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Pointwise density topologies with respect to a fixed sequence
Journal Pre-proof Pointwise density topologies with respect to a fixed sequence
Magdalena Górajska, Małgorzata Filipczak, Jacek Hejduk
PII:
S0166-8641(19)30352-9
DOI:
https://doi.org/10.1016/j.topol.2019.106941
Reference:
TOPOL 106941
To appear in:
Topology and its Applications
Received date:
5 August 2019
Revised date:
28 October 2019
Accepted date:
6 November 2019
Please cite this article as: M. Górajska et al., Pointwise density topologies with respect to a fixed sequence, Topol. Appl. (2019), 106941, doi: https://doi.org/10.1016/j.topol.2019.106941.
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Pointwise density topologies with respect to a fixed sequence Magdalena Górajskaa,∗, Małgorzata Filipczakb , Jacek Hejdukb a
Center of Mathematics and Physics, Łódź University of Technology, al. Politechniki 11, 90-924 Łódź, Poland b Faculty of Mathematics and Computer Sciences, Łódź University, Stefana Banacha 22, 90-238 Łódź, Poland
Abstract The paper presents a new family of density-type topologies. The pointwise density topology introduced in [4] is a motivation to consider topologies with respect to a fixed sequence. We study several properties of such topologies, in particular we discuss the separation axioms. Moreover, we apply our results to show that the pointwise density topology on the real line is not a regular space. Keywords: density point, density topology, pointwise convergence 2010 MSC: 54A10, 28A05
1. Introduction Through the paper we shall use standard notation: R, Q, N denote the sets of real numbers, rational numbers and positive integers, respectively. Let B be the σ-algebra of Borel sets on R. The Euclidean topology on R is denoted by Tnat . If T is a topology on R, then for any set A ⊂ R, clT A is the closure of A with respect to T . We will use the symbol λ to denote the Lebesgue measure on the real line. Let L be the family of Lebesgue measurable sets of R and let N be the family of Lebesgue null sets in R. We will use the symbol χA to denote the characteristic function of set A ⊂ R. It is well known definition that a point x ∈ R is a density point of a measurable set A 1 1 ,x+ n ]) λ(A∩[x− n = 1, or equivalently if limn→∞ = 1. It is sometimes if limh→0+ λ(A∩[x−h,x+h]) 2 2h n written in the form limn→∞ λ(n(A − x) ∩ [−1, 1]) = 2, where nA = {na : a ∈ A} and A − x = {a − x : a ∈ A} for n ∈ N. The last condition means that the sequence of characteristic functions {χn(A−x)∩[−1,1] }n∈N converges in measure λ to χ[−1,1] , (cf. [11]). For every A ∈ L, putting Φd (A) = {x ∈ R : x is a density point of A}, we get that Φd : L → L is the lower density operator (cf. [10] or [3]). Let us define Td = {A ∈ L : ∗
Corresponding author Email addresses: magdalena.gó[email protected] (Magdalena Górajska), [email protected] (Małgorzata Filipczak), [email protected] (Jacek Hejduk) Preprint submitted to Topology and its Applications
November 22, 2019
A ⊂ Φd (A)}. The family Td forms topology investigated in [3] and called the density topology. When we use various types of convergence of the sequence {χn(A−x)∩[−1,1] }n∈N we can obtain different density-type topologies. Thus, for example, if we consider the convergence everywhere the above sequence then we get the density topology which is called the pointwise density topology denoted by Tp and investigated in [4]. As we can see the sequence {n}n∈N plays a special role in the definition of Lebesgue density point. When we change this sequence in definition of the density point we get new density-type topologies. The motivation of this paper is generalization of the pointwise density topology by a modification of the definition of pointwise density point, replacing the sequence of natural numbers with the unbounded and nondecreasing sequence of positive real numbers. Also the problems posed in [4] and [5] will be solved. By S we denote the family of all unbounded and nondecreasing sequences of positive real numbers. Every sequence {sn }n∈N ∈ S will be denoted by s. First, we recall the definition of the s-density point for every measurable subset of R. Definition 1. ([2]) Let s ∈ S and A ∈ L. The point x ∈ R is a density point of a set A with respect to the sequence s ∈ S ( an s-density point of A) if λ A ∩ x − s1n , x + s1n lim = 1. (1) 2 n→∞
sn
The last condition means that the sequence of characteristic functions {χsn (A−x)∩[−1,1] }n∈N converges to χ[−1,1] with respect to measure λ. For any s ∈ S and A ∈ L, putting Φs (A) = {x ∈ R : x is s-density point of A}, we get that Φs : L → L is the lower density operator (cf. [2]). Let us define Ts = {A ∈ L : A ⊂ Φs (A)} for every sequence s ∈ S. Then the family Ts forms a topology on the real line investigated in [2]. It easy to see that Td ⊂ Ts for every s ∈ S. Let Ba and I be the σ-algebra of sets having the Baire property and the σ-ideal of first category sets on the real line with respect to the Euclidean topology, respectively. Let us recall the definition of an Is-density point of a set A ∈ Ba with respect to the sequence s ∈ S. Definition 2. ([6]) Let s ∈ S and A ∈ Ba. The point x ∈ R is an I-density point of a set A with respect to a sequence s ∈ S (an Is-density point of A) if for every subsequence p→∞ {snm }m∈N ⊂ s there exists a subsequence {snmp }p∈N such that χsnmp (A−x)∩[−1,1] −−−→ 1 I-a.e. For any s ∈ S and A ∈ Ba, putting ΦIs (A) = {x ∈ R : x is an Is-density point of A}, we get that ΦIs : Ba → Ba is a lower density operator (cf. [6]). Let us define TIs = {A ∈ Ba : A ⊂ ΦIs (A)}, for every sequence s ∈ S. The family TIs forms topology investigated in [6] and [8], which fulfils the inclusion TI ⊂ TIs , where TI denotes the I-density topology (cf. [11]). The rest of the paper is organized as follows. In the next section we introduce the concept of pointwise density point with respect to a fixed sequence. Section 3 contains the main results on pointwise density topology for invariant family of subset of 2
real line. In the final section we examine the separation axioms for s-pointwise density topology, in particular we present the proof that they do not satisfy the axiom of regularity. 2. Pointwise density points with respect to a fixed sequence The above definitions are motivation to introduce the concept of pointwise density point of a set A ⊂ R with respect to a sequence s ∈ S. Definition 3. Let A ⊂ R and s ∈ S. We shall say that (i) 0 is a pointwise density point of a set A with respect to a sequence s ∈ S ( an psdensity point of A) if and only if the sequence {χsn A∩[−1,1] }n∈N converges everywhere to the function χ[−1,1] , (ii) x ∈ R is a ps-density point of A if and only if 0 is a ps-density point of A − x, (iii) x ∈ R is a ps-dispersion point of A if only if x is a ps-density point of R \ A. Directly from the above definition we derive the following characterisation of ps-density points. Property 4. Let A ⊂ R and s ∈ S. Then (i) 0 is a ps-density point of A if and only if [−1, 1] ⊂ lim inf sn A, n→∞
(ii) x ∈ R is a ps-density point of A if and only if [−1, 1] ⊂ lim inf sn (A − x), n→∞
(iii) x ∈ R is a ps-dispersion point of A if and only if [−1, 1] ⊂ lim inf sn ((R \ A) − x). n→∞
Furthermore, for any s ∈ S and A ⊂ R let Φps (A) = {x ∈ R : x is a ps-density point of A}. The following properties are the consequence of the definition of an ps-density point. There are similar to the properties of pointwise density point describe in [4]. Theorem 5. Let A,B ⊂ R and s ∈ S. Then (i) (ii) (iii) (iv)
Φps (∅) = ∅ and Φps (R) = R, if A ⊂ B then Φps (A) ⊂ Φps (B), Φps (A ∩ B) = Φps (A) ∩ Φps (B), Φps (A) ⊂ A.
