One-Point Extensions of a Tychonoff Space X via Closed Ideals of $C_{B}(X)$

Abstract

For a Tychonoff space X, let $C_{B}(X)$ be the $C^{*}$-algebra of all bounded complex-valued continuous functions on X. In this paper, we mainly discuss Tychonoff one-point extensions of X arising from closed ideals of $C_{B}(X)$. We show that every closed ideal H of $C_{B}(X)$ produces a Tychonoff one-point extension $X(\infty _{H})$ of X. Moreover, every Tychonoff one-point extension of X can be obtained in this way. As an application, we study the partially ordered set of all Tychonoff one-point extensions of X. It is shown that the minimal unitization of a non-vanishing closed ideal H of $C_{B}(X)$ is isometrically $*$-isomorphic with the $C^{*}$-algebra $C_{B}\left( X(\infty _{H})\right)$. We provide a description for the Čech–Stone compactification of an arbitrary Tychonoff one-point extension of X as a quotient space of $\beta X$ via a closed ideal of $C_{B}(X)$. Then, we establish a characterization of closed ideals of $C_{B}(X)$ that have countable topological generators. Finally, an intrinsic characterization of the multiplier algebra of an arbitrary closed ideal of $C_{B}(X)$ is given.

1 Introduction

Throughout the paper, all topological spaces will be assumed Tychonoff, that is completely regular and Hausdorff. We use C(X) to denote the set of continuous complex-valued functions on a Tychonoff space X. Endowed with the usual pointwise addition, multiplication and multiplication by scalars, C(X) becomes an algebra over $\mathbb {C}$. Let $C_{B}(X)$ denote the set of all bounded elements of C(X). Equipped with the norm

$$\begin{aligned} \Vert f\Vert _{X}=\text {sup}\{|f(x)|: x\in X\},\quad \left( f\in C_{B}(X)\right) , \end{aligned}$$

the normed space $\left\langle C_{B}(X),\Vert \cdot \Vert _{X}\right\rangle$ is a commutative unital Banach algebra. For each $f\in C_{B}(X)$, the set $Z(f)=f^{-1}(\{0\})$ is called the zero-set of f, and the set $\text {coz}(f)=X{\setminus } Z(f)$ is called the cozero-set of f. Denote by $C_{0}(X)$ the set of all $f\in C_{B}(X)$ which vanish at infinity (that is, $|f|^{-1}\left( [\epsilon , +\infty )\right)$ is compact for each $\epsilon >0$). An ideal in $C_{B}(X)$ is a linear subspace I of $C_{B}(X)$ such that $fg\in I$ whenever $f\in C_{B}(X)$ and $g\in I$. An ideal of $C_{B}(X)$ which is closed with respect to the sup-norm $\Vert \cdot \Vert _{X}$ is called a closed ideal. An ideal I of $C_{B}(X)$ is non-vanishing if for each $x\in X$, there exists $f\in I$ such that $f(x)\ne 0$.

A family $\mathfrak {I}$ of subsets of a topological space X is called a set-ideal if it satisfies the following condition: if $A, B\in \mathfrak {I}$ and $C\subseteq A\cup B$, then $C\in \mathfrak {I}$. The set-ideal $\mathfrak {I}$ is proper provided $X\notin \mathfrak {I}$.

A topological space X is called locally null with respect to a set-ideal $\mathfrak {I}$ if each point of X has a neighborhood which is in $\mathfrak {I}$. A set-ideal $\mathfrak {I}$ is closed whenever $F\in \mathfrak {I}$, then $\text {cl}_{X}F\in \mathfrak {I}$.

Definition 1.1

For a topological space X and a subset H of $C_{B}(X)$, we let $\text {Coz}(H)=\bigcup _{h\in H}\text {coz}(h)$ and

$$\begin{aligned} \mathfrak {I}_{H}:=\left\{ A\subseteq X: (\exists h\in H) A\subseteq |h|^{-1}\left( [1,+\infty )\right) \right\} . \end{aligned}$$

Lemma 1.2

[16, Lemma 4.3] For every non-empty Tychonoff space X and every ideal H of $C_{B}(X)$, the following conditions are satisfied:

• (1) The family $\mathfrak {I}_{H}$ is a closed set-ideal in both X and $\text {Coz}(H)$. Moreover, the open set $\text {Coz}(H)$ is locally null with respect to $\mathfrak {I}_{H}$.

• (2) The space X is locally null with respect to $\mathfrak {I}_{H}$ if and only if H is a non-vanishing ideal.

We recall that a topological space Y is called a one-point extension of a topological space X if Y contains X as a dense subspace and $Y{\setminus } X$ is a singleton. The study of one-point extensions of a topological space X has been initiated by the Russian mathematician P. Alexandroff a century ago. He proved the well-known result that a locally compact non-compact Hausdorff topological space can be compactified by adding one point, the so-called point at infinity. His result motivated the study of one-point topological extensions. We recall that two one-point extensions $Y_{1}$ and $Y_{2}$ of X are called equivalent or identical if there exists a homeomorphism $\phi :Y_{1}\rightarrow Y_{2}$ which leaves X pointwise fixed. Equivalent extensions are treated as equal, so the class $\mathcal {E}(X)$ of all Tychonoff one-point extensions of X can be treated as a set. That one-point extensions $Y_1$, $Y_2$ of X are identical is denoted by $Y_1\equiv Y_2$. For any two $Y_{1}$ and $Y_{2}$ in $\mathcal {E}(X)$, define $Y_{1}\le Y_{2}$ to mean that there is a continuous map $\psi : Y_{2}\rightarrow Y_{1}$ which leaves X pointwise fixed. Note that $\le$ is a partial ordering on the set $\mathcal {E}(X)$. Extensive research (cf. [3] and [9,10,11,12,13,14]) has been done on various problems related to the partially ordered set of Tychonoff one-point extensions of a topological space.

One of the applications of set-ideals on a topological space X is to produce various one-point extensions of a topological space X (see, e.g., [4]). To be more precise, let $\langle X,\mathcal {T} \rangle$ be a topological space and let $\mathfrak {I}$ be a set-ideal on X. Given a point $\infty _{\mathfrak {I}}$ which does not belong to X, we denote the new set $X\cup \{\infty _{\mathfrak {I}}\}$ by $X(\infty _{\mathfrak {I}})$. Define

$$\begin{aligned} \mathcal {T}(\mathfrak {I}):=\mathcal {T} \cup \left\{ \{\infty _{\mathfrak {I}}\}\cup V: \text {cl}_{X}\left( X{\setminus } V\right) =X{\setminus } V\in \mathfrak {I} \right\} . \end{aligned}$$

Then $\mathcal {T}(\mathfrak {I})$ forms a topology on the set $X(\infty _{\mathfrak {I}})$. The topological space $\left\langle X(\infty _{\mathfrak {I}}),\mathcal {T}(\mathfrak {I})\right\rangle$ is called the one-point extension of X with respect to the set-ideal $\mathfrak {I}$. When no ambiguities arise, we may write $X(\infty _{\mathfrak {I}})$ to denote the topological space $\left\langle X(\infty _{\mathfrak {I}}),\mathcal {T}(\mathfrak {I})\right\rangle$.

In [4], under some mild assumption on a set-ideal $\mathfrak {I}$ on X, it is shown that the one-point extension $X(\infty _{\mathfrak {I}})$ is Tychonoff. We reassert this approach in more detail in the next section. Note that every one-point extension $X_{p}=X\cup \{p\}$ of a topological space X where $\{p\}$ is a closed subset of $X_{p}$ can be defined in a similar way, i.e., there exists a set-ideal $\mathfrak {I}$ in X such that X is locally null with respect to the set-ideal $\mathfrak {I}$ and $X_{p}=X(\infty _{\mathfrak {I}})$.

We remind the reader of a particular case that arises in traditional Banach spaces.

Example 1.3

(cf, e.g., [18], Example 9.4) Let X be a non-compact Tychonoff space. Then

$$\begin{aligned} \mathcal {K}_{X}:=\{A\subseteq X: \hbox { cl}_{X}A\hspace{0.2cm}\text {is a compact subset of}\ X\}, \end{aligned}$$

is a proper set-ideal in X. The space X is locally null with respect to $\mathcal {K}_{X}$ if and only if X is locally compact. It is also well known that $C_{0}(X)$ is a closed ideal of $C_{B}(X)$ which is non-vanishing if and only if X is locally compact. If X is non-compact, the closed ideal $C_{0}(X)$ can serve as an example of a commutative non-unital Banach algebra. It is known that whenever X is locally compact, the minimal unitization of $C_{0}(X)$ (i.e., the Banach algebra obtained from $C_{0}(X)$ by adjoining an identity) is isometrically $*$-isomorphic to $C_{B}(X(\infty _{_{\mathcal {K}_{X}}}))$, where $X(\infty _{_{\mathcal {K}_{X}}})$ is the one-point compactification of X. In particular whenever X is locally compact, then the topological spectrum of the unitization of $C_{0}(X)$ is homeomorphic to $X(\infty _{_{\mathcal {K}_{X}}})$.

