The family of I -density type topologies

The family of I -density type topologies

Gra˙zyna Horbaczewska

Abstract. We investigate a family of topologies introduced similarly as the I-density topology. In particular, we compare these topologies with respect to inclusion and we look for conditions under which these topologies are identical.

Keywords: I-density point, family of topologies Classification: 54A10

We use here a standard notation. LetN be the set of all positive integers,B the family of subsets of the real line having the Baire property andI theσ-ideal of meager sets. For every setA andx, t∈R, we setA+x={a+x;a∈A} and t·A = {t·a;a ∈ A}, where χ_A is the characteristic function of A and A^′ the complement ofA.

LetS be the family of all nondecreasing and unbounded sequences of positive real numbers. Every sequence{sn}n∈N∈S is denoted byhsi.

Let us recall the notion of an I-density point of a set A∈ B([PWW1]). The point 0 is an I-density point of a set A ∈ B if for every sequence {tn}n∈N∈ S there exists a subsequence {tnp}p∈N such that χ_{(tnp·A)∩[−1,1]p→∞}−→1 I-a.e. on [−1,1].

Based on the observation that starting from another fixed sequence different results can be obtained, the notion of anI-density point connected with a fixed sequence from the familyS has been introduced in [HH].

Definition 1. Lethsi ∈S. The point 0 is anhsi-I-density point of a setA∈ B if for every subsequence{snm}m∈N⊂ hsithere exists a subsequence{sn_mp}p∈N

such thatχ_{(snmp·A)∩[−1,1]p→∞}−→ 1 I-a.e. on [−1,1].

A point x∈R is anhsi-I-density point of A if 0 is an hsi-I-density point of the setA−x.

A pointx∈R is anhsi-I-dispersion point of A ifxis anhsi-I-density point ofA^′.

We can define one-sidedhsi-I-density points in the natural way.

For anyhsi ∈S andA∈ B, putting

Φ_hsiI(A) ={x∈R;x is anhsi-I-density point ofA}

we get that Φ_hsiI:B → B is a lower density operator (see [HH]).

Applying this operator we define for every fixed sequence hsi the topology ThsiI ={A∈ B;A⊂Φ_hsiI(A)}, which fulfils the inclusion: TI ⊂ ThsiI, whereTI

denotes theI-density topology ([HH]).

The main aim of this paper is to compare topologies connected with different sequences.

First of all, if hsi is the sequence of all natural numbers then ThsiI = TI

([PWW1]).

Now we state the main results.

LetS₀={hsi ∈ S: lim inf_n→∞s^sn

n+1 = 0}.

Theorem 1. Lethsi ∈S. ThenThsiI =TI if and only if hsi ∈S\S₀. Theorem 2. Lethsi,hti ∈ S₀ and lim_m→∞^s_t^m

m =α ∈(0,+∞). Then ThsiI = ThtiI if and only if α= 1.

Before presenting the proofs we need some properties of our topologies.

Properties.

(1) Lethsi,hti ∈ S. ThenThsiI =ThtiI if and only if Φ_hsiI(A) = Φ_htiI(A)for everyA∈ B.

(2) Lethsi ∈ S and 1≤α <∞. ThenThsiI ⊂ ThαsiI, wherehαsi={αsn}n∈N. (3) Lethsi ∈ S. Then for an arbitrary subsequencehs^′i ⊂ hsiwe have

ThsiI ⊂ Ths^′iI.

(4) Lethsi ∈ S. If for any subsequence of the sequence of all natural numbers hn^′i ⊂ {n}n∈Nthere exists a subsequencehn^′′i ⊂ hn^′isuch that

ThsiI ⊂ Thn^′′iI, thenThsiI⊂ TI.

(5) ∀ hsi ∈S ∀x∈R ∀A∈ B (A∈ ThsiI =⇒ A+x∈ ThsiI).

(6) ∀ hsi ∈S ∀A∈ B (A∈ ThsiI =⇒ −A∈ ThsiI).

(7) ∀ hsi ∈S ∀ |m| ≥1 ∀A∈ B (A∈ ThsiI =⇒ m·A∈ ThsiI).

