@ Appl. Gen. Topol. 22, no. 1 (2021), 139-147doi:10.4995/agt.2021.14332
© AGT, UPV, 2021

Remarks on the rings of functions which have

a finite number of discontinuities

M. R. Ahmadi Zand and Zahra Khosravi

Department of Mathematics, Yazd University, Iran (mahmadi@yazd.ac.ir, zahra.khosravi@stu.yazd.ac.ir)

Communicated by D. Georgiou

Abstract

Let X be an arbitrary topological space. F(X) denotes the set of all
real-valued functions on X and C(X)F denotes the set of all f ∈ F(X)
such that f is discontinuous at most on a finite set. It is proved that
if r is a positive real number, then for any f ∈ C(X)F which is not a
unit of C(X)F there exists g ∈ C(X)F such that g 6= 1 and f = g

r
f.

We show that every member of C(X)F is continuous on a dense open
subset of X if and only if every non-isolated point of X is nowhere
dense. It is shown that C(X)F is an Artinian ring if and only if the
space X is finite. We also provide examples to illustrate the results
presented herein.

2010 MSC: 54C40; 13C99.

Keywords: C(X)F ; Z-ultrafilter; completely separated; C(X)F -embedded;
Z-filter; over-rings of C(X); Artinian ring.

1. Introduction

Let X be a nonempty topological space and I(X) denote the set of isolated
points of X. The ring of all real-valued functions on X with pointwise addition
and multiplication is denoted by F(X), continuous members of F(X) is denoted
by C(X), the set of points at which f ∈ F(X) is continuous is denoted by C(f)
and bounded members of C(X) is denoted by C∗(X). For any f ∈ F(X), it
is well known and easy to prove that X \ C(f) is a countable union of closed
sets. The set of all f ∈ F(X) such that X \ C(f) is finite is a subring of F(X)

Received 12 September 2020 – Accepted 25 December 2020

http://dx.doi.org/10.4995/agt.2021.14332


M. R. Ahmadi Zand and Z. Khosravi

and it is denoted by C(X)F . The ring C(X)F where X is T1 was introduced
and studied in [5]. In this paper, topological spaces don’t have to satisfy any
separation axioms unless otherwise stated. Recall that lattice-ordered rings are
subdirect sums of totally ordered rings. Let f, g ∈ C(X)F , then

f ∨ g =
1

2
(f + g + |f − g|) ∈ C(X)F ,

and f ∧ g = −(−f ∨ −g) [5]. Thus, for any topological space X we observe
that C(X)F is a lattice-ordered subring of F(X) and C(X) is a lattice-ordered
subring of C(X)F . Also C

∗(X)F , consisting of all functions in C(X)F which
are bounded, is a lattice-ordered subring of C(X)F . Let S be a subset of X.
The characteristic function of S is denoted by χS. Let R be a commutative
ring. A nonzero ideal I in R is an essential ideal if I intersects every nonzero
ideal of R nontrivially. The socle of R denoted by Soc(R) is the sum of all
minimal ideals of R, or the intersection of all essential ideals of R. In [5], it
is shown that Soc(C(X)F ) consists of all functions which vanish everywhere
except on a finite subset of X. A subset of X is called a Gδ-set if it is a
countable intersection of open sets. As usual, clA and intA will denote the
closure and interior of a subset A in a space X, respectively.

In Section 2, we show that for any f ∈ C(X)F which is not a unit of
C(X)F there exists g ∈ C(X)F such that g 6= 1 and f = g

rf where r is a
positive number. Also, it is proved that if X is a T1-space, f ∈ C(X)F and
Z(f) ⊆ C(f), then Z(f) is Gδ. In section 3, It is shown that C(X)F is an
Artinian ring if and only if the space X is finite. We prove that every point
of X \ I(X) is nowhere dense and every dense open subset of X has finite
complement if and only if C(X)F = {f ∈ F(X)|f|D ∈ C(D) for some dense
open subset of X }. For undefined notations, the reader is referred to [4] and
[6].