One of our motivations to conduct the current research came from generalizing Example 1.3 by associating a closed ideal of $C_{B}(X)$ to each Tychonoff one-point extension of a topological space X. This paper builds upon the work of [10] and complements [11] whose primary focus is the partially ordered set of some special one-point extensions of a given locally compact space X. The topics of the present paper will explore structural properties of all Tychonoff one-point extensions of a given Tychonoff space.

Here is a brief outline of the paper. Following the preliminaries in Sect. 2, where we mainly recall some pertinent concepts and fix notation, we commence by recording some properties of the Čech–Stone compactification of a Tychonoff space X and closed ideals of the Banach algebra $C_B(X)$. Section 3 is devoted to the Tychonoff one-point extensions of a Tychonoff space X. We characterize set-ideals $\mathfrak {I}$ on X such that one-point extensions $X(\infty _{\mathfrak {I}})$ are Tychonoff. It turns out that there exists a non-vanishing closed ideal H in $C_{B}(X)$ such that the set-ideal $\mathfrak {I}$ is equal to $\mathfrak {I}_{H}$ (Proposition 3.2). Moreover, for every one-point extension $Y\in \mathcal {E}(X)$, we associate a non-vanishing closed ideal H of $C_{B}(X)$ such that Y is identical to $X(\infty _{\mathfrak {I}_{H}})$ (Proposition 3.5). In Sect. 4, we collect some properties of the partially ordered set $\langle \mathcal {E}(X), \le \rangle$. In this section, we examine the subsets of $\mathcal {E}(X)$ that have the least upper bound (Proposition 4.1 and Corollary 4.2). We apply the latter results to the proof of Theorem 4.5, which is one of the main results of this section. Theorem 4.5 asserts that every principal downset in $\mathcal {E}(X)$ is a distributive lattice. Further, we identify the maximal and minimal elements of the partially ordered set $\mathcal {E}(X)$ (Propositions 4.8 and 4.9). In Sect. 5, we apply Theorem 4.5 to study the partially ordered set of all Lindelöf one-point extensions of a non-Lindelöf, locally Lindelöf space. In Sect. 6, we study the relationship between minimal unitization of a non-vanishing closed ideal H of $C_{B}(X)$ (as a Banach subalgebra of $C_{B}(X)$) and the Tychonoff one-point extension $X(\infty _{\mathfrak {I}_{H}})$ of the space X. It is shown that for every non-vanishing closed ideal H of $C_{B}(X)$, the closed subalgebra $H+\mathbb {C}$ is isometrically $*$-isomorphic to the Banach algebra $C_{B}\left( X(\infty _{H})\right)$ (Theorem 6.2). Then, we develop this result to a closed ideal which is not generated by an idempotent (Theorem  6.4). In Sect. 7, we give a representation of the Čech–Stone compactification of one-point extensions of X. We provide a representation for the Čech–Stone compactification of a one-point extension of X via its associated closed ideal of $C_{B}(X)$. It is shown that for every $Y\in \mathcal {E}(X)$, there exists a closed non-vanishing ideal H of $C_{B}(X)$ such that the Čech–Stone compactification of Y is identical to the quotient space $\beta X/\mathfrak {Z}(H)$, where $\mathfrak {Z}(H)=\bigcap _{h\in H}Z(h^{\beta })$ (Theorem 7.1). Then, we obtain a characterization of closed ideals with countable topological generators. A list of topological characterizations for closed ideals having countable topological generators is given in Theorem 7.5. Finally, in Sect. 8, we study the multiplier algebras of closed ideals of $C_{B}(X)$. In fact in this section, for a closed ideal H of $C_{B}(X)$, we determine the largest unital $C^{*}$-algebra containing H as an essential ideal. It is stated in Proposition 8.1 that the multiplier algebra of a closed ideal H of $C_{B}(X)$ is isometrically $*$-isomorphic to the $C^{*}$-algebra $C_{B}\left( \text {Coz}(H)\right)$.

2 Preliminaries

Our primary references for background, terminology, and notation are the celebrated books [5] and [6]. We now review some well-known results that will be used in the sequel.

2.1 The Čech–Stone Compactification

For a Tychonoff space X, there exists a compact Hausdorff space $\beta X$, called the Čech–Stone compactification of X, and a mapping $\beta : X\rightarrow \beta X$ such that the following three properties hold:

• 1. The set $\beta (X)$ is dense in $\beta X$.

• 2. The mapping $\beta : X\rightarrow \beta (X)$ is a homeomorphism. (Hence, we can identify X with $\beta (X)$.)

• 3. Every $f\in C_{B}(X)$ admits a continuous extension $f^{\beta }\in C(\beta X)$ in the sense that $f^{\beta }\circ \beta =f$.

Note that $\Vert f\Vert _{X}=\Vert f^{\beta }\Vert _{\beta X}$, for every $f\in C_{B}(X)$ and hence the mapping

$$\begin{aligned} \beta : C_{B}(X)\rightarrow C(\beta X), \end{aligned}$$

defined by:

$$\begin{aligned} (\forall f\in C_{B}(X))\hspace{0.2cm} \beta (f)=f^{\beta } \end{aligned}$$

is an isometric $*$-isomorphism. The following theorem shows that the Čech–Stone compactification of a Tychonoff space X is uniquely determined as follows.

Theorem 2.1

Let X be a Tychonoff space. Let T be a compact Hausdorff space and let $\imath :X\rightarrow T$ be an embedding such that $\imath (X)$ is dense in T and for every $f\in C_{B}(X)$, there exists $F\in C(T)$ such that $F\circ \imath =f$. Then there exists a homeomorphism $\kappa :\beta X\rightarrow T$ such that $\kappa \circ \beta (X)=\imath (X)$.

2.2 The Spectrum of Closed Ideals of $C_{B}(X)$

Recall that given a commutative Banach algebra ${\textbf{A}}$ over the scalar field $\mathbb {C}$, the set of all multiplicative linear functionals on ${\textbf{A}}$ is denoted by $\mathfrak {M}({\textbf{A}})$. The topological spectrum (or just spectrum, if no confusion is possible) of ${\textbf{A}}$ is the set $\mathfrak {M}({\textbf{A}})$ endowed with the Gelfand topology $\mathfrak {g}$. It is well known that the topological space $(\mathfrak {M}({\textbf{A}}),\mathfrak {g})$ is locally compact and Hausdorff in case the Banach algebra ${\textbf{A}}$ is non-unital and it is compact and Hausdorff whenever the Banach algebra ${\textbf{A}}$ is unital.

If ${\textbf{A}}$ is a non-unital Banach algebra and ${\textbf{A}}^{*}={\textbf{A}}\oplus \mathbb {C}$ is its minimal unitization, then it is well known that the topological space $\mathfrak {M}({\textbf{A}})$ is homeomorphic to an open subset of the compact Hausdorff space $\mathfrak {M}({\textbf{A}}^{*})$. Moreover the space $\mathfrak {M}({\textbf{A}}^{*})$ is homeomorphic to the one-point compactification of the locally compact space $\mathfrak {M}({\textbf{A}})$ (cf, e.g., [18], Theorem 9.29).

In [15] (see also [7] and [8]), it was given an explicit topological description of the spectrum of a closed non-vanishing ideal of $C_{B}(X)$ as an open subset of $\beta X$ that contains the space X. Recently, this method has been extended to give a similar topological description for the spectrum of an arbitrary closed ideal of $C_{B}(X)$ (see [16]).

Recall that for a subset K of a topological space X, the quotient space which is obtained by collapsing K to a point is denoted by X/K (i.e., the quotient space that corresponds to the partition $\{K\}\cup \{\{x\}:x\in X{\setminus } K\}$ in X).

Definition 2.2

Let X be a Tychonoff space, and let H be a closed ideal of $C_{B}(X)$. The spectrum of H is denoted by $\mathfrak {s}\mathfrak {p}(H)$ and defined by:

$$\begin{aligned} \mathfrak {s}\mathfrak {p}(H)=\bigcup _{h\in H}\text {coz}(h^{\beta }). \end{aligned}$$

The complement of $\mathfrak {s}\mathfrak {p}(H)$ in $\beta X$ is denoted by $\mathfrak {Z}(H)$, that is,

$$\begin{aligned} \mathfrak {Z}(H)=\bigcap _{h\in H}Z(h^{\beta }). \end{aligned}$$

Note that $\mathfrak {s}\mathfrak {p}(H)$ is an open subset of $\beta X$, and hence, it is a locally compact space. Moreover, the quotient space $\beta X/\mathfrak {Z}(H)$ is homeomorphic to the one-point compactification of the space $\mathfrak {s}\mathfrak {p}(H)$. As a consequence of the celebrated Gelfand-Naimark representation theorem for commutative non-unital $C^{*}$-algebras, the topological spectrum of every closed ideal H of $C_{B}(X)$ is homeomorphic to the space $\mathfrak {s}\mathfrak {p}(H)$, (see [16] and [15]).

In the following, we give another topological description for $\mathfrak {s}\mathfrak {p}(H)$, where H is an ideal of $C_{B}(X)$, which is more suitable for the use in the forthcoming sections.

