Is there a minimum $\beta$ such that $\alpha\in \mathrm{V}_\beta$?

Question

Given some ordinal $\alpha$, is it known whether or not there is a minimum $\beta$ such that $\alpha\in\mathrm{V}_\beta$?

If so, then what is it? I've nailed it down for a couple ordinals:

$\omega:\omega+1$. This is because $\mathrm{V}_\omega$ does not contain $\omega$ (it is not hereditarily finite), but it is a set of hereditarily finite sets ($0,1,2,3,4...$), so $\mathrm{V}_{\omega+1} $ does.
$\omega+1:\omega+2$. This is because $\omega+1$ is not a set of hereditarily finite sets, (it contains $\omega$), but it is a set of sets of hereditarily finite sets.

This pattern of $\omega+\alpha:\omega+\alpha+1$ continues for all finite $\alpha$, but does it continue beyond $\omega\cdot2$? What about $\omega^2$?

Answer 1

When you see a pattern like this in the ordinals, the best thing to do is to see if transfinite induction can help: supposing that the pattern holds up to $\gamma$, can you show that it continues to hold at $\gamma$? At the same time, just like with normal induction, it's sometimes easier to prove a seemingly more ambitious statement.

In this case, try to prove the following:

$(*)\quad$ If $\alpha$ is an ordinal, then the ordinals in $V_\alpha$ are exactly those $<\alpha$.

(In symbols: "$Ord\cap V_\alpha=\alpha$." Remember that "$Ord$" denotes the class of ordinals itself, and an ordinal is exactly the set of ordinals smaller than itself.)

COMMENTS:

• Note that $(*)$ implies that for every ordinal $\alpha$, the first $\beta$ for which we have $V_\beta\ni\alpha$ is $\beta=\alpha+1$. So if you can prove $(*)$ you've answered your question.

• To get things started, $(*)$ is trivially true for $\alpha=0$: there are no ordinals $<0$, and there are no elements of $V_0$.

• For a limit ordinal $\lambda$, note that $Ord\cap V_{\lambda}=\bigcup_{\gamma<\lambda}(Ord\cap V_\gamma)$; do you see how to use this to show that if $(*)$ holds up to $\lambda$, it holds at $\lambda$ too?

• Finally, for the successor step: do you see how to show that if $Ord\cap V_\alpha=\alpha$, then $Ord\cap V_{\alpha+1}=\alpha+1=\alpha\cup\{\alpha\}$? This has two parts: first, show that $Ord\cap V_{\alpha+1}$ is at least as big as $\alpha\cup\{\alpha\}$, and secondly, showing that it is no bigger than $\alpha\cup\{\alpha\}$; try tackling each part separately, and remember how $V_{\alpha+1}$ is defined.