2
$\begingroup$

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.

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

$\endgroup$
2
$\begingroup$

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 $V_{\alpha + 4}$.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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