Table Of Content8 A model with no magic set
9
9
1
Krzysztof Ciesielski∗
n
a
J Department of Mathematics, West Virginia University,
5
Morgantown,WV 26506-6310,USA
1
[email protected]
]
O
Saharon Shelah†
L
. Institute of Mathematics, the Hebrew University of Jerusalem
h
t
a 91904 Jerusalem, Israel
m
and
[
Department of Mathematics, Rudgers University
1
v New Brunswick, NJ 08854,USA
4
5
1
1
0
Abstract
8
9
/ We will prove that there exists a model of ZFC+“c = ω2” in which
h
every M ⊆ R of cardinality less than continuum c is meager, and
t
a such that for every X ⊆ R of cardinality c there exists a continuous
m
function f:R→ R with f[X]= [0,1].
:
v In particular in this model there is no magic set, i.e., a set M ⊆ R
i
X such that the equation f[M] = g[M] implies f = g for every continu-
r ous nowhere constant functions f,g:R → R.
a
∗The authors wish to thank Professors Maxim Burke, Andrzej Ros lanowski and Jerzy
Wojciechowskifor reading preliminary versionsof this paper and helping in improving its
final version.
The work of the first author was partially supported by NSF Cooperative Reasearch
Grant INT-9600548, with its Polish part being financed by Polish Academy of Science
PAN, and 1996/97West Virginia University Senate Research Grant.
†This work was supported in part by a grant from NSF grant DMS 97-04477. Publi-
cation 653.
1
1 Introduction
The main goal of this paper is to prove the following theorem.
Theorem 1.1 There exists a model of ZFC in which c = ω ,
2
(⋆) for every X ⊆ R of cardinality c there exists a continuous function
f:R → R such that f[X] = [0,1], and
(⋆⋆) every M ⊆ R of cardinality less than c is meager.
Note that (⋆) of Theorem 1.1 is known to hold in the iterated perfect
set model. (See A. W. Miller [Mi].) This result was also generalized by
P. Corazza [Co] by finding another model leading to the following theorem.
Theorem 1.2 (Corazza) It is consistent with ZFC that (⋆) holds and
(⋆⋆′) every M ⊆ R of cardinality less than c is of strong (so Lebesgue)
measure zero.
Note that the condition (⋆⋆) is false in the iterated perfect set model and in
Corazza model. (See [BuCi].)
Corazza noticed also that Theorem 1.2 implies the following corollary
(since there exists a universal measure zero set of cardinality non(L), where
non(L) is the smallest cardinality of a nonmeasurable set).
Corollary 1.3 (Corazza [Co, Thm 0.3]) It is consistent with ZFC that (⋆)
holds and there is a universal measure zero set of cardinality c. In particular
in this model there are 2c many universal measure zero sets of cardinality c.
He asked also whether the similar statement is true with “always first-
category set” replacing “universal measure zero set.” The positive answer
easily follows from Theorem 1.1, since (in ZFC) there exists an always first-
categoryset ofcardinalitynon(M),wherenon(M)isthesmallest cardinality
of a nonmeager set.
Corollary 1.4 It is consistent with ZFC that (⋆) holds and there is an al-
ways first-category set of cardinality c. In particular in this model there are
2c many always first-category sets of cardinality c.
2
Clearly Theorem 1.1 can be viewed as dual to Theorem 1.2. However,
our original motivation for proving Theorem 1.1 comes from another source.
In [BeDi] A. Berarducci and D. Dikranjan proved that under the Continuum
Hypothesis (abbreviated as CH) there exists a set M ⊆ R, called a magic
set, such that for any two continuous nowhere constant functions f,g:R → R
if f[M] ⊆ g[M] then f = g. Different generalizations of a magic set were
also studied by M. R. Burke and K. Ciesielski in [BuCi]. In particular they
examined the sets of range uniqueness for the class C(R), i.e., sets which
definition is obtained from the definition of a magic set by replacing the
implication “if f[M] ⊆ g[M] then f = g” with “if f[M] = g[M] then f = g.”
They proved [BuCi, Cor. 5.15 and Thm. 5.6(5)] that if M ⊆ R is a set of
range uniqueness for C(R) then M is not meager and there is no continuous
function f:R → R for which f[M] = [0,1]. This and Theorem 1.1 imply
immediately the following corollary, which solves the problems from [BeDi]
and [BuCi].
