PERMITTED SETS ARE PERFECTLY MEAGER IN
TRANSITIVE SENSE
PETER ELIAŠ
Abstract. In this paper we generalize our previous result which says that
every set permitted for the family of Arbault sets is perfectly meager. We
extend this result to all sets permitted for families of N0 -sets and Nf0 -sets,
where f is a suitable non-negative function. We also prove that these sets are
perfectly meager in a stronger, transitive sense.
A set X ⊆ R is called an Arbault set if there exists an increasing sequence of
natural numbers {nk }k∈N such that for all x ∈ X,
lim sin πnk x = 0;
k→∞
it is an N0 -set if there exists an increasing sequence {nk }k∈N such that for all x ∈ X,
X
|sin πnk x| < ∞.
k∈N
We denote by A and N0 the families of all Arbault and N0 -sets, respectively. Sets
of this form were introduced by J. Arbault in his paper [1] where he studied the sets
of absolute convergence of trigonometric series. It is well known that Arbault sets,
and hence also N0 -sets, are both meager and measure zero, and that the families
A and N0 are closed under linear transformations and under unions with arbitrary
countable sets. For more information, we refer the reader to the book [4] and to
the survey paper [7].
In the definition of Arbault sets and N0 -sets it is convenient to use, instead of
the function |sin πx|, the distance of a real x to the nearest integer, i.e. the function
kxk = min{|x − k| : k ∈ Z}. The obtained notions remain equivalent to original
ones.
Z. Bukovská [5] introduced the following generalization. Let f be a non-negative
periodic function with the period 1. A set X ⊆ R is called an Af -set if there exists
an increasing sequence of natural numbers {nk }k∈N such that for all x ∈ X,
lim f (nk x) = 0;
k→∞
it is an N f0 -set if there exists an increasing sequence {nk }k∈N such that for all
x ∈ X,
X
f (nk x) < ∞.
k∈N
2000 Mathematics Subject Classification. Primary 03E15. Secondary 28A05, 43A46.
Key words and phrases. Arbault sets, N0 -sets, permitted sets, perfectly meager sets, perfectly
meager in transitive sense.
This research was supported by the VEGA grant No. 1/0427/03 of Slovak Grant Agency.
1
2
PETER ELIAŠ
We denote by Af and N0f the families of all Af -sets and Nf0 -sets, respectively.
Clearly N0f ⊆ Af , for every function f . Throughout this paper, it will be assumed
that f is continuous and f (0) = 0.
Relations between the families Af and N0f (and some other families) obtained
for various functions f were studied by Z. Bukovská and L. Bukovský in [6], where
the authors proved that the equality Af = A holds true if and only if the set
Z(f ) = {x ∈ [0, 1) : f (x) = 0} is a finite set of rationals. They also provided an
example of two functions f and g such that Z(f ) = Z(g) = {0} and no inclusion
beween N0f and N0g holds true.
For a family of sets F, we say that a set X is F-permitted if for every Y ∈ F,
also X ∪ Y ∈ F. We denote by Perm(F) the family of all F-permitted sets.
It is known (see e.g. [1], [4], [7]) that the families A and N0 are not closed under
unions. Hence the families Perm(A) and Perm(N0 ) are proper subfamilies of A and
N0 , respectively. The same was proved in [6] for the family N0f assuming that Z(f )
is a finite set of rationals. On the other side, every set of reals having cardinality
less than p is both A-permitted and N0 -permitted, see [7]. In [6], it was proved
that if f (x − y) ≤ f (x) + f (y) for all x, y ∈ R then every finite set of reals is
Nf0 -permitted. It is not known whether this result can be extended to infinite sets.
Also nothing is known about inclusions between families Perm(A), Perm(N0 ), and
Perm(N0f ).
For a ∈ NN , let us denote
n
o
A(a) = x : lim ka(n) xk = 0
n→∞
and
n X
N0 (a) = x :
n∈N
o
ka(n) xk < ∞ .
If f ∈ RR is a non-negative function, let us also denote
n X
o
f (a(n) x) < ∞ .
N0f (a) = x :
n∈N
n
J. Arbault [1, p. 288] proved that if a(n) = 22 for all n ∈ N and A(a) ⊆ A(b) for
Pjn
i
αi,n 22 where
some increasing sequence b ∈ NN then b(n)’s are of the form i=l
n
Pjn
|αi,n | are bounded. A similar result for the
αi,n ∈ Z, ln → ∞, and the sums i=l
n
Qn
case a(n) = i=0 pi where {pi }i∈N is an increasing sequence of natural numbers
was proved in [6]. A general version of this result, without restrictive assumptions
on the sequence a, was proved in [9]. Let us note that the result proved in [6] was
more general in a different sense: it was dealing with Af -sets and Nf0 -sets instead
of original Arbault sets.
