@
Appl. Gen. Topol. 20, no. 2 (2019), 363-377

doi:10.4995/agt.2019.11238

c© AGT, UPV, 2019

On ideal sequence covering maps

Sudip Kumar Pal a , Nayan Adhikary b and Upasana Samanta b

a Department of Mathematics, Diamond Harbour Women’s University, WB-743368, India

(sudipkmpal@yahoo.co.in)
b Department of Mathematics, Jadavpur University, Kol-32, WB, India (nayanadhikarysh@gmail.com,

samantaupasana@gmail.com)

Communicated by S. Garćıa-Ferreira

Abstract

In this paper we introduce the concept of ideal sequence covering map
which is a generalization of sequence covering map, and investigate
some of its properties. The present article contributes to the problem
of characterization to the certain images of metric spaces which was
posed by Y. Tanaka [22], in more general form. The entire investigation
is performed in the setting of ideal convergence extending the recent
results in [11, 15, 16].

2010 MSC: 54C10; 40A05; 54E35.

Keywords: sequence covering; sequentially quotient; sn-networks; boundary
compact map; ideal convergence.

1. Introduction

In the past 50 years, papers and surveys on discussion for theory of images
of metric spaces were published in large amounts. This issue has become a
typical research direction in developments of general topology, which has made
outstanding contributions for progress and prosperity of this subject. Taking
spaces as images of metric spaces, we deal with structures of metric spaces. In
1971, Swiec [20, 21] introduced the concept of sequence covering maps which is
closely related to the question about compact covering and s-images of metric
spaces (see also [4]). In [11] Lin discuss about sequence covering maps and its
properties also solved many open problem related to this concept.

Received 15 January 2019 – Accepted 11 June 2019

http://dx.doi.org/10.4995/agt.2019.11238


S. K. Pal, N. Adhikary and U. Samanta

The notion of statistical convergence, which is an extension of the idea of usual
convergence, was introduced by Fast [3] and Schoenberg [19] and its topolog-
ical consequences were studied first by Fridy [5] and Salat [6]. It seems more
appropriate to follow the more general approach of [6] where the notion of I-
convergence of sequence was introduced by using the ideas of ideal of the set
of positive integers. In [15, 16] Renukadevi studied some of the results of Lin
in statistical format.
As a natural consequences, in the present paper we continue the investigation
proposed in [11] and study similar problems in more general form. In Sec-
tion 3.1 we introduce the concepts of ideal sequence covering map and ideal
sequence covering compact map and study some of its properties. Also in Sec-
tion 3.2 we propose the concept of ideal sequentially quotient map and examine
related properties. The entire investigation is performed in the setting of ideal
convergence extending the recent results in [11, 14, 15, 16].

2. Preliminaries

We start by recalling the basic definition of ideals and filters. A family
I ⊂ 2Y of subsets of a non- empty set Y is said to be an ideal in Y if A,B ∈I
implies A∪B ∈I and A ∈I and B ⊂ A imply B ∈I. Further an admissible
ideal I of Y satisfies {x} ∈ Y for each x ∈ Y. Such ideals are called free
ideals. If I is a proper non-trivial ideal in Y (i.e. Y /∈ I , I 6= ∅ ), then
the family of sets F(I) = {M ⊂ Y : ∃A ∈ I : M = Y \ A} is called the
filter associated with the ideal I. Ifin = {A ⊂ N : A is finite}. It is an ideal
and Ifin-convergence implies original convergence. Now density of a subset
A of N is d(A) = limn→∞ 1n|{k : k ∈ A,k ≤ n}|, provided the limit exists.
Id = {A : A ⊂ N,d(A) = 0} is an ideal and Id−convergence implies the notion
of statistical convergence (see [7, 12, 17, 26]). Throughout all topological spaces
are assumed to be Hausdorff, all maps are onto and continuous and N stands
for the set of all natural number. Let X be a topological space and P ⊂ X. A
sequence {xn} converging to x in X is eventually in P if {xn : n > k}∪{x}⊂ P
for some k ∈ N; it is frequently in P if {xnk} is eventually in P for some
subsequence {xnk} of {xn}. Throughout by a space we will mean a topological
space, unless otherwise mentioned. Let us recall some definitions.

Definition 2.1 ([11]). Let X be a space and P ⊂ X.
(a) Let x ∈ P. Then P is called a sequential neighbourhood of x in X if

whenever {xn} is a sequence converging to x, then {xn} is eventually
in P.

(b) P is called a sequentially open subset in X if P is a sequential neigh-
bourhood of x in X for each x ∈ P.

Definition 2.2 ([9]). Let X be a space, and let P be a cover of X.
(1) P is a cs-cover of X, if for any convergent sequence S in X, there exists

P ∈P such that S is eventually in P.

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 364



On ideal sequence covering maps

(2) P is an sn-cover of X, if each element of P is a sequential neighbourhood
of some point of x and for each x ∈ X, there exists P ∈P such that P
is the sequential neighbourhood of x.

Definition 2.3 ([20]). A space X is strongly Fréchet if whenever {An|n ∈ N}
is a decreasing sequence of sets in X and x is a point which is in the closure of
each An, then for each n ∈ N there exists an element xn ∈ An such that the
sequence xn → x.

Definition 2.4 ([23]). A space X is said to have property ωD if every countably
infinite discrete subset has an infinite subset A such that there exists a discrete
open family {Ux|x ∈ A} with Ux ∩A = {x} for each x ∈ A.

Definition 2.5 ([1]). A class of mappings is said to be hereditary if whenever
f : X → Y is in the class, then for each subspace H of Y , the restriction of f
to f−1(H) is in the class.

