{0\})$. $\mathbb{Z}$ itself is a quotient set of $\mathbb{N} \times \mathbb{N}$. $\endgroup$ – Mike Nov 13 '14 at 5:45
  • $\begingroup$ I suppose it would be the equivalence class $[(-1,1)]$ with representative $(-1,1)$. If you have a set $A$, and you take a quotient set of it, what happens to the rank? $\endgroup$ – youngtableaux Nov 13 '14 at 5:54
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    $\begingroup$ Well $-1 \not\in \omega$, so how does that fit with your definition? $\endgroup$ – Mike Nov 13 '14 at 5:55
  • $\begingroup$ I see, thanks for pointing that out. What's the rank of $\mathbb Z$, then? It seems that $\mathbb Z$ has rank $\omega+2$ in this case (if $\mathbb Z$ is the quotient as you described). Does that not make $\mathbb Q$ have rank $\omega +4$, since $\mathbb Q\subseteq \mathcal P\mathcal P(\mathbb Z)$? $\endgroup$ – youngtableaux Nov 13 '14 at 6:02
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    $\begingroup$ An element of $\mathbb{Z}$ is a large set of pairs of natural numbers. Pairs of natural numbers belong to $V_{\omega}$, so $\mathbb{Z} \subseteq V_{\omega + 1}$. On the other hand, clearly $\mathbb{Z} \not\subseteq V_{\omega}$, since an equivalence class contains pairs containing arbitrarily large natural numbers. $\endgroup$ – Mike Nov 13 '14 at 6:08

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