Lemma 2.3

Let X be a Tychonoff space and let $f\in C_{B}(X)$. For every $\epsilon >0$,

$$\begin{aligned} |f^{\beta }|^{-1}\left( (\epsilon ,+\infty )\right) \subseteq \text {cl}_{\beta X}\left( |f|^{-1}\left( [\epsilon ,+\infty )\right) \right) \subseteq |f^{\beta }|^{-1}\left( [\epsilon ,+\infty )\right) . \end{aligned}$$

Proof

We only prove the first inclusion. Let $p\in \beta X$ be such that $|f^{\beta }(p)|>\epsilon$, and let $V_{p}$ be an arbitrary open neighborhood of p in $\beta X$. The set

$$\begin{aligned} W=V_{p}\cap |f^{\beta }|^{-1}\left( (\epsilon ,+\infty )\right) \end{aligned}$$

is also an open neighborhood of p in $\beta X$. Since X is dense in $\beta X$, the set $W\cap X$ is non-empty. Let $x\in W\cap X$. Note that $f^{\beta }(x)=f(x)$ and $|f(x)|>\epsilon$. Therefore, the set

$$\begin{aligned} V_{p}\cap |f|^{-1}\left( [\epsilon ,+\infty )\right) \end{aligned}$$

is non-empty and hence p belongs to $\text {cl}_{\beta X}\left( |f|^{-1}\left( [\epsilon ,+\infty )\right) \right)$. $\square$

The following corollary is an immediate consequence of Lemma 2.3.

Corollary 2.4

Let X be a Tychonoff space, and let H be an ideal of $C_{B}(X)$. Then,

$$\begin{aligned} \mathfrak {s}\mathfrak {p}(H)=\bigcup _{h\in H}\text {int}_{\beta X}\text {cl}_{\beta X} |h|^{-1}\left( [1,+\infty )\right) =\bigcup _{h\in H}\text {cl}_{\beta X} |h|^{-1}\left( [1,+\infty )\right) \end{aligned}$$

2.3 Set-Ideal Representations for Closed Ideals of $C_{B}(X)$

Let X be a Tychonoff space and let $\mathfrak {I}$ be a set-ideal in X. Motivated by the definition of the ideal $C_{0}(X)$ of $C_{B}(X)$, M. Reza Koushesh in [15] defined $C_{0}^{\mathfrak {I}}(X)$ as follows:

$$\begin{aligned} C_{0}^{\mathfrak {I}}(X)=\{f\in C_{B}(X): \hbox {} |f|^{-1}([\epsilon ,+\infty ))\in \mathfrak {I}, \hbox {for each} \epsilon >0\}. \end{aligned}$$

In [15], it was shown that $C_{0}^{\mathfrak {I}}(X)$ is a closed ideal of $C_{B}(X)$, for every set-ideal $\mathfrak {I}$ in X.

Example 2.5

Let X be a Tychonoff space and let $p\in \beta X$. It is well known that the set

$$\begin{aligned} M_{X}(p):=\left\{ f\in C_{B}(X): f^{\beta }(p)=0\right\} \end{aligned}$$

is a closed ideal of $C_{B}(X)$. Let $\mathcal {I}_{p}$ be defined as follows.

$$\begin{aligned} \mathcal {I}_{p}:=\left\{ A\subseteq X: p\notin \text {cl}_{\beta X}A\right\} . \end{aligned}$$

With an easy calculation one can observe that the set $\mathcal {I}_{p}$ is a proper set-ideal of X such that $M_{X}(p)=C_{0}^{\mathcal {I}_{p}}(X).$ Note that, for each $p\in \beta X$ the ideal $M_{X}(p)$ is a maximal ideal of $C_{B}(X)$. Moreover, if X is compact and Hausdorff, then every ideal of $C_{B}(X)$ is of the form $M_{X}(x)$, for some $x\in X$ (see, e.g., [6, Chapter 7]).

As one can expect, each closed ideal of $C_{B}(X)$ has a representation by a set-ideal (see [16]).

Lemma 2.6

[16, Lemma 4.4]. Let X be a Tychonoff space, and let H be a closed ideal of $C_{B}(X)$. Then $H=C_{0}^{\mathfrak {I}_{H}}(X)$.

The following lemma shows that the set-ideal $\mathfrak {I}_{H}$ is the smallest set-ideal for representing the closed ideal H of $C_{B}(X)$.

Lemma 2.7

[16, Lemma 4.6]. Let X be a Tychonoff space and let H be a closed ideal of $C_{B}(X)$. If $\mathfrak {I}$ is a set-ideal in X such that $H=C_{0}^{\mathfrak {I}}(X)$, then $\mathfrak {I}_{H}\subseteq \mathfrak {I}$.

Recall that in Lemma 1.2, it is mentioned that the family $\mathfrak {I}_{H}$ is also a set-ideal in $\text {Coz}(H)$. We also remark that $\text {Coz}(H)$ is locally null with respect to the ideal $\mathfrak {I}_{H}$. We conclude this subsection by the following representation theorem for closed ideals of $C_{B}(X)$.

Theorem 2.8

[16, Theorem 4.14]. Let X be a Tychonoff space and let H be a closed ideal of $C_{B}(X)$. Then the mapping

$$\begin{aligned} \Theta : H\rightarrow C_{0}^{\mathfrak {I}_{H}}\left( \text {Coz}(H)\right) \end{aligned}$$

defined by:

$$\begin{aligned} (\forall f\in H)\hspace{0.2cm} \Theta (f)=f\restriction _{\text {Coz}(H)} \end{aligned}$$

is an isometric $*$-isomorphism.

3 Tychonoff One-Point Extensions of a Topological Space X

Following Caterino et al. [4], a set-ideal $\mathfrak {I}$ in a topological space X is called functionally open if for each $A\in \mathfrak {I}$, there exists an open set $O\in \mathfrak {I}$ such that A and $X{\setminus } O$ are completely separated. It might be of some interest to know when $X(\infty _{\mathfrak {I}})$ is a Tychonoff space. This will be the context of the following result.

Theorem 3.1

[4, Theorem 3.3]. Let X be a Tychonoff space which is locally null with respect to a closed set-ideal $\mathfrak {I}$ in X. Then, $X(\infty _{\mathfrak {I}})$ is Tychonoff if and only if $\mathfrak {I}$ is functionally open.

We remark that functionally open set-ideals seem of natural interest and worth of additional study. The following result provides a connection between functionally open set-ideals on a Tychonoff space X and closed ideals of $C_{B}(X)$.

Proposition 3.2

Let X be a Tychonoff space, and let $\mathfrak {I}$ be a set-ideal in X. Then, $\mathfrak {I}$ is functionally open if and only if there exists a closed ideal H of $C_{B}(X)$ such that $\mathfrak {I}=\mathfrak {I}_{H}$.

Proof

Let H be a closed ideal of $C_{B}(X)$. To show that the set-ideal $\mathfrak {I}_{H}$ is functionally open, let us consider an arbitrary $A\in \mathfrak {I}_{H}$. There exists $h\in H$ such that $A\subseteq |h|^{-1}\left( [1,+\infty )\right)$. Let $O=|h|^{-1}\left( (\frac{1}{2},+\infty )\right)$. Then $A\subseteq O$ and O is open in X. Since $O\subseteq |2\,h|^{-1}\left( [1,+\infty )\right)$, we have $O\in \mathfrak {I}_{H}$. We notice that the sets $X{\setminus } O$ and $|h|^{-1}\left( [1,+\infty )\right)$ are disjoint zero-sets in X, so they are completely separated. Since $A\subseteq |h|^{-1}\left( [1,+\infty )\right)$, the sets A and $X{\setminus } O$ are completely separated. This proves that the set-ideal $\mathfrak {I}_{H}$ is functionally open.

Now we prove that every functionally open set-ideal in X is of the form $\mathfrak {I}_{H}$, for some closed ideal H of $C_{B}(X)$. Let $\mathfrak {I}$ be a functionally open set-ideal and set $H=C_{0}^{\mathfrak {I}}(X)$. Note that H is a closed ideal of $C_{B}(X)$. By Lemma 2.7, $\mathfrak {I}_{H}\subseteq \mathfrak {I}$. For the reverse inclusion, let $A\in \mathfrak {I}$. There exists an open set V in $\mathfrak {I}$ such that A and $X{\setminus } V$ are completely separated. Hence, there is $f\in C_{B}(X)$ such that $f(A)=\{2\}$ and f vanishes on $X{\setminus } V$. Since for each $\epsilon >0$,

$$\begin{aligned} |f|^{-1}\left( [\epsilon ,\infty )\right) \cap \left( X{\setminus } V\right) =\emptyset , \end{aligned}$$

the set $|f|^{-1}\left( [\epsilon ,\infty )\right)$ is included in V and hence it belongs to $\mathfrak {I}$. This means that $f\in C_{0}^{\mathfrak {I}}(X)=H$. Moreover, $A\subseteq |f|^{-1}\left( [1,\infty )\right)$ and thus $A\in \mathfrak {I}_{H}$. $\square$

Remark 3.3

It follows from Theorem 3.1 and Proposition 3.2 that, for every non-empty Tychonoff space X and every closed ideal H of $C_B(X)$, the extension $X(\infty _{\mathfrak {I}_{H}})$ is Tychonoff.

