In a paper I'm reading involving forcing, the following is used without proof, and it seems to be stated as if it is "obvious," but I don't see why. It's probably either easy or known, but I might not read the appropriate reference. Can anyone help?
Let $\overline{\alpha}$ be a $P$-name for an ordinal. Then for any $p\in P$ there is a $q\leq p$ and an $\alpha$ in the ground model such that $q\Vdash \overline{\alpha} = \check\alpha$.
(Here $\check\alpha$ is the canonical name).
On
This is true because forcing extensions do not add ordinals to the universe. So if we know that $\bar\alpha$ is a name which will be interpreted as an ordinal, it would have be interpreted as some ordinal from the ground model.
Which one may depend on the generic filter, but this really just means that at some point there will be a condition that will decide the value of that ordinal.
To see why forcing does not add ordinals, here is a Boolean-valued model approach Boolean-valued models have no essentially new ordinals.
Given any $p \in P$, let $G$ be a generic filter for $P$ containing $p$. Since forcing adds no new ordinals, and as $\overline{\alpha}$ is a $P$-name for an ordinal, there is an ordinal $\alpha$ in the ground model such that $\overline{\alpha}[G] = \alpha = \check{\alpha}[G]$. By the Forcing Theorem there is a $q \in G$ such that $q \Vdash \overline{\alpha} = \check{\alpha}$, and by extending if necessary we may assume $q \leq p$.
The above both gives the result you are looking for, and, equivalently, proves that the set $$\{ q \in P : ( \exists \alpha ) ( q \Vdash \overline{\alpha} = \check{\alpha} \}$$ is dense in $P$