Property 6. Let A ⊂ R and s ∈ S. Then (i) ∀a∈R
Φps (A) + a = Φps (A + a),
(ii) ∀m∈(0,∞)
mΦps (A) = Φp ms (mA).
Let J ⊂ 2R . We call J an invariant family of subsets (invariant for short) if J is closed under linear operation i.e. for each X ∈ J and all a, b ∈ R we obtain aX + b ∈ J . Property 7. Let s ∈ S and J be an invariant σ-ideal. If Tnat ∩ J = {∅} and A ∈ J , then Φps (A) = ∅. 3
In particular, the families N , I satisfies the assumption of Property 7, so we have the following corollary. Corollary 8. If A ∈ N or A ∈ I, then Φps (A) = ∅. It is known that the line can be decomposed into two complementary set such that one is of first category and the second is of measure zero. Using this fact we have the following property. Property 9. Let s ∈ S. Then (ii) there exists a set A ∈ L of full Lebesgue measure such that Φps (A) = ∅, (iii) there exists a residual set A ∈ Ba such that Φps (A) = ∅. A Let A ⊂ 2R be an arbitrary family of sets and Tps = {A ∈ A : A ⊂ Φps (A)}.
Proposition 10. If A ⊂ 2R is an invariant family of sets A ∈ A and s ∈ S then (i) ∀a∈R
A A A ∈ Tps ⇔ A + a ∈ Tps ,
(ii) ∀m∈(0,∞)
A A ∈ Tps ⇔ mA ∈ TpA s , m
A A (iii) if β ≥ 1 and A ∈ Tps then βA ∈ Tps , A A (iv) if β ≥ 1 then Tps ⊂ Tpβs , where βs = {βsn }n∈N . B = Following the concept of Theorem 9 in [5] we get that if A = B then the family Tps {A ∈ B : A ⊂ Φps (A)}, where s ⊂ Q is not a topology (see Theorem 9 in [5]). Observe A that if A = 2R , A = L or A = Ba, then the family Tps forms a topology. The proof that R 2 L ⊂ Ts Tps is a topology is straightforward. Then, for A = L or A = Ba we use the fact Tps Ba A and Tps ⊂ TIs . If Tps is a topology then we say that it is the topology generated by operator Φps in the space (R, A).
3. Pointwise topologies on the real line Hereafter we shall assume that a family A ⊂ 2R contain Tnat and co-countable sets. We A suppose that an operator Φps generates a topology Tps for every s ∈ S. Recall that A if and only if by property (iv) in Theorem 5 it follows that for every A ∈ A, A ∈ Tps Φps (A) = A. A A Let Tpt , Tps be topologies generated by sequences t, s ∈ S, respectively. For any t ∈ S we denote by At the set {x ∈ R : x = ttmn , n < m}. Theorem 11. Let s,t ∈ S. Then A A (i) if s is a subsequence of t then Tpt ⊂ Tps , 1 A , ∈ / Tps (ii) R \ ∞ n=1 sn
4
(iii) if At ∩ As is a finite set then R \
∞ 1 n=1
tn
A . ∈ Tps
Proof. Condition (i) is a consequence of definition of ps-density point. For the proof con- of ∞ 1 1 dition (ii) it is enough to observe that however R\ n=1 sn ∈ A but 1 ∈ / lim inf sn R \ ∞ . n=1 sn n→∞ 1 Next, we prove condition (iii). Let A = R\ ∞ i=1 ti . Obviously, A\{0} ∈ Tnat . It is sufficient to prove that 0 is an ps-density point of the set A, i.e. [−1, 1] ⊂ lim inf sn A. Clearly n→∞
[−1, 0] ⊂ lim inf sn A. Choose x ∈ (0, 1], n1 , n2 ∈ N such that n2 > n1 . Then n→∞
∞ 1 sn x ⇔ ∃ x = 1. ∈ / R\ i1 ∈N sn 1 ti t i1 i=1
Analogously ∞ 1 x sn ∈ / R\ ⇔ ∃ x = 2. i ∈N sn 2 t t i2 2 i i=1
1 ∞ 1 x / R\ ∞ ∈ / R\ Consequently, if snx ∈ and then i=1 ti i=1 ti sn2 1 N. As that At ∩ As is finite, it follows that the set
∞ 1 x n∈N: ∈ / R\ sn ti i=1
sn1 sn2
=
ti1 ti2
for some i1 , i2 ∈
1 A . . Therefore A ∈ Tps is finite. Hence x ∈ lim inf sn R\ ∞ i=1 ti n→∞
A A Corollary 12. Let s,t ∈ S. If At ∩ As is the finite set then the topologies Tpt , Tps are incomparable.
