forcings are equivalent if one is dense in the other
Suppose P and Q are forcing notions and that f:P→Q is a function such that:
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p1≤Pp2 implies f(p1)≤Qf(p2)
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If p1,p2∈P are incomparable then f(p1),f(p2) are incomparable
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f[P] is dense (http://planetmath.org/DenseInAPoset) in Q
then P and Q are equivalent.
Proof
We seek to provide two operations (computable in the appropriate universes) which convert between generic subsets of P and Q, and to prove that they are inverses.
F(G)=H where H is generic
Given a generic G⊆P, consider H={q∣f(p)≤q} for some p∈G.
If q1∈H and q1≤q2 then q2∈H by the definition of H. If q1,q2∈H then let p1,p2∈P be such that f(p1)≤q1 and f(p2)≤q2. Then there is some p3≤p1,p2 such that p3∈G, 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 d∈D there is some p∈P such that f(p)≤d. For each d∈D, assign (using the axiom of choice) some dp∈P such that f(dp)≤d, and call the set of these DP. This is dense in P, since for any p∈P there is some d∈D such that d≤f(p), and so some dp∈DP such that f(dp)≤d. If dp≤p then DP is dense, so suppose dp≰p. If dp≤p 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 p≤dp, and therefore f(p)=f(dp)=d, so p∈DP. Since DP is dense in P, there is some element p∈DP∩G. Since p∈DP, there is some d∈D such that f(p)≤d. But since p∈G, d∈H, so H intersects D.
G can be recovered from F(G)
Given H constructed as above, we can recover G as the set of p∈P 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 p1∈G such that f(p1)≤f(p). Take some dense D∈Q such that there is no d∈D 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 d1≤f(p)≤d). This set intersects f[G] in some q, so there is some p2∈G such that f(p2)≤q, and since G is directed, some p3∈G such that p3≤p2,p1. So f(p3)≤f(p1)≤f(p). If p3≰p then we would have p≤p3 and then f(p)≤f(p3)≤q, contradicting the definition of D, so p3≤p and p∈G 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={p∈P∣f(p)∈H}. If p1∈G and p1≤p2 then f(p1)∈H and f(p1)≤f(p2), so p2∈G since H is directed. If p1,p2∈G then f(p1),f(p2)∈H and there is some q∈H such that q≤f(p1),f(p2).
Consider D, the set of elements of Q which are f(p) for some p∈P and either f(p)≤q or there is no element greater than both f(p) and q. This is dense, since given any q1∈Q, if q1≤q then (since f[P] is dense) there is some p such that f(p)≤q1≤q. If q≤q1 then there is some p such that f(p)≤q≤q1. If neither of these and q there is some r≤q1,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)∈D∩H, and so p∈G. Since H is directed, there is some r≤f(p),q, so f(p)≤q≤f(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 q∈Q, there is some p∈P such that p≤q, and some d∈D such that d≤p≤q. So there is some f(p)∈f[D]∩H, and so p∈D∩G.
H can be recovered from F-1(H)
Finally, given G constructed by this method, H={q∣f(p)≤q} for some p∈G. To see this, if there is some f(p) for p∈G such that f(p)≤q then f(p)∈H so q∈H. On the other hand, if q∈H then the set of f(p) such that either f(p)≤q or there is no r∈Q such that r≤q,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 p∈G.
Title | forcings are equivalent if one is dense in the other |
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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 |