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$\mathbf{A.}$ Define a partial order (Domain,<,PO) as follows:

PosNeg := {1,-1, 2,-2, 3,-3 ....}

Domain := {FinSub$_i$ : Finsub$_i$ is a finite subset of PosNeg}

PO is ordered using "<" with the following rules:

FinSub$_i$ < FinSub$_j$ $\;$ iff $\;$ FinSub$_i$ $\subset$ FinSub$_j$ ...............................(1)

$\mathbf{B.}$ Define a Cohen Dense "D" subset of Domain having the following two properties :

(i) $\forall$ x $\in$ Domain $\exists$ y $\in$ D (x < y) .......................(2)

(ii) IF [(x $\in$ D) AND (x < y) AND (y $\in$ Domain) ] THEN y $\in$ D.......................(3)

$\mathbf{C.}$ $\mathbf{My}$ $\mathbf{Question}$ $\mathbf{is:}$

How to show that any 'consistent and complete' infinite subset "CC" of PosNeg will intersect every Cohen Dense set D (consistent means : if i/-i is in CC then -i/i is not in CC, and 'complete' means for every i, then i or -i is in CC)?

(Note this question is an attempt to 'modernise' the question Cohen Forcing in Set Theory - Proof that Forcing is Equivalent to intersection of Dense Sets).

My attempt at this proof showed that a Cohen Dense set "PerverseD" could always be chosen that would $\mathbf{stop}$ CC intersecting "PerverseD", making the question statement false (but Cohen indicates it is true).

My construction of this "PerverseD" is constructed by applying (2) first then (3), e.g. if CC:={1,2,...,n,.....} with all positive numbers:

If x={1,2,3,....,n} put {1,2,3,...,n,-(n+1)} $\in$ "PerverseD". Then take all possible supersets of these sets.

So as n tends to infinity there will be always be a finite part of CC that can be a subset of a set in "PerverseD", but none are ever exactly the same, for finite n.

  • 1
    This presentation is really hard to follow; why not use the modern language? "Complete consistent sets" are just maximal filters, Cohen forcing is more clearly described as the set of finite partial functions $\omega\rightarrow 2$ (or equivalently the set of finite binary sequences) ordered by reverse extension. You're just asking whether every maximal filter in Cohen forcing is fully generic. (Also, using TeX would help.) – Noah Schweber Dec 16 '19 at 15:49

1 Answers1

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In fact, no maximal filter (= complete consistent set) in any nontrivial forcing notion (you're looking at Cohen forcing in particular) meets every dense set: if $\mathbb{P}$ is a forcing notion and $G$ is a filter in $\mathbb{P}$ the set $\mathbb{P}\setminus G$ is a dense (by nontriviality) subset of $\mathbb{P}$ which $G$ does not meet.

(In your language: if $G$ is any complete consistent set, the collection of elements of the partial order which "disagree" with $G$ at some point is a dense set which $G$ doesn't meet.)

In particular, if $M$ is a model of ZFC and $\mathbb{P}$ is a nontrivial forcing notion in $M$, no filter which is $\mathbb{P}$-generic over $M$ will exist in $M$. Of course if $M$ is countable there will indeed be $\mathbb{P}$-generic filters of $M$, but they won't be elements of $M$.

What happens if we try to run the argument above inside a countable $M$? Well, we need to argue that for every filter $G$ the dense set $\mathbb{P}\setminus G$ is actually in $M$. But there's no reason to believe that it is. In fact, it never will be, since that would contradict the fact that generics over $M$ do exist.

And here's a proof of that: since $M$ is countable, the set $\mathcal{D}$ of dense subsets of Cohen forcing which are elements of $M$ is countable. We know that for any countable set of dense sets there is a filter meeting them all.