To formulate
the general theorem proved inª [9], we will need some notations. Let
©
Seq = a ∈ NN : a is increasing and a(0) = 1 . For a ∈ Seq, m ∈ Z, and z ∈ ZN ,
we say that z is a good expansion of m by a, if
¯P
¯ a(n)
X
¯
¯
z(n) a(n) and for all n ∈ N, ¯ j<n z(j) a(j)¯ ≤
m=
.
2
n∈N
Clearly if z is a good expansion then the set supp(z) = {n ∈ N : z(n) 6= 0} is finite.
A good expansion can be constructed as follows. Find some k such that |m| ≤
a(k)/2 and put z(n) = 0 for all n > k. Denote mk+1 = m. By induction on
n going from k to 0, let z(n) be the the nearest integer to mn+1 /a(n), and let
PERMITTED SETS ARE PERFECTLY MEAGER IN TRANSITIVE SENSE
3
mn = mn+1 − z(n) a(n). We end up with m0 = 0 since a(0) = 1. It is not difficult
to check that z is a good expansion of m by a. Let us note that a good expansion
of m by a is not necessarily unique.
¯P
¯
¯
¯
If z is a good expansion by a then for all n, |z(n) a(n)| ≤ ¯ j<n z(j) a(j)¯ +
µ
¶
¯P
¯
1
a(n + 1)
a(n) a(n + 1)
¯
¯
1+
. Thus
+
, and hence |z(n)| ≤
¯ j≤n z(j) a(j)¯ ≤
2
2
2
a(n)
all good expansions by a are bounded by a function depending only on a.
The mentioned generalization of Arbault’s theorem can now be formulated as
follows.
a(n)
= 0, and for all k ∈ N, let zk ∈ ZN be
a(n + 1)
a good expansion of b(k) by a. Then A(a) ⊆ A(b) if and only if
Theorem 1. Let a, b ∈ Seq, lim
n→∞
(1) ∀n ∈ N ∀∞ k zk (n)
P= 0, and
(2) ∃m ∈ N ∀k ∈ N
n∈N |zk (n)| ≤ m.
This theorem was proved in [9]; however, in one part of the proof we have used
a wrong argument. The corrected proof can be found in [10].
Let us denote
½
¾
a(n)
S = a ∈ Seq : lim
=0 .
n→∞ a(n + 1)
The condition limn→∞ a(n)/a(n + 1) = 0 in Theorem 1 is not restrictive since for
every sequence of positive reals {εn }n∈N and every Arbault set X it is easy to find
a ∈ Seq such that X ⊆ A(a) and a(n)/a(n + 1) ≤ εn for all n ∈ N. Thus every
Arbault set is included in a set of the form A(a) where a ∈ S. This allows the
following characterization of A-permitted sets (see [10], Theorem 1.2).
Theorem 2. A set X ⊆ R is A-permitted if and only if for every a ∈ S there
exists b ∈ S such that X ⊆ A(b) and the conditions (1) and (2) from Theorem 1
are satisfied, for any sequence {zk }k∈N of good expansions of b(k)’s by a.
This characterization was used in [10] to show that every A-permitted set is
perfectly meager. We say that a subset X of a Polish space is perfectly meager if
for every perfect set P , X ∩ P is meager in the relative topology of P , see [12]. A
set X is called universally meager if every Borel isomorphic image of X is meager,
see [15]. By [3], in a perfect Polish space, X is perfectly meager if and only if for
every perfect set P there exists an Fσ set F such that X ⊆ F and F ∩ P is meager
in P ; X is universally meager if and only if for every sequence {Pn }n∈N of perfect
sets there exists an Fσ set F such that X ⊆ F and F ∩ Pn is meager in Pn , for
each n. Every universally meager set is perfectly meager. However, the opposite
implication is known to be independent from the axioms of ZFC, see [2].
In accordance with [13] and [14], we say that a set X ⊆ R is perfectly meager in
transitive sense if for every perfect set P there exists an Fσ set F such that X ⊆ F
and for all t ∈ R, (F + t) ∩ P is meager in P . By the results of [13], the family
of all sets of reals which are perfectly meager in transitive sense is included in the
family of all universally meager sets, and this inclusion can be consistently strict.
