Cohen Forcing in Set Theory - Proof that Forcing is Equivalent to intersection of Dense Sets

Question

Cohen's book "Set Theory and the Continuum Hypothesis" on Page 126/127 (see below) shows that the existence of a completed new set a' is equivalent to its intersection with all dense subsets in M. I have the following questions :

(i) Is it possible to expand out the proof of the THEOREM into more detailed clear steps

and

(ii) Why in the Proof below is the set B "of forcing conditions that force either the statement or its negation" an element of M? Also how does the set of all these B i.e. {B : each element of B forces the same statement in the language} differ from the set of all dense sets ?

*********** Extract from Cohen p126/127 *************

We close with a theorem that restates the defining property of a complete sequence without using the notion of forcing. Even if we had defined complete sequences in this way, however, to show that N is a model we would still have to go through the same steps.

Definition. Let A be the set of all pairs where P and Q are disjoint finite subsets of ω. We write <P1,Q1> < <P2,Q2> if P1 ⊆ P2, Q1 ⊆ Q2. A subset B of A is called dense if i) for all x ∈ A, ∃y ∈ B and x < y, ii) x ∈ B, x < y implies y ∈ B. If {Pn} is an increasing sequence of forcing conditions, write a' = Lim Pn if a' = {n | ∃k (n ∈ a) ∈ Pk}.

THEOREM. If a' ⊆ ω, ∃ a complete sequence {Pn} with a' = Lim Pn if and only if for every dense subset B of A lying in M, ∃n such that if P = a' ∩ {0,...,n} and Q = {0,...,n} - P, then <P,Q> ∈ B.

Proof. Since whether {Pn} is a complete sequence depends only on Lim Pn, the theorem gives a characterization of complete sequences. It is clear from the definition of a dense set that if B is given then for any P, ∃ Q ⊇ P such that Q forces a' to have the property of the theorem. This implies that if {Pn} is complete a' has that property. Conversely, if a' satisfies our property let Pn be any sequence with a' = Lim Pn. Then, for any given statement let B be the set (which is in M) of forcing conditions which force either the statement or its negation. This is a dense set and hence by our condition some Pn forces either the statement or its negation.

Answer 1

Well, 7 years later I think I have cracked it:

The question's THEOREM is the origin of current methods of Forcing (a' $ \subseteq \omega$ is the forced new set, with dense defined as above - note its different from the current usual definition):

[ $\exists$ a negation complete sequence {$P_n$} with a' = Lim $P_n$ ] $\tag{1}$

if and only if

[ for every dense subset B of A lying in M, $\exists$ n such that if P = a' $\cap$ {0,...,n} and Q = {0,...,n} - P, then <P,Q> $\in$ B ]. $\tag{2}$

Cohen's objective is "Since whether {$P_n$} is a complete sequence depends only on Lim $P_n$, the theorem gives a characterization of complete sequences [independently of the particular {$P_n$} that produce a']". Indeed as a given a' could be generated in many ways, it would be very awkward if only particular {$P_n$} could be used to create a'. So the purpose of the theorem is to show that just (2) is needed to specify a negation complete a', but it is equivalent to (1) created using the Forcing concept.

Background to equation (1) : A little background is needed to show how (1) is constructed. There are just three properties that Forcing needs to have to create a consistent negation complete sequence a' (with $R \in A$):

Property 1 : $\forall R$ $\forall \varphi$ (we do not have $ R \Vdash \varphi$ AND $ R \Vdash \neg \varphi)$

Property 2 : $\forall R$ $\forall \varphi \text{ } (R \Vdash \varphi$ and $S \supseteq R$ then $S \Vdash \varphi)$

Property 3 : $ \forall R \text{ } \forall \varphi \text{ } (\exists \text{ } S \supseteq R$ : either $S \Vdash \varphi$ or $S \Vdash \neg \varphi \text{ but not both })$

Note that these properties are shared by the usual $\vdash$ (if it is associated with a consistent negation complete theory), so they are the 'typical' properties required by a negation complete and consistent deductive system, based upon the usual properties of $\vdash$.

Definition : Let A be the set of all pairs <P,Q>, where P and Q are disjoint finite subsets of $\omega$. We write

[ $<P_1,Q_1> \text{ } \leq \text{ } <P_2,Q_2>$ ] iff [ $P_1 \subseteq P_2 \text{ and } Q_1 \subseteq Q_2$].

If we wish to just use a' independently of the particular {$P_n$}, the above properties need to be represented as sets. As Property 1 can be derived from the other properties, then Properties 2 and 3 need to be used in set form.

