PARTITION FORCING AND INDEPENDENT FAMILIES

Abstract We show that Miller partition forcing preserves selective independent families and P-points, which implies the consistency of 
$\mbox {cof}(\mathcal {N})=\mathfrak {a}=\mathfrak {u}=\mathfrak {i}<\mathfrak {a}_T=\omega _2$
 . In addition, we show that Shelah’s poset for destroying the maximality of a given maximal ideal preserves tight mad families and so we establish the consistency of 
$\mbox {cof}(\mathcal {N})=\mathfrak {a}=\mathfrak {i}=\omega _1<\mathfrak {u}=\mathfrak {a}_T=\omega _2$
 .

It is well-known that the Baire space is not -compact (see [32]), which implies that a T is uncountable.Furthermore, d is the least size of a family of compact sets covering (see [3]), so it follows that d ≤ a T .It is known that the compact subspaces of the Baire space are in correspondence with the finitely branching subtrees of < .Using this correspondence and K önig's lemma, it is easy to prove that a T is equal to the least size of a maximal AD family of finitely branching subtrees of < .Džamonja, Hrušák, and Moore proved that 3 d implies that a T = 1 (see Theorem 7.6 of [43]).Thus, since 3 d holds in most of the natural models of d = 1 (see [28,43] for a precise formulation of this statement), a T = 1 also holds in these models.
On the other hand, given a partition C of into compact sets, Miller introduced a proper forcing notion Q(C) which has the Laver property and destroys C, that is, C no longer covers after forcing with Q(C).The forcing notion is known as Miller partition forcing and plays an important role in the current article (see Definition 2.1).Spinas showed that Q(C) is -bounding, which together with Miller's result establishes the Sacks property of Q(C) (see [49]).Thus, every partition of into compact sets can be destroyed with a proper forcing that has the Sacks property, which implies the consistency of cof (N ) < a T and in particular the consistency of d < a T .In [52, Proposition 4.1.31],Zapletal proved that Q(C) is forcing equivalent to the quotient of the Borel sets of modulo a -ideal generated by closed sets.In this way, the forcing Q(C) falls into the scope of the theory developed in [51,52].
In this article, we study the effect of Miller partition forcing on the independence number i and obtain the consistency of i < a T .The key argument is the fact that Miller partition forcing preserves selective independent families, a fact for which we provide two proofs: one building on the notion of fusion with witnesses (see Definition 2.5) and one building on Laflamme's filter games (see [34]).Both, the fusion with witnesses and the use of Laflamme's filter game in the context of Miller's partition forcing, are highly innovative and do not occur in earlier work on Q(C).
The notion of selective independent family was introduced by Shelah in his work on the consistency of i < u (see [46]).Selective independent families are families with very strong combinatorial properties, which resemble the combinatorial features of Ramsey ultrafilters.Studying the similarities and differences between selective independent families and Ramsey ultrafilters remains a very interesting line of research.For more recent work on maximal independent families see [13,14,17,18,45].
Employing our notion of fusion with witnesses, we show also that Q(C) preserves P-points.Together with the fact that Miller partition forcing and its iterations preserve tight mad families (see [25]), we obtain the consistency of the following constellation: The question if one can increase simultaneously u and a T , while preserving small witnesses to a and i becomes of interest.Further, we show that Shelah's poset Q I for destroying the maximality of a given maximal ideal from [46] strongly preserves tight MAD families.The following result appears as Corollary 5.14 in the current article: Finally, combining Miller partition forcing, Shelah's Q I , our preservation results, as well as the preservation results of [25,46], in Corollary 5. 15

we obtain:
Theorem.It is relatively consistent that i = a = 1 < u = a T = 2 .§2. Miller partition forcing.

