Mad families and non-meager filters
Haim Horowitz and Saharon Shelah
Abstract
We prove the consistency of ZF+DC+"there are no mad families"+"there
exists a non-meager filter on ω" relative to ZFC, answering
a question of Neeman and Norwood. We also introduce a weaker version
of madness, and we strengthen the result from [HwSh:1090] by showing
that no such families exist in our model.111Date: January 11, 2017
2000 Mathematics Subject Classification: 03E35, 03E15, 03E25
Keywords: mad families, non-meager filters, amalgamation
Publication 1103 of the second author
Partially supported by European Research Council grant 338821.
Introduction
This paper is a continuation of [HwSh:1090], which is part of
the ongoing effort to investigate the possible non-existence and definability
of mad families. In [HwSh:1090] we proved that ZF+DC+"there
are no mad families" is equiconsistent with ZFC (previous results
by Mathias and Toernquist established the consistency of that statement
relative to large cardinals, see [Ma1] and [To]). In this
paper we extend our results from [HwSh:1090] to address the following
question by Neeman and Norwood:
**Question ([NN]): **If there are no mad families, does it
follow that every filter is meager?
By a result of Mathias ([Ma2]), if every set of reals has the
Ramsey property, then every filter is meager.
We shall construct a model of ZF+DC where there are no mad families,
but there is a non-meager filter on ω. Our proof relies heavily
on [HwSh:1090], the main change is that now we’re dealing with
a class K2 consisting of pairs (P,A) such
that P is ccc and forces MAℵ1, and in addition,
P forces that A is independent (we shall require
more, see definition 2). In order to imitate the proof from [HwSh:1090],
we need to prove analogous amalgamation results for an appropriate
subclass of K2. As in [HwSh:1090], our final model is obtained
by forcing with P where (P,A) is a
“very large” object in a subclass of K2, and the non-meager
filter will be constructed from A, which should contain
many Cohen reals.
Finally, we consider the notion of nearly mad families (see definition
14), which was also introduced in [NN]. We introduce the notion
of a somewhat mad family, which includes both mad and nearly mad families,
and we prove that no somewhat mad families exist in our model.
A non-meager filter without mad families
**Hypothesis 1: **We fix μ and λ such that ℵ2≤μ,
λ=λ<μ, μ=cf(μ) and α<μ→∣α∣ℵ1<μ.
**Definition 2: A. **Let K2 be the class of k
such that:
a. Each k has the form (P,A)=(Pk,Ak).
b. P is a ccc forcing such that ⊩PMAℵ1.
c. A is a set of canonical P-names of subsets
of ω.
d. ⊩P"A is independent, i.e. every finite
non-trivial Boolean combination of elements of A is infinite".
B. For k∈K2 and a Pk-name ∼b,
let ⊩Pk"∼b∈pos(k)"
mean ⊩Pk"∼b∈[ω]ωand
there is no non-trivial Boolean combination of sets from Ak
that is almost disjoint to ∼b".
C. Let ≤1 be the following partial order on K2: k1≤1k2
if and only if:
a. Pk1⋖Pk2.
b. Ak1⊆Ak2.
D. Let ≤2 be the following partial order on K2:
k1≤2k2 if and only if:
a. As in B(a).
b. As in B(b).
c. If ∼b is a Pk1-name then
⊩Pk1"∼b∈pos(Ak1)"
implies ⊩Pk2"∼b∈pos(Ak2)".
E. Let K2+ be the class of k∈K2 such that ⊩Pk"Ak
is a maximal independent set everywhere", where A is
a maximal independent set everywhere if for every a0,...,an−1∈A
without repetition, b:=l<n∩alif l is even∈[ω]ω
and {a∩b:a∈A∖{a0,...,an−1}}
is a maximal independent set in [b]ω.
F. When we write "∼a0,...,∼an−1∈A",
we mean that (∼ai:i<n) is without repetition,
moreover, i<j<n→⊩P"∼ai=∼aj.
Observation 3: a. ≤1 and ≤2 are partial
orders, and if k1,k2∈K2+ then k1≤1k2→k1≤2k2.
b. If k1≤2k2 and ∼b is
a Pk1-name, then ⊩Pk1"∼b∈pos(Ak1)"
iff ⊩Pk2"∼b∈pos(Ak2)".