(8) ∀ hsi ∈S₀ ∃A∈ B ∀ |m|<1 (A∈ ThsiI∧m·A /∈ ThsiI).

The first four are simple consequences of the definitions and properties of lower densities. We want only to show one implication from (1) (the inverse is obvious).

Proof of (1): Lethsi,hti ∈S. We assume that ThsiI =ThtiI and there exists a setA∈ Bsuch that Φ_hsiI(A)6= Φ_htiI(A), for example Φ_hsiI(A)*Φ_htiI(A). Since Φ_htiI(A)∈ ThtiI=ThsiI, by definition ofThsiIwe have Φ_htiI(A)⊂Φ_hsiI(Φ_htiI(A)) which is equal to Φ_hsiI(A) because Φ_htiI(A) is equivalent to A (the Lebesgue Density Theorem works here), so we get a contradiction.

The next four properties have been already published ([HH], [H]). A justifica- tion of (5)–(7) is again easy so we can omit it. We want only to sketch the proof of the last one.

Proof of (8): Lethsi ∈S₀. Then there exists a subsequence{sn_k}k∈N ofhsi such that lim_k→∞s^snk

nk+1 = 0.

Put X = S∞ j=1[_s ¹

nj+1,√_s ¹

nj·s_nj+1]. Then 0 is an hsi-I-dispersion point of a setX. DefiningY =−X∪X we haveA={0} ∪(R\Y)∈ ThsiI.

Form= 0 it is obvious thatm·A /∈ ThsiI.

Now we want to show that 0 is not a righthsi-I-dispersion point of the setm·X form∈(−1,1)\ {0}. There is no loss of generality in assuming thatm∈(0,1).

We can findk₀ ∈Nsuch that for anyk > k₀we haveq _s

s_nknk₊₁ < m. Then 0 is not a righthsi-I-dispersion point of the set m·S_∞

j=k0[_s ¹

nj+1,√_s ¹

nj·s_nj+1], so neither of the setm·X. Hence m·A={0} ∪(R\m·Y)∈ T/ hsiI.

For details see [HH].

Proof of Theorem 1: Sufficiency. SinceTI⊂ ThsiI for every sequencehsi ∈ S, it is enough to show the inclusion: ThsiI ⊂ TI.

Lethsi ∈S\S0. We denote lim inf_k→∞sk+1^sk byλ, soλ >0.

Lethn^′i={n_j}j∈Ndenote an arbitrary sequence of natural numbers,hn^′i ∈S.

Then there exists j₀ ∈N such that for each j ≥j₀, j ∈ N, there exists k_j ∈N which fulfils the condition s_kj ≤n_j ≤ s_kj+1. There is no loss of generality in assuming thatj₀= 1. Now we choose a subsequence{nj_l}l∈Nfrom the sequence {nj}j∈N such that each interval [s_kjl, s_kjl+1] contains only one term of the se- quence{njl}l∈N. Sincesk_jl ≤njl≤skj+1 for eachl∈N, we have

1≤ nj_l

s_kjl ≤ s_kjl+1 s_kjl

and

1≤lim sup

l→∞

n_jl

s_kjl ≤lim sup

l→∞

sk_jl+1

s_kjl

= 1/lim inf

l→∞

s_kjl

sk_jl+1 ≤1/lim inf

k→∞

s_k s_k+1 = 1

λ <+∞. Therefore there exists a subsequence{sⁿ_kj^jlp

lp

}p∈N⊂ {sⁿ_kj^jl

l

}l∈Ntending toα, where 1≤α <∞. Then lim_{p→+∞α·s}^njlp

kjlp

= 1. Using the notation:

hn^′′i={n_jlp}p∈N and hs^′′i={s_k

jlp}p∈N

we obtain (by Theorem 2, which will be proved later) the equality of topologies Thn^′′iI=Thαs^′′iI.

Furthermore, by Properties (2) and (3), we have

ThsiI ⊂ Ths^′iI ⊂ Ths^′′iI⊂ Thαs^′iI=Thn^′′iI.