Definition 2.6 ([20]). Let f : X → Y be a mapping.
(a) f is a sequence covering map if for every convergent sequence {yn} in

Y , there is a convergent sequence {xn} in X with each xn ∈ f−1(yn).

(b) f is a 1-sequence covering map if for each y ∈ Y there exists x ∈ f−1(y)
such that whenever {yn} is a sequence converging to y, then there is a
sequence {xn} in X converging to x with each xn ∈ f−1(yn).

Clearly every 1-sequence covering map is a sequence covering map. But the
converse is not true, which is shown by the following example.

Example 2.7. Suppose {yn} is a convergent sequence of real numbers with
its limit y. Let Y = {yn : n ∈ N}∪{y} and

∧
= {α : α : N → N} where α

be an increasing mapping. Define Yα = {(yα(k),α) : k ∈ N}∪{(y,α)} for each
α ∈

∧
. The metric dα on Yα is defined by dα((p,α), (q,α)) = |p−q|. Take X =⊕

α∈
∧ Yα and the metric D on X is defined as D(x1,x2) = min{dα(x1,x2), 1}

when x1,x2 ∈ Yα for some α ∈
∧

and D(x1,x2) = 1 when x1 ∈ Yα,x2 ∈ Yβ for
distinct α,β ∈

∧
. Moreover, the Topology of Y is considered as: each {yn} is

open and a basic open set U containing y is of the form {y}∪{yn : n ≥ n0}
for some n0.
Now define a map f : X → Y by f(yα(k),α) = yα(k) and f(y,α) = y for each
α ∈

∧
. Clearly f is continuous. Any convergent sequence {zn} in Y is of the

form zs(k) = yα(k) for n = s(k),α ∈
∧

and zn = y otherwise. Then choose

rn = (yα(k),α) for n = s(k) and rn = (y,α) otherwise. Clearly rn ∈ f−1(zn)
and rn → (y,α) ∈ f−1(y). So f is a sequence covering map. But for y ∈ Y
choose any (y,α) ∈ f−1(y). Then there is yβ(k) → y, β 6= α but (yβ(k),β) does
not converge to (y,α). So f is not 1-sequence covering map.

Definition 2.8 ([2]). If X is a space that can be mapped onto a metric space
by a one-to-one mapping, then X is said to have weaker metric topology.

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 365



S. K. Pal, N. Adhikary and U. Samanta

Definition 2.9 ([6]). A sequence {xn} in a space X is said to be I−convergent
to x ∈ X (i.e I- limn→∞xn = x ) if for every neighbourhood U of x, {n ∈ N :
xn /∈ U}∈I.

Lemma 2.10 ([25]). Let X be a space with a weaker metric topology. Then
there is a sequence {Pi}i∈N of locally finite open covers of X such that ∩n∈Nst(K,Pi) =
K for each compact subset K of X.

Definition 2.11 ([10]). Let P = ∪{Px : x ∈ X} be a cover of a space X such
that for each x ∈ X, following conditions (a) and (b) are satisfied:

(a) Px is a network at x in X, i.e., x ∈ ∩Px and for each neighbourhood
U of x in X, P ⊂ U for some P ∈Px.

(b) If U,V ∈Px, then W ⊂ U ∩V for some W ∈Px.

(i) P is called a sn-network of X if each element of Px is a sequential
neighbourhood of x for each x ∈ X, where Px is called a sn-network at
x in X. A space Y is termed as an snf-countable if Y has a sn-network
P =

⋃
{Py : y ∈ Y} such that Py is countable and closed under finite

intersections for each y ∈ Y.

(ii) P is called a weak base of X if whenever G ⊂ X, G is open in X if and
only if for each x ∈ G, there exists a P ∈Px such that P ⊂ G.

Definition 2.12 ([13]). Let A ⊂ X and let O be a family of subsets of X.
Then O is an external base of A in X if for each x ∈ A and an open set U with
x ∈ U there is a V ∈O such that x ∈ V ⊂ U.

Definition 2.13 ([9]). Let f : X → Y be a map.
(a) f is a boundary compact map if ∂f−1(y) is compact in X for each y ∈ Y.
(b) f is called sequentially quotient if for each convergent sequence {yn}

in Y there is a convergent sequence {xk} in X with f(xk) = ynk for
each k, where {ynk} is a subsequence of {yn}. See also [14, 16] for more
details.

Definition 2.14 ([9]). Let X = {0}∪(N×N). For every n, m ∈ N and f ∈ NN,
let W(n,m) = {(n,k) ∈ N × N : k ≥ m}, and L(f) = ∪{W(n,f(n)) : n ∈ N}.
Then the set X with the following topology is called a sequential fan and
denoted briefly as Sω : for each x ∈ X, take Nx = {{x}}, if x ∈ N×N. Nx =
{{x}∪L(f) : f ∈ NN}, if x = 0 as a neighbourhood base of x.

Remark 2.15 ([9]). Sequential fan Sω is the space obtained by identifying the
limits of countably many convergent sequences.

Definition 2.16 ([18]). Let A,B be two non-empty collections of families of
subsets of an infinite set X. Then S1(A,B) is defined as: For each sequence
{An : n ∈ N} of elements of A, there is a sequence {bn : n ∈ N} such that
bn ∈ An for each n and {bn : n ∈ N}∈B.

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 366



On ideal sequence covering maps

3. Main results

3.1. (I,J)-sequence covering map. In this section we introduce a map
namely, (I,J)-sequence covering map and investigate some of its properties.

Definition 3.1. Let I and J be two admissible ideals. Then
(a) f : X → Y is said to be a (I,J)-sequence covering map if for any

given sequence {yn}, J -converging to y in Y , there exists a sequence
{xn}, I-convergent to x in X such that xn ∈ f−1(yn) for each n and
x ∈ f−1(y).

