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单词 ProofThatForcingNotionsAreEquivalentToTheirComposition
释义

proof that forcing notions are equivalent to their composition


This is a long and complicated proof, the more so because the meaning of Q shifts depending on what generic subset of P is being used. It is therefore broken into a number of steps. The core of the proof is to prove that, given any generic subset G of P and a generic subset H of Q^[G] there is a corresponding generic subset G*H of P*Q such that 𝔐[G][H]=𝔐[G*H], and conversely, given any generic subset G of P*Q we can find some generic GP of P and a generic GQ of Q^[GP] such that 𝔐[GP][GQ]=𝔐[G].

We do this by constructing functions using operationsMathworldPlanetmath which can be performed within the forced universesPlanetmathPlanetmath so that, for example, since 𝔐[G][H] has both G and H, G*H can be calculated, proving that it contains 𝔐[G*H]. To ensure equality, we will also have to ensure that our operations are inversesPlanetmathPlanetmathPlanetmathPlanetmath; that is, given G, GP*GH=G and given G and H, (G*H)P=P and (G*H)Q=H.

The remainder of the proof merely defines the precise operations, proves that they give generic sets, and proves that they are inverses.

Before beginning, we prove a lemma which comes up several times:

Lemma: If G is generic in P and D is dense above some pG then GD

Let D={pPpDp is incompatible with p}. This is dense, since if p0P then either p0 is incompatible with p, in which case p0D, or there is some p1 such that p1p,p0, and therefore there is some p2p1 such that p2D, and therefore p2p0. So G intersects D. But since a generic set is directed, no two elements are incompatible, so G must contain an element of D which is not incompatible with p, so it must contain an element of D.

G*H is a generic filter

First, given generic subsets G and H of P and Q^[G], we can define:

G*H={p,q^pGq^[G]H}

G*H is closed

Let p1,q^1G*H and let p1,q^1p2,q^2. Then we can conclude p1G, p1p2, q^1[G]H, and p1q^1q^2, so p2G (since G is closed) and q^2[G]H since p1G and p1 forces both q^1q^2 and that H is downward closed. So p2,q^2G*H.

G*H is directed

Suppose p1,q^1,p1,q^1G*H. So p1,p2G, and since G is directed, there is some p3p1,p2. Since q^1[G],q^2[G]H and H is directed, there is some q^3[G]q^1[G],q^2[G]. Therefore there is some p4p3, p4G, such that p4q^3q^1,q^2, so p4,q^3p1,q^1,p1,q^1 and p4,q^3G*H.

G*H is generic

Suppose D is a dense subset of P*Q^. We can project it into a dense subset of Q using G:

DQ={q^[G]p,q^D} for some pG

Lemma: DQ is dense in Q^[G]

Given any q^0Q^, take any p0G. Then we can define yet another dense subset, this one in G:

Dq^0={ppp0pq^q^0p,q^D} for some q^Q^

Lemma: Dq^0 is dense above p0 in P

Take any pP such that pp0. Then, since D is dense in P*Q^, we have some p1,q^1p,q^0 such that p1,q^1D. Then by definition p1p and p1Dq^0.


From this lemma, we can conclude that there is some p1p0 such that p1GDq^0, and therefore some q^1 such that p1q^1q^0 where p1,q^1D. So DQ is indeed dense in Q^[G].


Since DQ is dense in Q^[G], there is some q^ such that q^[G]DQH, and so some pG such that p,q^D. But since pG and q^H, p,q^G*H, so G*H is indeed generic.

GP is a generic filter

Given some generic subset G of P*Q^, let:

GP={pPppp,q^G} for some pP and some q^Q

GP is closed

Take any p1GP and any p2 such that p1p2. Then there is some pp1 satisfying the definition of GP, and also pp2, so p2GP.

GP is directed

Consider p1,p2GP. Then there is some p1 and some q^1 such that p1,q^1G and some p2 and some q^2 such that p2,q^2G. Since G is directed, there is some p3,q^3G such that p3,q^3p1,q^1,p2,q^2, and therefore p3GP, p3p1,p2.

GP is generic

Let D be a dense subset of P. Then D={p,q^pD}. Clearly this is dense, since if p,q^P*Q^ then there is some pp such that pD, so p,q^D and p,q^p,q^. So there is some p,q^DG, and therefore pDGP. So GP is generic.

GQ is a generic filter

Given a generic subset GP*Q^, define:

GQ={q^[GP]p,q^G} for some pP

(Notice that GQ is dependant on GP, and is a subset of Q^[GP], that is, the forcingMathworldPlanetmath notion inside 𝔐[GP], as opposed to the set of names Q which we’ve been primarily working with.)

GQ is closed

Suppose q^1[GP]GQ and q^1[GP]q^2[GP]. Then there is some p1GP such that p1q^1q^2. Since p1GP, there is some p2p1 such that for some q^3, p2,q^3G. By the definition of GQ, there is some p3 such that p3,q^1G, and since G is directed, there is some p4,q^4G and p4,q^4p3,q^1,p2,q^3. Since G is closed and p4,q^4p4,q^2, we have q^2[GP]GQ.

GQ is directed

Suppose q^1[GP],q^2[GP]GQ. Then for some p1,p2, p1,q^1,p2,q^2G, and since G is directed, there is some p3,q^3G such that p3,q^3p1,q^1,p2,q^2. Then q^3[GP]GQ and since p3G and p3q^3q^1,q^2, we have q^3[GP]q^1[GP],q^2[GP].

GQ is generic

Let D be a dense subset of Q[GP] (in 𝔐[GP]). Let D^ be a P-name for D, and let p1GP be a such that p1D^ is dense. By the definition of GP, there is some p2p1 such that p2,q^2G for some q2. Then D={p,q^pq^Dpp2}.

Lemma: D is dense (in G) above p2,q^2

Take any p,q^P*Q such that p,q^p2,q^2. Then pD^ is dense, and therefore there is some q^3 such that pq^3D^ and pq^3q^. So p,q^3p,q^ and p,q^3D.


Take any p3,q^3DG. Then p3GP, so q^3D, and by the definition of GQ, q^3GQ.

GP*GQ=G

If G is a generic subset of P*Q, observe that:

GP*GQ={p,q^ppp,q^Gp0,q^G} for some p,q^,p0

If p,q^G then obviously this holds, so GGP*GQ. Conversely, if p,q^GP*GQ then there exist p,q^ and p0 such that p,q^,p0,q^G, and since G is directed, some p1,q1^G such that p1,q^1p,q^,p0,q^. But then p1p and p1q^1q^, and since G is closed, p,q^G.

(G*H)P=G

Assume that G is generic in P and H is generic in Q[G].

Suppose p(G*H)P. Then there is some pP and some q^Q such that pp and p,q^G*H. By the definition of G*H, pG, and then since G is closed pG.

Conversely, suppose pG. Then (since H is non-trivial), p,q^G*H for some q^, and therefore p(G*H)P.

(G*H)Q=H

Assume that G is generic in P and H is generic in Q[G].

Given any qH, there is some q^Q such that q^[G]=q, and so there is some p such that p,q^G*H, and therefore q^[G]H.

On the other hand, if q(G*H)Q then there is some p,q^G*H, and therefore some q^[G]H.

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更新时间:2025/5/4 8:40:25