**Proof: **We shall prove the second claim of 3(a), everything
else should be clear. Suppose that ⊩Pk1"∼b∈pos(Ak1)",
but for some ∼a0,...,∼an−1∈Ak2
and p∈Pk2, p⊩Pk2"∼b∩(l<n∩∼all is even)
is finite". Let G⊆Pk1 be generic
over V such that p∈G and we shall work over V[G]. WLOG
there is n1<n such that ∼al∈Ak1
iff l<n1, and denote ∼a∗=l<n1∩∼all is even.
As ⊩Pk1"∼b∈pos(Ak1)",
it follows that ∼b∩∼a∗ is
infinite. It’s enough to show that for some Boolean combination ∼a∗∗
from Ak1, ∼a∗∗⊆∼a∗
and ∼a∗∗⊆∗∼b, as
then ∼a∗∗∩(n1≤l<n∩∼all is even)⊆∗∼b∩(l<n∩∼all is even),
and therefore it’s finite, contradicting the definition of Ak2.
As k1∈K2+, it follows that
{∼a∗∩∼c:∼c∈Ak1∖{∼al:l<n1}}
is a maximal independent set in [∼a∗]ω,
hence there are ∼c0,...,∼cm−1∈Ak1∖{∼al:l<n1}
such that (l<n1∩∼alif l is even)∩(k<m∩∼cmif m is even)⊆∗∼b,
so ∼a∗∗=(l<n1∩∼alif l is even)∩(k<m∩∼cmif m is even)
is as required. □
Observation 4: k1≤2k2 and k1∈K2+
when the following hold for some κ:
a. k2∈K2+.
b. k2∈H(κ).
c. M is a model such that k2∈M≺Lℵ2,ℵ2(H(κ),∈).
d. k1=k2M.
**Proof: **By observation 3, recalling that "P⊨ccc"
and "P⊨MAℵ1" are Lℵ2,ℵ2-expressible
.
**Claim 5: **For every k∈K2 there is k′∈K2+
such that k≤1k′. Moreover, if ∣Pk∣<μ
then we can find such k′ that satisfies ∣Pk′∣<μ.
**Proof: **If ∣Pk∣<μ, let λ∗=μ,
otherwise, let λ∗ be a regular cardinal greater than (2+∣Pk∣)ℵ1,
such that α<λ∗→∣α∣ℵ1<λ∗.
we try to choose a sequence (kα:α<λ∗)
by induction on α<λ∗ such that:
-
k0=k.
-
(kβ:β≤α) is an increasing continuous
sequence of members of K2 (with respect to ≤1).
-
∣Pkα∣<μ.
-
For every α<λ∗, if kα∈/K2+,
we choose a Boolean combination ∼aα from
Akα and ∼bα⊆∼aα
witnessing the failure of the condition from Definition 2(E). We then
define Pkα+1 as an extension (with respect
to ⋖) of Pkα⋆Cohen to
a ccc forcing that forces MAℵ1, we let ∼ηα
be the relevant Cohen generic real and we let Akα+1=Akα∪{∼bα∪∼ηα−1({1})}.
-
If α<λ∗ is a limit ordinal, we define kα
as in the proof of claim 7 below.
Why can we carry the induction at stage α+1 for α
as in (4)? We shall prove that for each α, ⊩Pkα+1"Akα+1
is independent". Let ∼aα∗=∼bα∪∼ηα−1({1}),
note that ⊩Pkα+1"∼aα∗∈/Akα",
as otherwise there are p∈Pkα+1, n<ω
and ∼a′∈Akα such
that p⊩Pkα+1"∼a′∖n=∼aα∗∖n",
and therefore p⊩Pkα+1"∼ηα−1({1})↾(ω∖∼bα∖n)=∼aα∗∖∼bα∖n=∼a′∖∼bα∖n∈VPkα",
a contradiction (as ∼ηα is Cohen and
ω∖∼bα∖n is infinite).
Now if ∼aα=l<n∩∼aα,lif l is even
and ∼c=∼aα∩(l<m∩∼dlif l is even)
where ∼d0,...,∼dm−1∈Akα∖{∼aα,0,...,∼aα,n−1},
then ∼c∖∼aα∗
and ∼c∩∼aα∗ are infinite
(as ∼ηα is Cohen and ∼c∖∼b
is infinite), so Akα+1 is forced to
be independent.
If for some α<λ∗, kα∈K2+, when
we’re done. Otherwise, by Fodor’s lemma, there are α<β<λ∗
such that (∼aα,∼bα)=(∼aβ,∼bβ),
a contradiction. □
**Definition 6: **We say that (kα:α<β)
is increasing continuous if α1<α1→kα1≤2kα2,
and for every limit δ<β, i<δ∪Pki⋖Pkδ.