In the next proposition, we give a description for Tychonoff one-point extensions of a space X in terms of closed ideals H of $C_{B}(X)$. This point of view plays an important role for the present article. We first need the following auxiliary result. For convenience in using symbols, we denote $\infty _{H}$ instead of using $\infty _{\mathfrak {I}_{H}}$.

Proposition 3.4

Let X be a Tychonoff space, and let H and K be two proper closed ideals of $C_{B}(X)$. Then, the following conditions are equivalent:

• (1) $H\subseteq K$.

• (2) $\mathfrak {I}_{H}\subseteq \mathfrak {I}_{K}$.

• (3) $X(\infty _{H})\le X(\infty _{K})$.

Proof

The implications $(1)\Rightarrow (2)\Rightarrow (3)$ are trivial. If $\mathfrak {I}_{H}\subseteq \mathfrak {I}_{K}$, then it follows from Lemma 2.6 that $H=C^{\mathfrak {I}_{H}}(X)\subseteq C^{\mathfrak {I}_{K}}(X)=K$. Hence, (2) implies (1). To complete the proof, it suffices to show that (3) implies (2).

Suppose that (3) holds. Let $\phi : X(\infty _{K})\rightarrow X(\infty _{H})$ be a continuous map such that, for every $x\in X$, $\phi (x)=x$. Since X is dense in both $X(\infty _{H})$ and $X(\infty _{K})$, we have $\phi (\infty _{K})=\infty _{H}$. Let $A\in \mathfrak {I}_{H}$. There exists $h\in H$ such that $A\subseteq |h|^{-1}\left( [1, +\infty )\right)$. Note that

$$\begin{aligned} W=|h|^{-1}\left( [0,1)\right) \cup \{\infty _{H}\}, \end{aligned}$$

is a basic open neighborhood of $\infty _{H}$ in $X(\infty _{H})$. Therefore,

$$\begin{aligned} \phi ^{-1}(W)=|h|^{-1}\left( [0,1)\right) \cup \{\infty _{K}\} \end{aligned}$$

is an open neighborhood of $\infty _{K}$ in $X(\infty _{K})$. This implies that

$$\begin{aligned} X{\setminus } |h|^{-1}\left( [0,1)\right) =|h|^{-1}\left( [1,+\infty )\right) \in \mathfrak {I}_{K}. \end{aligned}$$

Hence $A\in \mathfrak {I}_{K}$ and, in consequence, (3) implies (2). $\square$

Proposition 3.5

Let X be a Tychonoff space and let $Y=X\cup \{p\}\in \mathcal {E}(X)$. Define

$$\begin{aligned} H:=\{F\restriction _{X}: F\in C_{B}(Y)\hspace{0.1cm}\text {and}\hspace{0.1cm} F(p)=0\}. \end{aligned}$$

Then, the following statements hold:

• (1) H is a non-vanishing closed ideal of $C_{B}(X)$.

• (2) $Y\equiv X(\infty _{H})$.

Proof

(1) It is clear that H is a vector subspace of $C_{B}(X)$. Suppose that $g\in C_{B}(X)$ and $f=F\restriction _{X}\in H$. Define

$$\begin{aligned} h(t)={\left\{ \begin{array}{ll} f(t)g(t) &{}\quad \text {if } t\in X,\\ 0 &{} \quad \text {if } t=p. \end{array}\right. } \end{aligned}$$

Let us show that $h\in C_{B}(Y)$. To this aim, fix a positive real number M such that, for every $t\in X$, $|g(t)|\le M$. Take any real number $\epsilon >0$. There exists an open neighborhood U of p in Y such that, for every $x\in X$, $|F(x)|<\frac{\epsilon }{M}$. Then, for every $x\in X$, we have $|h(x)|=|F(x)|\cdot |g(x)|<\epsilon$. Hence, h is continuous at p. Since X is open in Y, it follows from the continuity of both F and g that h is continuous at every $x\in X$. Hence $h\in C_B(Y)$. Since, in addition, $h\restriction X= gf$, we have $gf\in H$. This shows that H is an ideal.

Claim 1

The ideal H is closed in $C_{B}(X)$.

Proof

Assume that $(f_{n}:n\in \mathbb {N})$ is a sequence in H which is uniformly convergent to $f\in C_{B}(X)$. There exists a sequence $(F_{n}:n\in \mathbb {N})$ in $C_{B}(Y)$ such that $F_{n}\restriction _{X}=f_{n}$ and $F_{n}(p)=0$, for each $n\in \mathbb {N}$. Define $F:Y\rightarrow \mathbb {C}$ as follows.

$$\begin{aligned} F(t)={\left\{ \begin{array}{ll} f(t) &{}\quad \text {if } t\in X,\\ 0 &{}\quad \text {if } t=p. \end{array}\right. } \end{aligned}$$

Since X is open in Y, it suffices to show that F is continuous at p. Let $\epsilon >0$. There exists $m\in \mathbb {N}$ such that

$$\begin{aligned} \Vert f_{m}-f\Vert _{X}<\frac{\epsilon }{2}. \end{aligned}$$

Since $F_{m}$ is continuous at p, there exists an open neighborhood V of p in Y such that $V\subseteq |F_{m}|^{-1}\left( [0,\frac{\epsilon }{2})\right)$. Therefore, $|f(x)|<\epsilon$, for each $x\in V\cap X$. This implies that $V\subseteq |F|^{-1}\left( [0,\epsilon )\right)$. Therefore, $F\in C_{B}(Y)$. Since $F\restriction _{X}=f$ and $F(p)=0$, the continuous mapping f belongs to H.

Claim 2

The ideal H is non-vanishing.

Proof

In view of part (2) of Lemma 1.2, it suffices to show that X is locally null with respect to $\mathfrak {I}_{H}$. Let $x\in X$. Note that Y is Tychonoff and $x\ne p$. Hence, there exists $F:Y\rightarrow [0,2]$ such that $F(x)=2$ and $F(p)=0$. Note that $f=F\restriction _{X}\in H$ and $x\in |f|^{-1}\left( (1,+\infty )\right)$. Therefore, X is locally null with respect to $\mathfrak {I}_{H}$.

(2) Define the mapping $\psi : Y\rightarrow X(\infty _{H})$ by

$$\begin{aligned} \psi (t)={\left\{ \begin{array}{ll} t &{} \quad \text {if } t\in X,\\ \infty _{H} &{}\quad \text {if } t=p. \end{array}\right. } \end{aligned}$$

To show that $\psi$ is continuous, it suffices to prove that $\psi$ is continuous at p. Let V be an open neighborhood of $\infty _{H}$ in $X(\infty _{H})$. Then, $V\cap X$ is open in X and $X{\setminus } V\in \mathfrak {I}_{H}$. Thus, there exists $h\in H$ such that $X{\setminus } V \subseteq |h|^{-1}\left( [1,+\infty )\right)$. Let ${\tilde{h}}:Y\rightarrow \mathbb {C}$ be defined as follows: ${\tilde{h}}(p)=0$ and, for every $x\in X$, ${\tilde{h}}(x)=h(x)$. Then ${\tilde{h}}\in C_B(Y)$ and $p\in |{\tilde{h}}|^{-1}\left( [0,1)\right) \subseteq \psi ^{-1}(V)$. Hence $\psi$ is continuous at p.

Now, we show that $\psi$ is an open mapping. Let W be an open neighborhood of p in Y. Since Y is a Tychonoff space, there exists $g\in C_{B}(Y)$, such that $g(X{\setminus } W)=\{1\}$ and $g(p)=0$. This implies that

$$\begin{aligned} p\in |g|^{-1}\left( [0,1)\right) \subseteq W. \end{aligned}$$

Note that $g\restriction _{X}\in H$ and

$$\begin{aligned} \psi \left( |g|^{-1}\left( [0,1)\right) \right) =|g\restriction _{X}|^{-1}\left( [0,1)\right) \cup \{\infty _{H}\}\subseteq \psi \left( W\right) . \end{aligned}$$

Therefore, $\infty _{H}$ is in the interior of $\psi \left( W\right)$. $\square$

4 The Partially Ordered Set $\mathcal {E}(X)$

M. Reza Koushesh [9,10,11] considered various problems related to the partially ordered set $\mathcal {E}(X)$. In this section, we will be interested in the lattice structure of some subsets of the partially ordered set $\mathcal {E}(X)$. We first determine the least upper bound or greatest lower bound of subsets of $\mathcal {E}(X)$, whenever they exist.

Proposition 4.1

Let X be a Tychonoff space. For every non-empty collection $\{H_i\}_{i\in {\textbf{I}}}$ of proper closed non-vanishing ideals of $C_B(X)$, the least upper bound $\bigvee _{i\in {\textbf{I}}}X(\infty _{H_i})$ exists in $\mathcal {E}(X)$ if and only if the ideal $\sum _{i\in {\textbf{I}}}H_i$ is proper in $C_B(X)$.