Corollary 1.5 There exists a model of ZFC in which there is no set of range
uniqueness for C(R). In particular there is no magic set in this model.
Finally, it is worthwhile to mention that for the class of nowhere constant
differentiable function the existence of a magic set is provable in ZFC, as
noticed by Burke and Ciesielski [BuCi2]. In the same paper [BuCi2, cor. 2.4]
it has been noticed that in the model constructed below there is also no set
or range uniquness for C(X) for any perfect Polish space X.
2 Preliminaries
Our terminology is standard and follows that from [BaJu], [Ci], or [Ku].
A model satisfying Theorem 1.1 will be obtained as a generic extension
of a model V satisfying CH. The forcing used to obtain such an extension
will be a countable support iteration P of length ω of a forcing notion P
ω2 2
defined below. Note that P, which is a finite level version of Laver forcing,1
is a version of a tree-forcing Qtree(K,Σ) from [RoSh 470, sec. 2.3] (for a 2-
1
big finitary local tree-creating pair (K,Σ); it is also a relative of the forcing
notion defined in [RoSh 470, 2.4.10]) and most of the results presented in
1Note that Theorem 1.1 is false in Laver model, since in this model there is a c-Lusin
set (there is a scale) and such a set cannot be mapped continuously onto [0,1].
3
this section is a variation of general facts proved in this paper. To define P,
we need the following terminology.
A subset T ⊆ ω<ω is a tree if t|`n ∈ T for every t ∈ T and n < ω. For a
tree T ⊆ ω<ω and t ∈ T we will write succ (t) for the set of all immediate
T
successors of t in T, i.e.,
succ (t) = {s ∈ T:t ⊆ s & |s| = |t|+1}.
T
We will use the symbol T to denote the set of all nonempty trees T ⊆ ω<ω
with no finite branches, i.e.,
T = {T ⊆ ω<ω: T 6= ∅ is a tree & succ (t) 6= ∅ for every t ∈ T}.
T
For T ∈ T we will write limT to denote the set of all branches of T, i.e.,
limT = {s ∈ ωω:s|`n ∈ T for every n < ω}.
Also if t ∈ T ∈ T then we define
Tt = {s ∈ T:s ⊆ t or t ⊆ s}.
Now define inductively the following “very fast increasing” sequences
hb ,n < ω:i < ωi by putting n = 1, and for i < ω
i i −1
b = (i+2)(ni−1!)i and n = (b )(bi)i.
i i i
In particular b = 2, n = 2, b = 9, n = 99, b = 4[(99)!]2, etc. (For the
0 0 1 1 2
purpose of our forcing any sequences that grows at least “as fast” would
suffice.) Also let
T⋆ = n = {s|`k:k < ω & s ∈ n }
i i
k<ωi<k i<ω
[ Y Y
and
T ⋆ = {T ∈ T :T ⊆ T⋆}.
Forcing P is defined as a family of all trees T ∈ T⋆ that have “a lot of
branching.” To define this last term more precisely we need the following
definition for every i < ω, T ∈ T and t ∈ T ∩ωi:
norm (t) = log log |succ (t)| ∈ [−∞,∞).
T bi bi T
4
Note that norm (t) = i for every t ∈ T⋆ ∩ωi. Now for T ∈ T ⋆ and k < ω
T⋆
let
norm (k) = inf{norm (t):t ∈ T & |t| ≥ k}
T T
and define
P = T ∈ T⋆: lim norm (k) = ∞ .
T
k→∞
n o
The order relation on P is standard. That is, T ∈ P is stronger than T ∈ P,
0 1
what we denote by T ≥ T , provided T ⊆ T . Note also that norm (k) ≤
0 1 0 1 T
norm (k) = k for every k < ω.
T⋆
In what follows for t ∈ T ∈ P we will also use the following notation
norm (t) = norm (|t|) = inf{norm (s):s ∈ T & t ⊆ s}.
T Tt T
It is easy to see that
norm (k) = min{norm (t):t ∈ T ∩ωk}.
T T
For n < ω define a partial order ≤ on P by putting T ≥ T if
n 0 n
T ≥ T & T |`ωk = T |`ωk & norm (k) ≥ n,
0 0 T0
where k = min{j < ω:norm (j) ≥ n}.