**Claim 7: **Every increasing continuous sequence in (K2+,≤2)
has an upper bound. Moreover, if the length of the sequence has cofinality
>ℵ1, then the union is an upper bound in K2+.
**Proof: **Given an increasing continuous sequence (kα:α<β),
we choose Pkβ as in [HwSh:1090] and
we let Akβ=α<β∪Akα.
This is enough for ≤1, so by claim 5 we’re done. □
**Claim 8: **A. If k1∈K2 then there are k2
and ∼a such that:
a. k1≤1k2.
b. Ak1∪{∼a}⊆Ak2.
c. ∼a is Cohen over VPk1.
d. ∣Pk2∣≤(2+∣Pk1∣)ℵ1.
B. Moreover, we may require that k2∈K2+.
**Proof: **A. Let P=Pk1⋆C
where C is Cohen forcing, now let Pk2
be a ccc forcing such that P⋖Pk2,
⊩Pk2"MAℵ1" and ∣Pk2∣≤(2+∣P∣)ℵ1.
Finally, let Ak2=Ak1∪{∼a}
where ∼a is a name for a Cohen real added by Pk2,
it’s easy to see that (Pk2,Ak2)
are as required.
B. By claim 5. □
**Definition 9: **We define the amalgamation property in the
context of K2+ as follows: K2+ has the amalgamation property
if A implies B where:
A. a. kl∈K2+ (l=0,1,2).
b. k0≤2kl (l=1,2).
c. Pk1∩Pk2=Pk0.
B. There exists k3=(Pk3,Ak3)∈K2+
such that kl≤2k3 (l=1,2).
**Claim 10: **a. (K2+,≤2) has the amalgamation property.
b. Suppose that k0,k1 and k2∈K2+,
g:Pk0→Pk1
is an embedding such that (g(Pk0),g(Ak0))≤2k1
and k0≤2k2, then there exist k,k2′∈K2+
and f such that:
-
k1≤2k and ∣Pk∣≤(2+∣Pk1∣+∣Pk2∣)ℵ1.
-
(g(Pk0),g(Ak0))≤2k2′≤2k.
-
f:Pk2→Pk2′
is an isomorphism mapping Ak2 to Ak2′
-
g⊆f.
**Proof: **a. We shall first prove that Ak1∩Ak2=Ak0.
Note that Ak0⊆Ak1∩Ak2
is true by the definition of ≤2, so suppose that ∼a∈Ak1∖Ak0,
we need to show that ∼a∈/Ak2.
As k0≤2k1, ∼a is not a
Pk0-name. Therefore, it’s not a Pk2-name,
hence ∼a∈/Ak2.
Now construct P as in [HwSh:1090], i.e. we take the
amalgamation P′=Pk1Pk0×Pk2
and then we take P∈K such that P′⋖P
and ∣P∣≤(2+∣P′∣)ℵ1. Now let A:=Ak1∪Ak2.
We need to show that A is as required, i.e. we need to
prove that (P,A) satisfy requirements (A)(d) and
(D)(c) in Definition 2 (in the end, we will use claim 5 for the requirement
in Definition (2)(D)(c)). By symmetry, it’s enough to show that if
⊩Pk1"∼b∈pos(Ak1)"
and ∼a0,...,∼an−1∈A,
then ⊩P"∼b∩(l<n∩∼alif (l is even))∈[ω]ω".
Let n2:=n, wlog there are n0<n1<n2 such that ∼al∈Ak0⟺l<n0,
∼al∈Ak1⟺l∈[0,n1)
and ∼al∈Ak2⟺l∈[0,n0)∪[n1,n2).
It’s enough to show that the last statement is forced by P′,
so let k<ω and p=(p1,p2)∈P′, we shall find
q∈P′ and m<ω such that p≤q, k≤m
and q⊩P′"m∈∼b∩(l<n∩∼alif (l is even))".
Let p0∈Pk0 witness "(p1,p2)∈P′",
i.e. p0⊩Pk0"l=1,2∧pl∈Pkl/Pk0".
Let ∼b∗={m:p1⊮Pk1/Pk0"m∈/∼b∩(∩{∼alif l is even:l∈[n0,n1)})"}
(so ∼b∗ is a Pk0-name)
and let p0∈G0⊆Pk0 be generic
over V,** **then ∼b∗=∼b∗[G0]∈V[G0]
and as p1⊩Pk1"∼b∈pos(Ak1)",
it follows that p0⊩Pk0"∼b∗∈[ω]ω",
moreover, p0⊩Pk0"∼b∗∈pos(Ak0)".