The converse statement is not true. Theorem 13. There exist sequences t, s ∈ S such that At ∩ As is an infinite set and A A the topologies Tpt , Tps are incomparable. Proof. Let as fix t, s ∈ S such that At ∩ As is an infinite set and the sequence s contains s ∈ S and the sequence t contains t ∈ S suchthat sets At ∩ As , At ∩ As ∞ 1 ∞ 1 A / are finite. Then by Theorem 11 we conclude that R\ i=1 t ∈ Tps and R\ i=1 t ∈ i i 1 1 A A A A A Tpt ∈ Tpt ∈ / Tps . Also R\ ∞ and R\ ∞ . Hence the topologies Tpt , Tps i=1 si i=1 si are incomparable. Theorem 14. For every sequence s ∈ S we have A . Tnat Tps 5
A Proof. Let s ∈ S. Obviously Tnat ⊂ Tps (see [4]). There exists a sequence t ∈ S such ∞ 1 A that At ∩ As = ∅. Let A = R\ i=1 ti . Obviously, A ∈ . / Tnat and A ∈ Tps A Theorem 15. Let T = s∈S Tps . Then T = Tnat .
Proof. The family T is a topology on the real line, as the intersection of the topology. Obviously, T contains natural topology. We now prove that T ⊂ Tnat . Let A ∈ T and x ∈ A. Without loss of generality, we can assume that x = 0. We prove that there exists δ > 0 such that (−δ, δ) ⊂ A. On the contrary, suppose that 1 1 1 ∀ ∃ ∈ − , \A. (2) k∈N tk ∈R tk k k Hence there exists a subsequence of natural numbers {kn }n∈N such that tkn > 0 for all n ∈ N or tkn < 0 for all n ∈ N and sequence {|tkn |}n∈N is unbounded and nondecreasing. Set A sn = |tkn |. Then, s ∈ S and A ∈ / Tps . Indeed, if tkn > 0 then 1 ∈ / lim inf sn A and if tkn < 0 n→∞
then (−1) ∈ / lim inf sn A, which is impossible. Therefore, 0 belongs to the set A with some n→∞ open neighborhood. A Property 16. There does not exist the largest topology Tps ∗ in the sense of inclusion such A A that Tps ⊂ Tps∗ for any s ∈ S.
Proof. We can see that for every sequence s ∈ S there exists a sequence t ∈ S such that A A As ∩ At = ∅. By Theorem 13 topologies Tps and Tpt are incomparable, which completes the proof. Lemma 17. Let s ∈ S. Then A (i) if s is a sequence of rational numbers then the set R \ Q ∈ Tps , A , (ii) there exists a sequence of irrational numbers t ∈ S such that R \ Q ∈ / Tpt A . (iii) there exists a countable dense set C such that R \ C ∈ Tps
Proof. The proof of the condition (i) is similar to Proposition 3.2 in [4]. Nevertheless, we include the proof for the completeness of the exposition and the convenience of the readers. Let {sn }n∈N ⊂ Q and y ∈ R\Q. Recall that y ∈ Φps (R\Q) is equivalent to the condition that for all x ∈ [−1, 1] there exists k ∈ N such that sxn + y ∈ R\Q for all n ≥ k. Fix x ∈ [−1, 1]. There exists at most one sn such that x ∈ / sn ((R\Q) − y). On the contrary, / sn1 ((R\Q) − y) suppose that there exist two different numbers sn1 , sn2 such that x ∈ and x ∈ / sn2 ((R\Q) − y). Hence x p1 x p2 +y = , + y = , where p1 , p2 ∈ Z q1 , q2 ∈ N. sn 1 q1 sn 2 q2 From the above equalities, we thus obtain y = y ∈ R\Q. Finally, we get R\Q ⊂ Φps (R\Q). 6
p2 p −sn1 q1 q2 1 sn2 −sn1
sn2
∈ Q, which is impossible, since
Now we prove condition (ii). Let y0 be a positive irrational number and {yn }n∈N ⊂ Q be a decreasing sequence convergent to y0 such that yn > y0 for all n ∈ N. Define tn =
1 for all n ∈ N. yn − y0
Obviously, t ∈ S and 1 ∈ / lim inf tn (R\Q − y0 ). Hence y0 ∈ / Φpt (R\Q). This implies that n→∞
A R\Q ∈ / Tpt . Finally we shall prove condition (iii). Let s ∈ S. Let A be the set of all elements of sequence s and B be given by formula
k i=1 ai , ai , bj , cj ∈ A, bj = cj , 1 ≤ i ≤ k, 1 ≤ j ≤ l, k, l ∈ N . B = x ∈ R : x = l (b − c ) j j j=1
Let C = SpanB, where SpanB is a linear space over Q generated by elements from a set A B. Obviously the set C is dense and countable. Now we prove that R \ C ∈ Tps . Let y ∈ R \ C. Recall that y ∈ Φps (R \ C) is equivalent to the condition that for all x ∈ [−1, 1] there exists k ∈ N such that sxn + y ∈ R \ C for all n ≥ k. Fix x ∈ [−1, 1]. There exists at most one sn such that sxn + y ∈ C. On the contrary, suppose that there exist two different numbers sn1 , sn2 such that x + y = c1 sn 1 Hence y=
∧
x + y = c2 for some c1 , c2 ∈ C. sn 2
s n2 sn1 c2 − c1 . sn 2 − sn 1 sn 2 − s n 1
Thus we obtain y ∈ C, which is impossible. 4. The separation axioms A Let s ∈ S and (R, Tps ) be a topological space assuming always that A contain Tnat A A and all co-countable sets. Since Tnat ⊂ Tps , we get that (R, Tps ) is a Hausdorff space. A Theorem 18. Let s ∈ S and A ⊂ Ba. Then the topological space (R, Tps ) is not regular.
Proof. Recall that the family TIs is a topology such that Tnat ⊂ TIs (cf. [6]). By Theorem 6 in [7] it follows that every Tnat -dense set can not be separated from any point by TIs -open A sets. Since Tps ⊂ TIs and the set C described in proof of Lemma 17 (iii) is Tnat -dense, A Tps -closed and countable we get that C and any point x ∈ / C can not be separated by A A -open sets. So that (R, Tps ) is not regular. Tps A Now we shall prove that for every s ∈ S the space (R, Tps ), where A ⊂ L is not regular. It is the answer to the question possed in [4]. The proof will be preceded by the following Lemmas. 7
Lemma 19. Let s ∈ S, A ⊂ R and J be an invariant σ-ideal such that J ∩ Tnat = {∅} . A If (a, b) \ A ∈ J , then the closure of the set A with respect to Tps contains (a, b). A (A). Suppose that B is not empty set. The set B is obviously Proof. Define B = (a, b)\clTps
A . Hence B = Φps (B). Since B ∈ J , we get from Property 7 in open with respect to Tps Section 2 that Φps (B) = ∅ and finally, B = ∅. This completes the proof.