(b) f : X → Y is said to be a (I,J)-1-sequence covering map if for each
y ∈ Y there exists x ∈ f−1(y) such that whenever {yn} is a sequence
J -converging to y, there exists a sequence {xn}, which is I-convergent
to x in X such that xn ∈ f−1(yn) for each n.

These are generalization of sequence covering map and 1-sequence cov-
ering map respectively. Clearly every (I,J)-1-sequence covering map is a
(I,J)-sequence covering map. But the converse is not true which is shown
by the next example.

Example 3.2. Y = {yn : n ∈ N}∪{y}. Topology of Y is defined by each {yn}
is open and a basic open set U containing y equals to {yn : n ≥ n0}∪{y} for
some n0 ∈ N. Suppose {pn} is a convergent sequence of real numbers with its
limit p. Define xn = pn if n ∈ M where M ∈ F(I) and xn = n, otherwise.
Clearly {xn}, I-converges to p. Let

∧
= {α : α : N −→ N is an increasing

mapping }. Take Xα = {(xk,α) : k ∈ N}∪{(p,α)} and metric dα on Xα is
defined by dα((m,α), (n,α)) = |m−n|. X =

⊕
α∈

∧ Xα and the metric D on
X is defined as D(m1,m2) = min{dα(m1,m2), 1} when m1,m2 ∈ Xα for some
α ∈

∧
and D(m1,m2) = 1 when m1 ∈ Yα,m2 ∈ Yβ for distinct α,β ∈

∧
.

Now define a map f : X → Y by f(xk,α) = yα(k) and f(p,α) = y for each
α ∈

∧
. Clearly f is continuous. Convergent sequence {zn} in Y is of the

form zs(k) = yα(k) α ∈
∧

and zn = y for n ∈ {s(k) : k ∈ N}c. Then choose
rs(k) = (xk,α) and rn = (p,α) for n ∈ {s(k) : k ∈ N}c. Clearly rn ∈ f−1(zn)
and {rn}I-converges to (p,α) ∈ f−1(y). So f is (I,J)-sequence covering map.
But for y ∈ Y choose any (p,α) ∈ f−1(y). Then there is {yβ(k)} → y, β 6= α
but {(xk,β)} does not converge to (p,α). So f is not (I,J)-1-sequence covering
map.

Proposition 3.3. Let f : X → Y and g : Y → Z be any two maps. Also let
I,J ,K be admissible ideals. Then the following hold:

(a) If f is (I,J)-sequence covering map and g is (J ,K)-sequence covering
map then g ◦f is (I,K)-sequence covering map.

(b) If g ◦f is (I,K)-sequence covering map then g is (I,K)-sequence cov-
ering map.

Proof. (a) Let z ∈ Z and {zn} be a sequence K-converging to z. Since g is
(J ,K)-sequence covering map so there exists a sequence {yn}, J -converging
to y in Y such that yn ∈ g−1(zn) and y ∈ g−1(z). Since f is (I,J)-sequence

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 367



S. K. Pal, N. Adhikary and U. Samanta

covering map so there exists a sequence {xn}, I-converging to x in X such that
xn ∈ f−1(yn) and x ∈ f−1(y). So we get xn ∈ (g◦f)−1(zn) and x ∈ (g◦f)−1(z).
Hence g ◦f is (I,K)-sequence covering map.
(b) Since f is continuous g is (I,K)-sequence covering map. �

Proposition 3.4. Let I,J be two admissible ideals. Then,
(a) Finite product of (I,J)-sequence covering mappings is (I,J)-sequence

covering map.
(b) (I,J)-sequence covering mappings are hereditarily (I,J)-sequence cov-

ering mappings.

Proof. (a) Let
∏N
i=1 fi :

∏N
i=1 Xi →

∏N
i=1 Yi be a map where each fi : Xi → Yi

is (I,J)-sequence covering map for i = 1, 2, ...N. Let {(yi,n)}n∈N be a se-
quence J -converging to (yi) in

∏N
i=1 Yi. Then {yi,n} J -converging to yi in Yi

for each i = 1, 2, ...N. Since each fi is (I,J)-sequence covering map there ex-
ists {xi,n} I-converges to xi in Xi such that fi(xi,n) = yi,n and fi(xi) = yi,
i = 1, 2...N. Now consider the sequence {(xi,n)}n∈N which is I-convergent to
(xi). So

∏N
i=1 fi is (I,J)-sequence covering map.

(b) Let f : X → Y be (I,J)-sequence covering map and H be subspace of
Y. Take g = f|f−1(H) such that g : f−1(H) → H is a map. Let {yn} be a
sequence J -converges to y in H. Then {yn}, J -converges to y in Y. Since f is
(I,J)-sequence covering map so there exists a sequence {xn}, I-converging to
x in X such that xn ∈ f−1(yn) ⊂ f−1(H) and x ∈ f−1(y) ⊂ f−1(H). Hence g
is (I,J)-sequence covering map. �

Lemma 3.5. Let
∧

be any index set and let X =
∏
α∈

∧ Xα has the product
topology. Then {(xα,i)}i∈N is I-converges to (xα) if and only if sequence of
each component {xα,i} I-converges to xα.

Proof. Suppose pα : X → Xα is a projection. So if {(xα,i)}i∈N I-converges
to (xα) then pα({(xα,i)}i∈N)= {xα,i}i∈N I-converges to xα. Conversely assume
that sequence of each component {xα,i} I-converges to xα. Let U be a basic
open set containing (xα) then pα(U) = Xα for all α ∈

∧
but finitely many.