Property (4) now yieldsThsiI⊂ TI which is the desired conclusion.

Necessity of the conditionhsi ∈ S\S0 has been already stated in [HH]. We repeat here the proof. We want to show that ifhsi ∈S₀ thenThsiI*TI.

From our assumption there exists a subsequence {snk}k∈N ⊂ {sn}n∈N such that lim_{k→∞ s}^snk

nk+1 = 0. We can assume that the sequence {s_nk^snk+1}k∈N is de- creasing (if necessary we can choose a subsequence).

Let

A= [∞ j=1

1

snj+1, 1

√snj·snj+1

.

We will show that 0 is a righthsi-I-dispersion point of the setA, it means that for each subsequence {snm}m∈N ⊂ {sn}n∈N there exists a subsequence {sn_mp}p∈N

such thatχ_{(snmp·A)∩[0,1]p→∞}−→ 0I-a.e. on [0,1]. Letj(l) = min{j∈N:l < nj+1}. We observe that

snm·

[∞ j=1

1

snj+1, 1

√snn·snj+1

∩[0,1]

=

snm· [∞ j=j(nm)

1

s_nj+1, 1

√snj·s_nj+1

∩[0,1]

⊂

snm·

0, 1

psn_j(nm)·sn_j(nm)+1

∩[0,1]

=

0, snm

psn_j(nm)·sn_j(nm)+1

∩[0,1]⊂

0, sn_j(nm)

psn_j(nm)·sn_j(nm)+1

∩[0,1]

=

0,

s sn_j(nm)

sn_j(nm)+1

∩[0,1].

Since lim sup_m 0,

r _s

nj(nm)

s_nj(nm)+1

={0}, we haveχ_{(snm·A)∩[0,1]m→∞}−→ 0 I-a.e. on [0,1], so 0 is anhsi-I-dispersion point of the setAe=−A∪A.

Let B = (0,_s¹

n1)\A and Be = −B∪B ∪ {0}. Then Be ∈ ThsiI. Of course B=S∞

j=1 √_s 1

nj·s_nj+1,_s¹

nj

.

We will show that 0 is not a rightI-density point of the setB, it means that there exists a sequence{tk}k∈N∈S such that for each subsequence{tkp}p∈N ⊂ {t_k}k∈N, the convergenceχ_(t

kp·B)∩[0,1]_p→∞−→ 1 I a.e. does not hold. Let t_k=√sn_k·s_nk+1 fork∈N. Observe that

(t_k·B)∩[0,1] =

√snk·snk+1· [∞ j=1

1

√snj·snj+1, 1 snj

∩[0,1]

=

√sn_k·s_nk+1· [∞ j=k+1

1

√snj·snj+1

, 1 snj

∩[0,1]

⊂

√snk·sn_k+1·

0, 1 sn_k+1

∩[0,1]⊂

0,

√sn_k·s_nk+1 sn_k+1

∩[0,1]

=

0, r snk

snk+1

∩[0,1].

Since lim sup_k[0,q _snk

s_nk+1] = {0}, we have χ_{tk·B∩[0,1]}(x) −→

k→∞ 0 for x ∈ (0,1].

ThereforeB /e∈ TI.

Corollary 1. For every sequencehsi ∈S\S₀ and for every sequence hti ∈S₀, ThsiI(ThtiI.

Now we can add one more property.

Corollary 2. For every sequence hsi ∈ S\S₀ and for every m ∈ R\ {0}, if A∈ ThsiI thenm·A∈ ThsiI.

For the proof of Theorem 2 we need two lemmas.

Lemma 1 ([PWW2]). Let A be an open set and let the sequences {in}n∈N

and {jn}n∈N have the following properties: in > 0, jn > 0 for each n ∈ N, lim_n→∞in= +∞,lim_n→∞jn= +∞, lim_n→∞^j_iⁿ

n = 1and let

χ_{(in·A)∩[−1,1]n→∞}−→ 0 I-a.e. on[−1,1]. Then also χ_{(jn·A)∩[−1,1] n→∞}−→ 0 I-a.e. on [−1,1].