T
Note that the sequence {≤ :n < ω} witnesses forcing P to satisfy the
n
axiom A. (In particular P is proper.) That is (see [BaJu, 7.1.1] or [RoSh 470,
2.3.7])
(i) T ≥ T implies T ≥ T for every n < ω and T ,T ∈ P;
0 n+1 1 0 n 1 0 1
(ii) if {T :n < ω} ⊆ P is such that T ≥ T for every n < ω then there
n n+1 n n
exists T ∈ P extending each T , namely T = T ∈ P; (such T is
n n<ω n
often called a fusion of a sequence hT :n < ωi;) and,
n
T
(iii) if A ⊆ P is an antichain, then for every T ∈ P and n < ω there exists
T ∈ P such that T ≥ T and the set {S ∈ A:S is compatible with T}
0 0 n
is at most countable.
In fact, in case of the forcing P the set {S ∈ A:S is compatible with T} from
(iii) is finite. Since this fact will be heavily used in Section 5 we will include
here its proof. (See Corollary 2.3.) However, this fact will not be used in the
next three sections so it can be skipped in the first reading.
The following definition is a modification of the similar one for the Laver
forcing. (See [BaJu, p. 353].)
5
Let D ⊆ P be dense below p ∈ P and n < ω. For t ∈ p with norm (t) ≥ n
p
we define the ordinal number rn(t) < ω as follows:
D
(1) rn(t) = 0 if there exists p′ ∈ D extending pt such that norm (t) ≥ n−1;
D p′
(2) if rn(t) 6= 0 and t ∈ ωi then
D
rn(t) = min α: ∃U ∈ [succ (t)]≥(bi)(bi)n−1 (∀s ∈ U)(rn(s) < α) .
D p D
n (cid:16) (cid:17) o
Lemma 2.1 Let D ⊆ P be dense below p ∈ P and n < ω. Then rn(t) is
D
well defined for every t ∈ p with norm (t) ≥ n.
p
Proof. By way of contradiction assume that there exists t ∈ p with
norm (t) ≥ n for which rn(t) is undefined. Then n > 1 (since otherwise we
p D
would have rn(t) = 0) and for any such t belonging to ωi the set
D
U = {s ∈ succ (t):rn(s) is defined}
p D
has cardinality less than (b )(bi)n−1. So
i
|{s ∈ succ (t):rn(s) is undefined}| = |succ (t)\U| ≥ |succ (t)|/2 (1)
p D p p
since |succ (t)|/2 = (b )(bi)normp(t)/2 ≥ (b )(bi)n/2 ≥ (b )(bi)n−1 > |U|. Con-
p i i i
struct a tree p ∈ T⋆ such that p ⊆ pt,
0 0
|succ (s)| ≥ |succ (s)|/2 (2)
p0 p
and rn(s) is undefined for every s ∈ p with t ⊆ s. The construction can
D 0
be easily done by induction on the levels of a tree, using (1) to make an
inductive step. But (2) implies that for every i < ω and s ∈ p ∩ ωi with
0
t ⊆ s
norm (s) = log log |succ (s)| ≥ log log |succ (s)|/2 ≥ norm (s)−1.
p0 bi bi p0 bi bi p p
So p ∈ P. Take p′ ∈ D with p′ ≥ p . We can find t ∈ p′ such that
0 0 1
norm (t ) ≥ norm (t ) ≥ n − 1. Then rn(t ) = 0, contradicting the fact
p 1 p′ 1 D 1
that rn(s) is undefined for every s ∈ p ⊇ p′.
D 0
6
Lemma 2.2 Let D ⊆ P be dense below p ∈ P and n < ω. Then for every
t ∈ p with norm (t) ≥ n there exist p ≥ pt and a finite set A ⊆ p such
p t n−1 t t
that p = (p )s and (p )s ∈ D for every s ∈ A .
t s∈At t t t
Proof.SThe proof is by induction on rn(t).
D
If rn(t) = 0 then p = p′ ∈ D will satisfy the lemma with A = {t}.
D t t
If rn(t) = α > 0 choose U ∈ [succ (t)]≥(bi)(bi)n−1 from the definition of
D p
rn(t). By the inductive assumption for every u ∈ U there exist q ≥ pu
D u n−1
and a finite set A ⊆ q such that q = (q )s and (q )s ∈ D for every
u u u s∈Au u u
s ∈ A . Then p = q and A = A satisfy the lemma.
u t u∈U u t uS∈U u
The next corollaSry canbe also founSd, in generalform, in[RoSh 470, 2.3.7,
3.1.1].