Take the finite set be {α1,α2, ...αk}. Since{xαj,i} I-converges to xαj so we get
Aαj = {i ∈ N : xαj,i /∈ pαj (U)} ∈ I for j = 1, 2, ..k. {i ∈ N : (xα,i) /∈ U} ⊂
∪kj=1Aαj ∈I. Hence {(xα,i)}i∈N is I-converges to (xα). �

From Lemma 3.5 we can say that if f : X =
∏
α∈

∧ Xα → Y = ∏α∈∧ Yα
be the infinite product of (I,J)-sequence covering mappings fα : Xα → Yα
and X,Y has the product topology then f is also (I,J)-sequence covering map.

The following example shows that the concept of ideal sequence covering map
generalizes the concept of statistical sequence covering map and hence sequence
covering map.

Example 3.6. Let F be the set of all increasing mappings from N to N. For
each f ∈ F consider Sf be a I-convergent sequence with its I-limit xf. i.e

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 368



On ideal sequence covering maps

Sf = {xf,n : n ∈ N}∪{xf}. Topology of Sf is defined as each {xf,n} is open
and a basic open set U containing xf is such that Au = {n ∈ N : xf,n /∈ U}∈I
and there exists at least one U such that d(Au) 6= 0. Consider the topological
sum X = ⊕f∈FSf . Let {yn}→ y and Y = {yn : n ∈ N}∪{y}. Topology of Y
is defined as each {yn} is open and a basic open set U containing y equals to
{yn : n ≥ n0}∪{y} for some n0 ∈ N. Let φ : X → Y defined as φ(xf,k) = yf(k),
for all k ∈ N and φ(xf ) = y. Consider any subsequence {yf(k))} of {yn}. Let
zk = yf(k) then consider corresponding xf,k in Sf . Then {xf,n}I-converging
to xf. Now φ(xf,k) = yf(k) = zk and φ(xf ) = y. Consider β ∈F and s : N → N
is an increasing function. Suppose zn = yβ(k),n = s(k) and zn = y, otherwise.
Then clearly zn → y. Consider ps(k) = xβ,k and pn = xβ, otherwise. Then
φ(ps(k)) = φ(xβ,k) = yβ(k) = zs(k) and φ(pn) = y otherwise. Let U be an open
set containing xf. Hence {n ∈ N : pn /∈ U} = {s(k) : ps(k) /∈ U,k ∈ N}. But
{k ∈ N : xβ,k /∈ U} ∈ I. If I has increasing function property i.e B ∈ I ⇒
f(B) ∈I for each increasing function f from N to N then {pn}I-converging to
xf. So φ is (I,Ifin)−sequence covering map but not sequence covering map.

Theorem 3.7. Let f : X → Y be a (I,J)−sequence covering compact map
where I,J be two admissible ideals and suppose there exists a sequence {Mn}
of disjoint infinite subset of N such that Mn /∈ I for each n. Then for each
y ∈ Y there exists x ∈ f−1(y) such that whenever U is an open neighbourhood
of x, f(U) is a sequential neighbourhood of y.

Proof. Suppose not, that is there exists y ∈ Y such that for every x ∈ f−1(y)
there exists an open neighbourhood Ux of x such that f(Ux) is not a sequential
neighbourhood of y. Since f−1(y) ⊂∪x∈f−1(y) Ux and f is a compact map, so
there exists x1,x2, . . . ,xn0 ∈ f−1(y) such that f−1(y) ⊂ ∪

n0
i=1 Uxi. Since each

f(Uxm ) is not a sequential neighbourhood of y, choose {ym,n}∞n=1 converging to
y such that ym,n /∈ f(Uxm ) for all m ∈{1, 2, ...n0},n ∈ N. Now define yk = ym,k
if k ∈ Mm,m ∈{1, 2, ...n0} and yk = y, otherwise. Then clearly yk → y, which
shows that {yk}, J -converging to y. Since f is (I,J)-sequence covering map so
there exists {xk}, I-converging to x such that xk ∈ f−1(yk) and x ∈ f−1(y).
Now x ∈ f−1(y) ⊂ ∪n0i=1 Uxi. So there exists m0 such that x ∈ Uxm0 and
{k ∈ N : xk /∈ Uxm0}∈I. Thus {k ∈ N : f(xk) /∈ f(Uxm0 )}∈I which shows
that {k ∈ N : yk /∈ f(Uxm0 )} ∈ I. But Mm0 ⊂{k ∈ N : yk /∈ f(Uxm0 )} ∈ I,
which contradicts that Mm0 6∈ I. Thus f(U) is a sequential neighbourhood of
y. �

Theorem 3.8. Let I be an admissible ideal and suppose there exists a sequence
{Mn} of disjoint infinite subset of N such that Mn /∈ I for each n. Then the
following conditions are equivalent for a space Y :

(a) Y is a (I,Ifin)-1-sequence covering compact image of a weaker metric
topology.

(b) Y is a (I,Ifin)-sequence covering compact image of a weaker metric
topology.

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 369



S. K. Pal, N. Adhikary and U. Samanta

(c) Y has a sequence {Fi}i∈N of point-finite sn-covers such that ∩i∈Nst(y,Fi) =
{y} for each y ∈ Y.

(d) Y has a sequence {Fi}i∈N of point-finite cs-covers such that ∩i∈Nst(y,Fi) =
{y} for each y ∈ Y.