Fusion with witnesses.
Recall that Sacks forcing S consists of all perfect trees in 2 < ordered by inclusion.That is, p ∈ S if and only if: We  Note that, Miller defined and used PM in [40] to show that cov(M) = 1 does not imply that a T = 1 .
Notice that if C is the partition of 2 into singletons, then Q(C) = S. Actually, it can be seen that if C is an analytic subset of K (2 ), where K (2 ) is the space of non-empty closed subsets of 2 equipped with the Vietoris topology, then Q(C) is forcing equivalent to Sacks forcing S. Theorem 2.2.Let C ⊆ K (2 ) be an uncountable analytic partition of 2 .Then Q(C) is forcing equivalent to Sacks forcing S.
, f| X is injective, and im(f| X ) ⊆ [p].This implies that there is q ∈ S such that [q] ⊆ im(f| X ).It is easy to see that q ⊆ p and |[q] ∩ K | ≤ 1 for every K ∈ C. Checking that q is as desired is straightforward.
A main difficulty in adapting Sacks fusion sequences to Q(C) is guaranteeing that the fusion is indeed an element of Q(C).To have a better control over fusion sequences in Q(C) we introduce the notion of a fusion with witnesses (see Definition 2.5).We begin with some auxiliary notions.Definition 2.3.Let C = {C α } α∈ 1 be an uncountable partition of 2 into closed sets.
(1) We say that x, y ∈ 2 are C-different if x, y belong to different elements of C.
(2) A tree p ⊆ 2 < is said to be C-branching if for any s ∈ p there are C-different branches in [p] extending s.
Note that, a C-branching tree is perfect.We will use the following notation: whenever C as above is given, for each x ∈ 2 we denote by α x the unique ordinal such that x ∈ C αx .
The equivalence of (a) and (c) in the lemma below can be found in [25].
Lemma 2.4.Let p ⊆ 2 < be a tree.The following are equivalent: (c) p is perfect and [p] contains a countable dense subset with C-different branches.
Proof.((a) ⇒ (c)) Let p ∈ Q(C).p is a perfect tree by the definition.Thus arrange split(p) and assign by induction to each splitting node s a real x from [p] extending s which was either already considered or belongs to a different set from C than all previously selected reals.This is possible since any s ∈ split(p) may be extended to t ∈ split(p) with [p(t)] being disjoint with finitely many sets from C containing all previously selected reals.The set of all assigned branches is the required dense set.
((b) ⇒ (a)) Let < 1 and s ∈ p.There are x, y ∈ [p] such that s ⊆ x, y and α x = α y .We take z ∈ {x, y} such that The particular enumeration constructed in Lemma 2.4 will be applied several times.Therefore we state explicitly that we may assume the dense set in Lemma 2.4 is enumerated as {x t : t ∈ p} such that s ⊆ x s , and if s ⊆ t ⊆ x s then x t = x s .

Definition 2.5 (Fusion sequence with witnesses).
(1) Let p be a condition in Q(C).We say that a set X ⊆ 2 is a p-witness for the n-th level if X ⊆ [p], for each s ∈ split n (p) there is x ∈ X extending s, and X has C-different elements.Note that if X is a p-witness for the (n + 1)-st level then each node from the n-th splitting level of p is contained in C-different branches in X. (2) Let (p, X ), (q, Y ) be pairs with p, q conditions in Q(C).Let X be a p-witness for the (n + 1)-st level and let Y be a q-witness for the n-th level.Then (p, X ) ≤ n (q, Y ) if and only if p ≤ q and X ⊇ Y.

a fusion sequence with witnesses then the fusion {p
Proof.We denote p = {p n : n ∈ }, X = {X n : n ∈ }, and we assume that we have s ∈ p.We take n ∈ and t ∈ split n (p) such that t extends s.Since split n (p) = split n (p n+1 ), the set X n+1 contains C-different branches extending t.Hence, X is dense in [p].One can easily see that X is contained in [p].Finally, by Lemma 2.4 we conclude that p ∈ Q(C).
Miller [40] and Spinas [49] applied separate fusion arguments in their proofs, while Miller [40] introduced the notion of a fusion even formally.The partial order Q(C) was recently used in [25], where the notion of a nice sequence was isolated from Spinas's fusion arguments.Our definition of fusion sequence covers both approaches.The sequence {X n } n∈ in our definition may be obtained as sets of leftmost branches in Miller's fusion argument, and as certain terms of nice sequence in Spinas's approach.
In addition to fusion sequences, we shall use two basic schemas to amalgamate conditions.Let us have a condition p ∈ Q(C), and for each s ∈ split n (p), i ∈ {0, 1}, a condition q(s, i) extending p(s i).Using Lemma 2.4, one can easily see that the tree is a condition in Q(C) as well.In the second amalgamation technique, we are given a decreasing sequence {q i } i∈ of extensions of p with strictly increasing stems s n = stem q n .We set x = i∈ s i and take q = i∈ q i (s i 1x(|s i |) ).Again, using Lemma 2.4, one can easily see that q is a condition in Q(C).
The proof of the fact that Q(C) is -bounding is underlying many fusion arguments associated with Q(C).For convenience of the reader, we repeat it here.We will make use of the following two lemmas.Lemma 2.7.Let ḟ be a Q(C)-name for a function in and let h be a function in ∩ V .The set of all conditions q satisfying the following property is dense in Q(C): There is a real x ∈ [q] and a sequence {f s } s∈x split(q) of functions in < such that for any s = x split n (q) we have q(s) One can construct a decreasing sequence {q i } i∈ of extensions of p with strictly increasing stems such that q n ḟ h(n) = f n for some We denote s n = stem(q n ) and we set x = i∈ s i .Finally, we take the amalgamation q = i∈ q i (s i 1x(|s i |) ).
Lemma 2.8.Let ḟ be a Q(C)-name for a function in .The set of all conditions q satisfying the following property is dense in Q(C): For all m ∈ , for all t ∈ split m (q) there is f t ∈ m+1 such that q(t) ḟ (m + 1) = ft .
Proof.Let p ∈ Q(C).We build a fusion sequence {(q n , X n )} n∈ with q 0 ≤ p such that its fusion q has the required property.Let the condition q 0 , branch x, and sequence {f s } s∈x split(q 0 ) be obtained from Lemma 2.7 for p and h(n) = n + 1.We set X 0 = {x}.
Let 0 ≤ n < .Suppose we have defined q n ∈ Q(C) and finite X n ⊆ [q n ].Let s ∈ split n (q n ).Take the unique branch x ∈ X n extending s, node r = x split n+1 (q n ), and number i = x(|s|) in {0, 1}.We set q(s, i) = q n (r).Let t ⊇ s 1i be such that [q n (t)] ∩ C αx = ∅ for all already considered branches x (i.e., all branches in X n and those assigned to previous nodes in some order of split n (q n )).Use Lemma 2.7 for q n (t) and h(j) = n + j + 2 to obtain q(s, 1i) ≤ q n (t), branch x, and sequence {f s } s∈x split(qn ) .
Finally, let X n+1 be the set of all considered branches in this step, and One can verify that the sequence {(q n , X n )} n∈ is a fusion sequence with witnesses.
As an application, we obtain a straightforward proof of the fact that Q(C) is -bounding.Since the poset has the Laver property (see [40]), this also gives the Sacks property of Q(C).Lemma 2.9 (Spinas [49]).The poset Q(C) has the Sacks property.
Proof.As explained above, by Miller's result it is sufficient to show that Q(C) is -bounding.Let ḟ be a Q(C)-name for a function in and let p ∈ Q(C).We will show that there is q ≤ p and g ∈ V ∩ such that q ḟ ≤ * ǧ.
By Lemma 2.8 we can assume that there is q ≤ p such that for all m ∈ , for all t ∈ split m (q) there is f t ∈ m+1 such that q(t) ḟ (m + 1) = ft .Define g ∈ as follows: Then q ∀n( ḟ(n) < g(n)).