Lemma 20. Let s ∈ S. If A ∈ Ba is a set of the second category, then the closure of A A with respect to Tps contains an interval. Proof. Let A be a set of the second category with the Baire property. Obviously there exists an interval (a, b) such that the set A is residual on (a, b). By Lemma 19, for J = I we A (A). obtain that (a, b) ⊂ clTps A Lemma 21. Let s ∈ S, A ⊂ L and A ∈ Tps . If A ∩ I = ∅ for every open interval I = ∅, A is equal to R. then the closure of the set A ⊂ R with respect to topology Tps A (A) and suppose that the set B is not empty. Since A, B are Proof. Define B = R \ clTps
A A topology, Tps ⊂ Ts then we get A, B ∈ Ts . Observe open set with respect to the Tps that Φs (A) ∩ Φs (B) = Φs (A ∩ B) = Φs (∅) = ∅.
Thus Φs (B) ⊂ R \ Φs (A) ⊂ R \ A. We divide the proof into two steps. 1. Firstly, assume that Φs (B) is the set of the second category. Since Φs (B) is a Borel set, this set is residual on some interval (a, b) (Proposition 1, p.188 in [2]). Thus A is the set of the first category on (a, b), then ∅ = Φps ((a, b) ∩ A) = Φps ((a, b)) ∩ Φps (A) ⊃ (a, b) ∩ Φps (A) = (a, b) ∩ A. It implies that the set A and (a, b) are disjoint. This contradicts our assumption. A 2. Here, we assume that Φs (B) is the set of the first category. Since B ∈ Tps we get that B ⊂ Φps (B). Hence B is the first category and thus Φps (B) = ∅. It implies that B = ∅.This contradiction ends the proof.
A Lemma 22. Let s ∈ S and A ⊂ L. If V ∈ Tps and V = ∅, then the closure of V with A respect to Tps contains an interval. A is nonempty and for every open interval I we have Proof. Let us assume that V ∈ Tps A (V ). It means that set R \ C meets the assumptions of that I \ C = ∅, where C = clTps A (R \ C) = R. It is a contradiction with the fact that V ∩ (R \ C) = ∅ Lemma 21, so that clTps and the proof is completed. 8
A Theorem 23. Let s ∈ S and A ⊂ L. The topological space (R, Tps ) is not regular.
Proof. Recall that a topological space (X, T ) is regular if and only if for every set U ∈ T and every point x ∈ U , there exists a neighbourhood V ∈ T such that x ∈ V ⊂ clT (V ) ⊂ U (see Theorem 1.5.5 in [1]). Let U = R \ C be the set defined in Lemma 17. Obviously, U is A A not empty and U ∈ Tps . Let V ∈ Tps be a non empty subset of U . From Lemma 22 we A (V ) contains some interval. Therefore clT A (V ) \ U = ∅. In this way we obtain that clTps ps A ) is not regular. have proved that the space (R, Tps
Theorem 23 is the answer to the question formulated in [4], if the topological space L (R, Tps ) is regular. Now, we are solving Problem 8 formulated in [5]: Ba L
= Tps , where s = {n}n∈N ? Is it true that Tps Lemma 24. Let H be a Hamel base and s = {n}n∈N . Then, every point x ∈ R \ H is a ps-density point of set R \ H. Proof. Let x ∈ R. Recall that x ∈ Φp (R \ H) is equivalent to the condition that for all a ∈ [−1, 1] there exists k ∈ N such that na + x ∈ / H for all n ≥ k. Fix a ∈ [−1, 1]. We prove that there exist at most two different number sn1 , sn2 , such that na1 + x ∈ H and na2 + x ∈ H. On the contrary, suppose that there exist three different numbers sn1 , sn2 , sn3 such that h1 =
a a a + x ∈ H and h2 = + x ∈ H and h3 + x ∈ H. n1 n2 n3
From the above equalities, we obtain that x=
n 1 h1 − n 2 h2 n 1 h1 − n 3 h3 and x = . n1 − n2 n1 − n3
Therefore, h1 n1 (n3 − n2 ) + h2 n2 (n1 − n3 )h2 + h3 n3 (n2 − n1 ) = 0.
(3)
which is a contradiction with the assumption that h1 , h2 , h3 are linearly independent. Finally, we get x ∈ Φp (R \ H). Corollary 25. Let H be a Hamel base and s = {n}n∈N . Then the set R \ H is open set with respect to topology Tps = {A ∈ R : A ⊂ Φp (A)}. Note that Lemma 24 is true for any s ⊂ Q. From Lemma 24 and Corollary 25 we obtain the following theorem. L Ba Theorem 26. The topology Tps and Tps for s ⊂ Q are not comparable.
Proof. Recall that there exist a Hamel base H1 of measure zero without the Baire property and a Hamel base H2 of the first category and Lebesgue nonmeasurable (see Corollary 11.4.3 in [9]). Then L Ba R \ H1 ∈ Tps and R \ H1 ∈ / Tps 9
and
Ba L R \ H2 ∈ Tps and R \ H2 ∈ / Tps , L Ba Ba L \ Tps
= ∅ and Tps \ Tps
= ∅. Tps
Hence
L Ba
= Tps . Tps
Bibliografia References [1] Engelking R., General topology, Polish Scientific Publishers, Warsaw (2007). [2] Filipczak M., Hejduk J., On topologies associated with the Lebesgue measure, Tatra Mt. Math. Publ. 28(2), (2004) pp. 187-197. [3] Goffman C., Neugebauer C. J., Nishura T., The density topology and approximate continuity, Duke Math. J. 28,(1961), pp. 497–506. [4] Górajska M., Pointwise density topology, Open Mathematics, vol. 13, no. 1, (2015), pp. 75—82. [5] Górajska M., Hejduk J., Pointwise density topology with respect to admissible σ-algebras, Tatra Mt. Math. Publ. 55(1), (2013), pp. 77–83. [6] Hejduk J., Horbaczewska G., On I-density topologies with respect to a fixed sequence, Reports on Real Analysis, Conference at Rowy, (2003), pp. 78—85. [7] Hejduk J., On the regularity of topologies in the family of sets having the Baire property, Filomat 27(7), (2013), pp. 1291-1295. [8] Horbaczewska G., The family of I-density type topologies, Comment. Math. Univ. Carolinae 46(4), (2005), pp. 735–745. [9] Kuczma M., An introduction to the theory of functional equations and inequalities: Cauchy’s equation and Jensen’s inequality, Springer Science and Business Media (2009). [10] Oxtoby J. C., Measure and category, Springer Verlag, New York-Heidelberg-Berlin (1980). [11] Poreda W., Wagner-Bojakowska E., Wilczyński W., A category analogue of the density topology, Fund. Math. 125(2), (1985), pp.167–173.