Proof. It is clear that (a) ⇒ (b), (c) ⇐⇒ (d), (c) ⇒ (a).[15]
(b) ⇒ (c) Suppose f : X → Y is a (I,If )−sequence covering compact mapping.
As X being a space with a weaker metric topology, there is a sequence {Pi}i∈I
of locally finite open covers of X such that ∩i∈Nst(K,Pi) = K for each compact
subset K ⊂ X, by Lemma 3.5. For each i ∈ N, put Fi = f(Pi). Then Fi is
a point finite cover of Y, since f is compact. By Theorem 3.7 for each y ∈ Y
there exists x ∈ f−1(y) such that for every open neighbourhood Ux of x, f(Ux)
is a sequential neighbourhood of y. Since each Pi is an open cover of X, there
exists P ∈Pi such that x ∈ P and so F = f(P) is a sequential neighbourhood
of y. Choose F′i ⊂ Fi which are sequential neighbourhood of y. Thus F

′
i is a

point finite sn-cover of Y. For each y ∈ Y, f−1(y) is compact subset of X and
∩i∈Nst(f−1(y),Pi) = f−1(y). Thus ∩i∈Nst(y,Fi) = {y}. �

Theorem 3.9. Let X be a strongly Fréchet space with property ωD. If f : X →
Y is a closed and (I,J)−sequence covering map, then Y is strongly Fréchet.

Proof. Clearly Y is a Fréchet space, since it is a closed image of a strongly
Fréchet. Suppose Y is not strongly Fréchet. Then Y contains a homomorphic
copy of the sequential fan Sω and the copy can be closed in Y [9]. Consider
Sω ⊂ Y as a closed set where Sω = {y} ∪ {ym,n : m,n ∈ ω}. Take each
Sm = {ym,n}n∈ω is a sequence converges to y. Take A ∈F(I). For each m ∈ N,
choose ymk = y0,k if k ∈ A

c and ymk = ym,k, if k ∈ A. Then the sequence {ymk}
converges to y. Since f is (I,J)-sequence covering map there exists xm ∈
f−1(y) and a sequence Qm, I-converging to xm such that f(Qm) = {ymk}.
Now for each k ∈ ω take Tk = ∪{f−1(Sm) : m ≥ k}. Suppose that there exists
z ∈ X such that for every open neighbourhood U of z, {n ∈ N : xn ∈ U} is
infinite. Then z ∈∩k∈NTk. Since X is strongly Fréchet, there exists a sequence
{zk} converges to z where zk ∈ Tk. But {f(zk)}k∈N does not converges to y,
which is a contradiction. So for every z ∈ X there is an open neighbourhood Uz
of z such that {n ∈ N : xn ∈ Uz} is finite. So z cannot be a limit point of {xn}.
Hence {xn} is closed. Suppose the set {xn} is finite. Let it be z = xn, n ∈ N′
where N′ is a infinite subset of N. Then for every open neighbourhood U of
z, {n ∈ N : xn ∈ U} is infinite which is a contradiction. Therefore the set
{xn}n∈N is infinite, closed and discrete in X.
Since X has the property ωD there exists an infinite subset {xnj}j∈N and a
discrete open family {Uj}j∈ω such that Uj ∩{xnj}j∈ω = {xnj}. Recall that
Qnj, I-converges to xnj and f(Qnj ) = {ynj}. Therefore we can take uj ∈
Uj∩Qnj such that {f(uj)}j∈ω is infinite and contained in {y0,n : n ∈ N}. Since
{uj}j∈ω is closed in X, {f(uj)}j∈ω is closed in Sw, which is a contradiction.
Thus Y is strongly Fréchet. �

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 370



On ideal sequence covering maps

Corollary 3.10. Every closed and (I,J)-sequence covering image of a metric
space is metrizable.

Proof. Suppose f : X → Y is closed and a (I,J)-sequence covering map. Let
X be a metric space. Since every metrizable space has ωD property [13], Y is
strongly Fréchet by Theorem 3.9. Also every strongly Fréchet space which is a
closed image of a metric space is metrizable [8, 24]. Hence Y is metrizable. �

3.2. (I,J ,L)- sequentially quotient map. In this section we introduce the
concept of ideal sequentially quotient map and investigate some of its proper-
ties. Also we investigate under what condition the mapping will coincide with
the ideal sequence covering map.

Definition 3.11. Let I,J ,L be three admissible proper ideals of N. A map
f : X → Y is said to be (I,J ,L)- sequentially quotient map if {yn} is a
sequence, J -converging to y there is a sequence {xk}, I-converging to x such
that xk ∈ f−1(ynk ), x ∈ f

−1(y) and {n ∈ N : xk 6∈ f−1(yn) for all k ∈ N}∈L.

Remark 3.12. If I = Ifin = J and L = Id then each (I,J ,L)-sequentially
quotient map is a statistically sequentially quotient map.

We observe following implications.
sequence covering map=⇒ statistically sequentially quotient map and (I,J ,L)-
sequentially quotient map=⇒ sequentially quotient map.
The reverse implications need not be true. We consider the following examples:

Example 3.13. Let f = {A ⊂ N : N\A is infinite and d(A) < 1}. Let I be an
admissible maximal ideal of N then F(I) is an ultrafilter on N. For each α ∈ f
consider a sequence {xα,i : i ∈ N} converging to xα and let {yn : n ∈ N} be a
sequence converging to y. Define Sα = {xα,i : i ∈ α}∪{xα} and Y = {yn : n ∈
N}. Topologies of Sα, α ∈ f and Y are defined below. Each {xα,i} is open and
basic open set containing xα is of the form {xα,i : i ≥ n0, i ∈ α}∪{xα}. Each
{yn} is open and basic open set containing y is of the form {yn : n ≥ n0}∪{y}.
Let X =