Preservation of P-points.
Next, we show that Miller partition forcing preserves P-points.We will make use of the following notation: Given Theorem 2.10.The forcing notion Q(C) preserves P-points and Ramsey ultrafilters.
Proof.We prove just the first part.The second claim follows from the first one and the fact that the forcing notion Q(C) is -bounding (see [26,Lemma 21.12]).Note that a filter base G generates an ultrafilter on if and only if P( ) = G up ∪ G * dn .Let U be a P-point in V. We shall prove that the family U generates an ultrafilter in By Lemma 2.8 we can assume that for all m ∈ , for all t ∈ split m (p) there is u t ∈ m+1 2 such that Note that the latter property remains true for any stronger condition q, since t in the m-th level of q is an extension of some s in the m-th level of p.
Claim.We can assume that The sets U 0 , U 1 are disjoint and their union is p.We may distinguish two cases: (i) There is s ∈ p such that p(s) ⊆ U 0 .In this case, just take p(s).
(ii) For each s ∈ p there is t ∈ p(s) such that t ∈ U 1 .We build a fusion sequence {(p n , X n )} n∈ such that the fusion has the required properties.Taking s ∈ split 0 (p) there is t ∈ p(s) such that t ∈ U 1 .We set p 0 = p(t) and X 0 = {x t }.Let 0 ≤ n < .Suppose we have defined

Finally, let
and let X n+1 be the set of all x t 's for t ∈ split n+1 (p n+1 ).One can verify that the sequence {(p n , X n )} n∈ is a fusion sequence with witnesses.Moreover the fusion q of this sequence satisfies that ( \Y s ) ∈ U for all s ∈ q.We assume that Y 0 ⊆ U, the other case may be handled analogously.We take a pseudointersection Z of Y 0 in U , with Z ⊆ Y ∅ .We shall simultaneously build two fusion sequences with witnesses, namely , and a partition of Z into two sets Z 0 , Z 1 such that for their respective fusions q 0 , q 1 ≤ p we obtain q 0 Ž0 ⊆ Ẏ and n and the choice of number k 2n+3 are done similarly, and leads to p Since Z ∈ U, Z 0 or Z 1 is in U , and so q 0 Ẏ ∈ U up or q 1 Ẏ ∈ U up .§3. Selective independence.

