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:

  1. $\Vdash_\nu \dot{\alpha} \le \nu$

  2. 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:

  1. If $\dot{\alpha}$ is prompt $\eta \le \nu$ and if $p \Vdash_\nu \eta \le \dot{\alpha}$ then $p \upharpoonright \eta ^\smallfrown 1_\nu \upharpoonright [\eta, \nu) \Vdash_\nu \eta \le \dot{\alpha}$
  2. 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!

  • $\begingroup$ Cross posted from math.SE: math.stackexchange.com/questions/107134/… $\endgroup$ – Asaf Karagila Feb 9 '12 at 10:26
  • $\begingroup$ Presumably both occurrences of $\xi$ in conclusion 1 should be $\eta$ (or all the $\eta$'s should be $\xi$'s). Also, hypothesis 2 is presumably intended to be assumed for all $\xi$, not for some particular (unspecified) $\xi$. $\endgroup$ – Andreas Blass Feb 9 '12 at 14:36
  • $\begingroup$ Yes, you are right. I corrected it. $\endgroup$ – Stefan Hoffelner Feb 9 '12 at 14:44

Here's proof for conclusion 1; I think 2 will then follow fairly easily. For brevity, I'll write things like $p\upharpoonright\eta$ when I really mean its extension by 1 to domain $\nu$. Suppose $\eta$ were a counterexample to 1. Since $p\upharpoonright\eta$ fails to force $\eta\leq\dot\alpha$, it must have an extension $q$ forcing $\dot\alpha$ to have some specific value $\xi<\eta$. Apply promptness to infer that $q\upharpoonright\xi$ already forces this value $\xi$ for $\dot\alpha$. But, since $\xi<\eta$ and $q$ extends $p\upharpoonright\eta$, the conditions $q\upharpoonright\xi$ and $p$ are compatible. Since the former forces $\dot\alpha$ to have value $\xi$ while the latter forces $\eta\leq\dot\alpha$, this is a contradiction.

EDIT: I think I was too optimistic in expecting conclusion 2 to "follow fairly easily", so I'm adding information about that. Notice first that what I wrote above remains correct if we replace the inequalities $\eta\leq\dot\alpha$ and $\xi<\eta$ by $\eta<\dot\alpha$ and $\xi\leq\eta$, respectively. (Of course, $\eta<\dot\alpha$ is equivalent to $\eta+1\leq\dot\alpha$, so I could apply the preceding paragraph directly, but then I'd get a conclusion about $p\upharpoonright(\eta+1)$, whereas I really want $p\upharpoonright\eta$.)

Next, let me establish a "dual" (i.e., order-reversed) version of conclusion 1, namely that if $\dot\alpha$ is prompt and $p$ forces $\dot\alpha\leq\eta$, then $p\upharpoonright\eta$ already forces the same. To prove this, suppose it fails, and let $q$ be an extension of $p\upharpoonright\eta$ forcing $\eta<\dot\alpha$. By what I proved above, $q\upharpoonright\eta$ suffices to force $\eta<\dot\alpha$. But $q\upharpoonright\eta$ is an extension of $p\upharpoonright\eta$ and is therefore compatible with $p$. That's absurd since compatible conditions can't force contradictory things about the ordering of $\eta$ and $\dot\alpha$.

Of course, this dual version of conclusion 1 also has an analog for conditions $p$ that force $\dot\alpha<\eta$.

At last, I'm ready to prove the part of conclusion 2 that deals with the supremum, say $\dot\beta$, of some prompt names $\dot\alpha_i$. So suppose, toward a contradiction, that $p$ forces $\dot\beta=\xi$ but $p\upharpoonright\xi$ doesn't force this. There are two possibilities: Either some extension of $p\upharpoonright\xi$ forces some $\dot\alpha_i$ to be strictly above $\xi$, or some extension of $p\upharpoonright\xi$ forces some ordinal $\eta<\xi$ to be an upper bound for all the $\dot\alpha_i$'s. In either case, let $q$ be such an extension of $p\upharpoonright\xi$.

Consider the first case: $q$ forces $\xi<\dot\alpha_i$ for a certain index $i$. Then, by one of the versions of conclusion 1, $q\upharpoonright\xi$ already forces the same inequality. But $q\upharpoonright\xi$ is an extension of $p\upharpoonright\xi$ and is therefore compatible with $p$, which forces the opposite. So this case cannot occur.

There remains the case that, for a certain $\eta<\xi$, the condition $q$ forces $\dot\alpha_i\leq\eta$ for every index $i$. By a dual version of conclusion 1, as proved above, $q\upharpoonright\eta$ forces all these inequalities and therefore forces $\dot\beta\leq\eta$. But $q\upharpoonright\eta$ is compatible with $p$, which forces the opposite. So this case is also impossible, and the proof of conclusion 2 for the supremum is complete.

Finally, since I was (I hope) careful to use, in this proof for the supremum, only versions of conclusion 1 for which the dual is also available, we can dualize this whole proof to get the result for the infimum, which is of course the minimum since we're dealing with a well-rodering.


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.

  • $\begingroup$ Thank you again! Actually this approach was my first idea of how to prove the claims but I falsely thought that the Boolean algebra $B_\eta$ is isomorphic to the product of the $(B_\xi)_{\xi < \eta}$ (which is as I belive only true for inverse limits?). After realizing my error I didn't continue to think in that direction as I couldn't figure out how the embeddings $B_{\xi} \rightarrow B_{\eta}$ actually look like. But anyway thank you a lot! $\endgroup$ – Stefan Hoffelner Feb 11 '12 at 18:58

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