⊕
α∈f Sα. Define a map φ : X → Y by φ(xα,i) = yi and φ(xα) = y for

all α ∈ f.
Now there is β ∈F(I) with d(β) < 1. Suppose β = {nk : k ∈ N} and let xk =
xβ,nk. Then {xk} converges to xβ and φ(xk) = ynk. Also {nk : k ∈ N}∈F(I).
Consider a subsequence {ymi} of {yn}. Let A = {mi : i ∈ N}. Now two cases
may appear.
Case 1 : Let N\A be infinite. Name zi = ymi. Consider a set, B which is in F(I)
and of density less than 1. Let MB be the set of all indices of corresponding
elements of {ymi}. Clearly d(MB) < 1 and N \ MB is infinite. Then there is
a sequence converging to xMB. Define a sequence {rj} where rj = xMB,mj for
j ∈ B and rj = xMB for j 6∈ B then rj → xMB. Also B ⊂{nk : φ(rj) = znk}.
Case 2 : Let N\A be finite. Consider a set B ∈F(I) and of density less than 1.
Let MB be the set of all corresponding elements of {ymi}. Clearly d(MB) < 1
and N \ MB is infinite. Then there is a sequence converging to xMB. Define
a sequence {rj} where rj = xMB,mj for j ∈ B and rj = xMB for j 6∈ B then

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 371



S. K. Pal, N. Adhikary and U. Samanta

rj → xMB. Also φ(rj) = ymj = zj for j ∈ B and φ(rj) = xMB, j 6∈ B.
Next consider a sequence {zi} where znk = yk, k ∈ N and zi = y, i 6∈ {nk : k ∈
N}. Then zi → y. As F(I) is an ultrafilter on N then either {nk ∈ N} ∈F(I)
or its complement belongs to F(I). Let F = {nk ∈ N} ∈ F(I) and consider
a proper subset of F, F′ that belongs to F(I) and whose density is less than
1. Let F′ = {nki : i ∈ N} and let C = {ki : i ∈ N}. Also d(C) < 1. Define
rnki = xC,ki and rj = xC for j 6= nki. Then F

′ ⊂{nk : φ(rk) = znk}. Consider
a sequence {zj} where znk = yα(k) and zj = y for j 6= α(k) where α : N → N is
an increasing function. Let F = {nk ∈ N}∈F(I). Consider an infinite subset
B of α(N) where N \ MB is infinite. Then there is an infinite subset, F′ of
the set of all indices of corresponding elements of znk belonging to F(I). Let
F′ = {nki : i ∈ N} and C = {α(ki) : i ∈ N}. Consider a sequence {rj : j ∈ N}
where rj = xC,α(ki), j = nki and rj = xC. Therefore rj → xC and F

′ ⊂ {nk :
φ(rk) = ynk}. Again let N \F = F

{ ∈ F(I), as F(I) is an ultrafilter on N
there is an infinite subset of F{, F{,′, which is a member of F(I) and of density
less than 1. Let us construct a sequence {rj} where rj = xF{,′ if j ∈ F

{,′ and
put elements of {xF{,′,i} in rest of the places. But as there is no set in f with
density equals to 1 and so φ cannot be a statistically sequentially quotient map
but it is an (Ifin,Ifin,I)-sequentially quotient map.

Example 3.14. Let there be an infinite set A ⊂ N with A /∈I and N\A /∈I.
Also let Y be a convergent sequence with its limit {y}. Consider Xα = {xi :
i ∈ A}∪{xα} and Xβ = {zi : i ∈ N \A}∪ (xβ) where xi → xα and zi → xβ.
Topology of Xα is defined as follows: {xi} are open and open set containing xα
is of the form {xi : i ≥ i0, i ∈ A}∪{xα}. Similarly Topologies of Xβ and Y are
defined. Define a map f : Xα

⊕
Xβ → Y by f(xi) = yi,f(zi) = yi,f(xα) =

f(xβ) = y. Then f is a sequentially quotient map but f is not an (Ifin,Ifin,I)-
sequentially quotient map.

Proposition 3.15.

(1) Let I,J ,K,L be four ideals of N and let F(L) satisfy the property:
If {nk : k ∈ N} ∈ F(L) and {ki : i ∈ N} ∈ F(L) then {nki : i ∈
N} ∈ F(L). f : X → Y and g : Y → Z be (I,J ,L)- and (J ,K,L)-
sequentially quotient maps. Then g ◦ f : X → Z is an (I,K,L)-
sequentially quotient map.

(2) If g ◦f is (I,J ,L)- sequentially quotient map then g is so.

Proof. (1) Let {zn} be a sequence, K-converging to z in Z. So there is a se-
quence {yk}, J -converging to y so that for each k ∈ N, g(yk) = znk, g(y) = z
and {nk : g(yk) = znk}∈F(L). Now as f : X → Y is an (I,J ,L)-sequentially
quotient map there is a sequence {xi}, I-converging to x such that f(xi) = yki
for each i, y = f(x) and {ki ∈ N : yki = f(xi)}∈F(L). Now (g◦f)(x) = z for
each i. Thus (g◦f)(xi) = znki and {nki : i ∈ N}∈F(L)(by our assumptions).
Therefore g ◦f : X → Z is an (I,K,L)- sequentially quotient map.
(2) Let {zn} be a sequence J -converging to z. Then there is a sequence
{xk}, I-converging to x. Now (g ◦ f)(xk) = znk and (g ◦ f)(x) = z. Also

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 372



On ideal sequence covering maps

{nk ∈ N : (g ◦ f)(xk) = znk} ∈ F(L). By continuity of f, it follows that
{f(xk)}, I-converges to f(x) and hence g is (I,J ,L)- sequentially quotient
map. �

Proposition 3.16.