In Lemma 2 we state an equivalent condition for being anhsi-I-dispersion point of an open set. The idea was motivated by [ L].

Lemma 2. Let hsi ∈ S. The point 0 is a right-handhsi-I-dispersion point of an open set G if and only if, for every natural numbern, there exist a natural

number k and a real numberδ >0 such that for each m∈Nsuch that _s¹

m < δ and for eachi∈ {1, . . . , n}, there exists a natural numberj∈ {1, . . . , k}such that

G∩

i−1

n +j−1 nk

· 1 sm,

i−1

n + j

nk

· 1 sm

=∅.

Proof: We shall first prove the necessity forhsi-I-dispersion. Assume that 0 is a right-handhsi-I-dispersion point of the open setGand suppose the assertion of the lemma is false. Then we could find a natural numbern₀ such that, for each k∈N and δ_k = ¹_k, there exist m_k ∈N such thatk < smk andi_k ∈ {1, . . . , n₀} such that, for eachj∈ {1, . . . , k}

G∩

i_k−1

n₀ +j−1 n₀k

· 1 sm_k,

i_k−1 n₀ + j

n₀k

· 1 sm_k

6

=∅.

Since i_k is chosen from a finite set, there exists a subsequence {sm_kl}l∈N ⊂ {smk}k∈Nsuch that the numberi_kl is common for alll. For simplicity we denote it byi₀ and the chosen subsequence by {smk}k∈N. Let {sm_kz}z∈Nbe any subse- quence of{smk}k∈N. For every natural numberp∈Nthe setS_∞

z=p((sm_kz ·G)∩ (ⁱ⁰_n⁻¹₀ ,_nⁱ⁰₀)) is open and dense on [ⁱ⁰_n⁻¹₀ ,_nⁱ⁰₀], so

\∞ p=1

[∞ z=p

sm_kz ·G

∩

i₀−1 n₀ , i₀

n₀

is residual on [ⁱ⁰_n⁻¹₀ ,_nⁱ⁰₀]. Consequently

lim sup

z sm_kz ·G

∩[−1,1]

⊃

\∞ p=1

[∞ z=p

sm_kz ·G

∩

i₀−1 n₀ , i₀

n₀

∈/I.

Hence there exists a sequence{smk}k∈Nsuch that for each subsequence

{sm_kz}z∈N⊂ {smk}k∈N, lim sup_z((sm_kz·G)∩[−1,1]) is a not a meager set. This contradicts our assumption that 0 is anhsi-I-dispersion point ofG.

Now assume that the condition from our lemma is true and our goal is to show that 0 is a right-handhsi-I-dispersion point ofG.

Let{smp}p∈Nbe an arbitrary subsequence ofhsi. The subsequence of {smp}p∈N will be defined by induction. Forn= 1 there exist k1 ∈Nandδ1 >0 such that for each m ∈ N for which _s¹

m < δ₁ and for i = 1 there exists j = j(sm,1)∈ {1, . . . , k₁}such that

G∩ j−1

k₁ · 1 sm

, j k₁ · 1

sm

=∅.

Let{sm_α1(z)}z∈N be a subsequence of{smp}p∈N such that for eachz∈Nwe have _s ¹

mα1(z) < δ₁ and the numberj(sm_{α1 (z)},1) =j₁₁ is common for all z ∈N.

Putsmp1 =sm_α1(1).

Assume the sequence{sm_αn−1(z)}z∈N and sm_pn−1 =sm_{αn−1 (1)} to be defined.

For a natural numbern there existkn andδn>0 such that for eachm∈Nfor which _s¹

m < δn and for i∈ {1. . . n} there existsj =j(sm, i)∈ {1, . . . , kn} such that

G∩

i−1

n + j−1 n·kn

· 1 sm,

i−1

n + j

n·kn

· 1 sm

=∅.

Let{sm_αn(z)}z∈Nbe a subsequence of{sm_αn−1(z)}z∈Nsuch that for eachz∈N we have _s ¹

mαn(z) < δn and j = (sm_αn(z),1) = j_n1,..., j(sm_αn(z), n) = jnn are common for allz∈N. Putsm_pn =sm_αn(1). We proceed by induction.