Let us denote
¾
½
X
a(n)
<∞ .
S0 = a ∈ Seq :
n∈N a(n + 1)
4
PETER ELIAŠ
If f is a uniformly continuous non-negative function and ε ≥ 0, let
f ∗ (ε) = sup {|f (x) − f (y)| : |x − y| ≤ ε} .
P
∗
Clearly f ∗Pis continuous, non-decreasing,
f
(0)
=
0,
and
n∈N f (xn + yn ) < ∞
P
∗
whenever n∈N f (xn ) < ∞ and n∈N f (|yn |) < ∞. We denote
½
µ
¶
¾
X
a(n)
f
∗
f
<∞ .
S0 = a ∈ Seq :
n∈N
a(n + 1)
Note that every continuous periodic function is uniformly continuous. We will need
the following generalization of Lemma 2.6 from [10].
Lemma 3. Let f ∈ RR be a continuous non-negative periodic function with the
period 1 such that f (0) = 0. Let a ∈ S0f , b ∈ Seq, and for all k ∈ N, let zk ∈ ZN
be a good expansion of b(k) by a. Assume that the set {|zk (n)| : k, n ∈ N} is
unbounded. Then there exists x ∈ N0f (a) such that x ∈
/ A(b).
In the proof of Lemma 3, we will use the following technical results. Here, by an
interval we mean a closed bounded interval with nonempty interior. For the proofs
see [9] (Lemma 6, 7) or [10] (Lemma 2.4, 2.5).
Lemma 4. Let a ∈ Seq, n ∈ N, and let a(n)/a(n + 1) ≤ 1/4. Let z ∈ ZN be a good
expansion by a such that |z(n)| ≥ 2. Then for every interval I such that diam(I) =
4/(3a(n)) there exists an interval J ⊆ I such that diam(J) = 4/(3a(n + 1)) and for
all x ∈ J,
°P
° 1
2a(n)
1
°
°
ka(n) xk ≤
and
+
z(j)
a(j)
x
°
°≥ .
j≤n
3a(n + 1) |z(n)| − 12
6
Lemma 5. Let a ∈ Seq, n ∈ N, and let a(n)/a(n + 1) ≤ 1/8. Let z ∈ ZN be a
good expansion
°P by a. If I is° an interval such that diam(I) = 4/(3a(n)) and for
°
°
all x ∈ I, ° j<n z(j) a(j) x° ≥ 1/6, then there exists an interval J ⊆ I such that
diam(J) = 4/(3a(n + 1)) and for all x ∈ J,
ka(n) xk ≤
4a(n)
3a(n + 1)
and
°P
° 1
°
°
° j≤n z(j) a(j) x° ≥ .
6
Proof of Lemma 3. We will find increasing sequences of natural numbers {ni }i∈N ,
{ki }i∈N such that
(i) for all n ≥ n0 , a(n)/a(n + 1) ≤ 1/8,
(ii) for all µi ∈ N, |zki (ni )|¶≥ 2,
X
1
(iii)
f∗
< ∞,
1
|z
(n
)|
−
k
i
i
2
i∈N
(iv) for all i ∈ N, ni+1 > max supp (zki ).
The sequences {ni }i∈N , P
{ki }i∈N can be defined as follows. Fix a sequence of positive
reals {εi }i∈N such that i∈N f ∗ (εi ) < ∞ and let m0 be such that a(n)/a(n + 1) ≤
1/8 for all n ≥ m0 . We know that the set {|zk (n)| : k ∈ N} is bounded for
every n. If mi is already defined, we can find ni and ki such that ni ≥ mi and
|zki (ni )| ≥ 2 + 1/εi . Let us take mi+1 = max supp (zki ) + 1 and continue by
induction. It can be easily checked that the sequences {ni }i∈N and {ki }i∈N are
increasing and satisfy the conditions (i)–(iv).
Let us now define a sequence of intervals {In }n≥n0 . We can start with arbitrary
interval In0 such that diam (In0 ) = 4/(3a(n0 )).