2. CF -embedded subsets of X

Let f ∈ C(X)F . The set f
−1(0) = {x ∈ X|f(x) = 0} denoted by Z(f) is

called the Zero-set of f and CoZ(f) = X \ Z(f) is called a Co-Zero-set in X
[5]. The collection of all Zero-sets in X is denoted by Z[C(X)F ] or Z(X).

Proposition 2.1 ([5]). For a topological space X, f ∈ C(X)F is a unit element
if and only if Z(f) = ∅.

Units in C∗(X)F is characterized by the following result.

Lemma 2.2. A function g in C∗(X)F is a unit of C
∗(X)F if and only if g is

bounded away from zero.

Proof. Necessity. If there is a function h in C∗(X)F such that gh = 1, then
|h| < n for some n ∈ N and so |g| > 1

n
. Hence g is bounded away from zero.

Sufficiency. Suppose that g is bounded away from zero. So there is r > 0 such
that |g(x)| > r for every x ∈ X. Thus for any x ∈ X, 0 < 1

|g(x)|
< 1

r
and so

1
g
∈ C∗(X)F , i.e., g is a unit of C

∗(X)F . �

© AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 1 140



Rings of functions which have a finite number of discontinuities

Proposition 2.3. If X is an arbitrary topological space and r is a positive real
number, then, for any f ∈ C(X)F which is not a unit of C(X)F there exists
g ∈ C(X)F such that g 6= 1 and f = g

rf.

Proof. Since f is not a unit of C(X)F , Proposition 2.1 implies that there exists
a ∈ Z(f). If g = 1 − χ{a}, then 1 6= g ∈ C(X)F and f = g

rf. �

Definition 2.4 ([5]). A nonempty subset F of Z(X) is said to be a Z-filter
on X, if it satisfies the following conditions:

(1) ∅ /∈ F;
(2) if Z1, Z2 ∈ F, then Z1 ∩ Z2 ∈ F; and
(3) if Z ∈ F, Z′ ∈ Z(X) and Z′ ⊇ Z, then Z′ ∈ F.

Let I be a proper ideal in C(X)F . Then Z[I] = {Z(f)|f ∈ I} is a Z-filter
on X [5].

Definition 2.5. A Z-filter U on X is called a Z-ultrafilter on X, if there isn’t
any Z- filter F on X, such that U ( F.

Definition 2.6 ([5]). An ideal I in a subring of C(X)F is called fixed ( resp.,
free), if the intersection of all members of Z[I] is non-empty (resp., empty).

In [5], it is noted that every Z-filter F is of the form Z[IF] for some ideal
IF in C(X)F and F is called fixed ( resp., free), if IF is fixed ( resp., free).

Definition 2.7. A Z-filter F on X is called a prime Z-filter, if the union of
two Zero-sets belongs to F, then at least one of them belongs to F.

Theorem 2.8. The following statements are correct.

(1) If I is a prime ideal in C(X)F , then Z[I] = {Z(f)|f ∈ I} is a prime
Z-filter on X.

(2) If F is a prime Z-filter on X, then Z−1[F] = {f ∈ C(X)F |Z(f) ∈ F}
is a prime ideal in C(X)F .

Proof. The proof is similar to [6, Theorem 2.12]. �

Lemma 2.9. Let f ∈ C(X)F . Then there is a positive unit u of C(X)F such
that (−1 ∨ f) ∧ 1 = uf.

Proof. It’s straightforward. �

Theorem 2.10. Let X be a topological space which is T1. Then the following
statements are equivalent.

(1) For any f ∈ C(X)F , there is a unit u of C(X)F such that f = u|f|.
(2) For any g ∈ C∗(X)F , there exists a unit v of C

∗(X)F such that g =
v|g|.