(1) (I,J ,L)- sequentially quotient maps is preserved by finite products.
(2) (I,J ,L)- sequentially quotient maps are hereditarily (I,J ,L)- sequen-

tially quotient maps.

Proof. (1) Let fi : Xi → Yi be (I,J ,L)- sequentially quotient map for i =
1, 2, · · · ,N. We define a map f :

∏N
i=1 Xi →

∏N
i=1 Yi by f(x1,x2, · · · ,xN ) =

(f1(x1),f2(x2), · · · ,fN (xN )). Then f is continuous and onto. Let {(yi,n) : n ∈
N} be a sequence J -converge to (yi) in

∏N
i=1 Yi. Then {yi,n}, J -converging

to yi in Yi for i = 1, 2, · · · ,N. Now there is a sequence {xi,k}, I-converging
to xi such that fi(xi,k) = yi,nk, k ∈ N and {n ∈ N : xi,k 6∈ f

−1
i (yi,n) for all

k ∈ N} ∈ L for i = 1, 2, · · · ,N, . Put x = (xi) ∈
∏N
i=1 Xi. Then {(xi,k) : k ∈

N} I-converges to (xi). Let Ni = {n ∈ N : xi,k ∈ f−1i (yi,n) for all k ∈ N} ∈

F(L). Put N =
N⋂
i=1

Ni ∈ F(L). Then N ⊂ {n ∈ N : (xi,k) ∈ f−1(yi,n) for all

k ∈ N}. Therefore f is an (I,J ,L)- sequentially quotient map.
(2) Let f : X → Y be an (I,J ,L)- sequentially quotient map and let H be a
subspace of Y. Suppose that {yn} is a sequence in H, J -converging to y in H.
Then there is a sequence {xn}, I-converging to x in X where xk ∈ f−1(ynk ).
Clearly xk ∈ f−1(H) for each k ∈ N and x ∈ f−1(H). �

Recall that there is a class of ideals of N, β ‘say’ that satisfies: if A ∈ I
then for every strictly increasing function f : N → N, f(A) ∈I. Several known
ideals are β ideals, for example

(1) The (analytic P-ideal) Id of natural density zero sets,
(2) The (Fσ P-ideal) summable ideal I1

n
= {A ⊂ N : Σ

n∈A
1
n
< ∞},

(3) The (analytic P-ideal)

Ilog = {A ⊂ N : lim
n→∞

( Σ
i∈A∩{1,2,··· ,n}

1

i
)/( Σ

i∈{1,2,··· ,n}

1

i
) = 0}

of logarithmic density zero sets.

Theorem 3.17. Let f : X → Y be an (I,J ,L)- sequentially quotient and
boundary compact map where I ∈ β. Also let there be an infinite countable
partition {Mi : i ∈ N} of N such that Mi /∈ L for each i ∈ N and I ⊂ L.
Moreover, let Y be snf-countable. Then for each non-isolated point y ∈ Y,
there is a point xy ∈ ∂f−1(y) such that whenever U is an open subset with
xy ∈ U, there exists a P ∈Py satisfying P ⊂ f(U).

Proof. Let it be false. Then there exists a non-isolated point y ∈ Y such that
for each x ∈ ∂f−1(y) there exists an open neighbourhood Ux of x so that

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 373



S. K. Pal, N. Adhikary and U. Samanta

for all P ∈ Py,P 6⊂ f(Ux). Therefore ∂f−1(y) ⊂ {Ux : x ∈ ∂f−1(y)}. Since
∂f−1(y) is compact, there exists a finite subfamily U of {Ux : x ∈ ∂f−1(y)}
that covers ∂f−1(y). Name U = {Ui : 1 6 i 6 n0}. Let Py = {Pn : n ∈ N}

and Wy = {Fn =
n⋂
i=1

Pi : n ∈ N}. Then Wy ⊂ Py. Also Fn+1 ⊂ Fn for all

n ∈ N. Now for each 1 ≤ m ≤ n0, n ∈ N there exists xn,m ∈ Fn \ f(Um).
Put yn = xn,1, n ∈ M1, yn = xn,2, n ∈ M2, · · · ,yn = xn,n0, n ∈ Mn0 and
yn = y, n ∈ N\ (

⋃
i=1,2,··· ,n0

Mi).

Then {yn} converges to y. As f is (I,J ,L)- sequential quotient map there
is a sequence {xk}, I-converging to x ∈ ∂f−1(y) in X such that for each
k ∈ N, f(xk) = ynk and {n ∈ N : xk 6∈ f

−1(yn) for all k} ∈ L. Now there
is m0 ∈ {1, 2, · · · ,n0} such that x ∈ Um0 as x ∈ ∂f−1(y) ⊂ ∪U. Hence {k ∈
N : xk 6∈ Um0} ∈ I which shows that {nk ∈ N : ynk 6∈ f(Um0 )} ∈ I. Thus
{nk ∈ N : ynk 6∈ f(Um0 )} ∈ L as I ⊂ L. Therefore {nk ∈ N : f(xk) =
ynk} ∈ F(L) and hence {nk ∈ N : ynk ∈ f(Um0 )} ∈ F(L) which shows
that {n ∈ N : yn ∈ f(Um0 )} ∈ F(L). Thus {n ∈ N : yn 6∈ f(Um0 )} ∈ L.
But Mm0 /∈ L and for each n ∈ Mm0,yn = xn,m0 6∈ f(Um0 ), which is a
contradiction. Hence the theorem. �

The following result gives the relation between ideal sequentially quotient
map and ideal sequence covering map.

Theorem 3.18. Let f : X → Y be an (I,J ,L)- sequentially quotient and
boundary compact map where X is first countable. Suppose J ⊂ I and X
satisfies S1(F(I),F(I)). Then f is (I,J)-sequence covering map provided Y
is snf-countable.