Conflicts of Interest The authors declare no conflict of interest. This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.
10
Magdalena Górajska, Małgorzata Filipczak, Jacek Hejduk
PII:
S0166-8641(19)30352-9
DOI:
https://doi.org/10.1016/j.topol.2019.106941
Reference:
TOPOL 106941
To appear in:
Topology and its Applications
Received date:
5 August 2019
Revised date:
28 October 2019
Accepted date:
6 November 2019
Please cite this article as: M. Górajska et al., Pointwise density topologies with respect to a fixed sequence, Topol. Appl. (2019), 106941, doi: https://doi.org/10.1016/j.topol.2019.106941.
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Pointwise density topologies with respect to a fixed sequence Magdalena Górajskaa,∗, Małgorzata Filipczakb , Jacek Hejdukb a
Center of Mathematics and Physics, Łódź University of Technology, al. Politechniki 11, 90-924 Łódź, Poland b Faculty of Mathematics and Computer Sciences, Łódź University, Stefana Banacha 22, 90-238 Łódź, Poland
Abstract The paper presents a new family of density-type topologies. The pointwise density topology introduced in [4] is a motivation to consider topologies with respect to a fixed sequence. We study several properties of such topologies, in particular we discuss the separation axioms. Moreover, we apply our results to show that the pointwise density topology on the real line is not a regular space. Keywords: density point, density topology, pointwise convergence 2010 MSC: 54A10, 28A05
1. Introduction Through the paper we shall use standard notation: R, Q, N denote the sets of real numbers, rational numbers and positive integers, respectively. Let B be the σ-algebra of Borel sets on R. The Euclidean topology on R is denoted by Tnat . If T is a topology on R, then for any set A ⊂ R, clT A is the closure of A with respect to T . We will use the symbol λ to denote the Lebesgue measure on the real line. Let L be the family of Lebesgue measurable sets of R and let N be the family of Lebesgue null sets in R. We will use the symbol χA to denote the characteristic function of set A ⊂ R. It is well known definition that a point x ∈ R is a density point of a measurable set A 1 1 ,x+ n ]) λ(A∩[x− n = 1, or equivalently if limn→∞ = 1. It is sometimes if limh→0+ λ(A∩[x−h,x+h]) 2 2h n written in the form limn→∞ λ(n(A − x) ∩ [−1, 1]) = 2, where nA = {na : a ∈ A} and A − x = {a − x : a ∈ A} for n ∈ N. The last condition means that the sequence of characteristic functions {χn(A−x)∩[−1,1] }n∈N converges in measure λ to χ[−1,1] , (cf. [11]). For every A ∈ L, putting Φd (A) = {x ∈ R : x is a density point of A}, we get that Φd : L → L is the lower density operator (cf. [10] or [3]). Let us define Td = {A ∈ L : ∗
Corresponding author Email addresses: magdalena.gó[email protected] (Magdalena Górajska), [email protected] (Małgorzata Filipczak), [email protected] (Jacek Hejduk) Preprint submitted to Topology and its Applications
November 22, 2019
A ⊂ Φd (A)}. The family Td forms topology investigated in [3] and called the density topology. When we use various types of convergence of the sequence {χn(A−x)∩[−1,1] }n∈N we can obtain different density-type topologies. Thus, for example, if we consider the convergence everywhere the above sequence then we get the density topology which is called the pointwise density topology denoted by Tp and investigated in [4]. As we can see the sequence {n}n∈N plays a special role in the definition of Lebesgue density point. When we change this sequence in definition of the density point we get new density-type topologies. The motivation of this paper is generalization of the pointwise density topology by a modification of the definition of pointwise density point, replacing the sequence of natural numbers with the unbounded and nondecreasing sequence of positive real numbers. Also the problems posed in [4] and [5] will be solved. By S we denote the family of all unbounded and nondecreasing sequences of positive real numbers. Every sequence {sn }n∈N ∈ S will be denoted by s. First, we recall the definition of the s-density point for every measurable subset of R. Definition 1. ([2]) Let s ∈ S and A ∈ L. The point x ∈ R is a density point of a set A with respect to the sequence s ∈ S ( an s-density point of A) if λ A ∩ x − s1n , x + s1n lim = 1. (1) 2 n→∞
sn
The last condition means that the sequence of characteristic functions {χsn (A−x)∩[−1,1] }n∈N converges to χ[−1,1] with respect to measure λ. For any s ∈ S and A ∈ L, putting Φs (A) = {x ∈ R : x is s-density point of A}, we get that Φs : L → L is the lower density operator (cf. [2]). Let us define Ts = {A ∈ L : A ⊂ Φs (A)} for every sequence s ∈ S. Then the family Ts forms a topology on the real line investigated in [2]. It easy to see that Td ⊂ Ts for every s ∈ S. Let Ba and I be the σ-algebra of sets having the Baire property and the σ-ideal of first category sets on the real line with respect to the Euclidean topology, respectively. Let us recall the definition of an Is-density point of a set A ∈ Ba with respect to the sequence s ∈ S. Definition 2. ([6]) Let s ∈ S and A ∈ Ba. The point x ∈ R is an I-density point of a set A with respect to a sequence s ∈ S (an Is-density point of A) if for every subsequence p→∞ {snm }m∈N ⊂ s there exists a subsequence {snmp }p∈N such that χsnmp (A−x)∩[−1,1] −−−→ 1 I-a.e. For any s ∈ S and A ∈ Ba, putting ΦIs (A) = {x ∈ R : x is an Is-density point of A}, we get that ΦIs : Ba → Ba is a lower density operator (cf. [6]). Let us define TIs = {A ∈ Ba : A ⊂ ΦIs (A)}, for every sequence s ∈ S. The family TIs forms topology investigated in [6] and [8], which fulfils the inclusion TI ⊂ TIs , where TI denotes the I-density topology (cf. [11]). The rest of the paper is organized as follows. In the next section we introduce the concept of pointwise density point with respect to a fixed sequence. Section 3 contains the main results on pointwise density topology for invariant family of subset of 2
real line. In the final section we examine the separation axioms for s-pointwise density topology, in particular we present the proof that they do not satisfy the axiom of regularity. 2. Pointwise density points with respect to a fixed sequence The above definitions are motivation to introduce the concept of pointwise density point of a set A ⊂ R with respect to a sequence s ∈ S. Definition 3. Let A ⊂ R and s ∈ S. We shall say that (i) 0 is a pointwise density point of a set A with respect to a sequence s ∈ S ( an psdensity point of A) if and only if the sequence {χsn A∩[−1,1] }n∈N converges everywhere to the function χ[−1,1] , (ii) x ∈ R is a ps-density point of A if and only if 0 is a ps-density point of A − x, (iii) x ∈ R is a ps-dispersion point of A if only if x is a ps-density point of R \ A. Directly from the above definition we derive the following characterisation of ps-density points. Property 4. Let A ⊂ R and s ∈ S. Then (i) 0 is a ps-density point of A if and only if [−1, 1] ⊂ lim inf sn A, n→∞
(ii) x ∈ R is a ps-density point of A if and only if [−1, 1] ⊂ lim inf sn (A − x), n→∞
(iii) x ∈ R is a ps-dispersion point of A if and only if [−1, 1] ⊂ lim inf sn ((R \ A) − x). n→∞
Furthermore, for any s ∈ S and A ⊂ R let Φps (A) = {x ∈ R : x is a ps-density point of A}. The following properties are the consequence of the definition of an ps-density point. There are similar to the properties of pointwise density point describe in [4]. Theorem 5. Let A,B ⊂ R and s ∈ S. Then (i) (ii) (iii) (iv)
Φps (∅) = ∅ and Φps (R) = R, if A ⊂ B then Φps (A) ⊂ Φps (B), Φps (A ∩ B) = Φps (A) ∩ Φps (B), Φps (A) ⊂ A.