PERMITTED SETS ARE PERFECTLY MEAGER IN TRANSITIVE SENSE
5
For n ≥ n0 , let In be an interval of the length 4/3(a(n)). If n = ni for some i,
then by Lemma 4 there exists an interval In+1 ⊆ In of the length 4/(3a(n + 1))
such that for all x ∈ In+1 ,
°P
° 1
2a(n)
1
°
°
and ° j≤n zki (j) a(j) x° ≥ .
ka(n) xk ≤
+
3a(n + 1) |zki (ni )| + 12
6
°P
°
°
°
Otherwise we have ni < n < ni+1 for some i, and ° j<n zki (j) a(j) x° ≥ 1/6 for all
x ∈ In . By Lemma 5 there exists an interval In+1 ⊆ In of the length 4/(3a(n + 1))
such that for all x ∈ In+1 ,
°P
° 1
4a(n)
°
°
ka(n) xk ≤
and ° j≤n zki (j) a(j) x° ≥ .
3a(n + 1)
6
T
Let us take x ∈ n≥n0 In . For all n ≥ n0 such that n ∈
/ {ni : i ∈ N} we have
µ
¶
4a(n)
a(n)
, hence f (a(n) x) ≤ 2f ∗
. If n = ni for some i,
ka(n) xk ≤
3a(n + 1)
a(n + 1)
2a(n)
1
then ka(n) xk ≤
+
and thus
3a(n + 1) |zki (ni )| − 12
µ
¶
µ
¶
a(n)
1
∗
∗
f (a(n) x) ≤ f
+f
.
a(n + 1)
|zki (ni )| − 12
We obtain that
¶
µ
¶ X µ
X
X
a(n)
1
∗
∗
< ∞,
f (a(n) x) ≤ 2
f
+
f
a(n + 1)
|zki (ni )| − 21
n≥n0
n≥n0
i∈N
hence x ∈ N0f (a). For all i, by the condition (iv) we have
°P
° 1
°
°
kb(ki ) xk = ° j<ni+1 zki (j) a(j) x° ≥ ,
6
therefore x ∈
/ A(b).
¤
For a, b ∈ Seq, we say that b is bounded by a if for any sequence {zk }k∈N of good
expansions of b(k)’s by a, the set {|zk (n)| : k, n ∈ N} is bounded. Let us denote
Bd (a) = {b ∈ Seq : b is bounded by a}.
Corollary 6. Let f be a continuous non-negative periodic function with the period
1 such that f (0) = 0 and Z(f ) is a finite set of rationals. If X ∈ Perm(N0f ) then
for every a ∈ S0f there exists b ∈ Bd (a) such that X ⊆ A(b).
Proof. If X is Nf0 -permitted then for every a ∈ S0f , X ∪ N0f (a) is an Nf0 -set, hence
there exists b0 ∈ S0f such that X ∪N0f (a) ⊆ N0f (b0 ). If Z(f ) is a finite set of rationals
then there exists a positive integer p such that pu ∈ Z for every u ∈ Z(f ). Put
b(0) = 1 and b(k) = p b0 (k) for every k ≥ 1. Clearly b ∈ Seq and for every real x, if
f (b0 (k) x) → 0 then kb(k) xk → 0. Hence N0f (a) ⊆ N0f (b0 ) ⊆ A(b) and X ⊆ A(b).
By Lemma 3, the set {|zk (n)| : k, n ∈ N} is bounded for any sequence {zk }k∈N of
good expansions of b(k)’s by a, hence b is bounded by a.
¤
We will use the following classical result of harmonic analysis, known as Kronecker’s theorem. For the proof, see e.g. [8] or [11].
6
PETER ELIAŠ
Theorem 7. Let x1 , . . . , xn ∈ R be linearly independent over Q, i.e. if q1 x1 + · · · +
qn xn = 0 for some rational q1 , . . . , qn then q1 = · · · = qn = 0. Let y1 , . . . , yn ∈ R
and ε > 0. Then there exist infinitely many natural numbers m such that for all i,
kmxi − yi k < ε.
The following lemma is a generalization of Lemma 1.5 from [10].
Lemma 8. Let P ⊆ R be a perfect set and let f be a continuous non-negative
periodic function with the period 1 such that f (0) = 0. Then there exists a ∈ S0f
such that for any b ∈ Bd (a) and any sequence of reals {yk }k∈N , the set
½
¾
x ∈ P : lim inf kb(k) x − yk k = 0
k→∞
is dense in P .
