forcings are equivalent if one is dense in the other

Suppose P and Q are forcingMathworldPlanetmath notions and that f:PQ is a function such that:

  • p1Pp2 implies f(p1)Qf(p2)

  • If p1,p2P are incomparable then f(p1),f(p2) are incomparable

  • f[P] is dense ( in Q

then P and Q are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath.


We seek to provide two operationsMathworldPlanetmath (computable in the appropriate universesPlanetmathPlanetmath) which convert between genericPlanetmathPlanetmathPlanetmath subsets of P and Q, and to prove that they are inversesPlanetmathPlanetmathPlanetmath.

F(G)=H where H is generic

Given a generic GP, consider H={qf(p)q} for some pG.

If q1H and q1q2 then q2H by the definition of H. If q1,q2H then let p1,p2P be such that f(p1)q1 and f(p2)q2. Then there is some p3p1,p2 such that p3G, and since f is order preseving f(p3)f(p1)q1 and f(p3)f(p2)q2.

Suppose D is a dense subset of Q. Since f[P] is dense in Q, for any dD there is some pP such that f(p)d. For each dD, assign (using the axiom of choiceMathworldPlanetmath) some dpP such that f(dp)d, and call the set of these DP. This is dense in P, since for any pP there is some dD such that df(p), and so some dpDP such that f(dp)d. If dpp then DP is dense, so suppose dpp. If dpp then this provides a member of DP less than p; alternatively, since f(dp) and f(p) are compatible, dp and p are compatible, so pdp, and therefore f(p)=f(dp)=d, so pDP. Since DP is dense in P, there is some element pDPG. Since pDP, there is some dD such that f(p)d. But since pG, dH, so H intersects D.

G can be recovered from F(G)

Given H constructed as above, we can recover G as the set of pP such that f(p)H. Obviously every element from G is included in the new set, so consider some p such that f(p)H. By definition, there is some p1G such that f(p1)f(p). Take some dense DQ such that there is no dD such that f(p)d (this can be done easily be taking any dense subset and removing all such elements; the resulting set is still dense since there is some d1 such that d1f(p)d). This set intersects f[G] in some q, so there is some p2G such that f(p2)q, and since G is directed, some p3G such that p3p2,p1. So f(p3)f(p1)f(p). If p3p then we would have pp3 and then f(p)f(p3)q, contradicting the definition of D, so p3p and pG since G is directed.

F-1(H)=G where G is generic

Given any generic H in Q, we define a corresponding G as above: G={pPf(p)H}. If p1G and p1p2 then f(p1)H and f(p1)f(p2), so p2G since H is directed. If p1,p2G then f(p1),f(p2)H and there is some qH such that qf(p1),f(p2).

Consider D, the set of elements of Q which are f(p) for some pP and either f(p)q or there is no element greater than both f(p) and q. This is dense, since given any q1Q, if q1q then (since f[P] is dense) there is some p such that f(p)q1q. If qq1 then there is some p such that f(p)qq1. If neither of these and q there is some rq1,q then any p such that f(p)r suffices, and if there is no such r then any p such that f(p)q suffices.

There is some f(p)DH, and so pG. Since H is directed, there is some rf(p),q, so f(p)qf(p1),f(p2). If it is not the case that f(p)f(p1) then f(p)=f(p1)=f(p2). In either case, we confirm that H is directed.

Finally, let D be a dense subset of P. f[D] is dense in Q, since given any qQ, there is some pP such that pq, and some dD such that dpq. So there is some f(p)f[D]H, and so pDG.

H can be recovered from F-1(H)

Finally, given G constructed by this method, H={qf(p)q} for some pG. To see this, if there is some f(p) for pG such that f(p)q then f(p)H so qH. On the other hand, if qH then the set of f(p) such that either f(p)q or there is no rQ such that rq,f(p) is dense (as shown above), and so intersects H. But since H is directed, it must be that there is some f(p)H such that f(p)q, and therefore pG.

Title forcings are equivalent if one is dense in the other
Canonical name ForcingsAreEquivalentIfOneIsDenseInTheOther
Date of creation 2013-03-22 12:54:43
Last modified on 2013-03-22 12:54:43
Owner Henry (455)
Last modified by Henry (455)
Numerical id 6
Author Henry (455)
Entry type Result
Classification msc 03E35
Classification msc 03E40