# potential theory and forcing[精选推荐pdf]

arXivmath/0502394v1 [math.LO] 17 Feb 2005Potential theory and forcing∗Jindˇ rich Zapletal†University of FloridaFebruary 1, 2008AbstractWe isolate a property of capacities which leads to construction of proper forcings, and prove that among others, the Newtonian capacity enjoys this property.1IntroductionWe will be concerned with outer regular subadditive capacities on Polish spaces. These are functions c PX → R∪ {∞} satisfying the following demands1. c0 0,A ⊂ B → cA ≤ cB,cA ∪ B ≤ cA cB2. cA inf{cO O ⊂ X is open and A ⊂ O} for every set A ⊂ X3. cSA sup{cAn n ∈ ω} whenever An n ∈ ω is an inclusion- increasing sequence of subsets of X4. cK cA.∗2000 AMS subject classification 03E40, 31C15. †Partially supported by GAˇCR grant 201-03-0933 and NSF grant DMS 0300201.1It turns out that very many capacities share this property. The set˜A is always one that has been long known and studied–for the Newtonian capacity [1] it is˜A A ∪ {x ∈ R3 the potential of A is ≥ 1 at x}; for the Stepr¯ ans capacities [10] it is the set˜A A ∪ {x ∈ X the set A has upper density 1 at x}. We include a list of examples in Section 5.Theorem 1.2. Suppose that c is an outer regular subadditive stable capacity on some Polish space X. Then1. the forcing PIcis proper2. ZFAD the capacity c is continuous in increasing wellordered unions, and every set has a Borel subset of the same capacity.The finer forcing properties of the posets PIcare shrouded in mystery except for a couple of general observations the forcings PIcare capacitable and therefore bounding [2] 7.13 the ideal Icis generated by Gδsets and therefore the forcing PIcmakes the ground model reals meager [2] 2.17 The poset PIc, where c is the Newtonian capacity, is nowhere c.c.c. [4] Theorem 4.6 If c is strongly subadditive then the forcing PIcpreserves Lebesgue outer measure [12].An interesting feature of the proofs is that they can be combined in the following sense. If {cm m ∈ n} is a finite collection of countably subadditivesubmeasures on some Polish space X let b be their join, the submeasure defined by bA inf{ΣmcmBm A S mBm}.This is the largest submeasure smaller than all of the submeasures {cm m ∈ n}. We do not know if a join of a collection of capacities must be a capacity. However, we do know that the σ-ideal Ibis generated by the union of the ideals Icm m ∈ n andTheorem 1.3. Suppose that {cm m ∈ n} is a finite collection of outer regular strongly subadditive stable capacities, and b is their join. Then1. the forcing PIbis proper2. ZFAD the submeasure b is continuous in increasing wellordered unionsof uncountable cofinality, in particular the ideal Ibis closed under wellordered unions. Every set has a Borel subset of the same submeasure.Note the additional assumption of strong subadditivity on the capacities con- cerned. Not all stable capacities are strongly subadditive–the Stepr¯ ans capaci- ties are not, while the Newtonian capacity is. Another possible combination is the following.2Theorem 1.4. Suppose that {cm m ∈ n} is a finite collection of outer regular strongly subadditive stable capacities on a compact Polish space X with a metric d, and let s 0 be a real number. Let I be the σ-ideal generated by the ideals Icm m ∈ n and the sets of finite s-dimensional Hausdorffmeasure. Then1. the forcing PIis proper2. ZFAD the ideal I is closed under wellordered unions, and every I- positive set has a Borel I-positive subset.The corresponding result for just the ideal of σ-finite Hausdorffmeasuresets was proved in [11]. Further variations are possible, but not effortless–forexample we can further adjoin the ideal of sets of σ-finite s-dimensional packingmeasure, but we do not know how to adjoin ideals of sets of σ-finite measurefor two different Hausdorffmeasures and preserve properness and closure under wellordered unions. We have not studied the properties of the resulting forcings. We do not know if they are bounding. We do not know if there are ideals I,J such that the forcings PIand PKare proper while the forcing PKis not, where K is the ideal generated by I ∪ J. Perhaps the most important open problem regards the relationship between stability and other properties of capacities.Question 1.5. Is every outer regularsubadditive capacity stable Is every outer regular strongly subadditive capacity stable Is the factor forcing PIcproper for every outer regular subaditive capacityIs every outer regular capacity continuous in increasing wellordered unions under ADOur notation follows the set theoretic standard of [6]. AD denotes the use of the Axiom of Determinacy, AD is a technical strengthening of AD due to W. Hugh Woodin. Special thanks go to Murali Rao of University of Florida, who showed that the capacities encountered in potential theory are stable. Without this result I would have never considered writing the