TITLE: Question about prompt names of ordinals
QUESTION [5 upvotes]: I asked this question first on math SE and was told that it would better fit here. So:    
The following concept is due to Shelah and I have some issues with a claim using this notion:
Suppose that $\nu$ is a limit ordinal and that $P_\nu$ is an iteration of forcing notions. We say that a $P_\nu$ name $\dot{\alpha}$ of an ordinal is $prompt$ iff the following two things hold:

$\Vdash_\nu \dot{\alpha} \le \nu$
If $p \Vdash_\nu "\dot{\alpha} = \xi"$ then even $p \upharpoonright \xi ^\smallfrown 1_\nu \upharpoonright [\xi, \nu) \Vdash_\nu \dot{\alpha} = \xi$  ( $1_\nu$ should be the largest element of the iteration, and $\xi$ is the hacek name of an ordinal though I refused to write the hacek)

Then the following two things should hold:

If $\dot{\alpha}$ is prompt $\eta \le \nu$ and if $p \Vdash_\nu \eta \le \dot{\alpha}$ then $p \upharpoonright \eta ^\smallfrown 1_\nu \upharpoonright [\eta, \nu) \Vdash_\nu \eta \le \dot{\alpha}$
If $\dot{\alpha_i}$ are prompt then so is the supremum $sup$ and the minimum $min$

I have problems proving those two assertions so any help would be highly appreciated. Thank you!

REPLY [6 votes]: I'm giving a second answer rather than editing the first, partly because the first is already unpleasantly long, and partly because I think the following is a better way to view the situation.  Let me first recall some general information about iterated forcing.  I'll write  $P_\xi$ for the forcing resulting from the first $\xi$ steps of the iteration, and I'll write $B_\xi$ for its regular open algebra, i.e., the complete Boolean algebra that has $P_\xi$ as a dense subset.  If $\xi<\eta$, then the obvious embedding $P_\xi\to P_\eta$, appending $(\eta-\xi)$ 1's to any condition, extends to a complete embedding of Boolean algebras $B_\xi\to B_\eta$.  The completeness of these embeddings seems to be what underlies the results in the question.  
I'll simplify notation by identifying all the algebras $B_\xi$ with their images in $B_\nu$ (where $\nu$ is, as in the question, the length of the iteration). To say that $\dot\alpha$ is prompt is just to say that, for each $\xi<\nu$ the Boolean truth value $\Vert\dot\alpha=\xi\Vert$ is in $B_\xi$.  This implies, thanks to completeness of the subalgebras, that the following is in $B_\eta$ (for any $\eta\leq\nu$):
$$\Vert\eta\leq\dot\alpha\Vert=
-\bigvee_{\xi<\eta}\Vert\dot\alpha=\xi\Vert,$$
and this, when translated back from Boolean-algebra language to forcing language, gives conclusion 1 of the question.  Similarly, for conclusion 2, using the assumption that $\Vert\dot\alpha_i=\xi\Vert\in B_\xi$ for all $i$ and $\xi$, we find that $B_\eta$ contains
$$\Vert\sup_i\ \dot\alpha_i=\eta\Vert=
\left(\bigwedge_i\bigvee_{\xi\leq\eta}\Vert\dot\alpha_i=\xi\Vert\right)\land
\left(\bigwedge_{\gamma<\eta}\bigvee_i\Vert\gamma+1\leq\dot\alpha_i\Vert\right)$$
and
$$\Vert\min_i\ \dot\alpha_i=\eta\Vert=
\left(\bigvee_i\Vert\dot\alpha_i=\eta\Vert\right)\land
\left(\bigwedge_i\Vert\eta\leq\dot\alpha_i\Vert\right).
$$
As far as I can see, all the work in my previous answer --- extending conditions, truncating them, and observing compatibility --- was just repeating (too many times) the argument for why each $B_\eta$ is a complete subalgebra of $B_\nu$ (and why promptness is equivalent to its Boolean formulation).  The moral of the story is that the Boolean-valued  viewpoint sometimes makes things considerably easier and clearer than the forcing viewpoint.