Proof. For a given perfect set P , fix a sequence of open intervals {Hn }n∈N such
that P ∩ Hn 6= ∅ for all n, and for every open set U such that P ∩ U 6= ∅ there exists
n such that Hn ⊆ U . Such sequence can be selected from a base of topology on real
line. P
Fix sequences of positive reals {εn }n∈N , {ηn }n∈N such that limn→∞ εn = 0
and n∈N f ∗ (ηn ) < ∞. For each n, fix a finite set Tn ⊆ R such that for every
y ∈ R there exists t ∈ Tn with kt − yk ≤ εn .
We will define a ∈ S0f and a sequence of finite sets of open intervals {In }n∈N
such that for every n, P ∩ I 6= ∅ for all I ∈ In .
Put a(0) = 1 and I0 = {H0 }. If a(n) and In are defined, denote
©
ª
Rn = (r, s) ∈ Z2 : |r| ≤ (n + 1) a(n) and 1 ≤ |s| ≤ n .
For every I ∈ In , t ∈ Tn , and (r, s) ∈ Rn , let us select a point xI,t
r,s ∈ P ∩ I in such
© I,t
ª
a way that the set xr,s : I ∈ In , t ∈ Tn , (r, s) ∈ Rn will be linearly independent
over Q. This is possible since the vector space over Q generated by a finite set is
countable, while P ∩ I is uncountable for every I ∈ In . By Theorem 7, there exists
a(n + 1) > a(n) such that a(n)/a(n + 1) ≤ ηn and for all I ∈ In , t ∈ Tn , and
(r, s) ∈ Rn ,
°
°
°(r + sa(n + 1))xI,t
°
r,s − t ≤ εn .
I,t
I,t
I,t
Choose an open interval Jr,s
⊆ I such that xI,t
r,s ∈ Jr,s and for all x ∈ Jr,s ,
k(r + sa(n + 1))x − tk ≤ 2εn . Put
© I,t
ª
In+1 = {Hn+1 } ∪ Jr,s
: I ∈ In , t ∈ Tn , (r, s) ∈ Rn .
We clearly obtain a ∈ S0f . To prove the rest, let us assume that b ∈ Bd (a),
{yk }k∈N is a sequence of reals, and U is an open set such that P ∩ U 6= ∅. Denote
½
¾
Q = x ∈ P : lim inf kb(k) x − yk k = 0 .
k→∞
We have to show that Q ∩ U is nonempty.
We will define increasing sequences of natural numbers {ni }i∈N , {ki }i∈N , and a
sequence of intervals {Jn }n≥n0 such that Jn ∈ In for all n ≥ n0 . Let {zk }k∈N be a
sequence of good expansions of b(k)’s by a. Since b ∈ Bd (a), there exists m such
that |zk (n)| ≤ m for all n, k ∈ N. Let us take n0 ≥ m such that Hn0 ⊆ U .
Put k0 = 0 and Jn0 = Hn0 . If ni , ki , and Jni ∈ I¡ni are¢ defined, let us find
ki+1 > ki and ni+1 > ni such that ni+1 = max supp zki+1 . Denote k = ki+1 ,
PERMITTED SETS ARE PERFECTLY MEAGER IN TRANSITIVE SENSE
7
P
n = ni+1 − 1, r = j≤n zk (j) a(j), and s = zk (n + 1). Clearly b(k) = r + sa(n + 1).
Since zk is a good expansion, we have
¢
¡
|r| ≤ |zk (n)| + 12 a(n) ≤ (m + 1)a(n) and 1 ≤ |s| ≤ m,
and since m ≤ n, we obtain (r, s) ∈ Rn . Let t ∈ Tn be such that kt − yk k ≤ εn . For
every j such that ni ≤ j < n, let Jj+1 ∈ Ij+1 be any interval such that Jj+1 ⊆ Jj .
Let Jn+1 ∈ In+1 be such that for all x ∈ Jn , k(r + sa(n + 1))x − tk ≤ 2εn . We
obtain that for all x ∈ Jni+1 , and thus also for all x ∈ cl(Jni+1 ), kb(k) x − yk k ≤
kb(k) x − tk + kt − yT
k k ≤ 3εn .
Let us take x ∈ n≥n0 cl(Jn ). Since for each n we have cl(In+1 ) ⊆ cl(In ) and
P ∩ cl(In ) 6= ∅, we obtain x ∈ P . Moreover, for all i we have
°
°
°b(ki+1 )x − yk ° ≤ 3εn −1 ,
i+1
i+1
and thus x ∈ Q. Finally, x ∈ U follows from the choice of Jn0 .
¤
We can now prove the main result of our paper.