Proof

Let us first assume that the family $\left\{ X(\infty _{H_i})\right\} _{i\in {\textbf{I}}}$ has a least upper bound, say Y, in $\mathcal {E}(X)$. Take a proper closed non-vanishing ideal K of $C_B(X)$ such that $Y\equiv X(\infty _{K})$. Using Proposition 3.4, $H_i\subseteq K$, for all $i\in {\textbf{I}}$. Since the closure of the proper ideal $\sum _{i\in {\textbf{I}}}H_i$ is contained in K, it is a proper ideal of $C_B(X)$.

Conversely, assume that H is the closure of the proper ideal $\sum _{i\in {\textbf{I}}}H_i$ in $C_B(X)$. So, H is a proper ideal of $C_B(X)$ as well, (see [6, Exercise 2 M.4]). Applying Proposition 3.4, we obtain that $X(\infty _{H})$ is an upper bound for the family $\left\{ X(\infty _{H_i})\right\} _{i\in {\textbf{I}}}$. Let $Z\in \mathcal {E}(X)$ be an arbitrary upper bound for the family $\left\{ X(\infty _{H_i})\right\} _{i\in {\textbf{I}}}$. Take a proper closed non-vanishing ideal K of $C_B(X)$ such that $Z\equiv X(\infty _{K})$. Thus, $H\subseteq K$ and hence $X(\infty _{H})\le X(\infty _{K})$. This means that $X(\infty _{H})$ is the least upper bound of the family $\left\{ X(\infty _{H_i})\right\} _{i\in {\textbf{I}}}$. $\square$

The following corollary is an immediate consequence of Propositions 3.5 and 4.1.

Corollary 4.2

Let X be a Tychonoff space. If a non-empty subfamily $\{Y_i\}_{i\in {\textbf{I}}}$ of $\mathcal {E}(X)$ has an upper bound, then the least upper bound $\bigvee _{i\in {\textbf{I}}}Y_i$ exists in $\mathcal {E}(X)$.

Corollary 4.3

Let X be a Tychonoff space, and let H and K be two proper non-vanishing closed ideals of $C_B(X)$. The following statements hold:

• (1) $X(\infty _{H})\wedge X(\infty _{K})=X(\infty _{H\cap K})$.

• (2) $X(\infty _{H})\vee X(\infty _{K})=X(\infty _{H+K})$, whenever $H+K$ is a proper ideal of $C_B(X)$.

Proof

• (1) It is easy to check that $H\cap K$ is a non-vanishing closed ideal of $C_B(X)$. Now, using Propositions 3.4 and 3.5, the statement holds trivially.

• (2) Using Theorem 3.8 of [16], $H+K$ is a closed ideal. Now, assuming that it is a proper ideal of $C_B(X)$, the equality follows from Proposition 4.1.

$\square$

Notation 4.4

For an element x of a partially ordered set $(P,\le )$, the principal downset generated by x is denoted by $\downarrow x$ and is defined as follows:

$$\begin{aligned} \downarrow x=\{y\in P: y\le x\}. \end{aligned}$$

Theorem 4.5

Let X be a Tychonoff space and let $Y\in \mathcal {E}(X)$. Then, the principal downset $\downarrow Y$ is a distributive lattice. Furthermore, every non-empty subset of $\downarrow Y$ has a least upper bound.

Proof

Let Z and T be in $\downarrow Y$. By part (1) of Corollary 4.3 the meet $Z\wedge T$ exists in $\downarrow Y$. Proposition 4.1 implies that the join $Z\vee T$ exists in $\downarrow Y$, as well. Therefore $\downarrow Y$ is a lattice. Next, we show that $\downarrow Y$ is a distributive lattice. For some proper non-vanishing closed ideals $H, K_1, K_2$ of $C_B(X)$, we have:

$$\begin{aligned} \begin{aligned} Z\wedge \left( T_1\vee T_2\right)&\equiv X(\infty _{H})\wedge \left( X(\infty _{K_1})\vee X(\infty _{K_2})\right) \\&\equiv X(\infty _{H})\wedge \left( X(\infty _{K_1+K_2})\right) \\&\equiv X(\infty _{H\cap (K_1+K_2)}).\\ \end{aligned} \end{aligned}$$

Note that the set of all closed ideals of $C_B(X)$ is a distributive lattice (see Proposition 4.10 of [16]). Hence we have $H\cap \left( K_1+K_2\right) =\left( H\cap K_1\right) +\left( H\cap K_2\right)$. Thus

$$\begin{aligned} \begin{aligned} X(\infty _{H\cap (K_1+K_2)})&\equiv X(\infty _{\left( H\cap K_1\right) +\left( H\cap K_2\right) }) \\&\equiv X(\infty _{\left( H\cap K_1\right) })\vee X(\infty _{\left( H\cap K_2\right) }) \\&\equiv \left( X(\infty _{H})\wedge X(\infty _{K_1} )\right) \vee \left( X(\infty _{H})\wedge X(\infty _{K_2})\right) .\\ \end{aligned} \end{aligned}$$

Therefore, we conclude that

$$\begin{aligned} Z\wedge \left( T_1\vee T_2\right) \equiv \left( Z\wedge T_1\right) \vee \left( Z\wedge T_2\right) . \end{aligned}$$

$\square$

Now we turn our attention to the maximal elements of the partially ordered set $\mathcal {E}(X)$.

Notation 4.6

Let X be a Tychonoff space and let $p\in \beta X{\setminus } X$. The subspace $X\cup \{p\}$ of $\beta X$ is denoted by $\Lambda _{_X}(p)$.

Lemma 4.7

Let X be a Tychonoff space and let $p\in \beta X{\setminus } X$. Then $\Lambda _{X}(p)\equiv X(\infty _{H})$, where H equals the maximal ideal $M_X(p)$ of $C_{B}(X)$.

Proof

Suppose that $Y=\Lambda _{_X}(p)$. Note that X is $C^{*}$-embedded in Y since

$$\begin{aligned} X\subseteq Y=\Lambda _{_X}(p)\subseteq \beta X. \end{aligned}$$

We can easily observe that

$$\begin{aligned} M_X(p)=\{F\restriction _{X}: F\in C\left( Y\right) \hspace{0.1cm}\text {and}\hspace{0.1cm} F(p)=0\}. \end{aligned}$$

To complete the proof, one can use similar arguments to the ones in the proof of part (2) in Proposition 3.5. $\square$

Now we are ready to characterize the maximal elements of $\mathcal {E}(X)$.

Proposition 4.8

Let X be a Tychonoff space and let $Y\in \mathcal {E}(X)$. The following statements are equivalent.

• (1) Y is maximal in the partially ordered set $\mathcal {E}(X)$.

• (2) There exists $p\in \beta X{\setminus } X$ such that $Y\equiv \Lambda _{_X}(p)$.

• (3) X is $C^{*}$-embedded in Y.

Proof

(1) implies (2). Let Y be a maximal element of the partially ordered set $\mathcal {E}(X)$. By part (1) of Proposition 3.5, there exists a non-vanishing closed ideal H of $C_{B}(X)$ such that $Y\equiv X(\infty _{H})$. There exists a maximal ideal $M=M(p)$ of $C_{B}(X)$, for some $p\in \beta X{\setminus } X$, such that $H\subseteq M$. By Proposition 3.4, $Y\equiv X(\infty _{H})\le X(\infty _{M})$. Since Y is maximal, $Y\equiv X(\infty _{M})$. Now Lemma 4.7 implies that $Y\equiv \Lambda _{_X}(p)$.

(2) implies (3). Note that $X\subseteq \Lambda _{_X}(p)\subseteq \beta X$, and since X is $C^{*}$-embedded in $\beta X$, therefore X is $C^{*}$-embedded in $\Lambda _{_X}(p)$, as well.

(3) implies (1). Let X be a $C^{*}$-embedded subset of Y and let $Y{\setminus } X=\{q\}$. By Proposition 3.5,

$$\begin{aligned} H=\{f\restriction _{X}: f\in C(Y)\hspace{0.1cm}\text {and}\hspace{0.1cm} f(q)=0\} \end{aligned}$$

is a closed ideal of $C_{B}(X)$ such that $Y\equiv X(\infty _{H})$. To show that H is a maximal ideal, consider any proper ideal M of $C_B(X)$ such that $H\subseteq M$. Suppose that $f\in M{\setminus } H$. There exists ${\tilde{f}}\in C_B(Y)$ such that $f={\tilde{f}}\restriction X$ and ${\tilde{f}}(p)=c\ne 0$. Then $g=f-{\textbf{c}}\in H$. Since $f,g\in M$, we obtain that a nonzero constant function on X belongs to M. This is impossible because M is proper. The contradiction obtained shows that $H=M$, so H is a maximal ideal. By part (2) of Proposition 3.5, $Y\equiv X(\infty _{H})$. Proposition 3.4, taken together with Proposition 3.5, implies that Y is maximal in the partially ordered set $\mathcal {E}(X)$. $\square$

Our final subject in this section is a characterization of minimal elements in $\mathcal {E}(X)$, for every non-compact Tychonoff space X. Surprisingly, although the partially ordered set $\mathcal {E}(X)$ always has a maximal element and more precisely, the number of the maximal elements equals the cardinality of $\beta X{\setminus } X$, whenever it has a minimal element, it is unique and the space X is locally compact.

