# Omega, well ordering and the cumulative hierarchy $V$.

Hi I need help with a problem of set theory. I'm not sure how to prove that the well ordering on $\omega$ isomorphic to $\omega+\omega$ belongs to the level $V_{\omega+\omega}$ in the hierarchy $V$.

any help?

thanks!

-

Any relation on $\omega$ is a subset of $\mathcal P (\mathcal P ( \mathcal P(\omega)))$.

$V_{\omega + \omega} = \bigcup_{\beta < \omega + \omega} \mathcal P (V_\beta)$ and $\omega \in V_{\omega + 1}$.

Now combine these two to obtain the desired.

-

Note that $\omega$ is a very particular set, it belongs $V_{\omega+1}$. It is not hard to calculate and see that $\omega\times\omega$ belongs to $V_{\omega+5}$ (or even less), so every subset of it would be in $V_{\omega+6}$ and in particular the subsets which are well-orders of any order type.

-
Still having fun with Igor, I see. It's simply astounding that some people don't realise that science advances precisely because it is incorrect, and is willing to adjust itself. Otherwise we'd still be adhering to Ptolemy's conception of a geo-centric universe, since that was science (and math)! – Arthur Fischer Apr 22 '13 at 10:34
@Arthur: If I won't do that, I'll have to resort to actually working! :-) – Asaf Karagila Apr 22 '13 at 10:40
Any pair $(n,m)$ is in $V_\omega$, so $\omega\times\omega$ and all its subsets are in $V_{\omega+1}$. – Andrés Caicedo Apr 22 '13 at 18:01
@Andres: Ah, yes. That is a good observation. I have a vague memory of this being pointed out before in the comments to my answer on this site... – Asaf Karagila Apr 22 '13 at 18:12
This is a test$\vphantom{@Arthur}$. Did you get pinged? :-) – Asaf Karagila May 5 '13 at 22:28