Proof. (1) ⇒ (2). Let g ∈ C∗(X)F . Then by (1), there exists a unit u of
C(X)F such that g = u|g|, and so |u(x)| = 1 for any x ∈ X such that g(x) 6= 0.
Now, Proposition 2.1 implies that Z(u) = ∅. Let A = {x ∈ X|u(x) > 0} and
B = {x ∈ X|u(x) < 0}. Then X is a disjoint union of A and B. If u = 1 or

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M. R. Ahmadi Zand and Z. Khosravi

u = −1, then u ∈ C∗(X). Let u 6= 1 and u 6= −1, then there is a finite subset F
of X such that C(u) = X \ F . Since X is a T1-space, C(u) is open in X. A \ F
and B \ F are disjoint open sets in C(u) and so A \ F and B \ F are disjoint
open sets in X. Therefore, there is a unite v of C∗(X)F such that v|A = 1 and
v|B = −1. Clearly, g = v|g| and so (1) implies (2).
(2) ⇒ (1). Let f ∈ C(X)F . Then, there is g ∈ C

∗(X)F such that g =
(f ∧1)∨−1, so according to (2) there is a unit v of C∗(X)F such that v|g| = g.
From Z(f) = Z(g) it follows that v|f| = f and hence, (2) implies (1). �

Theorem 2.11. Let X be an arbitrary topological space, f be in F(X) and H
be a finite closed subset of X such that f ∈ C(X \ H). If H ∩f−1(0) = ∅, then
f−1(0) is a Gδ-set in X.

Proof. Since by our assumption f−1(0) ∩ H = ∅, there is m ∈ N such that
f−1(− 1

m
, 1
m
)∩H = ∅. Thus for every n ≥ m, f−1(− 1

n
, 1
n
) is open in X \H and

so it is open in X. Hence, f−1(0) =
⋂∞

n=m f
−1(− 1

n
, 1
n
) is a Gδ-set in X. �

By the above theorem we note that if X is a T1-space, f ∈ C(X)F and
Z(f) ⊆ C(f), then Z(f) is Gδ. The following example shows that Theorem
2.11 is not true in general case.

Example 2.12. Let βN be the Stone-Čech compactification of N and y ∈ βN\
N. It is well known that {y} is not a Gδ-set in βN. Thus, f = 1−χ{y} ∈ C(βN)F
and Z(f) = {y} is not a Gδ-set.

Definition 2.13 ([5]). Two nonempty subsets A and B of a topological space
X are said to be F-completely separated in X if there is a function f ∈ C∗(X)F
such that f[A] = {0} and f[B] = {1}.

Theorem 2.14 ([5]). Let X be a topological space. Then two nonempty subsets
A and B are F-completely separated in X if and only if they are contained in
disjoint Zero-sets.

Definition 2.15. Let S be a nonempty subset of X. We say that S is CF -
embedded in X if for any function f ∈ C(S)F , there exists g ∈ C(X)F called
an extension of f such that g|S = f.
In the same way, S is called C∗F -embedded if every f ∈ C

∗(S)F can be extended
to g ∈ C∗(X)F , i.e.,

∃g ∈ C∗(X)F such that g|S = f.

Example 2.16. Let X = R and S1 = R \ {0}. Then, f ∈ C
∗(S1) with value

1 for all positive r, and -1 for negative r, has no continuous extension, i.e.,
S1 is not C

∗-embedded in X. Clearly, S1 is C
∗
F -embedded and CF -embedded.

Let S = R \ Z and h be the restriction of the bracket function to S. Then,
g ∈ C(S)F since it is in C(S). Since there is no g ∈ C(R)F such that g|S = h,
S is not CF -embedded in X.

We observe that if S is a subset of a topological space X with finite comple-
ment, then S is C∗F -embedded and CF -embedded in X.

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Rings of functions which have a finite number of discontinuities

Proposition 2.17. Let η : S → X be a one-one mapping and K be the set
of discontinuity points of η. If K is a finite set , then φ : C(X)F → C(S)F
defined by φ(g) = g ◦ η is a ring homomorphism.

Proof. If g ∈ C(X)F , then H = X \ C(g) is a finite set. Thus, S \ C(g ◦ η) is a
finite set since it is a subset of η−1(H) ∪ K and η is continuous at every point
of X \ K and one-one so φ(g) = g ◦ η ∈ C(S)F , i.e., φ is well defined. Clearly,
φ is a ring homomorphism. �

The following result is an immediate consequence of Proposition 2.17.