Proposition 4.9

Let X be a Tychonoff space. The following statements are equivalent.

• (1) $\mathcal {E}(X)$ has the first element.

• (2) $\mathcal {E}(X)$ has a minimal element.

• (3) The space X is locally compact and not compact.

Proof

(1) trivially implies (2).

(2) implies (3). Assume on the contrary that X is not locally compact. Thus, X is not compact and hence $\beta X{\setminus } X$ is a nonempty set which is not closed in $\beta X$. Let $Y\in \mathcal {E}(X)$. By Proposition 3.5, there exists a non-vanishing closed ideal H of $C_{B}(X)$ such that $Y\equiv X(\infty _{H})$. Since H is non-vanishing, $\mathfrak {Z}(H)\subseteq \beta X{\setminus } X$. We can pick $x\in \beta X{\setminus } X$ that $x\notin \mathfrak {Z}(H)$. Let us consider $K=\mathfrak {Z}(H)\cup \{x\}$ and define

$$\begin{aligned} I:=\{f\in C_{B}(X): K\subseteq Z\left( f^{\beta }\right) \}. \end{aligned}$$

Note that $I=\bigcap _{p\in K}M_{X}(p)$ and hence I is closed, since $M_{X}(p)$ is a closed ideal, for each $p\in K$. We claim that H properly contains I. Assume that f belongs to I. Note that $f^{\beta }$ vanishes on K, and hence, for every real number $\epsilon >0$,

$$\begin{aligned} \text {cl}_{\beta X} |f|^{-1}\left( [\epsilon ,+\infty )\right) \subseteq \mathfrak {s}\mathfrak {p}(H). \end{aligned}$$

Note that $\text {cl}_{\beta X} |f|^{-1}\left( [\epsilon ,+\infty )\right)$ is a compact set and hence there exists $h\in H$ such that

$$\begin{aligned} \text {cl}_{\beta X} |f|^{-1}\left( [\epsilon ,+\infty )\right) \subseteq \text {cl}_{\beta X} |h|^{-1}\left( [1,+\infty )\right) \end{aligned}$$

(Corollary 2.4). This implies that

$$\begin{aligned} |f|^{-1}\left( [\epsilon ,+\infty )\right) \subseteq |h|^{-1}\left( [1,+\infty )\right) , \end{aligned}$$

and hence we can infer that $|f|^{-1}\left( [\epsilon ,+\infty )\right) \in \mathfrak {I}_{H}$, for every $\epsilon >0$. Lemma 2.7 implies that $f\in C_{0}^{\mathfrak {I}_{H}}(X)=H$.

Now we claim that H properly contains I. Since $\mathfrak {Z}(H)$ is compact and $p\notin \mathfrak {Z}(H)$, there exists $f\in C_{B}(X)$ such that $f^{\beta }$ vanishes on $\mathfrak {Z}(H)$ and $f^{\beta }(p)=1$. By the same argument as in the first part, f belongs to H, but clearly $f\notin I$. This implies that $\mathfrak {I}_{I}$ is properly contained in $\mathfrak {I}_{H}$. Now, by applying Proposition 3.4, $X(\infty _{I})\lvertneqq X(\infty _{H})\equiv Y$, which contradicts the minimality of Y in $\mathcal {E}(X)$.

(3) implies (1). If X is a locally compact, non-compact Hausdorff space, then the Alexandroff one-point compactification of X is the first element of $\mathcal {E}(X)$. $\square$

5 Application to Lindelöf One-Point Extensions

This section is devoted to the partially ordered set of all Lindelöf one-point extensions of a Tychonoff space X. Let us denote this partially ordered set by $\mathcal {E}^{L}(X)$. As we obtain below, the study of this ordered set will be meaningful when X is a locally Lindelöf space. Recall that a topological space is locally Lindelöf whenever each point of which has a Lindelöf neighborhood.

Notation 5.1

The set of all subsets of a topological space whose closures are Lindelöf is denoted by $\mathcal {L}$. Note that $\mathcal {L}$ is a set-ideal on X.

Lemma 5.2

Let X be a Tychonoff space. The closed ideal $C_{\circ }^{\mathcal {L}}(X)$ is proper and non-vanishing if and only if X is non-Lindelöf and locally Lindelöf.

Proof

It is clear that the ideal $C_{\circ }^{\mathcal {L}}(X)$ is proper if and only if X is non-Lindelöf. Assume that $C_{\circ }^{\mathcal {L}}(X)$ is a non-vanishing ideal of $C_B(X)$. Let $x\in X$. There exists $f\in C_{\circ }^{\mathcal {L}}(X)$ such that $f(x)\ne 0$. There exists $n\in \mathbb {N}$ such that x is in $|f|^{-1}\left( (1/n,+\infty )\right)$. Since $|f|^{-1}\left( [1/2n,+\infty )\right)$ is Lindelöf, we infer that x has a Lindelöf neighborhood. Conversely, assume that X is locally Lindelöf and let $x\in X$. Hence, there is a Lindelöf neighborhood $U_x$ of x. Since X is Tychonoff, there is $f\in C_B(X)$ such that $\text {coz}(f)\subseteq U_x$. Note that, for each $n\in \mathbb {N}$, $|f|^{-1}\left( [1/n,+\infty )\right)$ is a closed subset of the Lindelöf space $U_x$ and hence it is Lindelöf. Therefore, $f\in C_{\circ }^{\mathcal {L}}(X)$, and we conclude that the ideal is non-vanishing. $\square$

Lemma 5.3

Let X be a Tychonoff, non-Lindelöf and locally Lindelöf space. Then, the one-point extension $X(\infty _{L})$, where $L=C_{\circ }^{\mathcal {L}}(X)$, is Lindelöf.

Proof

Since the ideal L is non-vanishing and closed, the extension $X(\infty _{L})$ is in $\mathcal {E}(X)$. To show that $X(\infty _{L})$ is Lindelöf, let $\mathcal {U}$ be an open covering of $X(\infty _{L})$. The covering $\mathcal {U}$ must contain an open set $U_0$ which contains $\infty _{L}$. Thus, there are $f\in L$ and $n\in \mathbb {N}$ such that $|f|^{-1}\left( [0,1/n)\right) \cup \{\infty _{K}\}$ is contained in $U_0$. The set $X{\setminus } U_0$ is a closed subset of the Lindelöf space $|f|^{-1}\left( [1/n,+\infty )\right)$. So there exists a countable subset $\mathcal {V}$ of $\mathcal {U}$ which covers $X{\setminus } U_0$. Therefore $\mathcal {V}\cup \{U_0\}$ is countable covering of $X(\infty _{L})$. $\square$

Lemma 5.4

Let X be a Tychonoff non-Lindelöf space. Then the set $\mathcal {E}^{L}(X)$ is non-empty if and only if X is locally Lindelöf.

Proposition 5.5

Let X be a Tychonoff, non-Lindelöf and locally Lindelöf space. Then, $\mathcal {E}^{L}(X)=\downarrow X(\infty _{L})$, where $L=C_{\circ }^{\mathcal {L}}(X)$. Furthermore, the partially ordered set $\mathcal {E}^{L}(X)$ is a distributive lattice in which the least upper bound of each non-empty set exists.

Proof

Assume that Y is in $\downarrow X(\infty _{L})$. It means that Y is a continuous image of the Lindelöf space $X(\infty _{L})$. Hence, $Y\in \mathcal {E}^{L}(X)$. Conversely, assume that $T\in \mathcal {E}^{L}(X)$. By Proposition 3.5, there is a non-vanishing ideal H of $C_B(X)$ such that $T\equiv X(\infty _{H})$. Now let f be in H. The set $|f|^{-1}\left( [1/n,+\infty )\right)$ is a closed subset of $X(\infty _{H})$, for each $n\in \mathbb {N}$. Therefore, we can infer that $f\in L=C_{\circ }^{\mathcal {L}}(X)$. Since $H\subseteq L$, we conclude that $T\le X(\infty _{L})$. The second part follows immediately from Corollary 4.2. $\square$

6 The Minimal Unitization of Closed Ideals of $C_{B}(X)$ via One-Point Extensions of X

In this section, for a non-vanishing closed ideal H of $C_B(X)$, we focus our attention on the continuous functions on X which have continuous extensions over $X(\infty _{H})$. In particular, we try to generalize Example 1.3 and give a representation for the minimal unitization of every non-vanishing closed ideal of $C_B(X)$. We conclude the section by discussing a consequence of this result.

Definition 6.1

For every Tychonoff space X and for every proper closed ideal H of $C_B(X)$, the minimal unital subalgebra of $C_B(X)$ generated by H is the set

$$\begin{aligned} H+\mathbb {C}=\{h+{\textbf{c}}:h\in H, \hspace{0.1cm} c\in \mathbb {C}\}. \end{aligned}$$

Theorem 6.2

Let X be a Tychonoff space and let H be a proper non-vanishing closed ideal of $C_{B}(X)$. Then the algebra $H+\mathbb {C}$ is isometrically $*$-isomorphic with $C_{B}\left( X(\infty _{H})\right)$.