present paper.2Stable capacitiesOnce and for all fix a Polish space X with a countable basis B for its topology,closed under finite unions. Let c be an outer regular subadditive stable capacity with the A 7→˜A operation as indicated in Definition 1.1. Let P be a forcing adding a point ˙ x ∈˙X.Consider an infinite game G between Players I and II. In the beginning, Player I indicates an initial condition pini∈ P and then produces a sequence Dk k ∈ ω of open dense subsets of the forcing P as well as a c-null set A. Player II produces a sequence pini≥ p0≥ p1≥ ... such that pk∈ Dkand pk decides the membership of the point x in the k-th basic open subset of the spaceX in some fixed enumeration. Player II wins if, writing g ⊂ P for the filter his conditions generate, the point ˙ x/g falls out of the set A.3In order to complete the description of the game, we have to describe the exact schedule for both players. At round k ∈ ω, Player I indicates the open dense set Dk⊂ PIcand sets Ak,l ∈ B for l ∈ k so that cAk,l ≤ 2−land l ∈ k0∈ k1implies Ak0,l ⊂ Ak1,l and cAk1,l − cAk0,l ≤ 2−k0. In the end, let Al S kAk,l and recover the set A ⊂ X as A T lAl. The continuity of the capacity in increasing unions shows that cAl ≤ 2−land so cA 0. Note that apart from the open dense sets, Player I has only countably many moves at his disposal. Still, he can produce a superset of any given c-nullset as his final set A ⊂ X. Player II is allowed to tread water, that is, to waitfor an arbitrary finite number of rounds place trivial moves before placing the next condition pkon his sequence.Lemma 2.1. Player II has a winning strategy in the game G if and only if P ˙ x falls out of all ground model coded c-null sets.Proof. The key point is that the payoffset of the game G is Borel in the large tree of all legal plays, and therefore the game is determined by [8]. A careful computation will show that the winning condition for Player I is in fact a union of an Fσand a Gδset. For the left-to-right direction, if there is some condition p ∈ P and a c-null Borel set B ⊂ X such that p ˙ x ∈˙B, then Player I can win by indicating pini p, producing some null set A ⊃ B, and on the side producing an increasing sequence Mi i ∈ ω of countable elementary submodels of some large structure and playing in such a way that the sets Dk k ∈ ω enumerate all open dense subsets of the poset P in the model M S iMi, and {pk k ∈ ω} ⊂ M. Inthe end, this must bring success this way, Player II’s filter g ⊂ P is M-generic containing the condition p, by the forcing theorem applied in the model M, M[g] | ˙ x/g ∈ A, and by Borel absoluteness ˙ x/g ∈ A as desired. The right-to-left direction is harder. Suppose that P ˙ x falls out of all ground model coded c-null sets, and σ is a strategy for Player I. By the de-terminacy of the game G, it will be enough to find a counterplay against the strategy σ winning for Player II. The following claim will be used repeatedly.Claim 2.2. Suppose that p ∈ P is a condition. There is a real number ǫp 0 such that for every set Borel B ⊂ X there is a condition q ≤ p forcing the point ˙ x out of˙B.Proof. If this failed for some condition p ∈ P, then for each number i ∈ ω there would be a Borel set of capacity ≤ 2−isuch that p ˙ x ∈˙Bi. But then, the set B T iBiis a Borel set of zero capacity and p ˙ x ∈˙B. This is a contradiction.Let pini∈ P be the initial condition indicated by the strategy σ, and let l ∈ ω be a number such that 2−l¯k the condition * will besatisfied after Player II places the move r at the round k. Then for every nuber k ¯k there is a Borel set Bk ⊃ Ak,l such that cBk ≤ cAk,l 2−k such that r ˙ x ∈˙Bk.Claim 2.3. cTkBk ∪ B cB.Proof. Note B S kAk,l is an increasing union.If the claim failed, by the continuity of the capacity in increasing unions there would have to be a number i ¯k such that cTkBk ∪ Ai,l cB 2−i. However, Bi⊃T kBk∪Ai,l and cBi ≤ cAi,l2−i≤ cB2−i, contradiction.By the properties of the tilde operation, it must be the case that cTkBk\ ˜B 0. At the same time, r ˙ x ∈T kBk\˜B. This contradicts the assumption that P ˙ x falls out of all ground model coded c-null setsCorollary 2.4. The forcing PIcis proper.Proof. Note that the forcing PIcforces the generic point ˙ xgen∈˙X to fall out of all ground model coded c-null sets. Let σ be the corresponding winning strategy for Player II in the game G. Let M be a countable elementary submodel of a large structure containing the strategy σ, and let B ∈ PIc∩ M be an arbitrary condition. We must prove [11] that the set {x ∈ B x is M-generic} is Ic- positive. Suppose that A ∈ Icis a c-null set, and simulate a play of the game G in which Player I indicates B pini, enumerates all open dense subsets of PIcin the model M and produces the set A or some of its c-null supersets, and Player II follows his strategy σ. By elementarity, all the moves in this play arein