Corollary 2.18. Let T be a nonempty subset of a topological space X. Then,
the restriction function φ : C(X)F → C(T )F (resp. φ : C

∗(X)F → C
∗(T )F )

defined by φ(f) = f|T is an onto ring homomorphism if and only if T is CF -
embedded (resp. C∗F -embedded).

Theorem 2.19. Each CF -embedded subset is a C
∗
F -embedded subset of X.

Proof. Let X be a topological space and S ⊆ X be a CF -embedded subset of
X. Now suppose that f ∈ C∗(S)F . Since C

∗(S)F ⊆ C(S)F , there exists an
extension g ∈ C(X)F such that g|S = f. Thus there is a positive integer m
such that for every x ∈ S, |f(x)| ≤ m. We put

h := (g ∨ −m) ∧ m,

obviously h ∈ C∗(X)F and for each s ∈ S, we have

h(s) = (g(s) ∨ −m) ∧ m = (f(s) ∨ −m) ∧ m = f(s).

Hence, S is a C∗F -embedded subset of X. �

Theorem 2.20. Let S and X be two subsets of a topological space Y such that
∅ 6= S ⊆ X ⊆ Y . If X is a CF -embedded in Y then S is a CF -embedded in X
if and only if S is CF -embedded in Y .

Proof. Necessity, assume that S is a CF -embedded in X and f ∈ C(S). So f
has an extension g ∈ C(X)F such that g|S = f and according to assumption,
X is CF -embedded in Y , so there is h ∈ C(Y )F such that h|X = g. Hence, S
is CF -embedded in Y .
Sufficiency, assume that S is CF -embedded in Y and put f ∈ C(S)F . Since S
is CF -embedded in Y , there is an extension g ∈ C(Y )F such that g|S = f and
so g|X ∈ C(X)F is an extension of f which completes the proof. �

Theorem 2.21. A C∗F -embedded subset S of X is CF -embedded in X if and
only if it is F-completely separated from every Zero-set disjoint from it.

Proof. Let S be a C∗F -embedded in X. Assume that h ∈ C(X)F and Z(h)∩S =
∅ and let f(s) = 1

h(s)
for any s ∈ S, then f ∈ C(S)F . Thus by the hypothesis,

there is g ∈ C(X)F , such that g|S = f. Therefore, gh ∈ C(X)F . From
gh[S] = {1} and gh[Z(h)] = {0} it follows that Z(h) and S are F-completely
separated in X.
Conversely, suppose that f ∈ C(S)F . So arctan◦f ∈ C

∗(X)F and there is

© AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 1 143



M. R. Ahmadi Zand and Z. Khosravi

g ∈ C(X)F such that g|S = arctan◦f. Therefore Z = {x ∈ X : |g(x)| ≥
π
2
}

belongs to Z(X) and Z ∩ S = ∅. So according to the assumption there exists
h ∈ C∗F (X) such that h[S] = {1} and h[Z] = {0}. It’s obvious that for every
x ∈ X, |g(x)h(x)| < π

2
and gh|S = arctan ◦f. Hence tan ◦(gh) ∈ C(X)F is an

extension of f and so S is CF -embedded in X. �

By the above theorem and Theorem 2.14 we have the following result.

Theorem 2.22. Each C∗F -embedded Zero-set is CF -embedded.

Proof. It is straigthforward. �

3. Some algebraic aspects of C(X)F

Recall that T
′

(X) is the ring of all f ∈ F(X) where for each f there is an
open dense subset D of X such that f|D ∈ C(D) [1]. Let X be a topological

space which is T1. Then C(X)F is a subring of T
′

(X). The following example
shows that it is not true in general.

Example 3.1. Let X = {0, 1, 2} and τ = {∅, {0}, {1, 2}, X}. Then, C(X)F =
F(X) and (X, τ) is not a discrete space and so C(X) 6= F(X). Since X is the

only dense open subset of the space, C(X) = T
′

(X). Thus, C(X)F is not a

subring of T
′

(X).