The task is now to show that{x:χ_{(smpn·G)∩[0,1]}90} ∈I. Let (a, b)⊂[0,1].

Then there exist a natural numbern₀ andi₀∈ {1, . . . , n₀}such that [ⁱ⁰_n⁻¹₀ ,_nⁱ⁰₀]⊂ (a, b).

We shall consider a sequence{sm_αn

0 (z)}z∈Nand a natural number kn0 corre- sponding ton₀. Then for each n≥n₀ sm_pn ∈ {sm_αn

0 (z)}z∈N. Hence for each n≥n₀ there existsj=jn0i₀ such that

G∩

i0−1 n₀ , j−1

n₀kn0

· 1 sm_pn,

i0−1 n₀ + j

n₀kn0

· 1 sm_pn

=∅. Let

(c, d) =

i₀−1

n₀ + j−1 n₀kn0

,i₀−1 n₀ + j

n₀kn0

.

Then (c, d)⊂(a, b) and for eachn≥n₀ we have

∅=G∩

c· 1

sm_pn, d· 1 sm_pn

= 1

sm_pn sm_pn ·G

∩(c, d) , so

(c, d)⊂[0,1]\ sm_pn ·G

∩[0,1]

. Therefore

(c, d)⊂ [∞ n=1

\∞ n=r

[0,1]\ sm_pr·G

∩[0,1]

and lim sup_r((sm_pr ·G)∩[0,1]) is nowhere dense. Thus χ_{(smpr·G)∩[0,1]r→∞}−→ 0 I a.e.

which completes the proof.

Proof of Theorem 2: Let hsi,hti ∈ S and limm→∞ sm

tm = 1. Then using Lemma 1 we get immediately the equality of topologies.

Now, let hsi,hti ∈ S₀ and lim_m→∞^s_t^m

m =α ∈(0,+∞). Let us suppose that 0< α <1. We can assume that ^s_t^m

m > ¹₂αfor allm∈N. We want to show that ThsiI6=ThtiI.

From the proof of Property (8) it follows that there exists a setY, which is a countable sum of closed intervals, such that {0} ∪(R\Y)∈ ThtiI and 0 is not a hti-I-density point of the setR\αY, which is equivalent to the fact that 0 is not anhαti-I-dispersion point of the setY, so neither of the set G= intY since Y \intY ∈I.

It suffices to show that 0 is not anhsi-I-dispersion point of the setG, because it means that 0 is not anhsi-I-dispersion point ofY, so{0} ∪(R\Y)∈ T/ hsiI.

For convenience we restrict our consideration to the right-hand case and sup- pose, contrary to our claim, that 0 is a right-handhsi-I-dispersion point of the open setG. By Lemma 2 we know that

(∗) for every natural numbernthere exist a natural numberkand a real number δ >0 such that for every naturalmsatisfying _s¹_m < δ and for each

i∈ {1, . . . , n} there exists a natural numberj ∈ {1, . . . , k} such that

G∩

i−1n +^j−1_nk

· s¹m,

i−1n +_nk^j

·s¹m

=∅. We shall show that

for every natural number N there exist a natural number K and a real number ∆>0 such that for every naturalmsatisfying the inequality

αt1m < ∆ and for each ˜i ∈ {1, . . . , N} there exists a natural number ˜j ∈ {1, . . . , K}such thatY ∩_˜

i−1N +^˜j−1_{N K}

·αt¹m, _˜

i−1N +_{N K}^˜j

·αt¹m

=∅.

Consider an arbitrary natural numberN. Applying (∗) for n=N we choose k∈N and δ >0 satisfying (∗). Since, by assumption, _αt^sn_n tends to 1, it follows that

(∗∗) for everyǫ >0 there exists a natural numbernǫ such that for everyn > nǫ

we have an inequality|^snαt^−αtnⁿ|< ǫ.