Noah Schweber
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  • So on p127 of Cohens book where he characterizes a complete sequence as one that intersects each "Cohen Dense" set in M must not be what he is doing ? though strictly he says Exists n : P= G intersects {0,..,n} and Q ={0,...,n}-P, forming <P,Q> is in the dense set - in Cohen's notation. I'm now mystified what Cohen is saying. –  Dec 16 '19 at 16:30
  • @LittleCheese Cohen's complete sequence is not in $M$. Per the last sentence of my answer, since $M$ is a countable model of ZFC there do exist $\mathbb{P}$-generic filters over $M$, they're just not in $M$ itself. (Similarly, there is a bijection between $\omega_{17}^M$ and $\omega$, but no such bijection in $M$ itself.) – Noah Schweber Dec 16 '19 at 16:31
  • True, though on p127 he 'looks' to be intersecting it with all dense sets in M (even though he doesnt say it).? –  Dec 16 '19 at 16:33
  • @LittleCheese So? The filter isn't in $M$, the dense sets are. We can talk about how a thing not in $M$ relates to things which are in $M$. – Noah Schweber Dec 16 '19 at 16:34
  • Again, modern presentations make all of this much clearer - it's so much easier to just talk about filters. What we're saying here is: if $M$ is a countable model of ZFC and $\mathbb{P}$ is a nontrivial forcing notion with $\mathbb{P}\in M$, then there does exist a $\mathbb{P}$-generic filter over $M$ but there does not exist a $\mathbb{P}$-generic filter over $M$ which is an element of $M$. – Noah Schweber Dec 16 '19 at 16:35
  • I agree - my question relates to intersecting all dense sets in M by the Cohen complete consistent sequence. My 'proof' that its not possible must therefore be incorrect ? –  Dec 16 '19 at 16:36
  • @LittleCheese To be honest I can't really follow your argument, but I think it may actually be right. Note that you're not talking about $V$ vs. $M$, you're just talking about $V$, and the statement you want to prove (which is indeed true) is "There is no maximal filter meeting all dense sets in Cohen forcing." But that doesn't contradict what Cohen's saying, since he's saying "There is a maximal filter meeting all dense sets in $M$ in Cohen forcing." – Noah Schweber Dec 16 '19 at 16:40
  • (Also, you really should start using TeX - it will make things much easier to follow, and write for that matter.) – Noah Schweber Dec 16 '19 at 16:41
  • How do I use TeX in comments - is there a link ? –  Dec 16 '19 at 16:43
  • Cohen is saying within M, but how does that make my construction in the question incorrect ? –  Dec 16 '19 at 16:44
  • @LittleCheese I mean both in your comments and questions. See https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference. – Noah Schweber Dec 16 '19 at 16:46
  • @LittleCheese I didn't say the construction in your question is incorrect - I said "to be honest I can't really follow your argument, but I think it may actually be right." My comment re: you vs. Cohen was explaining why there isn't any tension ("that doesn't contradict what Cohen's saying"). – Noah Schweber Dec 16 '19 at 16:47
  • I thought I was reasoning using Dense sets in M ! Maybe the dense set I constructed isn't in M because I was using the knowledge about what G was (which I did) ? I'm mystified as to what changes by just reasoning using Cohen Dense subsets in M. –  Dec 16 '19 at 16:53
  • @LittleCheese You didn't mention a "small model" in your question anywhere, so as written you're doing everything in $V$. What changes ... is, as I've said above, the existence of generics. Again, there is no fully generic filter in Cohen forcing; however, for every countable $M$ there is a filter which is generic over $M$. The point is that by only looking at dense sets in $M$, Cohen has only countably many "requirements" to meet (but of course that's taking place in $V$). – Noah Schweber Dec 16 '19 at 16:56
  • "Maybe the dense set I constructed isn't in M because I was using the knowledge about what G was (which I did) ?" Once we bring $M$ into the picture (and again, $M$ isn't present in your question as written) that's exactly right. And this is the same point I made about expressibility in my answer to your other question - $M$ can't "see" $G$, so in general questions which make reference to $G$ (as opposed to the name in $M$ for the generic) can't be asked in $M$ and things defined using $G$ (again, as opposed to the name) don't exist in $M$. – Noah Schweber Dec 16 '19 at 16:58
  • OK, so Cohen is working in M. The Cohen Dense set concept is derived from the formulae like (1) & (2) in the question, but generalises them to make equations (1) & (2). There will be a Cohen Dense set for each formula that's Forced by the conditions in the Cohen Dense set. But as Cohen has generalised the concept to be any set that satisfies (1) & (2), what do these other dense sets represent (if indeed there are any dense sets that don't force a formua to be true/false? Also how do they intersect with the final Cohen G ? –  Dec 16 '19 at 17:13
  • @LittleCheese " how do they intersect with the final Cohen G" Well, by definition $G$ - which is not an element of the countable model $M$ - meets all the dense sets which are in $M$. So ... – Noah Schweber Dec 16 '19 at 17:18
  • "as Cohen has generalised the concept to be any set that satisfies (1) & (2), what do these other dense sets represent (if indeed there are any dense sets that don't force a formua to be true/false?" In fact, every dense set does "make decisions:" if $D$ is dense, pick $p\in D$ and consider the sentence $\varphi_p$ = "$G$ contains an element compatible with $D$." For any $q\in D$, either $q\Vdash\varphi_p$ (iff $q$ is compatible with $p$) or $q\Vdash\neg\varphi_p$ (iff $q$ is not compatible with $p$). – Noah Schweber Dec 16 '19 at 17:21
  • OK thanks...... –  Dec 16 '19 at 17:32