In the following result we show that every non-isolated point of a topological
space X is nowhere dense if and only if C(X)F is a subring of T

′

(X).

Lemma 3.2. Let X be an arbitrary topological space. Then, C(X)F is a

subring of T
′

(X) if and only if for any x ∈ X \ I(X), intcl{x} = ∅, i.e., {x}
is nowhere dense.

Proof. Let C(X)F be a subring of T
′

(X). If x ∈ X \ I(X), then χ{x} ∈

T
′

(X) since χ{x} ∈ C(X)F . Thus, there is a dense open subset D of X such
that χ{x} ∈ C(D). Clearly, x /∈ D and so int(cl{x}) = ∅. Conversely, let
int(cl{x}) = ∅ for every x ∈ X \ I(X). If f ∈ C(X)F , then F = X \ C(f) is a
finite subset of X \ I(X) and X \ clF = X \ (∪x∈F cl{x}) = ∩x∈F (X \ cl{x}) is
an open dense subset of X contained in C(f). Thus, f ∈ T ′(X), i.e., C(X)F ⊆

T
′

(X). �

Remark 3.3. Recall that a subset A of a topological space X is said to be a
generalized closed (briefly g-closed) set if clA ⊆ U whenever A ⊆ U and U is
open in X. If the set of all closed sets and g-closed sets in X are coincide,
then X is called a T 1

2

-space [7]. It is well known that X is a T 1
2

-space if and

only if for each x ∈ X , {x} is either closed or open [3]. Thus, if X is a T 1
2

-

space then by Lemma 3.2, C(X)F is a subring of T
′(X). It is noted that all

topological spaces in the paper [5] is assume to be T1 but some of its results
hold in the class of T 1

2

-topological spaces, for example if we borrow the proof

of [5, Proposition 3.1], word-by-word, then we observe that Proposition 3.1 in
[5] holds for T 1

2

-spaces.

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Rings of functions which have a finite number of discontinuities

Lemma 3.4. Let X be an arbitrary topological space. Then, T ′(X) ⊆ C(X)F
if and only if every dense open subset of X has finite complement.

Proof. ⇒. Let D be a dense open subset of X and T ′(X) ⊆ C(X)F . Then,
χD ∈ C(X)F since χD ∈ T

′(X). If X \ D is infinite, then χD is not continuous
at x for any x ∈ X \ D and so X \ D ⊂ X \ C(χD). Thus, χD /∈ C(X)F which
is a contradiction.
⇐. It is obvious. �

The following results is a direct consequence of the above lemma and Lemma
3.2.

Theorem 3.5. Let X be an arbitrary topological space. Then, T ′(X) = C(X)F
if and only if every point of X \ I(X) is nowhere dense and every dense open
subset of X has finite complement.

Corollary 3.6. Let X be a topological space which is T 1
2

. Then, T ′(X) =

C(X)F if and only if every dense open subset of X has finite complement.

The above corollary shows that [5, Proposition 5.4 ] holds for topological
spaces which are T 1

2

.

Remark 3.7. Recall that if the intersection of any two nonempty open sets in
a topological space X is nonempty, then X is called hyperconnected. Thus,
if f ∈ C(X)F and X is T1, then there is a finite closed subset H of X such
that f ∈ C(X \ H). Since X is a hyperconnected T1-space and H is a finite
subset of X, we have X \ H is hyperconnected. Therefore C(X \ H) consists
of constant functions. Hence, the image of the function f is a finite set. Also,
we have the following:

R ≃ C∗(X) = C(X)  C(X)F ⊆ T
′(X) ( F(X).

In particular if the topology of X is cofinite topology and Y is the direct sum
of X with itself, then by Theorem 3.5 we have the following:

C(Y )  C(Y )F = T
′(Y ) ( F(Y ).

Let F be a free ultrafilter on N. The set N with topology F ∪ {∅} become a
topological space denoted by X. From [1, Example 3.4] it follows that

C(X)  T ′(X)  F(X).