Definition : A Dense set B : $ B \subseteq A$ has the following two properties :

(i) Based upon Property 3, [$\forall x \in A \text{ } \exists y \in B : x \leq y$]

(ii) Based upon Property 2, [$ \forall x \text{ } \forall y \text{ } (y \in B \text{ and } x \in A \text{ and } (x \geq y)) \text{ } \implies x \in B $]

Note that (i) & (ii) have effectively fixed $\varphi$ from Properties 3 & 2, and generalised B from relating to a particular $\varphi$, by removing $\Vdash \varphi$ from the definition of B. The approach is to ensure that the Dense sets are in M, but do not have any explicit reference to $\Vdash$ and a', since the intention is that a' is not in M. Since the B are not defined in terms of just $\varphi$, the objective is for a' to intersect all B, to ensure a complete set of $\varphi$ is captured. This intersection can't be done in the base model M, as otherwise it would define a Truth function within M contrary to the Tarski Truth definability theorem.

Definition : We write a' = Lim $P_n$ iff {$P_n$} is any increasing sequence of forcing conditions and a' = {n | $\exists$ k (n $\in $ a) $\in $ $P_k$}. So this extracts a' from the {$P_n$} used to construct it.

Background to equation (2) : To see what equation (2) is doing, we need to unpack the meaning of [$\exists$ n such that if P = a' $\cap$ {0,...,n} and Q = {0,...,n} - P, then <P,Q> $\in$ B]. Since the objective is to use a' and not specific {$P_n$}, we need a type of <P,Q> that is in any B and works just with a'. So if a' is ordered in $\omega$ so that it is an increasing set of elements of $\omega$, for finite n $\in \omega$ then {$x : x \in a' \land x \leq n$} will be the elements of {0..n} in a', i.e. its equal to P = a' $\cap$ {0,...,n}. Similarly for those elements not in a' but $\leq n$, Q={0..n} - P. So the question is can such a <P,Q> be found in any dense set B, for some n ? Remarkably it can.

Proof Equation (1) implies Equation (2) :

Since B is dense, for any x $\in$ A, there exists a $y \geq x$ : $y \in B$. Due to (ii) B also contains all supersets of y. From Equation (1) & Property 3, for any fixed $\varphi$ there exists an $S \in A : S \supseteq$ y that gives $S \Vdash \varphi \text{ or } S \Vdash \neg \varphi$. As B is dense it contains all supersets of y, so includes S. If S = <P,Q> then the maximum n in P or Q can be found. As B is dense, it contains all supersets of S. Although B does not know a', or the elements from 0..n that are in a', since every superset of S is produced, one of them S' = <P',Q'> will only include exactly which elements of a' $\leq$ n are in a'. Hence (2) applies.

Proof Equation (2) implies Equation (1) :

Although the definition of a dense set dropped the reference to $\varphi$ and $\Vdash$, in fact if ALL B $\in$ M are considered (as $\Vdash$ is absolute and definable in a base countable model M of ZFC) then there will be, for each $\varphi_i$ one $B_i$ that includes only those b : $b \Vdash \varphi$ (or $b_i \Vdash \neg \varphi$). Let {$P_n$} be any set that generates a'. We know that $B_i$ will include a <P,Q> that just includes the [0..r] elements in a'. Even though <P,Q> may not be in {$P_n$} used to create a', we know as the $P_n$ are finite, whatever the {$P_n$}, there will be a first one $P_k$ that first fully describes what elements from a' are in the first [0..r] elements from $\omega$, but may of course also describe elements from a' that are greater than r. However this doesn't matter since $B_i$ is dense then it will include all supersets of <P,Q>, which will therefore include the $P_k$ $\in$ {$P_n$}. So there will be some $P_k$ that forces $\varphi$ (or $\neg \varphi$) in $B_i$. Since all $B_i$ and therefore $\varphi_i$ are in the set of all dense B, then the sequence {$P_n$} will be complete. Hence (1) applies.

Note 1 :

To illustrate that Properties 1, 2 & 3 arise naturally using $\vdash$ :

Property 1 : $\vdash$ is consistent : (we do not have $ P \vdash \varphi$ AND $ P \vdash \neg \varphi)$

Property 2 : $P \vdash \varphi \text{ implies } P \land Q \vdash \varphi$

Property 3 : If $\vdash$ is associated with some axioms that make the theory T negation complete (suppressing mention of T):

Assume for each $\varphi$, $\exists \; B$ ($B \vdash \varphi$ or $B \vdash \neg \varphi$)

Then for any C, $\exists \; Q = B \land C \; $ : ($Q \vdash \varphi$ or $Q \vdash \neg \varphi$)

Hence $\forall \varphi \; \forall C$ ($\exists \; Q = B \land C$ : $Q \vdash \varphi$ or $Q \vdash \neg \varphi)$