Dense maximality.
Recall the definition: Definition 3.1.A family A ⊆ P( ) is an independent family if for every distinct A 0 , ... , A n ∈ A and h : {A 0 , ... , A n } −→ 2, the set It is maximal independent, if it is independent and maximal under inclusion.
We will be exclusively interested in infinite independent families.For an independent family A let FF(A) be the set of all finite partial functions from A to 2 and order it by inclusion.For h ∈ FF(A), we let where A 0 = \A and A 1 = A for A ⊆ .The density ideal of A, denoted id(A) is the set of all X ⊆ such that for all h ∈ FF(A) there is h ⊇ h in FF(A) such that A h ∩ X is finite (or equivalently empty).Dual to the density ideal of A is the density filter of A denoted fil(A) and consisting of all X ⊆ such that for all h ∈ FF(A) there is h ⊇ h in FF(A) such that A h \X is finite (or equivalently empty). 3emma 3.2.Let A be an infinite independent family.The following are equivalent: (1) For all X ∈ P( ) and all h ∈ FF(A) there is Proof.First we show that (1) implies (2).Let h ∈ FF(A), let X ⊆ A h , and suppose A h \X / ∈ id(A).Thus, there is h ∈ FF(A) such that for all h ⊇ h the set A h ∩ (A h \X ) is non-empty.Note that if h and h are incompatible, then A h ∩ (A h \X ) = ∅, which is a contradiction.Therefore h and h are compatible and without loss of generality, we can assume that h ⊇ h.Thus, we have that for all h ⊇ h , the set Next, we show that (2) implies (3).Thus, consider any X ∈ P( )\ fil(A).Then, in particular \X / ∈ id(A) and so there is h ∈ FF(A) such that for all h ⊇ h, To see that (3) implies (2), consider any h ∈ FF(A) and To see that (2) implies (1) consider any X ∈ [ ] and let h ∈ FF(A).We want to show that there is h ⊇ h such that either An independent family A is said to be densely maximal if any one of the above three properties holds.Note that if an independent family is densely maximal, then it is also maximal.The notion of dense maximality of independent families appears (to the best knowledge of the authors) for the first time in [24].In particular, we obtain: Corollary 3.3.Let A be an infinite independent family.Then, A is densely maximal iff 4The fact that partial orders has the Sacks property implies in particular: Lemma 3.4 [14].Let W be a P-generic extension of V, where P has the Sacks property.If A ∈ V is an independent family, then in W, fil(A) is generated by fil(A) V .

Selectivity.
Recall the following definitions.Let F ⊆ P( ).Then F is centered if for every finite subfamily H, H ∈ F; F is a P-set, if every countable subfamily has a pseudo-intersection in F; F is a Q-set, if for every partition E of into finite sets, there is an X ∈ F meeting each element of the partition in at most one point, i.e., |X ∩ E| ≤ 1 for each E ∈ E. Definition 3.5.Let F be a filter over containing the Fréchet filter.We say that F is a selective filter if and only if for every partition {X i } i∈ of into elements of F * , where F * is the dual ideal of F, there exists Note that a filter F ⊆ P( ) is selective (also called Ramsey) if and only if F extends the Fréchet filter and is both a P-set and a Q-set.Definition 3.6.An independent family is said to be selective if it is densely maximal and fil(A) is a selective filter.
Selective independent families exist under CH, a result which is due to Shelah (see [46]).Further studies of selective independent families can be found in [14,17].
Of particular interest for us will be the following combinatorial characterization of Q-filters which is similar to a characterization of happy families (see Proposition 0.7 in the work of Mathias [37] or Proposition 11.6 in [26]), as well as the fact that Q-filters are preserved by -bounding forcing notions (see Lemma 3.8).
Lemma 3.7.Let F be a filter.The following are equivalent: Proof.((a) ⇒ (b)) Note that the partition relations occurring in this proof, and so implicitly the proof itself, can be found in [46].Inductively, choose a sequence {n(l )} l ∈ such that n(0) = 0 and We consider the partition Note that E 1 is clearly reflexive and symmetric.Transitivity holds, since no three pairwise distinct elements of . However, on the other hand by the definition of sequence {n(l )} l ∈ we have f(m 1 ) ≤ n(3l 2 + 1) < n(3l 3 ), a contradiction.Thus, E 1 is an equivalence relation on C 1 and each E 1 -equivalence class has at most two elements.
Extend E 1 to an equivalence relation E 2 on by defining ) Let E be a partition of into finite sets.We set There is {k(n In particular, we get the following, which we state for completeness.
Proof.If P is an -bounding forcing notion, F a Q-filter in V, then we use part (2) of Lemma 3.7 for f ∈ V ∩ dominating function g ∈ V P ∩ .
Selective filter has a property similar to Mathias' notion of a happy family [37] (see [26] as well).Note that Mathias [37, Proposition 0.10] has shown that an ultrafilter G is Ramsey if and only if G is happy (see Proposition 11.7 in [26] as well).
Lemma 3.9.Let F be a filter over containing the Fréchet filter.The following are equivalent: Proof.((a) ⇒ (b))F is a P-set and therefore there is Let us take {k(n): n ∈ } ∈ F from Lemma 3.7 such that C = {k(n + 1): n ∈ } ⊆ C 0 .Hence, we have k(n + 1) ∈ C 0 \ f(k(n)), and so k(n + 1) ∈ G k(n) .((b) ⇒ (a)) First we shall show that F is a P-set.Let {G i } i∈ be a sequence in F. We set We shall show that F is a Q-set using Lemma 3.7.Indeed, let the function f ∈ be increasing.We consider sets G i = {(f(i), +∞)}, and we take a ∈ F such that a(n + 1) ∈ G a(n) .Hence, a(n + 1) > f(a(n)).
The following preservation theorem will be central to the proof that iterations of Miller partition forcing, as well as other partial orders which are of interest for this article, preserve selective independent families.Lemma 3.10 [46, Lemma 3.2].Let F be a selective filter and let H ⊆ P( )\F be cofinal in P( )\F with respect to ⊆ * .If P α , Qα | α < is a countable support iteration of -bounding proper forcing notions such that for all α < , we have 1 Pα "H is cofinal in P( )\ F , " then the same holds for .
The forcing iterations, that we will be interested in, all have the Sacks property.Thus, in the corresponding generic extensions fil(A) is generated by fil(A) ∩ V , where V denotes the ground model (see Lemma 3.4).Thus, if fil(A) is selective in the ground model, then it will remain selective in the desired generic extensions.Thus, the above preservation theorem implies that in order to guarantee that a given selective independent family remains selective (in our desired generic extensions), it is sufficient to guarantee that each iterand preserves the dense maximality of the family.Note that the fact that the density filter is selective will play a crucial role in this preservation arguments.That is, our techniques do not imply that densely maximal independent families are preserved, but only selective ones.Before giving detailed proofs of these crucial preservation properties of Miller partition forcing (see Theorem 4.1 and Corollary 4.10) which are also the most technical arguments in the paper, we state our main result: Theorem 3.11.Assume CH.There is a cardinals preserving generic extension in which Proof.Let V denote the ground model.We assume that A is a selective independent family in V, U is a P-point in V, and E is a tight MAD family in V (according to [25]).Using an appropriate bookkeeping device define a countable support iteration P α , Q : α ≤ 2 , < 2 of posets such that for each α, P α forces that Q α = Q(C) for some uncountable partition C of 2 into compact sets and such that V P 2 a T = 2 .P 2 has the Sacks property and therefore cof(N ) = 1 .By Shelah's preservation Theorem 3.10 and Corollary 4.10, or alternatively by Theorem 4.1, the family A remains selective independent in V P 2 and so a witness to i = 1 .Similarly, U generates a P-point in V P 2 , so u = 1 as well.And finally, a = 1 since E is a tight MAD family (see [25]).§4.Miller partition forcing preserves selective independent families.We proceed with taking care of the successor stages of our forcing construction, i.e., the fact that Miller partition forcing preserves selective independent families.In Section 4.1 using the technique of fusion with witnesses, we give a proof of this fact, while in Section 4.2 we give yet one more proof using Laflamme's filter games.