Property 6. Let A ⊂ R and s ∈ S. Then (i) ∀a∈R
Φps (A) + a = Φps (A + a),
(ii) ∀m∈(0,∞)
mΦps (A) = Φp ms (mA).
Let J ⊂ 2R . We call J an invariant family of subsets (invariant for short) if J is closed under linear operation i.e. for each X ∈ J and all a, b ∈ R we obtain aX + b ∈ J . Property 7. Let s ∈ S and J be an invariant σ-ideal. If Tnat ∩ J = {∅} and A ∈ J , then Φps (A) = ∅. 3
In particular, the families N , I satisfies the assumption of Property 7, so we have the following corollary. Corollary 8. If A ∈ N or A ∈ I, then Φps (A) = ∅. It is known that the line can be decomposed into two complementary set such that one is of first category and the second is of measure zero. Using this fact we have the following property. Property 9. Let s ∈ S. Then (ii) there exists a set A ∈ L of full Lebesgue measure such that Φps (A) = ∅, (iii) there exists a residual set A ∈ Ba such that Φps (A) = ∅. A Let A ⊂ 2R be an arbitrary family of sets and Tps = {A ∈ A : A ⊂ Φps (A)}.
Proposition 10. If A ⊂ 2R is an invariant family of sets A ∈ A and s ∈ S then (i) ∀a∈R
A A A ∈ Tps ⇔ A + a ∈ Tps ,
(ii) ∀m∈(0,∞)
A A ∈ Tps ⇔ mA ∈ TpA s , m
A A (iii) if β ≥ 1 and A ∈ Tps then βA ∈ Tps , A A (iv) if β ≥ 1 then Tps ⊂ Tpβs , where βs = {βsn }n∈N . B = Following the concept of Theorem 9 in [5] we get that if A = B then the family Tps {A ∈ B : A ⊂ Φps (A)}, where s ⊂ Q is not a topology (see Theorem 9 in [5]). Observe A that if A = 2R , A = L or A = Ba, then the family Tps forms a topology. The proof that R 2 L ⊂ Ts Tps is a topology is straightforward. Then, for A = L or A = Ba we use the fact Tps Ba A and Tps ⊂ TIs . If Tps is a topology then we say that it is the topology generated by operator Φps in the space (R, A).
3. Pointwise topologies on the real line Hereafter we shall assume that a family A ⊂ 2R contain Tnat and co-countable sets. We A suppose that an operator Φps generates a topology Tps for every s ∈ S. Recall that A if and only if by property (iv) in Theorem 5 it follows that for every A ∈ A, A ∈ Tps Φps (A) = A. A A Let Tpt , Tps be topologies generated by sequences t, s ∈ S, respectively. For any t ∈ S we denote by At the set {x ∈ R : x = ttmn , n < m}. Theorem 11. Let s,t ∈ S. Then A A (i) if s is a subsequence of t then Tpt ⊂ Tps , 1 A , ∈ / Tps (ii) R \ ∞ n=1 sn
4
(iii) if At ∩ As is a finite set then R \
∞ 1 n=1
tn
A . ∈ Tps
Proof. Condition (i) is a consequence of definition of ps-density point. For the proof con- of ∞ 1 1 dition (ii) it is enough to observe that however R\ n=1 sn ∈ A but 1 ∈ / lim inf sn R \ ∞ . n=1 sn n→∞ 1 Next, we prove condition (iii). Let A = R\ ∞ i=1 ti . Obviously, A\{0} ∈ Tnat . It is sufficient to prove that 0 is an ps-density point of the set A, i.e. [−1, 1] ⊂ lim inf sn A. Clearly n→∞
[−1, 0] ⊂ lim inf sn A. Choose x ∈ (0, 1], n1 , n2 ∈ N such that n2 > n1 . Then n→∞
∞ 1 sn x ⇔ ∃ x = 1. ∈ / R\ i1 ∈N sn 1 ti t i1 i=1
Analogously ∞ 1 x sn ∈ / R\ ⇔ ∃ x = 2. i ∈N sn 2 t t i2 2 i i=1
1 ∞ 1 x / R\ ∞ ∈ / R\ Consequently, if snx ∈ and then i=1 ti i=1 ti sn2 1 N. As that At ∩ As is finite, it follows that the set
∞ 1 x n∈N: ∈ / R\ sn ti i=1
sn1 sn2
=
ti1 ti2
for some i1 , i2 ∈
1 A . . Therefore A ∈ Tps is finite. Hence x ∈ lim inf sn R\ ∞ i=1 ti n→∞
A A Corollary 12. Let s,t ∈ S. If At ∩ As is the finite set then the topologies Tpt , Tps are incomparable.
The converse statement is not true. Theorem 13. There exist sequences t, s ∈ S such that At ∩ As is an infinite set and A A the topologies Tpt , Tps are incomparable. Proof. Let as fix t, s ∈ S such that At ∩ As is an infinite set and the sequence s contains s ∈ S and the sequence t contains t ∈ S suchthat sets At ∩ As , At ∩ As ∞ 1 ∞ 1 A / are finite. Then by Theorem 11 we conclude that R\ i=1 t ∈ Tps and R\ i=1 t ∈ i i 1 1 A A A A A Tpt ∈ Tpt ∈ / Tps . Also R\ ∞ and R\ ∞ . Hence the topologies Tpt , Tps i=1 si i=1 si are incomparable. Theorem 14. For every sequence s ∈ S we have A . Tnat Tps 5
A Proof. Let s ∈ S. Obviously Tnat ⊂ Tps (see [4]). There exists a sequence t ∈ S such ∞ 1 A that At ∩ As = ∅. Let A = R\ i=1 ti . Obviously, A ∈ . / Tnat and A ∈ Tps A Theorem 15. Let T = s∈S Tps . Then T = Tnat .