Theorem 9. Let f be a continuous non-negative periodic function with the period
1 such that f (0) = 0 and Z(f ) is a finite set of rationals. If X ∈ Perm(A) or
X ∈ Perm(N0f ) then X is perfectly meager in transitive sense. Specially, if X ∈ N0
then X is perfectly meager in transitive sense.
Proof. Let P be a perfect set. By Lemma 8 there exists a ∈ S0f such that for every
b ∈ Bd (a) and for every sequence of reals {yk }k∈N , the set
½
¾
x ∈ P : lim inf kb(k) x − yk k = 0
k→∞
is dense in P . If X is A-permitted or Nf0 -permitted then by Theorem 2 or by
Corollary 6, there exists b ∈ Bd (a) such that X ⊆ A(b). For j ∈ N, denote
½
¾
1
Fj = x ∈ R : ∀k ≥ j kb(k) xk ≤
4
S
and put F = j∈N Fj . Clearly F is an Fσ set and A(b) ⊆ F . It remains to show
that (F + t) ∩ P is meager in P , for any t ∈ R.
Fix t ∈ R and put yk = b(k) t + 1/2, for all k ∈ N. Denote
½
¾
Q = x ∈ P : lim inf kb(k) x − yk k = 0 .
k→∞
If x ∈ Q then for infinitely many k’s we have kb(k) x − yk k < 1/4, and thus
kb(k)(x − t)k > 1/4. It follows that x ∈
/ Fj + t, for any j ∈ N. Therefore the
open set R \ (Fj + t) contains Q, and since Q is dense in P , (Fj + t) ∩ P is nowhere
dense in P , and (F + t) ∩ P is meager in P .
¤
Let us finish with some open problems.
Problem 10. Does any inclusion between the families Perm(A) and Perm(N0f )
for various functions f hold true?
Problem 11. Are permitted sets universally null?
Problem 12. Can one in ZFC prove the existence of an uncountable permitted
set?
8
PETER ELIAŠ
References
[1] Arbault J., Sur l’ensemble de convergence absolue d’une série trigonométrique, Bull. Soc.
Math. France 80 (1952), 253–317.
[2] Bartoszyński T., On perfectly meager sets, Proc. Amer. Math. Soc. 130 (2001), 1189–1195.
[3] Bartoszyński T., Remarks on small sets of reals, Proc. Amer. Math. Soc. 131 (2002), 625–630.
[4] Bary N. K., A Treatise on Trigonometric Series, The Macmillan Co., New York, 1964.
[5] Bukovská Z., Thin sets defined by a sequence of continuous functions, Math. Slovaca 49
(1999), 323–344.
[6] Bukovská Z., Bukovský L., Comparing families of thin sets, Real Anal. Exchange 27 (2001–
2002), 609–625.
[7] Bukovský L., Kholshchevnikova N. N., Repický M., Thin sets of harmonic analysis and
infinite combinatorics, Real Anal. Exchange 20 (1994–1995), 454–509.
[8] Cassels J. W. S., An Introduction to Diophantine Approximation, Cambridge Tracts in Mathematics and Mathematical Physics, No. 45, Cambridge University Press, New York, 1957.
[9] Eliaš P., On inclusions between Arbault sets, Acta Univ. Carol. Math. Phys. 44 (2003), 65–72.
[10] Eliaš P., Arbault permitted sets are perfectly meager, to appear in Tatra Mt. Math. Publ.
[11] Hewitt E., Ross K. A., Abstract Harmonic Analysis, Die Grundlehren der Mathematischen
Wissenschaften, Bd. 115, Academic Press, Inc., New York; Springer-Verlag, Berlin-GöttingenHeidelberg, 1963.
[12] Miller A. W., Special subsets of the real line. In: Handbook of Set-Theoretic Topology, K.
Kunen and J. E. Vaughan, eds., North-Holland, Amsterdam, 1984, pp. 201–233.
[13] Nowik A., Weiss T., Not every Q-set is perfectly meager in the transitive sense, Proc. Amer.
Math. Soc. 128 (2000), 3017–3024.
[14] Nowik A., Weiss T., Some remarks on totally imperfect sets, Proc. Amer. Math. Soc. 132
(2003), 231–237.
[15] Zakrzewski P., Universally meager sets, Proc. Amer. Math. Soc. 129 (2001), 1793–1798.
Mathematical Institute, Slovak Academy of Sciences, Grešákova 6, 040 01 Košice,
Slovakia
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