the model M, therefore the filter g ⊂ M ∩PIcPlayer II created is M-generic.5Since the strategy σ is winning, the generic point ˙ xgen/g falls out of the set A. Thus the set of all generic points in the set B is Ic-positive as desired. In fact, a second look will show that the collection of generic points of the set B has the same capacity as the set B itself.Corollary 2.5. In the choiceless Solovay model, the ideal Icis closed under wellordered unions.Proof. Let κ be an inaccessible cardinal and let G ⊂ Collω, 0. Then in N there must be a point x ∈ B which falls out of all ground model coded c-null Borel sets, and in V there must be a forcing P of size 0. This is more or less the same as the previous proof. Use Lemma 2.1 in theground model to find a winning strategy σ for Player II in the game G associated with the name ˙ x. Apply a wellfoundedness argument to see that this strategy is still winning in the model N. Now given a c-null set D ⊂ X in the modelN, find a play of the game G in which Player I indicates the initial condition p pini, enumerates all the dense subsets of the forcing P in the ground model, and produces some c-null superset of the set D. The resulting point x ˙ x/g falls into the set C \ D, showing that the set C cannot be c-null.A little bit of extra work will show that actually the capacity c is continuous in increasing wellordered unions in the choiceless Solovay model. There is a related integer game. Suppose that c is an outer regular stablecapacity, B ⊂ X is a set and ǫ 0 is a real number. The infinite game HB,ǫ is played between Players I and II. Player I creates an open set A such that cA ≤ ǫ and Player II creates a point x ∈ X. Player II wins if x ∈ B \ A. The precise schedule of the two players is similar to the game G. At round k ∈ ω, Player I plays a basic open set Ak ⊂ X such that cAk ≤ ǫ and k0∈ k1 implies Ak0 ⊂ Ak1 and cAk1 − cAk0 ≤ 2−k0. In the end, the set A is recovered asS kAk. For Player II, fix some Borel bijection π 2ω→ X. Player II can tread water for an arbitrary number of steps before playing a nontrivial move, a bit 0 or 1. Let y ∈ 2ωbe the sequence of bits he got in the end; the point x ∈ X is recovered as x πy.Lemma 2.6. cB 0 be a real number and argue that ΣkbAkǫ ≥ bA.For every number k find sets Bmk m ∈ n such that AkS m∈nBm kand Σm∈ncmBmk lAm k. It is clear from the last expression and the subadditivity of the submeasure cmthat cmBm 0. Thus we expressed the set A as a union of sets of respective zero submeasures as desired.We will now prove Theorem 1.3. Note the extra assumption of strong sub- additivity for the capacities. We do not know if it is necessary, however our economical proofs do use it in one small, absolutely critical point.For the record let us stateDefinition 3.2. A capacity c is strongly subadditive if cA ∪ B cA ∩ B ≤ cA cB for all sets A,B.Claim 3.3. Suppose that c is a strongly subadditive capacity and B,An,Bn n ∈ ω are sets such that An⊂ Bn∩ B and cBn − cAn ≤ ǫnfor all n and some real numbers ǫn. Then cSnBn∪ B − cB ≤ Σnǫn.Proof. First note that for every number n ∈ ω, cBn∪B−cB ≤ ǫn. Namely, by the strong subadditivity cBn∪BcBn∩B ≤ cBncB and therefore cBn∪ B ≤ cBn cB − cBn∩ B ≤ cBn cB − cAn ≤ cB ǫn. Now argue that cB0∪ B1∪ B − cB ≤ ǫ0 ǫ1, the rest follows by the continuity of the capacity under increasing wellordered unions. But this is just like the situation in the previous paragraph cB0∪ B ∪ B1∪ B cB0∩ B1 ∪ B ≤ cB0∪ B cB1∪ B and cB0∪ B1∪ B ≤ cB0∪ B cB1∪ B − cB0∩ B1 ∪ B ≤ cB ǫ0 cB ǫ1− cB cB ǫ0 ǫ1.Suppose that {cm m ∈ n} are outer regular strongly subadditive stable capacities on a Polish space X and let b be their join, with the associated ideal Ib. Each of them has the associated tilde operation. We will abuse the notation to use the same tilde to denote this operation for any of the capacities. Which capacity is concerned will be always clear from the index of the set˜Bm denotes the cmtilde of the set Bm. Suppose that P is a forcing adding a point ˙ x ∈˙X. Consider the infinite game G between Players I and II. In the beginning, Player I indicates an initial condition pini∈ P and then produces a sequence Dk k ∈ ω of open dense subsets of the forcing P as well as a b-null set A. Player II produces a sequence pini≥ p0≥ p1≥ ... such that pk∈ Dkand the condition pkdecides the membership of the point ˙ x in the k-th basic open subset of the space X in somefixed enumeration, generating some filter g ⊂ P. Player II wins if the realization ˙ x/g falls out of the set A. In order to complete the description of the game, we have to describe the exact schedule for both players. At round k ∈ ω, Player I indicates the open dense set Dk⊂ PIcand sets Ak,l,m ∈ B for l ∈ k and m ∈ n so that9cmAk,l,m ≤ 2−land l ∈ k0∈ k1implies Ak0,l,m ⊂ Ak1,l,m and cmAk1,l,m−cAk0,l,m ≤ 2−k0. In the end, let Al,m SkAk,l,m,Am T lAl,m and recov