Proof

Suppose that H is a closed non-vanishing ideal of $C_{B}(X)$. Clearly X is locally $\mathfrak {I}_{H}$-null. Let $g\in H+\mathbb {C}$. There exist $h\in H$ and $c\in \mathbb {C}$ such that $g=h+{\textbf{c}}$. To see that g has a unique extension to $X(\infty _{H})$, define a mapping ${\hat{h}}$ from $X(\infty _{H})$ to $\mathbb {C}$ as follows.

$$\begin{aligned} {\hat{h}}(x)={\left\{ \begin{array}{ll} h(x) &{} \quad \text {if } t\in X, \\ 0 &{}\quad \text {if } t=\infty _{H}. \end{array}\right. } \end{aligned}$$

Note that $X(\infty _{H})$ is a Tychonoff extension of X, so every singleton of $X(\infty _{H})$ is closed. Therefore, $X=X(\infty _{H}){\setminus } \{\infty _{H}\}$ is open in $X(\infty _{H})$. Hence, it is enough to show that ${\hat{h}}$ is continuous at $\infty _{H}$. As for the continuity at $\infty _{H}$, let $\epsilon >0$ and define $K_{\epsilon }=|h|^{-1}\left( [\epsilon , +\infty )\right)$. Note that $K_{\epsilon }\in \mathfrak {I}_{H}$ and hence $U_{\epsilon }=\{\infty _{H}\}\cup \left( X{\setminus } K_{\epsilon }\right)$ is open in $X(\infty _{H})$. Moreover, $U_{\epsilon }$ is a subset of $|h|^{-1}\left( [0,\epsilon )\right)$. Therefore ${\hat{h}}$ is continuous at $\infty _{H}$ and hence it belongs to $C_{B}(X(\infty _{H}))$. Now define ${\hat{g}}={\hat{h}}+{\textbf{c}}$. Clearly, ${\hat{g}}$ is the unique extension of g to $X(\infty _{H})$.

Define a mapping $\theta$ from $H+\mathbb {C}$ to $C_{B}\left( X(\infty _{H})\right)$ by the rule $\theta (g)={\hat{g}}$, where ${\hat{g}}$ is the unique extension of g over $X(\infty _{H})$. Since H is non-unital, $\mathfrak {I}_{H}$ is proper and hence X is dense in $X(\infty _{H})$. Therefore $\theta$ is well-defined, it preserves algebraic operations and it is injective.

To show that $\theta$ is onto, let $f\in C_{B}\left( X(\infty _{H})\right)$. To see that the restriction $f\restriction _{X}$ belongs to $H+\mathbb {C}$, assume that $f(\infty _{H})=c$ and set $k=f-{\textbf{c}}$. Since k is a continuous function, $|k|^{-1}\left( [0,\epsilon )\right)$ is an open neighborhood of $\infty _{H}$. Hence there exists $E\in \mathfrak {I}_{H}$ such that

$$\begin{aligned} |k|^{-1}\left( [0,\epsilon )\right) =\{\infty _{H}\}\cup \left( X{\setminus } E\right) . \end{aligned}$$

Note that

$$\begin{aligned} |k|^{-1}\left( [\epsilon , +\infty )\right) = |k\restriction _{X}|^{-1}\left( [\epsilon ,+\infty )\right) , \end{aligned}$$

and thus $k\restriction _{X}$ belongs to $C_{0}^{\mathfrak {I}_{H}}(X)=H$. In consequence, $k\restriction _{X}+{\textbf{c}}=f\restriction _{X}$ is an element of $H+\mathbb {C}$ and therefore $\theta (f\restriction _{X})=f$.

Finally, we show that $\theta$ is an isometry. Note that for each $f\in H+\mathbb {C}$, we have the following inequality.

$$\begin{aligned} \Vert f\Vert _{X}\le \Vert {\hat{f}}\Vert _{X(\infty _{H})}=\Vert \theta (f)\Vert _{X(\infty _{H})}. \end{aligned}$$

For the reverse inequality, let $\epsilon >0$. There exists $t\in X(\infty _{H})$ such that

$$\begin{aligned} \Vert {\hat{f}}\Vert _{X(\infty _{H})}-\frac{\epsilon }{2}\le |{\hat{f}}(t)|. \end{aligned}$$

If $t\ne \infty _{H}$, then ${\hat{f}}(t)=f(t)$ and hence $\Vert {\hat{f}}\Vert _{X(\infty _{H})}-\epsilon \le \Vert f\Vert _{X}.$ On the other hand if $t=\infty _{H}$, since ${\hat{f}}$ is continuous on $X(\infty _{H})$ and $\infty _{H}$ is in the closure of X, there exists $x\in X$ such that

$$\begin{aligned} |{\hat{f}}(t)-f(x)|<\frac{\epsilon }{2}. \end{aligned}$$

It follows that

$$\begin{aligned} \Vert {\hat{f}}\Vert _{X(\infty _{H})}-\epsilon <|f(x)|\le \Vert f\Vert _{X}. \end{aligned}$$

Both cases imply that

$$\begin{aligned} \Vert {\hat{f}}\Vert _{X(\infty _{H})}\le \Vert f\Vert _{X}+\epsilon . \end{aligned}$$

Since $\epsilon$ is arbitrary, we deduce that $\Vert {\hat{f}}\Vert _{X(\infty _{H})}\le \Vert f\Vert _{X}.$ $\square$

Lemma 6.3

Let H be a closed ideal of $C_{B}(X)$. Then, H is generated by an idempotent if and only if $\text {Coz}(H)$ belongs to $\mathfrak {I}_{H}$.

Proof

Assume that H is a closed ideal of $C_B(X)$. If H is generated by an idempotent e, then clearly $\text {Coz}(H)=X{\setminus } Z(e)=\vert e\vert ^{-1}(1)$ and hence $\text {Coz}(H)\in \mathfrak {I}_{H}$.

Now, assume that $\text {Coz}(H)\in \mathfrak {I}_{H}$. Hence, there exists $h\in H$ such that $\text {Coz}(H)=|h|^{-1}\left( [1,+\infty )\right)$. Therefore $\text {Coz}(H)$ is clopen in X and $Z(h)=|h|^{-1}\left( [0,1)\right)$. Now, consider the mapping $e:X\rightarrow \{0,1\}$ defined as follows: for every $x\in \text {Coz}(H), e(x)=1$, and for every $x\in X{\setminus } \text {Coz}(H)$, $e(x)=0$. Note that e is continuous and idempotent. Since $Z(e)=Z(h)$, by [6, Exercise 1D.1], e is a multiple of h and h is a multiple of e. Thus, the principal ideals generated by e and h are equal, i.e., $\langle e\rangle = \langle h\rangle$ and hence $\langle e\rangle \subseteq H$. Next, assume that $f\in H$. Since $X{\setminus } Z(f)\subseteq \text {Coz}(H)$, therefore $Z(e)\subseteq Z(f)$. Using [6, Exercise 1D.1], we infer that f is a multiple of e, i.e., $f\in \langle e\rangle$. This shows that $H\subseteq \langle e\rangle$. $\square$

Theorem 6.4

Let X be a Tychonoff space and let H be a closed ideal of $C_{B}(X)$ which is not generated by an idempotent. Then $H+\mathbb {C}$ is isometrically $*$-isomorphic with $C_{B}\left( Y(\infty _{K})\right)$, where $Y=\text {Coz}(H)$ and $K=C_{0}^{\mathfrak {I}_{H}}(Y)$.

Proof

Since $Y=\text {Coz}(H)$ is a locally $\mathfrak {I}_{H}$-null space, by Lemma 1.2, the ideal $K=C_{0}^{\mathfrak {I}_{H}}(Y)$ of $C_{B}\left( Y\right)$ is non-vanishing. Regarding Theorem 6.2, $C_{0}^{\mathfrak {I}_{H}}(Y)+\mathbb {C}$ is isometrically $*$-isomorphic with $C_{B}\left( Y(\infty _{K})\right)$. So, it is enough to show that $H+\mathbb {C}$ is isometrically $*$-isomorphic with $C_{0}^{\mathfrak {I}_{H}}(Y)+\mathbb {C}$. It follows from Theorem 2.8 that the mapping

$$\begin{aligned} \Theta : H\rightarrow C_{0}^{\mathfrak {I}_{H}}(Y), \end{aligned}$$

defined by $\Theta (h)=h\restriction _{\text {Coz}(H)}$, where $h\in H$, is an isometric $*$-isomorphism. Define the mapping

$$\begin{aligned} {\bar{\Theta }}: H+\mathbb {C}\rightarrow C_{0}^{\mathfrak {I}_{H}}(Y)+\mathbb {C}, \end{aligned}$$

by the rule ${\bar{\Theta }}(h+{\textbf{c}})=\Theta (h)+{\textbf{c}}$, where $h\in H$ and $c\in \mathbb {C}$. We first show that ${\bar{\Theta }}$ is well-defined. To see this, let $h_{1}+\mathbf {c_{1}}=h_{2}+\mathbf {c_{2}}$, where $h_{1}$ and $h_{2}$ are in H and $c_{1}$ and $c_{2}$ are in $\mathbb {C}$. Since H is not generated by an idempotent, H is proper and thus $\mathbf {c_{1}}=\mathbf {c_{2}}$, and hence, $h_{1}=h_{2}$. Therefore, $\Theta (h_{1})+\mathbf {c_{1}}=\Theta (h_{2})+\mathbf {c_{2}}$. Since $\Theta$ is onto, it is clear to observe that ${\bar{\Theta }}$ is onto and it preserves addition and multiplication. Now, we show that ${\bar{\Theta }}$ is injective. Let $h_{1}$ and $h_{2}$ be in H and let $c_{1}$ and $c_{1}$ be in $\mathbb {C}$, such that $\Theta (h_{1})+\mathbf {c_{1}}=\Theta (h_{2})+\mathbf {c_{2}}$. Therefore, we have

$$\begin{aligned} h_{1}\restriction _{\text {Coz}(H)}+\mathbf {c_{1}}=h_{2}\restriction _{\text {Coz}(H)}+\mathbf {c_{2}}. \end{aligned}$$