Corollary 2.3 Let A ⊆ P be an antichain. Then for every p ∈ P and n < ω
there exists q ∈ P such that q ≥ p and the set
n
A = {r ∈ A:r is compatible with q}
0
is finite.
Proof. Extending A, if necessary, we can assume that A is a maximal
antichain. Thus D = {q ∈ P:(∃p ∈ A)(q ≥ p)} is dense in P.
Leti < ω besuchthatnorm (i) ≥ n+1. ByLemma2.2foreveryt ∈ p∩ωi
p
there exists p ≥ pt and a finite set A ⊆ p such that p = (p )s and
t n t t t s∈At t
(p )s ∈ D for every s ∈ A . Put q = p . Then it satisfies the corollary.
t t t∈p∩ωi t S
S
3 Proof of the theorem
For α ≤ ω let P be a countable support iteration of forcing P defined in
2 α
the previous section. Thus P is obtained from a sequence hhP ,Q˙ i:β < αi,
α β β
whereeachP forcesthatQ˙ isaP -nameforforcingP. Alsowewillconsider
β β β
elements of P as functions p which domains are countable subset of α. In
α
particular if p ∈ P and 0 ∈ dom(p) then p(0) is an element of P as defined
α
in V.
Now let V be a model of ZFC+CH and let G be a V-generic filter in P .
ω2
We will show that the conclusion of Theorem 1.1 holds in V[G].
7
In what follows for α ≤ ω we will use the symbol G to denote G∩P .
2 α α
In particular each G is a V-generic filter in P and V[G ] ⊆ V[G ] = V[G].
α α α ω2
Since CH holds in V, forcing P is ω -cc in V. Thus since P satisfies
ω2 2
the axiom A, we conclude that P preserves cardinal numbers and indeed
ω2
c = ω holds in V[G].
2
To prove that (⋆⋆) holds in V[G] consider n = limT⋆ with the
i<ω i
product topology. Since n is homeomorphic to the Cantor set 2ω it
i<ω i Q
is enough to show that every subset S of n of cardinality less than
cV[G] = ω is meager in Q n . But every xi<∈ωSibelongs already to some
2 i<ω i Q
intermediate model V[G ] with α < ω , since P satisfies the axiom A (so
α 2
Q
is proper), and the iteration is with countable support. In particular there
exists an α < ω such that S ⊆ V[G ]. So it is enough to prove that
2 α
n ∩V[G ] is meager in n .
i<ω i α i<ω i
Since V[G ] is obtained from V[G ] as a generic extension via forcing
α+1 α
(cid:0)Q (cid:1) Q
P (in V[G ]) our claim concerning (⋆⋆) in V[G] follows immediately from the
α
following lemma. (See also [RoSh 470, 3.2.8].)
Lemma 3.1 Let V be a model of ZFC+CH and H be a V-generic filter in
P. Then in V[H] the set n ∩V is a meager subset of n .
i<ω i i<ω i
(cid:0)Q (cid:1) Q
Proof. Let r ∈ n be such that {r} = {limT:T ∈ H} and put
i<ω i
M = M where
j<ω j Q T
S
M = s ∈ n :s(k) 6= r(k) for every j ≤ k < ω .
j i
( )
i<ω
Y
Since clearly every M is closed nowhere dense it is enough to show that
j
n ∩V ⊆ M. For this pick s ∈ n ∩V and consider a subset
i<ω i i<ω i
D = D ∈ V of P, where
(cid:0)Q j<(cid:1)ω j (cid:0)Q (cid:1)
S D = {p ∈ P:(∀t ∈ p)(∀k ∈ dom(t)\j)(s(k) 6= t(k))}.
j
It is enough to prove that D is dense in P, since H ∩ D 6= ∅ implies that
j
s ∈ M .
j
So let p ∈ P and let j < ω be such that p ∩ωj−1 ⊆ p (1) and define
0 0 0
p = {t ∈ p :(∀k ∈ dom(t)\j)(s(k) 6= t(k))}.
