Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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 $V_{\alpha + 4}$.

share|improve this answer
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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.