Proof. The family T is a topology on the real line, as the intersection of the topology. Obviously, T contains natural topology. We now prove that T ⊂ Tnat . Let A ∈ T and x ∈ A. Without loss of generality, we can assume that x = 0. We prove that there exists δ > 0 such that (−δ, δ) ⊂ A. On the contrary, suppose that 1 1 1 ∀ ∃ ∈ − , \A. (2) k∈N tk ∈R tk k k Hence there exists a subsequence of natural numbers {kn }n∈N such that tkn > 0 for all n ∈ N or tkn < 0 for all n ∈ N and sequence {|tkn |}n∈N is unbounded and nondecreasing. Set A sn = |tkn |. Then, s ∈ S and A ∈ / Tps . Indeed, if tkn > 0 then 1 ∈ / lim inf sn A and if tkn < 0 n→∞
then (−1) ∈ / lim inf sn A, which is impossible. Therefore, 0 belongs to the set A with some n→∞ open neighborhood. A Property 16. There does not exist the largest topology Tps ∗ in the sense of inclusion such A A that Tps ⊂ Tps∗ for any s ∈ S.
Proof. We can see that for every sequence s ∈ S there exists a sequence t ∈ S such that A A As ∩ At = ∅. By Theorem 13 topologies Tps and Tpt are incomparable, which completes the proof. Lemma 17. Let s ∈ S. Then A (i) if s is a sequence of rational numbers then the set R \ Q ∈ Tps , A , (ii) there exists a sequence of irrational numbers t ∈ S such that R \ Q ∈ / Tpt A . (iii) there exists a countable dense set C such that R \ C ∈ Tps
Proof. The proof of the condition (i) is similar to Proposition 3.2 in [4]. Nevertheless, we include the proof for the completeness of the exposition and the convenience of the readers. Let {sn }n∈N ⊂ Q and y ∈ R\Q. Recall that y ∈ Φps (R\Q) is equivalent to the condition that for all x ∈ [−1, 1] there exists k ∈ N such that sxn + y ∈ R\Q for all n ≥ k. Fix x ∈ [−1, 1]. There exists at most one sn such that x ∈ / sn ((R\Q) − y). On the contrary, / sn1 ((R\Q) − y) suppose that there exist two different numbers sn1 , sn2 such that x ∈ and x ∈ / sn2 ((R\Q) − y). Hence x p1 x p2 +y = , + y = , where p1 , p2 ∈ Z q1 , q2 ∈ N. sn 1 q1 sn 2 q2 From the above equalities, we thus obtain y = y ∈ R\Q. Finally, we get R\Q ⊂ Φps (R\Q). 6
p2 p −sn1 q1 q2 1 sn2 −sn1
sn2
∈ Q, which is impossible, since
Now we prove condition (ii). Let y0 be a positive irrational number and {yn }n∈N ⊂ Q be a decreasing sequence convergent to y0 such that yn > y0 for all n ∈ N. Define tn =
1 for all n ∈ N. yn − y0
Obviously, t ∈ S and 1 ∈ / lim inf tn (R\Q − y0 ). Hence y0 ∈ / Φpt (R\Q). This implies that n→∞
A R\Q ∈ / Tpt . Finally we shall prove condition (iii). Let s ∈ S. Let A be the set of all elements of sequence s and B be given by formula
k i=1 ai , ai , bj , cj ∈ A, bj = cj , 1 ≤ i ≤ k, 1 ≤ j ≤ l, k, l ∈ N . B = x ∈ R : x = l (b − c ) j j j=1
Let C = SpanB, where SpanB is a linear space over Q generated by elements from a set A B. Obviously the set C is dense and countable. Now we prove that R \ C ∈ Tps . Let y ∈ R \ C. Recall that y ∈ Φps (R \ C) is equivalent to the condition that for all x ∈ [−1, 1] there exists k ∈ N such that sxn + y ∈ R \ C for all n ≥ k. Fix x ∈ [−1, 1]. There exists at most one sn such that sxn + y ∈ C. On the contrary, suppose that there exist two different numbers sn1 , sn2 such that x + y = c1 sn 1 Hence y=
∧
x + y = c2 for some c1 , c2 ∈ C. sn 2
s n2 sn1 c2 − c1 . sn 2 − sn 1 sn 2 − s n 1
Thus we obtain y ∈ C, which is impossible. 4. The separation axioms A Let s ∈ S and (R, Tps ) be a topological space assuming always that A contain Tnat A A and all co-countable sets. Since Tnat ⊂ Tps , we get that (R, Tps ) is a Hausdorff space. A Theorem 18. Let s ∈ S and A ⊂ Ba. Then the topological space (R, Tps ) is not regular.
Proof. Recall that the family TIs is a topology such that Tnat ⊂ TIs (cf. [6]). By Theorem 6 in [7] it follows that every Tnat -dense set can not be separated from any point by TIs -open A sets. Since Tps ⊂ TIs and the set C described in proof of Lemma 17 (iii) is Tnat -dense, A Tps -closed and countable we get that C and any point x ∈ / C can not be separated by A A -open sets. So that (R, Tps ) is not regular. Tps A Now we shall prove that for every s ∈ S the space (R, Tps ), where A ⊂ L is not regular. It is the answer to the question possed in [4]. The proof will be preceded by the following Lemmas. 7
Lemma 19. Let s ∈ S, A ⊂ R and J be an invariant σ-ideal such that J ∩ Tnat = {∅} . A If (a, b) \ A ∈ J , then the closure of the set A with respect to Tps contains (a, b). A (A). Suppose that B is not empty set. The set B is obviously Proof. Define B = (a, b)\clTps
A . Hence B = Φps (B). Since B ∈ J , we get from Property 7 in open with respect to Tps Section 2 that Φps (B) = ∅ and finally, B = ∅. This completes the proof.