0
8
Clearly p is a tree. It is enough to show thatp ∈ P, since then p ∈ D extends
j
p . But if t ∈ p ∩ωk for some k ≥ j and t ∈ p is an immediate predecessor
0 0 0
of t then
|succ (t )| ≥ |succ (t )|−1 = (b )(bk−1)normp0(t0) −1 > 0
p 0 p0 0 k−1
so succ (t ) is nonempty and for every s ∈ limp
p 0
lim norm (s|`i) ≥ lim(norm (s|`i)−1) = ∞.
i→∞ p i→∞ p0
This finishes the proof of Lemma 3.1.
To show that (⋆) holds in V[G] we will use the following two propositions.
The first of them is an easy modification of the Factor Theorem from [BaJu,
Thm 1.5.10]. For the case of Sacks forcing this has been proved in [BaLa,
Thm 2.5].
Proposition 3.2 Let β < α ≤ ω and γ be such that β +γ = α. If P⋆ is a
2 γ
P -name for the iteration P of P (as constructed in VPβ) then forcings P
β γ α
and P ⋆P⋆ are equivalent.
β γ
The analog of the next proposition for the iteration of Sacks forcing can
be found in an implicit form in [Mi].
Proposition 3.3 Suppose that p k− “τ ∈ 2ω \V” for some p ∈ P . Then
ω2
(in V) there exists a continuous function f:2ω → 2ω with the property that
• for every r ∈ 2ω there exists q ≥ p such that
r
q k− f(τ) = r.
r
The proof of Proposition 3.3 will be postponed to the next section. The
proof of (⋆) based on Proposition 3.3 and presented below is an elaboration
of the proof from [Mi] that (⋆) holds in the iterated Sacks model.
First note (compare [Co]) that to prove (⋆) it is enough to show that
(◦) for every X ⊆ 2ω of cardinality c there exists a continuous function
f:2ω → 2ω such that f[X] = 2ω.
9
Indeed if X ⊆ R has cardinality c andthere isno zero-dimensional perfect
set P ⊂ R such that |X ∩ P| = c then X is a c-Lusin subset of R. Then
there is c-Lusin subset of 2ω as well, and such a set would contradict (◦) since
it cannot be mapped continuously onto [0,1] (so onto 2ω as well). (See e.g.
[Mi, Sec. 2].)
So there are a,b ∈ R and a zero-dimensional perfect set P ⊂ [a,b] with
|X∩P| = c. But P and 2ω are homeomorphic. Therefore, by (◦), there exists
a continuous f:P → P ⊂ [a,b] such that f[X ∩P] = P. Then a continuous
extension F:R → [a,b] of f, which exists by Tietze Extension theorem, has
a property that F[X] ⊇ P. Now if g:R → [0,1] is continuous and such that
g[P] = [0,1] then f = g ◦F satisfies (⋆).
We will prove (◦) in V[G] by contraposition. So let X ⊆ 2ω be such
that f[X] 6= 2ω for every continuous f:2ω → 2ω. Thus for any such f there
exists an F (f) ∈ 2ω such that F (f) ∈/ f[X]. We will prove that this implies
0 0
|X| < c by showing that X ⊆ V[G ] for some α < ω . This is enough, since
α 2
V[G ] satisfies CH.
α
Now let D = 2<ω ∈ V. Since D is dense in 2ω any continuous f:2ω → 2ω
is uniquely determined by f |`D. Let F:(2ω)D → 2ω, F ∈ V[G], be such that
F(f |`D) = F (f) for every continuous f:2ω → 2ω. Thus
0
F(f |`D) ∈/ f[X]
for every continuous f:2ω → 2ω. We claim that there exists an α < ω of
2
cofinality ω such that
1
F |` (2ω)D ∩V[G ] ∈ V[G ]. (3)
α α
(cid:0) (cid:1)
To show (3) first recall that for every real number r and every α ≤ ω of
2
uncountable cofinality if r ∈ V[G ] then r ∈ V[G ] for some β < α. This is
α β
a general property of a countable support iteration of forcings satisfying the
axiom A (and, more generally, proper forcings). In particular
(2ω)D ∩V[G ] = (2ω)D ∩V[G ] (4)
α β
β<α
[ (cid:0) (cid:1)
for every α < ω of cofinality ω .
2 1
Now let hf :α < ω i ∈ V[G] be a one-to-one enumeration of (2ω)D, and
α 2
put y = F(f ). Then there exists a sequence S = hhϕ ,η i:α < ω i ∈ V
α α α α 2
such that ϕ and η are the P -names for f and y , respectively. Moreover,
α α ω2 α α
10