Let b∗∗ be the Pk0-name defined as b∗
if p0 is in the generic set, and as ω otherwise. As k0≤2k2,
it follows that p2⊩Pk2/G0"∼b∗∗∩(l∈[0,n0)∪[n1,n2)∩∼alif (l is even))∈[ω]ω".
Therefore, in V[G0] there are (p2′,m) such that:
a. p2≤p2′∈Pk2/G0.
b. m>k.
c. p2′⊩Pk2/G0"m∈∼b∗∗∩(l∈[0,n0)∪[n1.n2)∩∼alif (l is even))".
Note that as ∼b∗∗∈V[G0], V[G0]⊨"m∈∼b∗∗[G0]=∼b∗[G0]".
Therefore, by the definitions of ∼b and ∼b∗,
there is p1′∈Pk1/G0 above p1 such
that p1′⊩Pk1/G0"m∈∼b∩(∩{∼alif l is even:l∈[n0,n1)})".
Therefore, there is p0≤p0′∈G0 forcing (in Pk0)
all of the aforementioned statements about (p1′,p2′) in V[G0].
Now it’s easy to check that q=(p1′,p2′) is as required. Finally,
extend (P,A) (with respect to ≤1) to a
member of K2+. By observation 3, we’re done.
b. Follows from (a) by changing names. □
**Claim 11: **There exists k=(Pk,Ak)=(P,A)∈K2+
such that ∣Pk∣=λ and:
-
For every X⊆P of cardinality <μ, there
exists k′=(Q,A′)∈K2+ such that
X⊆Q, k′≤2k and ∣Q∣<μ.
-
If k1,k2∈K2+, ∣Pk1∣,∣Pk2∣<μ,
k1≤2k2 and f1:Pk1→P
is a complete embedding such that (f1(Pk1),f1(Ak1))≤2(P,A),
then there is a complete embedding f2 such that f1⊆f2
and (f2(Pk2),f2(Ak2))≤2(P,A).
**Proof: **The first property is satisfied by every k∈K2+
by observation 4. The proof of (2) is as in [HwSh:1090]. □
**Claim 12: **A implies B where:
A. a. k0,k1,k2∈K2+, k0≤2kl
(l=1,2) and Pk0=Pk1∩Pk2.
b. ∼D is a Pk0-name of a
nonprincipal ultrafilter on ω.
c. For l=1,2, ∼al and ∼bl
are canonical Pkl-names of a member of [ω]ω.
d. For l=1,2, ⊩Pkl"∼al∩∼bl
is infinite and ∼al contains no members of ∼D
from VPk0.
e. Pk1∩Pk2=Pk0.
f. For l=1,2, ⊩Pkl"∼bl is a pseudo intersection of ∼D".
B. There is k∈K2+ such that:
a. ∣Pk∣≤(2+∣Pk1∣+∣Pk2∣)ℵ1.
b. kl≤2k (l=1,2).
c. ⊩Pk"∼a2∖∼a1
and ∼a1∖∼a2 are infinite".
**Proof: **Let k∈K2+ be the object constructed
by the proof of claim 10. We need to prove that k satisfies
clause (B)(c). For l=0,1,2, let Pl=Pkl
and let P′ be as in the proof of claim 10, so it suffices
to prove that ⊩P′"∼a2∖∼a1
and ∼a1∖∼a2 are infinite".
Let p=(p1,p2)∈P′, k<ω and let p0∈P0
be a witness of "p=(p1,p2)∈P′". Now let G0⊆P′
be generic over V such that p0∈G0 and let D=∼D[G0].
By the assumptions, for l=1,2, pl⊩P1/G0"∼al∩∼bl
is an infinite pseudo intersection of D". In V[G0], let bl∗={m:pl⊮Pl/G0"m∈/∼al∩∼bl"},
then pl⊩Pl/G0"∼al∩∼bl⊆∼bl∗,
hence bl∗ infinite". As pl⊩Pl/G0"∼al∩∼bl
is a pseudo intersection of D", necessarily V[G0]⊨bl∗∈D.
Let al∗={m:pl⊩Pl/G0m∈∼al},
so al∗∈V[G0] and pl⊩Pl/G0"al∗⊆∼al".
