Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
    
Cross posted from math.SE: math.stackexchange.com/questions/107134/… –  Asaf Karagila Feb 9 '12 at 10:26
    
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) $\xi$. –  Andreas Blass Feb 9 '12 at 14:36
    
Yes, you are right. I corrected it. –  Stefan Hoffelner Feb 9 '12 at 14:44
add comment

2 Answers

up vote 7 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
    
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! –  Stefan Hoffelner Feb 11 '12 at 18:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.