SetK= 3kand we fix ∆>0 such that (1) ∆< ^δ₂

and

(2) for everym∈N, if _s¹

m <2∆ thenm > nǫ, whereǫ=_{2N K}¹ . Therefore for every m ∈ N such that _αt¹

m < ∆ we have _s¹

m < 2∆ < δ (since

αt1m >_2s¹

m), so by (2) and (∗∗) the following inequality holds:

sm−αtm

αtm

< 1

2N K .

Fix an arbitrary ˜i ∈ {1, . . . , N}. From (∗) for i = ˜i there exists a natural numberj ∈ {1, . . . , k}such that

Y ∩

i−1

n +j−1 nk

· 1 sm,

i−1

n + j

nk

· 1 sm

=∅.

To obtain a contradiction, suppose that for every ˜j ∈ {1, . . . , K}the setY has common points with the interval ((ⁱ⁻¹_N +^˜j−1_{N K})· αt¹m,(ⁱ⁻¹_N +_{N K}^˜j )·αt¹m), so for every ˜j∈ {1, . . . , K}there existsy∈Gsuch thaty ∈(ⁱ⁻¹_n +^˜j−1_3nk,ⁱ⁻¹_n +_3nk^˜j )·αt¹m, it meansy∈(0,_αt¹_m) andy·αtm ∈(ⁱ⁻¹_n +^˜j−1_3nk,ⁱ⁻¹_n +_3nk^˜j ). From (∗) we see that there exists a numberj∈ {1, . . . , n}such that for anyy∈Y the pointy·smdoes not belong to the interval (ⁱ⁻¹_n +^j−1_nk ,ⁱ⁻¹_n +_nk^j ). But for ˜j= 3j−1 there exists a point y ∈ Y such that y·αtm ∈ (ⁱ⁻¹_n +^3j−2_3nk ,ⁱ⁻¹_n +^3j−1_3nk ). Simultaneously

|y·αtm−y·sm| = |y·(αtm−sm)| < _αt¹_m|αtm−sm| < _{2N K}¹ = _6nk¹ , hence y·sm ∈ {ⁱ⁻¹_n + ^3j−3_3nk ,ⁱ⁻¹_n + _3nk^3j ) = (ⁱ⁻¹_n + ^j−1_nk ,ⁱ⁻¹_n +_nk^j ). This contradiction

completes the proof.

By Theorem 1 it is obvious that for sequences belonging toS\S₀ we can have the same topology even if the sequences considered do not satisfy the condition limn→∞ sn

tn = 1.

The following theorems show more properties of the family ofI-density type topologies.

Theorem 3. For every sequence hti ∈S₀ there exists a sequencehsi ∈S₀ such thatThsiI(ThtiI.

Proof: Let hti ∈S₀. Then set α∈ (0,1) and let hsi = hαti. Then hsi ∈ S₀ and limn→∞sn

tn = α6= 1, so by Theorem 2 ThtiI 6= ThsiI and by Property (2)

ThsiI⊆ ThtiI.

Theorem 4. For every sequence hti ∈ S there exists a sequence hsi ∈ S such thatThtiI(ThsiI.

Proof: If hti ∈ S\S₀ then ThtiI =TI and it is sufficient to take an arbitrary sequence hsi ∈ S₀. Let us assume that hti ∈ S₀. We define hsi = hαti, where α∈R and α >1. Then by Property (2), ThtiI ⊂ ThsiI and from Theorem 2 it

follows thatThtiI6=ThsiI.

Theorem 5. There exist sequenceshsi,hti ∈S₀ such thatThsiI\ ThtiI 6=∅ and ThtiI\ ThsiI6=∅.

Proof: Lethsi={(2n−1)!}n∈N,hti={(2n)!}n∈N. Of coursehsi,hti ∈S₀. Set Y₁ = S_∞

k=1(_(2k)!¹ ,_(2k−1)!¹ ), Y₂ =S_∞

k=2(_(2k−1)!¹ ,_(2k−2)!¹ ). We have Y₁ ∩Y₂ = ∅ and [0,1]\(Y₁∪Y₂)∈I. Moreover

(tn·Y₁)∩[0,1] =

(2n)!· [∞ k=1

1

(2k)!, 1 (2k−1)!

