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I guess that amounts to if there is a continuous $f$ with $\mathbb{P}(\mathbb{L}_{f(\alpha)}) \cap \mathbb{L} = \mathbb{L}_{f(\alpha+1)}$

I seem to remember reading that it is, but I forget where or when I read it, why it's true, what f is, and I can't find it now.

Thanks for any info.

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I'm pretty sure that you know who read it and how... – t.b. Jul 28 '11 at 21:49
First, even if it is in the title, you should re-ask your question in the body of the post. Second, what is your notation? Beyond the fact that your tag says set-theory, I have no way of knowing if I am capable of answering your question. Can you give a little context? Even if the notation is standard, there are generally lots of different places where similar notation is used for very different things. – Aaron Jul 28 '11 at 21:52
Is $V_{\omega+1}$ an $L_\alpha$? Which one? – Andrés E. Caicedo Jul 28 '11 at 21:54
@Aaron: This notation is as basic in set theory as $\int$ signs are in analysis :-) – Asaf Karagila Jul 28 '11 at 21:55
@Aaron: $V = L$ is the abbreviation for the Axiom of Constructibility. The indices are explained on the pages on the von Neumann universe and the Gödel (constructible) universe. – t.b. Jul 28 '11 at 22:02
up vote 6 down vote accepted

The answer is no.

  1. It is a theorem of $ZFC$ that $V\models\alpha>\omega\rightarrow |L_\alpha|=|\alpha|$, therefore it is consistent with $V=L$ as well, so $L\models |L_\alpha|=|\alpha|$ for infinite $\alpha$.
  2. On the other hand, $V\models V_{\alpha+1}=\mathcal P(V_\alpha)$, which for $\alpha>0$ is not the case that $|V_\alpha|=|\alpha|$, but rather much larger. (Of course for inaccessible cardinals there is an equality but this requires a consistency strength greater than $Con(ZFC)$).
  3. Now consider the following case: $L\models 2^\omega=\omega_1$, since $L_{\omega+1}$ is countable it cannot have all the subsets of $\omega$. On the other hand $\mathcal P(\omega)\subseteq V_{\omega+1}$ so clearly $L_{\omega+1}\neq V_{\omega+1}$.
  4. And finally $\beta$ is the set of all ordinals in both $L_\beta$ and $V_\beta$. If for some $\beta$ we had $V_{\omega+1}=L_\beta$ then $\beta=\omega+1$ but we have seen that this is not the case.
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Huh, it's not only wrong but obviously wrong. Not sure what I was remembering. Thanks! – where when why what who how Jul 28 '11 at 22:06

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