Fusion sequences and selectivity.
Theorem 4.1.Let A be a selective independent family and let G be a Q(C)-generic filter.Then, in V [G], A is still selective independent.
Proof.Recall that by Lemmas 3.4 and 3.8, fil(A) remains a selective filter.Thus, it is sufficient to show that A remains densely maximal in the generic extension.In V Q(C) , take any

Thus, for all h ∈ FF(A), Y ⊆ \A h and so for all h ∈ FF(A), |Y ∩ A
By Lemma 2.8 we can assume that for all m ∈ , for all t ∈ split m (p) there is u t ∈ m+1 2 such that p(t) Ẏ (m + 1) = ǔt .Now, in V for each t ∈ p, let m / ∈ Ẏ there is a condition q ≤ p(s) such that q m ∈ Ẏ .However, by our assumption on p due to Lemma 2.
is called the degree of acceptability of x.Note that for each acceptable branch x, y x = {k(i + 1): i ∈ i(x)} ∈ fil(A).Moreover, each s ∈ p can be extended to an acceptable branch.Indeed, if s ∈ split l (p), take j such that l ≤ k(j).Thus, we assume that k(j + 1) ∈ Y s .By part (iv) of Claim 4.2 there is t ∈ split k(j+1) (p) extending s such that p(t) ǩ(j + 1) ∈ Ẏ .Repeat the procedure recursively to build an acceptable branch extending s.In the following, we continue using a fusion argument.We build a fusion sequence {(p n , X n )} n∈ .
To define p 0 , take some acceptable branch x extending some node in split k(0)+1 (p) with degree of acceptability at most 0, and a node s = x split k(1) (p).We set p 0 = p(s) and X 0 = {x}.
Let us assume that p n and X n are defined, and consider s ∈ split k(n) (p) ∩ split(p n ).Take the unique acceptable branch x ∈ X n extending s.Define i = x(|s|) ∈ {0, 1} and s i = x split k(n+1) (p).Then we set s 1-i to be an extension of s 1i such that: (i) [p(s 1-i )] ∩ C αx = ∅ for all already considered acceptable branches x (i.e., all branches in X n and those assigned to previous nodes in some order of split k(n) (p) ∩ split(p n )).This can be easily achieved since each C αx is nowhere dense in with degree of acceptability at most j.Let us recall that each node can be extended to an acceptable branch.
Finally, let X n+1 be the set of all considered acceptable branches in this step, and One can see that the sequence {(p n , X n )} n∈ is a fusion sequence with witnesses.Let q = {p n : n ∈ } and let X = {X n : n ∈ }.
We shall show that the family {y x : x ∈ X } possesses the desired properties.Indeed, let x ∈ X .For each n ∈ we have y x (n) = k(i(x)(n) + 1).Due to construction of q we have x split n (q) = x split k(jn +1) (p) for some increasing sequence {j i } i∈ , and if t = x split n (q) then p(t) ǩ(j n + 1) ∈ Ẏ .Thus k(j n + 1) ∈ y x and consequently j n ≥ i(x)(n) for each n.Let us now fix n and consider t = x split n (q).The definition of y x guarantees that p(s) ǩ(i(x)(n) + 1) ∈ Ẏ for s = x split k(i(x)(n)+1) (p).Thus we have q(s) yx (n) ∈ Ẏ .On the other hand, The last part of our proof resembles the proof of the previous claim.Let x s for s ∈ split(p) be the branch in X extending s such that if s ⊆ t ⊆ x s then x t = x s .The corresponding y xs is denoted by y s .The set y s belongs to fil(A) ∩ V .By Lemma 3.9 for G n being the family of all y t 's with t ∈ split ≤n (p), we obtain {l (n Let us denote C = {l (n + 1): n ∈ }.We shall construct a condition q * ≤ p such that q * Č ⊆ Ẏ .Then q * Ẏ ∈ fil(A) which is a contradiction.We build a fusion sequence {(p n , X n )} n∈ .Let p 0 = p, X 0 = {x t } for t ∈ split 0 (p), and suppose we have defined p n .For each t ∈ split n (p n ) ⊆ split l (n) (p) and each i ∈ {0, 1} take w * (t, i) ∈ split l (n)+1 (p) such that w * (t, i) end-extends t i.Then and so for each t, i we take w(t, i) = x w * (t,i) split l (n+1) (p).Note that by Claim 4.3 and the fact that l (n + 1) ≥ j for l (n + 1) = y w * (t,i) (j) we obtain Finally, let q * be the fusion of {(p n , X n )} n∈ .Then, since q * l (n + 1) ∈ Ẏ for all n ∈ , q * is as required.