Proof. Let y be a non-isolated point in Y. As Y is snf-countable by Theorem
3.17, there exists a point xy ∈ ∂f−1(y) such that whenever U is an open
neighbourhood of xy there exists P ∈ Py, P ⊂ f(U). Let {Bn : n ∈ N} be a
countable neighbourhood base at xy : Bn+1 ⊂ Bn, n ∈ N. Now for each Bn,
there exists a Pn ∈ Py : Pn ⊂ f(Bn) which shows that f(Bn) is a sequential
neighbourhood of y ∈ Y as each P ∈Py is a sequential neighbourhood of y. Let
{yi} be a sequence, J -converging to y in Y, i.e. {i ∈ N : yi 6∈ Pn} ∈ J . Now
For each n, let An = {i ∈ N : yi ∈ f(Bn)}. Choose in ∈ An, n ∈ N such that
{in : n ∈ N}∈F(I). Set xj = f−1(yj) if j 6= in, n ∈ N and xj = f−1(yj)∩Bn if
j = in. Let U be an open neighbourhood of xy. Then there is Bn ⊂ U. Therefore
{xj}, I-converging to xy and for each j ∈ N, f(xj) = yj, y = f(xy).

�

The following Lemma exhibits the nature of the image of ideal sequentially
quotient boundary compact map.

Lemma 3.19. Let Ω be the set of all topological spaces such that each compact
subset K ⊂ X is metrizable and has a countable neighborhood base in X and

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 374



On ideal sequence covering maps

f : X → Y be an (I,J ,L)- sequentially quotient and boundary compact map.
If X ∈ Ω then Y is snf-countable.

Proof. Consider a non-isolated point y ∈ Y. ∂f−1(y) 6= ∅, compact and since
X ∈ Ω there is a countable external base U for ∂f−1(y) in X. Let V = {∪U′ :
U′ ⊂ U is finite and ∂f−1(y) ⊂ ∪U′}. Now f(V) is countable as V is so. Now
we have to show that f(V) is a sn-network at y.
Let N be a neighbourhood of y. Then clearly ∂f−1(y) ⊂ f−1(N). Now for
each x ∈ ∂f−1(y) there is Ux ∈U : x ∈ Ux ⊂ f−1(N). So {Ux : x ∈ ∂f−1(y)}
covers ∂f−1(y). There is a finite subfamily K of {Ux : x ∈ ∂f−1(y)} that covers
∂f−1(y) and ∂f−1(y) ⊂∪K⊂ f−1(N). Clearly ∪K∈V and y ∈ f(∪K) ⊂N .
Let U1 = f(U

′
1) and U2 = f(U

′
2) where U

′
1 and U

′
2 are elements of V. Now

∂f−1(y) ⊂ U′1 ∩ U′2 then as in above we get U′ ∈ V such that ∂f−1(y) ⊂
U′ ⊂ U′1 ∩ U′2. Name f(U′) = V ′. Then V ′ ⊂ U1 ∩ U2. Next we show that
each V ∈ f(V) is a sequential neighbourhood of y. Let {yn} be a sequence
converging to y in Y. Then there is a sequence {xk}, I-converging to x such
that x ∈ ∂f−1(y) and f(xk) = ynk, k ∈ N. Now V = f(U) ∈ f(V). So
{k ∈ N : xk 6∈ U} ∈ I which shows that {k ∈ N : ynk 6∈ V} ∈ I. Suppose
that B = {n ∈ N : yn 6∈ V} is an infinite set. Then {y′n : n ∈ B} is a
subsequence of {yn} converging to y. There is a sequence {x′n}, I-converging
to x′ ∈ ∂f−1(y) ⊂ U and its image under f is a subsequence of {y′n : n ∈ B},
hence there is a m ∈ B so that ym ∈ V , which is a contradiction. So B must
be a finite set and hence the result. �

Theorem 3.20. Let I, J and L be three ideals of N that satisfy the following
conditions.

(1) I ∈ β and I satisfies S1(F(I),F(I)).
(2) J ⊂I
(3) There is a countable infinite partition {Mi : i ∈ N} of N such that

Mi /∈L for each i.
Then each (I,J ,L)- sequentially quotient and boundary compact map f : X →
Y, is an (I,J)-sequence covering map if X ∈ Ω.

Proof. Let f : X → Y be an (I,J ,L)- sequentially quotient and boundary
compact map and let X ∈ Ω. By Lemma 3.19, it follows that Y is snf-countable.
As ∂f−1(y) is compact hence applying Theorem 3.18, we have f is an (I,J)-
sequence covering map. �

Corollary 3.21. Let I, J and L be three ideals of N that satisfy the following
conditions.

(1) I ∈ β and I satisfies S1(F(I),F(I)).
(2) J ⊂I
(3) There is a countable infinite partition {Mi : i ∈ N} of N such that

Mi /∈L for each i.

c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 375



S. K. Pal, N. Adhikary and U. Samanta

Then each (I,J ,L)- sequentially quotient and boundary compact map f : X →
Y, is an (I,J)-sequence covering map if at least one of the following conditions
holds.

(1) X has a point-countable base.
(2) X is a developable space.

Proof. It is known that X ∈ Ω either if X has a point-countable base or if X
is a developable space. Then the result follows from Theorem 3.20 �

Acknowledgements. The work of N. Adhikary has been supported by UGC(Ref:
1127/(CSIR-UGC NET DEC. 2017)), India. The authors are grateful to Prof.
Pratulananda Das for his valuable suggestions to improve the quality of the
paper.

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