Clearly, C(X) 6= C(X)F and there is a dense open subset D in X such that
X \ D is infinite and so χD ∈ T

′(X) \ C(X)F .

Let X be an arbitrary topological space and R be a subring of F(X) such
that χ{x} ∈ R for some x ∈ X. If I is the ideal generated by χ{x} in R, then
I is ring isomorphic with the field of real numbers and so I is a minimal ideal
of R. Note that for any x ∈ X we have χ{x} ∈ C(X)F . If an ideal of C(X)F
contains a nonzero function f and r = f(x) 6= 0 for some x ∈ X, then it
contains χ{x} =

1
r
fχ{x} and so we have the following well known result.

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M. R. Ahmadi Zand and Z. Khosravi

Lemma 3.8 ([5]). Let X be an arbitrary topological space, then Soc(C(X)F )
which is equal to the ideal generated by χ{x}’s is a free ideal both in C(X)F
and in C∗(X)F .

Proposition 3.9. Let X be an arbitrary topological space. Then, Soc(C(X)F )
is the intersection of all the free ideals in C(X)F , and of all the free ideals in
C∗(X)F .

Proof. If I is a free ideal in C(X)F or C
∗(X)F and x ∈ X, then there exists

g ∈ I such that a = g(x) 6= 0 and so by the comment before Lemma 3.8, χ{x} ∈
I. Thus, Soc(C(X)F ) ⊆ I and so by Lemma 3.8 the proof is complete. �

An algebraic characterization of finite topological spaces is given in the fol-
lowing result.

Theorem 3.10. Let X be a topological space. Then the following are equiva-
lent.

(1) X is a finite set.
(2) C(X)F = Soc(C(X)F ).

Proof. (1) ⇒ (2). Let X be a finite set. Then 1 =
∑

x∈X χ{x} ∈ Soc(C(X)F )
by Lemma 3.8 . Thus, C(X)F = Soc(C(X)F ).
(2) ⇒ (1). If C(X)F = Soc(C(X)F ), then 1 ∈ Soc(C(X)F ) and so by Lemma
3.8 , there are xi ∈ X and ci ∈ R where 1 ≤ i ≤ n and n ∈ N such that
1 =

∑n
i=1 ciχ{xi}. Thus, X is a finite set. �

Recall that a commutative ring is called Artinian if it satisfies the descending
chain condition on ideals and it is called semisimple if the intersection of all
the maximal ideals called Jacobson radical is zero. It is well known that a ring
is Artinian semisimple if and only if it is equal to the sum of its minimal ideals
[2]. Let X be a topological space, then C(X)F is semisimple since by [5] its
Jacobson radical is zero. Thus by the above theorem we have the following
result.

Corollary 3.11. A topological space X is finite if and only if C(X)F is an
Artinian ring.

Theorem 3.12. Let Φ : C(X)F → C(X)F be a C(X)F -module homomorphism
which is one to one. Then Φ is a C(X)F -module isomorphism.

Proof. First we claim that Φ(1) is a unit of C(X)F . Assume to the contrary
that Φ(1) is not a unit of C(X)F . Thus by Proposition 2.3, there is 1 6= g ∈
C(X)F such that Φ(1) = gΦ(1) and so

Φ(1 − g) = (1 − g)Φ(1) = 0,

which is a contradiction since Φ is one to one. Therefore, there exists h ∈
C(X)F such that hΦ(1) = 1. Now for any f ∈ C(X)F , we have f = fΦ(h) =
Φ(fh) ∈ Φ(C(X)F ) which completes the proof. �

Corollary 3.13. Let f : X → X be a bijection. If f is continuous, then
Φ : C(X)F → C(X)F defined by Φ(g) = g◦f is a C(X)F -module isomorphism.

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Rings of functions which have a finite number of discontinuities

Proof. It is straightforward. �

Acknowledgements. We record our pleasure to the anonymous referee for

his or her constructive report and many helpful suggestions on the main results

of the earlier version of the manuscript which improved the presentation of the

paper.

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