Lemma 20. Let s ∈ S. If A ∈ Ba is a set of the second category, then the closure of A A with respect to Tps contains an interval. Proof. Let A be a set of the second category with the Baire property. Obviously there exists an interval (a, b) such that the set A is residual on (a, b). By Lemma 19, for J = I we A (A). obtain that (a, b) ⊂ clTps A Lemma 21. Let s ∈ S, A ⊂ L and A ∈ Tps . If A ∩ I = ∅ for every open interval I = ∅, A is equal to R. then the closure of the set A ⊂ R with respect to topology Tps A (A) and suppose that the set B is not empty. Since A, B are Proof. Define B = R \ clTps
A A topology, Tps ⊂ Ts then we get A, B ∈ Ts . Observe open set with respect to the Tps that Φs (A) ∩ Φs (B) = Φs (A ∩ B) = Φs (∅) = ∅.
Thus Φs (B) ⊂ R \ Φs (A) ⊂ R \ A. We divide the proof into two steps. 1. Firstly, assume that Φs (B) is the set of the second category. Since Φs (B) is a Borel set, this set is residual on some interval (a, b) (Proposition 1, p.188 in [2]). Thus A is the set of the first category on (a, b), then ∅ = Φps ((a, b) ∩ A) = Φps ((a, b)) ∩ Φps (A) ⊃ (a, b) ∩ Φps (A) = (a, b) ∩ A. It implies that the set A and (a, b) are disjoint. This contradicts our assumption. A 2. Here, we assume that Φs (B) is the set of the first category. Since B ∈ Tps we get that B ⊂ Φps (B). Hence B is the first category and thus Φps (B) = ∅. It implies that B = ∅.This contradiction ends the proof.
A Lemma 22. Let s ∈ S and A ⊂ L. If V ∈ Tps and V = ∅, then the closure of V with A respect to Tps contains an interval. A is nonempty and for every open interval I we have Proof. Let us assume that V ∈ Tps A (V ). It means that set R \ C meets the assumptions of that I \ C = ∅, where C = clTps A (R \ C) = R. It is a contradiction with the fact that V ∩ (R \ C) = ∅ Lemma 21, so that clTps and the proof is completed. 8
A Theorem 23. Let s ∈ S and A ⊂ L. The topological space (R, Tps ) is not regular.
Proof. Recall that a topological space (X, T ) is regular if and only if for every set U ∈ T and every point x ∈ U , there exists a neighbourhood V ∈ T such that x ∈ V ⊂ clT (V ) ⊂ U (see Theorem 1.5.5 in [1]). Let U = R \ C be the set defined in Lemma 17. Obviously, U is A A not empty and U ∈ Tps . Let V ∈ Tps be a non empty subset of U . From Lemma 22 we A (V ) contains some interval. Therefore clT A (V ) \ U = ∅. In this way we obtain that clTps ps A ) is not regular. have proved that the space (R, Tps
Theorem 23 is the answer to the question formulated in [4], if the topological space L (R, Tps ) is regular. Now, we are solving Problem 8 formulated in [5]: Ba L
= Tps , where s = {n}n∈N ? Is it true that Tps Lemma 24. Let H be a Hamel base and s = {n}n∈N . Then, every point x ∈ R \ H is a ps-density point of set R \ H. Proof. Let x ∈ R. Recall that x ∈ Φp (R \ H) is equivalent to the condition that for all a ∈ [−1, 1] there exists k ∈ N such that na + x ∈ / H for all n ≥ k. Fix a ∈ [−1, 1]. We prove that there exist at most two different number sn1 , sn2 , such that na1 + x ∈ H and na2 + x ∈ H. On the contrary, suppose that there exist three different numbers sn1 , sn2 , sn3 such that h1 =
a a a + x ∈ H and h2 = + x ∈ H and h3 + x ∈ H. n1 n2 n3
From the above equalities, we obtain that x=
n 1 h1 − n 2 h2 n 1 h1 − n 3 h3 and x = . n1 − n2 n1 − n3
Therefore, h1 n1 (n3 − n2 ) + h2 n2 (n1 − n3 )h2 + h3 n3 (n2 − n1 ) = 0.
(3)
which is a contradiction with the assumption that h1 , h2 , h3 are linearly independent. Finally, we get x ∈ Φp (R \ H). Corollary 25. Let H be a Hamel base and s = {n}n∈N . Then the set R \ H is open set with respect to topology Tps = {A ∈ R : A ⊂ Φp (A)}. Note that Lemma 24 is true for any s ⊂ Q. From Lemma 24 and Corollary 25 we obtain the following theorem. L Ba Theorem 26. The topology Tps and Tps for s ⊂ Q are not comparable.
Proof. Recall that there exist a Hamel base H1 of measure zero without the Baire property and a Hamel base H2 of the first category and Lebesgue nonmeasurable (see Corollary 11.4.3 in [9]). Then L Ba R \ H1 ∈ Tps and R \ H1 ∈ / Tps 9
and
Ba L R \ H2 ∈ Tps and R \ H2 ∈ / Tps , L Ba Ba L \ Tps
= ∅ and Tps \ Tps
= ∅. Tps
Hence
L Ba
= Tps . Tps
Bibliografia References [1] Engelking R., General topology, Polish Scientific Publishers, Warsaw (2007). [2] Filipczak M., Hejduk J., On topologies associated with the Lebesgue measure, Tatra Mt. Math. Publ. 28(2), (2004) pp. 187-197. [3] Goffman C., Neugebauer C. J., Nishura T., The density topology and approximate continuity, Duke Math. J. 28,(1961), pp. 497–506. [4] Górajska M., Pointwise density topology, Open Mathematics, vol. 13, no. 1, (2015), pp. 75—82. [5] Górajska M., Hejduk J., Pointwise density topology with respect to admissible σ-algebras, Tatra Mt. Math. Publ. 55(1), (2013), pp. 77–83. [6] Hejduk J., Horbaczewska G., On I-density topologies with respect to a fixed sequence, Reports on Real Analysis, Conference at Rowy, (2003), pp. 78—85. [7] Hejduk J., On the regularity of topologies in the family of sets having the Baire property, Filomat 27(7), (2013), pp. 1291-1295. [8] Horbaczewska G., The family of I-density type topologies, Comment. Math. Univ. Carolinae 46(4), (2005), pp. 735–745. [9] Kuczma M., An introduction to the theory of functional equations and inequalities: Cauchy’s equation and Jensen’s inequality, Springer Science and Business Media (2009). [10] Oxtoby J. C., Measure and category, Springer Verlag, New York-Heidelberg-Berlin (1980). [11] Poreda W., Wagner-Bojakowska E., Wilczyński W., A category analogue of the density topology, Fund. Math. 125(2), (1985), pp.167–173.
Conflicts of Interest The authors declare no conflict of interest. This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.
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