Laflamme's filter game.
We give yet one more proof of the fact that Miller partition forcing preserves selective independent families, which is based on Laflamme's filter game.We will make use of the following property, which can be deduced from [52, Propositions 4.1.2and 4.1.31].
Definition 4.5 (Laflamme [34]).Let F be a filter over .The game G(F, , F) is defined as follows.On the n turn, Player I plays some U n ∈ F and Player II responds with some a n ∈ U n .After turns, Player II wins if the sequence {a n } n∈ belongs to F. Otherwise, Player I wins.
It is not hard to prove that Player II never has a winning strategy in this game.On the other hand, we have the following theorem of Laflamme, see [34] (as well as [35]).

Theorem 4.6 (Laflamme). A filter F is not selective if and only if Player I does have a winning strategy for the game G(F, , F).
Lemma 4.7.Let F be a selective filter, p ∈ S, and H : [p] −→ P( ) be a continuous function such that for every s ∈ p, Then there are q ∈ S, Y ∈ F such that q ⊆ p and for every f ∈ [q], Y ⊆ H (f).
Proof.For every q ∈ S such that q ⊆ p, let L(q) = H [q] .Now consider the game G(F, , F). Player I will play the following strategy, while constructing a sequence {t } ∈2 < ⊆ p such that: (a) ∀ ∈ 2 < ∀i ∈ 2 t t i .(b) ∀ , ∈ 2 n = → (t and t are incomparable ) .On the first turn Player I defines t ∅ = ∅ and plays U 0 = L(p(t ∅ )).As the rules dictate, Player II responds with some ).As the rules dictate Player II responds with some a 1 ∈ U 1 .
In general, suppose that it is the n + 1 turn and that Player I has constructed {t } ∈2 ≤n and for every m Since H is continuous, there is k ∈ such that for every ∈ 2 n and every g ∈