∩[0,1]

=

(2n)!· [∞ k=n+1

1

(2k)!, 1 (2k−1)!

∩[0,1]

⊂

(2n)!·

0, 1

(2n+ 1)!

∩[0,1]

=

0, (2n)!

(2n+ 1)!

∩[0,1] =

0, 1 2n+ 1

and, of course, for any subsequence{tnp}p∈N⊂ hti, (tnp·Y1)∩[0,1]⊂[0,_2np¹₊₁).

It follows that lim sup_p(tnp·Y1)∩[0,1] = {0} ∈I, hence 0 is a right-hand hti- I-dispersion point of Y₁, which gives that it is a right-hand hti-I-density point ofY₂. FinallyZ₂= (−Y₂)∪ {0} ∪Y₂∈ ThtiI.

In the same manner we can see that (sn·Y₂)∩[0,1]⊂[0,_2n¹ ) and conclude that Z₁= (−Y₁)∪{0}∪Y₁∈ ThsiI. We thus getZ₁∈ ThsiI\ThtiI andZ₂∈ ThtiI\ThsiI. Theorem 6. LetT^∗ be a topology generated byS

hsi∈SThsiI. Then S

hsi∈SThsiI6=T^∗= 2^R.

Proof: It is immediate thatS

hsi∈SThsiI 6= 2^R because S

hsi∈SThsi ⊂ B. Our proof starts with the observation that if for every x ∈ A, where A ∈ B, there exists a sequencehsi ∈S such thatx∈Φ_hsiI(A) thenA∈ T^∗. Indeed, letA∈ B, x∈Aandhsi ∈S be a sequence such thatx∈Φ_hsiI(A). Since (Φ_I(A)△A)∈I, we havex∈Φ_hsiI(A∩ΦI(A)). SimultaneouslyA∩ΦI(A)∈ TI⊂ ThsiI. Therefore (A∩Φ_I(A))∪ {x} ∈ ThsiI⊂ T^∗ and finallyA=S

x∈A((A∩Φ_I(A))∪ {x})∈ T^∗. We next show that singletons are open in T^∗. LetE =S_∞

n=1(_a¹

n,_b¹

n) where an= (2n+1)!,bn= (2n)! forn∈N. Thenhai,hbi ∈S. We claim that 0 is a right- handhai-I-dispersion point of the setE, because (an·E)∩[0,1]⊂(0,_2n+2¹ ) and henceχ_{(an·E)∩[0,1]n→∞}−→ 0Ia.e. on [0,1] and so does each subsequence. Similarly 0 is a right-handhbi-I-density point of the setE, because (bn·E)∩[0,1]⊃(_2n+1¹ ,1) and henceχ_{(bn·E)∩[0,1]n→∞}−→ 1 I-a.e. on [0,1] and so does each subsequence.

Putting A = E∪ {0} ∪(−E) we obtain 0 ∈ Φ_hbiI(A) and for the set B = S_∞

n=1((_b ¹

n+1,_a¹

n)∪(−a¹n,−b_n+1¹ ))∪ {0} we have 0∈Φ_haiI(B), so by the above A, B ∈ T^∗. Therefore{0} = A∩B ∈ T^∗. Since the topologies considered are invariant under translations, we have{x}= (A+x)∩(B+x)∈ T^∗for anyx∈R,

and finallyT^∗= 2^R.

Theorem 7. Let T = {T_hsiI;hsi ∈ S} = {TI} ∪ {ThsiI;hsi ∈ S₀}. Then card(T) =c.

Proof: Obviously card(T)≤c.

Ifhsi ∈S₀ then for everyα >0 a sequencehαsi ∈S₀. By Theorem 2 for every α, β >0,α6=β we haveThαsiI 6=ThβsiI so card(T)≥c.

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University of L´od´z, Faculty of Mathematics, ul. Banacha 22, PL-90-238 L ´od´z, Poland

E-mail: [email protected]

(Received December 16, 2004,revised July 12, 2005)