# Why $\mathrm{rank}(x^y) < \alpha+\omega$, if $x$, $y$ have rank $\le$ $\alpha$?

This question is from Set Theory, Jech(2006), Page 70, 6.5.

Rank function is defined as on Page 64:

• $V_0=\emptyset$,
• $V_{\alpha+1}=P(V_{\alpha})$,
• $V_{\alpha}=\bigcup_{\beta<\alpha}V_\beta$, if $\alpha$ is a limit ordinal.

$\mathrm{rank}(x)=\operatorname{min}\{\alpha \in \mathrm{Ord}:x \in V_{\alpha+1}\}$

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As Asaf hints, you really have to get your hands dirty for this one.

If $x , y \in V_{\alpha + 1}$, then for each $u \in x$ and $v \in y$ we have that $u , v \in V_\alpha$ and so $\{ v \} , \{ v , u \} \in V_{\alpha + 1}$ and so $\langle v , u \rangle = \{ \{ v \} , \{ v , u \} \} \in V_{\alpha + 2}$. Thus $y \times x \subseteq V_{\alpha + 2}$, and so $y \times x \in V_{\alpha + 3}$. Also, every subset of $y \times x$ belongs to $V_{\alpha + 3}$, in particular every function $y \to x$ belongs to $V_{\alpha + 3}$, and so the set of all those functions belongs to $V_{\alpha + 4}$.

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Oh snap. I was correct about the $+5$ or so? Cool. –  Asaf Karagila Dec 6 '12 at 14:26
Thank you for your elaborate answer. –  Metta World Peace Dec 6 '12 at 14:30

Show that the rank of $y\times x$ is below $\alpha+5$ or so, and the conclusion should follow.

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Thank you very much for your hints which help highlight a gap in my understanding. –  Metta World Peace Dec 6 '12 at 14:29
No problem. I would have written a longer answer but I am on the train and he iPhone is not fun for long answers :-) –  Asaf Karagila Dec 6 '12 at 14:33