L(p(t )). As the rules dictate,
Player II responds with some a n+1 ∈ U n+1 .
Since F is a selective filter, this is not a winning strategy for Player I. Therefore there is a match where Player I plays by the above strategy, but Player II wins.Let {a n } n∈ and {t } ∈2 < be the sequences associated with one of these matches and let q = { ∈ p | ∃ ∈ 2 < ⊆ t }.It is straightforward that q and Y = {a n } n∈ are the objects we are looking for.Theorem 4.8.Let F be a selective filter and let G be a Q(C)-generic filter.In V [G], for every X ∈ P( ) one of the following statements occurs: Proof.Let Ẋ be a name for a subset of , p ∈ Q(C), and suppose no condition below p forces (b).By Lemma 4.4 there is a continuous H : [p] −→ P( ) such that p H ( ṙgen ) = Ẋ .For every q ∈ S such that q ⊆ p, let L(q) = H [q] and note that if q ∈ Q(C) then L(q) ∈ F. This holds, because q Ẋ ⊆ L(q) and q does not force (b).We say that a condition q ∈ S, which is not necessarily in Q(C), is special if q ⊆ p and for every s ∈ q we have that L(q(s)) ∈ F.
We will make use of the following notion: Given s ∈ p we say that the pair (q, T ) is s-special if q ∈ S is special, T ∈ C, [q] ⊆ [p(s)] ∩ T .We divide the proof in cases.
Case 1.For every s ∈ p, there is an s-special pair (q, T ).
In this case, consider the game G(F, , F). Player I will play by the following strategy, while recursively constructing sequences {q } ∈2 < ⊆ S, {s } ∈2 < ⊆ p, and {T } ∈2 < ⊆ C such that: On the first turn, Player I defines s ∅ = ∅, an s ∅ -special pair (q ∅ , T ∅ ) and plays U 0 = L(q ∅ ).As the rules dictate, Player II responds with some a 0 ∈ U 0 .Since a 0 ∈ H [q ∅ ] , there is some f ∈ [q ∅ ] such that a 0 ∈ H (f) and since H is continuous there is k ∈ such that for every g ∈ [p(f| k )], a 0 ∈ H (g). Notice that since (q ∅ , T ∅ ) is s ∅ -special, we have that f| k is compatible with s ∅ and moreover we can extend f| k to incomparable s 0 , s 1 ∈ p such that s ∅ s 0 , s 1 , s 0 ∈ q ∅ and [p(s 1 )] ∩ T ∅ = ∅.Now Player I defines q 0 = q ∅ (s 0 ), T 0 = T ∅ , an s 1 -special pair (q 1 , T 1 ) and plays U 1 = L(q 0 ) ∩ L(q 1 ).As the rules dictate, Player II responds with some a 1 ∈ U 1 .
In general, suppose that it is the n + 1 turn and Player I has constructed q , s , and T for every ∈ 2 ≤n .Moreover, suppose that for every m ≤ n Player I has played such that a n ∈ H (f ).As H is continuous, there is k ∈ such that for every ∈ 2 n and every g ∈ [p(f | k )], a n ∈ H (g). As each (q , T ) is s -special, we have that f | k is compatible with s and that ∈2 n T ∩ [p] is nowhere dense in [p].Then we can extend each f | k to incomparable s 0 , s 1 ∈ p such that s s 0 , s 1 , s 0 ∈ q and [p(s 1 )] ∩ T = ∅.Now Player I defines q 0 = q (s 0 ), T 0 = T , an s 1special pair (q 1 , T 1 ) and plays U n+1 = ∈2 n+1 L(q ).As the rules dictate, Player II responds with some a n+1 ∈ U n+1 .Since F is a selective filter, the above is not a winning strategy for Player I and so, there is a match where Player I follows the strategy, but Player II wins.Let {a n } n∈ , {q } ∈2 , {s } ∈2 , and {T } ∈2 be the sequences associated with one of these matches.To finish Case 1, define . But then, by Lemma 2.4, q ∈ Q(C).Since for every n ∈ , every ∈ 2 n+1 , and every g ∈ [p(s )], we have that a n ∈ H (g) and every g ∈ [q] satisfies this condition for some ∈ 2 n+1 , we obtain that for every g ∈ [q], {a n } n∈ ⊆ H (g). In particular we have that q {a n } n∈ ⊆ Ẋ .Since {a n } n∈ ∈ F, we are done.
Case 2. There is s 0 ∈ p for which there is no s 0 -special pair (q, T ).That is, every ordered pair (q, T ) does not satisfy one of the following conditions: q ∈ S is special, In this case, we use Lemma 4.7 to find q ∈ S and Y ∈ F such that q ⊆ p(s 0 ) and for every f ∈ [q], Y ⊆ H (f). Notice that q is special.Suppose towards a contradiction that q / ∈ Q(C).Since every element of C is closed, this means that there is some T ∈ C such that T ∩ [q] has non-empty interior in [q] and so we can find ∈ q such that [q( )] ⊆ T .Then (q( ), T ) is s 0 -special, which is a contradiction.Therefore q ∈ Q(C).To finish this case, just note that as before q Y ⊆ Ẋ .Suppose that C is the partition of 2 in singletons.Then Q(C) = S and so Case 1 of Theorem 4.8 never occurs.Therefore Lemma 4.7 actually yields a complete proof of Theorem 4.8 for Sacks forcing.Additionally, we obtain once again: Corollary 4.9.The poset Q(C) preserves selective ultrafilters.Corollary 4.10.Let A be a selective independent family and let G be a Q(C)generic filter.Then, in V [G], A is still selective independent.
Proof.Since Q(C) is proper and has the Sacks property, fil(A) V is a selective filter in V [G], but by Lemma 3.4 we know that fil(A) V [G] = fil(A) V .To show that A remains densely maximal in V [G], note that by Theorem 4.8, the family P( V and so we are done.§5.No small ultrafilter bases and tightness.5.1.The poset Q I .For a maximal ideal I on , Q I denotes the forcing notion introduced by Shelah in [46] for obtaining the consistency of i < u.In [46] it is shown that Q I is proper [46,Claim 1.13], -bounding [46, Claim 1.12], and even has the Sacks property [46,Claim 1.12].In the Q I -generic extension, I is no longer a maximal ideal [46,Claim 1.5].For completeness of the presentation we repeat below the definition and some of the key properties of Q I . and E 1 -equivalence classes are unions of E 2 -equivalence classes.(3) Let A ⊆ .A function g is A-n-determined if g : A {0, 1} → {0, 1} and there is w ⊆ A ∩ (n + 1) such that for any , ∈ A {0, 1} with w = w we have g( ) = g( ).
For i ∈ A, by g i we denote the function from A {0, 1} to {0, 1} which maps ∈ A {0, 1} to (i).