Now recall that ⊩Pkl"∼al
contains no member of ∼D from VPk0",
therefore pl⊩Pl/G0"al∗∈/∼D".
Hence in V[G0] (recalling D is an ultrafilter), b:=(b1∗∩b2∗)∖(a1∗∪a2∗)∈D.
Let m∈b be such that k<m. By the definition of bl∗,
there is pl′∈Pl/G0 above pl such that pl′⊩Pl/G0"m∈∼al∩∼bl".
By the definition of al∗, there is pl′′∈Pl/G0
above pl such that pl′′⊩Pl/G0"m∈/∼al".
Let p0′∈G0 be a condition above p0 forcing the above
statements, so p0′ is witnessing the fact that (p1′,p2′′),(p1′′,p2′)∈P′
are above p=(p1,p2). Now m>k, (p1′,p2′′)⊩"m∈∼a1∖∼a2"
and (p1′′,p2′)⊩"m∈∼a2∖∼a1",
which completes the proof. □
**Definition 13: **Let P=Pk be the
forcing from claim 11, let G⊆P be generic over
V and in V[G], let V1=HOD(R<μ∪{Ak}).
**Definition 14 ([NN]): **A family F⊆[ω]ω
is nearly mad if ∣A∩B∣<ℵ0 or ∣AΔB∣<ℵ0
for every A=B∈F, and F is maximal
with respect to this property.
Theorem 15: V1⊨ZF+DC<μ+"there are no mad families"+"there are no nearly mad families"+"there exists a non-meager filter on ω".
**Proof: **1. In order to see that there exists a non-meagre
filter in V1, let ∼D be the filter generated
by Ak and the cofinite sets. By claim 8 and
the choice of k, ∼D contains many Cohen
reals and therefore is non-meager.
-
The proof of the non-existence of mad families is exactly as in
[HwSh:1090], where (K2+,≤2) here replaces (K,⋖)
there, and claim 10 is used for the amalgamation arguments. Alternatively,
see the proof of (3) below.
-
The non-existence of a nearly mad family in V1 will follow
from the proofs below. □
Somewhat mad families
**Definition 16: **A family F⊆[ω]ω
is somewhat mad if:
a. For every a1,a2∈F, ∣a1∩a2∣<ℵ0
or a1⊆∗a2 or a2⊆∗a1.
b. If b∈[ω]ω then for some a∈F,
∣a∩b∣=ℵ0.
**Observation 17: **Nearly mad families are somewhat mad. □
**Definition 18: **Let Pr(k1,k2,∼D,∼b2)
mean:
a. k1≤2k2, ∼b2 is a Pk2-name
and ∼D is a Pk1-name such
that ⊩Pk2"∼b2∈[ω]ω"
and ⊩Pk1"∼D is a nonprincipal
ultrafilter on ω".
b. If G1⊆Pk1 is generic over V,
p1∈Pk2/G1, b0∗={n:p1⊩Pk1G1"n∈∼b2"}
and b1∗={n:p1⊮Pk1/G1"n∈/∼b2"},
then V[G1]⊨b1∗∖b0∗∈D.
**Claim 19: **(A) implies (B) where:
A. a. k1∈K2+.
b. ⊩Pk1"∼D is a nonprincipal
ultrafilter on ω".
c. ⊩Pk1"∼S1 is somewhat
mad".
d. ⊩Pk1"∼S1∩∼D=∅".
B. There is k2 such that:
a. k2∈K2+.
b. k1≤2k2.
c. ∣Pk2∣≤(2+∣Pk1∣)ℵ1.
d. (α) implies (β) where:
α. (k3,∼S2) satisfy the following
properties:
-
k2≤2k3∈K2+.
-
⊩Pk3"∼S2 is somewhat
mad and ∼S1⊆∼S2".
-
⊩Pk3"no member of ∼S2∖∼S1
contains a member of ∼D".
β. For some Pk3-name ∼a,
⊩Pk3"∼a∈∼S2"
and Pr(k1,k3,∼D,∼a).
**Proof: **Using Mathias forcing restricted to ∼D,
it’s easy to see that there is k2 and a Pk2
name ∼b such that k1≤2k2∈K2+,
∣Pk2∣≤(2+∣Pk1∣)ℵ1
and ⊩Pk2"∼b is a pseudo
intersection of ∼D". Therefore, ⊩Pk2"∼b∈[ω]ω
is almost disjoint to every ∼a∈∼S1"
(by the fact that ⊩Pk1"∼S1∩∼D=∅").
