# A question about the proof that forcing extensions don't add ordinals

I've been reading Halbeisen's Combinatorial Set Theory: with a gentle introduction to forcing (well, more like skimming for now...) and I've stumbled into an apparent problem with a proof of the theorem:

Let ${\bf V}\models ZFC$, let ${\mathbb P}$ be a forcing notion in $\bf V$, and let $G$ be $\mathbb P$-generic over $\bf V$; then $\Omega^{\mathbf V[G]}=\Omega^{\bf V}$

The proof is by contradiction; we take the least $\gamma$ which is an ordinal in $\mathbf V[G]\setminus \mathbf V$, then choose a $\mathbb P$-name $\tilde \gamma$ for $\gamma$ (the book uses undertilde, but I don't think it's supported here). The proof then continues with:

Then $\{ \tilde x:\exists p(\langle \tilde x,p\rangle\in\tilde\gamma)\}$ is a set in $\bf V$

That's all right (assuming $p$ ranges over $\mathbb P$, otherwise there's the same problem as stated below). But then the offending statement follows:

hence, the collection of all ordinals $\alpha\in\gamma$ is in fact a set in $\bf V$.

Why why would it be so? It seems to me that the collection of all ordinals $\alpha\in\gamma$ can be defined as a set using axiom schema of comprehension, but only using $G$ as a parameter, but we can't do this in $\bf V$, because $G\notin \bf V$.

Am I being overly cautious? If not, how do I see that it is a set?

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"Am I being overly cautious?" Not at all, there are a few issues with the sketch in Lorenz's book. To start with: If $\gamma$ is minimal, you probably want to fix a $p$ forcing this. This requires that you are able to express that $\gamma$ is minimal, which means you need to be able to say that it is not in $\mathbf V$. This is the trickiest part, actually. The rest follows from minimality in a straightforward fashion (any element of $\gamma$ comes from a name in $\hat \gamma$, and for some $\alpha\in\mathbf V$, some $r$ below $p$ forces that it equals $\alpha$, etc). –  Andres Caicedo Sep 11 '12 at 1:55
@AndresCaicedo: thanks for the suggestion. I'll try to work it out after I get some sleep. :) –  tomasz Sep 11 '12 at 2:05
Another approach would be to show that if $\tilde a$ is a name that denotes (relative to some generic $G$) an ordinal $\alpha$, then $\alpha\leq\rho$, where $\rho$ is the rank of the name $\tilde a$ in the cumulative hierarchy. –  Andreas Blass Nov 11 '12 at 1:42