For an I-equivalence relation E we denote
For a condition q ∈ Q I , let A q be A(E q ) in the following.
Finally, p ≤ n q if p ≤ q and A p contains the first n elements of A q .
is equal to H pn for large enough n.In order to guarantee that dom E q ∈ I * , it is sufficient to choose a play with the first player using the described strategy, but loosing, which by Lemma 5.9 exists.The other properties for q ∈ Q I are satisfied by the definition of q.
Finally q is (M, Q I )-generic, q ≤ n p n for all n and so q "(∀ Ż ∈ I(A) ∩ M [ Ġ]) | Ż ∩ B| = ." As a corollary we obtain that in Shelah's model of i < u (see [46]), also the almost disjointness number is small.Corollary 5.14.It is relatively consistent that a = i < u.Moreover, using the preservation results of the current article, together with the preservation results of [25], as well as the fact that Q I has the Sacks property, we obtain: Corollary 5.15.It is relatively consistent that cof(N ) = i = a = 1 < a T = u = 2 .
Proof.Work over a model of CH.Let A 0 be Shelah's selective independent family and let A 1 be a tight mad family.Using an appropriate bookkeeping device define a countable support iteration P α , Q : α ≤ 2 , < 2 of posets such that for even α, P α forces that Q α = Q(C) for some uncountable partition C of 2 into compact sets, for odd α, P α forces that Q α = Q I for some maximal ideal I on , and such that V P 2 a T = u = 2 .The iteration P 2 has the Sacks property and therefore cof(N ) = 1 .By the indestructibility of selective independence the family A 0 remains maximal independent in V P 2 and so a witness to i = 1 .Moreover, by the preservation properties of tight MAD families (see [25]), and the above preservation theorems, A 1 is a witness to a = 1 in the final model.§6.Appendix: The problem of Vaughan.We conclude the paper with an overview of the problem of Vaughan and point many of the difficulties surrounding a possible solution of it, in particular the fact that the most common forcing methods do not seem to help with the problem: (1) Finite support iteration of ccc forcings of length a regular cardinal over a model of CH.This approach cannot work since in the models obtained in this way, the size of the continuum is equal to cov (M) and it is known that cov (M) ≤ i. 5(2) Countable support iteration of definable proper forcings of length 2 over a model of CH.It follows by the results of Džamonja, Hrušák, and Moore in [43] that in all of these models the equality b = a will hold, so in particular we will have that a ≤ i. (3) Countable support iteration of non-definable proper forcings of length 2 over a model of CH.This approach could work; however a model of i < a obtained by this method will also be a model of 1 = d < a (since d ≤ i), thus solving the problem of Roitman, which is considered to be one of the hardest problems in the theory of cardinal invariants.
(4) Forcing with ultrapowers and iterating along a template.The method of forcing with ultrapowers and iterating along a template was introduced by Shelah in [48] to build models of d < a and u < a.This is a very powerful method that has been very useful and has been successfully applied to this day.Unfortunately, it seems that all forcings obtained using this method tend to increase i for the same reason they increase a.To learn more about this powerful method, see [8-10, 16, 20, 22, 39].(5) Short finite support iterations over models of MA.Performing a finite support iteration of length 1 over a model of MA (for example) is a powerful method to add "small witnesses" of some cardinal invariants while keeping others large.Models obtained in this way are often called "dual models" (see [12] for several interesting results and applications of these methods).In [2] a dual model was constructed to add a small maximal independent family in order to build a model of i <non (N ).Unfortunately, it is not clear how one could avoid adding a small MAD family with this method.Moreover, it seems likely that the principle 3 d of Hrušák will hold in these models 6 (see [28]).In principle, it could be possible to construct a model of i < a using matrix iterations (see [7,11,38] to learn more about this method), but one would need to be very careful in order to avoid problems like in the points 1 and 5 above.

Definition 5 . 1 .
Let I be an ideal on .(1)An equivalence relation E on a subset of is an I-equivalence relation if dom E ∈ I * and each E-equivalence class is in I.
Claim 4.3.We can assume that a dense set X ⊆ [p] with C-different elements has the associated family {y x : x ∈ X } of sets in fil(A) such that if t = x split n (p) then p(t) y x (n) ∈ Ẏ .Proof.By Claim 4.2(iii), Y t ∈ fil(A) ∩ V for each t ∈ split(p).By Lemma 3.9 for G n being the family of all Y t 's with t ∈ split ≤n (p), we obtain {k(n): n ∈ } ∈ 8, for any t ∈ split m (p) we have either p(t) m ∈ Ẏ or p(t) m / ∈ Ẏ .Since {p(t): t ∈ split m (p), t ⊇ s} is pre-dense in p(s), there is t ∈ split m (p) extending s such that p(t) m ∈ Ẏ .t : t ∈ split ≤k(n) (p)}.Moreover, by part (iv) of Claim 4.2 for any s ∈ split k(n) (p) there is t ∈ split k(n+1) (p) extending s such that p(t) ǩ(n + 1) ∈ Ẏ .For each branch x ∈ [p] we consider the set