We shall now prove that k2 satisfies (B)(d). Suppose that
(k3,∼S2) are as there. By the somewhat
madness of ∼S2, there is ∼a
such that ⊩Pk3"∼a∈∼S2
and ∣∼a∩∼b∣=ℵ0". Therefore,
⊩Pk3"∼a∩∼b∈[ω]ω
is a pseudo intersection of ∼D". Now let G1⊆Pk1
be generic over V. If p1∈Pk3/G1 then
b∗={n:p1⊮Pk3/G1"n∈/∼a∩∼b"}∈∼D[G1]
by the fact that ∼a∩∼b is
a pseudo intersection of ∼D. Now let a∗={n:p1⊩Pk3/G1"n∈∼a"},
then p1⊩PK3/G1"a∗⊆∼a is infinite".
If a∗∈∼D[G1], then p1 forces that ∼a
(which belongs to ∼S2∖∼S1)
contains a member of ∼D[G1], contradicting (α)(3).
Therefore, a∗∈/∼D[G1], and ∼a
is as required in the definition of Pr(k1,k3,∼D,∼a).
□
**Claim 20: **There is no somewhat mad family in V1.
**Proof: **Suppose towards contradiction that ∼S
is a P-name of a somewhat mad family. As in [HwSh:1090],
let ∼D be a P-name of a Ramsey ultrafilter
on ω such that ⊩P"∼S∩∼D=∅".
By claim 11(a), there is k1≤2k such that k1∈K2+,
∣Pk1∣<μ and ∼S is definable
using a Pk1−name. Let KP+ be
the set of k′∈K2+ such that k′≤2k,
∣Pk′∣<μ, ∼S↾Pk′
is a canonical Pk′-name of a somewhat mad family
in VPk′ and ∼D↾Pk′
is a Pk′-name of a Ramsey ultrafilter on ω.
As in [HwSh:1090], KP+ is ≤2-dense in K2+,
so there exists k2∈KP+ such that k1≤k2.
Let k3∈K2+ be as in claim 19 for (k2,∼S↾Pk2),
wlog k3≤2k (see claim 11). Choose k4∈KP+
such that k3≤2k4 and let ∼S4:=∼S↾Pk4.
Let ∼a be a Pk4-name such
that Pr(k3,k4,∼D,∼a)
holds, as guaranteed by claim 19.
As in [HwSh:1090], there are k5,k6,∈KP+
and an isomorphism f from k4 to k5 over k2
such that Pk5 adds a generic for M∼D↾Pk4
(Mathias forcing restricted to the ultrafilter ∼D↾Pk4)
and (k2,k4,k5,k6) here are as (k0,k1,k2,k3)
in claim 10, and wlog k6≤2k. By the choice
of f, ⊩P"∼a,f(∼a)∈∼S".
By claim 12, with (k2,k4,k5,k6,∼a,f(∼a))
standing for (k0,k1,k2,k,∼a1,∼a2)
there, it’s forced by Pk5, and hence by P,
that ∼a∖f(∼a) and f(∼a)∖∼a
are infinite. As in [HwSh:1090], ⊩P"∣∼a∩f(∼a)∣=ℵ0".
As ⊩P"∼a,f(∼a)∈∼S",
we get a contradiction. □
References
[HwSh:1090] Haim Horowitz and Saharon Shelah, Can you take Toernquist’s
inaccessible away? arXiv:1605.02419
[Ma1] A. R. D. Mathias, Happy families, Ann. Math. Logic **12
**(1977), no. 1, 59-111.
[Ma2] A. R. D.. Mathias, A remark on rare filters.
[NN] Itay Neeman and Zach Norwood, Happy and mad families in L(R),
preprint.
[To] Asger Toernquist, Definability and almost disjoint families,
arXiv:1503.07577.
(Haim Horowitz) Einstein Institute of Mathematics
Edmond J. Safra campus,
The Hebrew University of Jerusalem.
Givat Ram, Jerusalem, 91904, Israel.
E-mail address: [email protected]
(Saharon Shelah) Einstein Institute of Mathematics
Edmond J. Safra campus,
The Hebrew University of Jerusalem.
Givat Ram, Jerusalem, 91904, Israel.
Department of Mathematics
Hill Center - Busch Campus,
Rutgers, The State University of New Jersey.
110 Frelinghuysen road, Piscataway, NJ 08854-8019 USA
E-mail address: [email protected]