Note that $h=h_{1}-h_{2}$ is equal to $\mathbf {c_{2}}-\mathbf {c_{1}}$ on $\text {Coz}(H)$. Since H is not generated by an idempotent, $\text {Coz}(H){\setminus } |h|^{-1}\left( [\epsilon , +\infty )\right)$ is non-empty, for every $\epsilon >0$. Choose t in $\text {Coz}(H){\setminus } |h|^{-1}\left( [\epsilon , +\infty )\right)$. Then $|h(t)|<\epsilon$ and hence $|c_{2}-c_{1}|<\epsilon$. Since $\epsilon$ is an arbitrary positive number, we have $c_{1}=c_{2}$. Moreover, $h_{1}\restriction _{\text {Coz}(H)}=h_{2}\restriction _{\text {Coz}(H)}$, and whence $h_{1}=h_{2}$. Therefore, ${\bar{\Theta }}$ is injective.

Finally, we show that ${\bar{\Theta }}$ is an isometry. It is enough to show that ${\bar{\Theta }}$ is norm-preserving. For each $h\in H$ and each $c\in \mathbb {C}$, we have

$$\begin{aligned} \Vert \Theta (h)+{\textbf{c}}\Vert _{Y}\le \Vert h+{\textbf{c}}\Vert _{X}. \end{aligned}$$

Now, we prove the reverse inequality. Let $\epsilon$ be a positive real number. There exists $x_{0}\in X$ such that

$$\begin{aligned} \Vert h+\textbf{c}\Vert _{X}\le |h(x_{0})+c|+\frac{\epsilon }{2}. \end{aligned}$$

If $x_{0}\in Y= \text {Coz}(H)$, then we can automatically observe that

$$\begin{aligned} \Vert h+{\textbf{c}}\Vert _{X}\le \Vert \Theta (h)+{\textbf{c}}\Vert _{Y}+\epsilon . \end{aligned}$$

If $x_{0}\notin Y=\text {Coz}(H)$, then $h(x_{0})=0$ and hence $|h(x_{0})+c|=|c|$. Since H is not generated by an idempotent, the set

$$\begin{aligned} \text {Coz}(H){\setminus } |h|^{-1}\left( [\epsilon , +\infty )\right) \end{aligned}$$

is non-empty. Choose p in $\text {Coz}(H){\setminus } |h|^{-1}\left( [\frac{\epsilon }{2}, +\infty \right)$. Then $|h(p)|<\frac{\epsilon }{2}$ and thus

$$\begin{aligned} |c|\le |h(p)+c|+\frac{\epsilon }{2}. \end{aligned}$$

The last inequality implies that

$$\begin{aligned} \Vert h+{\textbf{c}}\Vert _{X}\le |c|+\frac{\epsilon }{2}\le \Vert \Theta (h)+{\textbf{c}}\Vert _{Y}+\epsilon . \end{aligned}$$

Thus, we conclude that for each $\epsilon >0$,

$$\begin{aligned} \Vert h+{\textbf{c}}\Vert _{X}\le \Vert \Theta (h)+{\textbf{c}}\Vert _{Y}+\epsilon . \end{aligned}$$

Since $\epsilon$ is arbitrary, we have $\Vert h+{\textbf{c}}\Vert _{X}\le \Vert \Theta (h)+{\textbf{c}}\Vert _{Y}$. $\square$

7 The Čech–Stone Compactification of One-Point Extensions of X via Closed Ideals of $C_{B}(X)$

In this section, as a consequence of Theorem 7.1, we will observe that for each non-vanishing closed ideal H of $C_{B}(X)$, the one-point extension $X(\infty _{H})$ can be embedded in the quotient space $\beta X/\mathfrak {Z}(H)$. In addition, it is shown that the Čech–Stone compactification of $X(\infty _{H})$ is identical to $\beta X/\mathfrak {Z}(H)$. Such a description helps us to precisely identify closed ideals with countable topological generators.

Theorem 7.1

Let H be a closed ideal of $C_{B}(X)$ which is not generated by an idempotent and let $Y=\text {Coz}(H)$. Then $Y(\infty _{H})$ is embedded in $\beta X/\mathfrak {Z}(H)$. Moreover, the Čech–Stone compactification of $Y(\infty _{H})$ is equal to $\beta X/\mathfrak {Z}(H)$ (see also Lemma 1 in [17]).

Proof

Assume that $\pi : \beta X\rightarrow \beta X/\mathfrak {Z}(H)$ is the corresponding quotient mapping of the quotient space $\beta X/\mathfrak {Z}(H)$. Now define

$$\begin{aligned} \rho (t)={\left\{ \begin{array}{ll} \pi (t)=\{t\} &{}\quad \text {if } t\in \text {Coz}(H),\\ \mathfrak {Z}(H) &{}\quad \text {if } t=\infty _{H}.\end{array}\right. } \end{aligned}$$

Note that $\rho$ is well-defined, one-one and onto the subspace $\left( X\cup \mathfrak {Z}(H)\right) /\mathfrak {Z}(H)$ of $\beta X/\mathfrak {Z}(H)$.

Claim 1 $\rho$ is continuous.

Proof

Let ${\widetilde{V}}$ be an open subspace of $\beta X/\mathfrak {Z}(H)$. Note that the set $V=\bigcup {\widetilde{V}}$ is open in $\beta X$. We have to verify that $\rho ^{-1}({\widetilde{V}})$ is open in $Y(\infty _{H})$. Let $p\in \rho ^{-1}({\widetilde{V}})$. If $p\in Y=\text {Coz}(H)$, then $p\notin \mathfrak {Z}(H)$. Note that $\mathfrak {Z}(H)$ is a compact subset of $\beta X$. Thus there exists an open neighborhood $O_{p}$ of p in $\beta X$ such that $O_{p}\cap \mathfrak {Z}(H)=\emptyset$. Hence, $O_{p}\cap V\cap X$ is an open neighborhood of p in X. Note that

$$\begin{aligned} O_{p}\cap V\cap X=O_{p}\cap V\cap \text {Coz}(H). \end{aligned}$$

Therefore, $O_{p}\cap V\cap X\subseteq \rho ^{-1}({\widetilde{V}})$ and hence p is in the interior of $\rho ^{-1}({\widetilde{V}})$. Now suppose that $p=\infty _{H}$. Then $\mathfrak {Z}(H)\subseteq V$ and hence

$$\begin{aligned} \beta X{\setminus } V\subseteq \bigcup _{h\in H}\text {Coz}\left( h^{\beta }\right) . \end{aligned}$$

Since $\beta X{\setminus } V$ is compact, there exist $h_{1},\ldots ,h_{n}\in H$ such that

$$\begin{aligned} \beta X{\setminus } V\subseteq \text {Coz}(h_{1}^{\beta })\cup \cdots \cup \text {Coz}\left( h_{n}^{\beta }\right) . \end{aligned}$$

Now define

$$\begin{aligned} h=h_{1}\bar{h_{1}}+\cdots +h_{n}\bar{h_{n}}. \end{aligned}$$

Since H is an ideal of $C_{B}(X)$, the continuous function h belongs to H and

$$\begin{aligned} \beta X{\setminus } V\subseteq \text {Coz}\left( h^{\beta }\right) . \end{aligned}$$

Since $\beta X{\setminus } V$ is a compact subset of $\beta X$, there exists $m\in \mathbb {N}$, such that

$$\begin{aligned} \beta X{\setminus } V\subseteq |h^{\beta }|^{-1}\left( [1/m,+\infty )\right) . \end{aligned}$$

Now define

$$\begin{aligned} \mathcal {V}:=\left( \text {Coz}(H){\setminus } |mh|^{-1}\left( [1,+\infty )\right) \right) \cup \{\infty _{H}\}. \end{aligned}$$

The set $\mathcal {V}$ is an open neighborhood of $\infty _{H}$ in the space $Y(\infty _{H})$ and clearly

$$\begin{aligned} \mathcal {V}\subseteq \rho ^{-1}({\widetilde{V}}). \end{aligned}$$

Claim 2 $\rho : Y(\infty _{H})\rightarrow \rho \left( Y(\infty _{H})\right)$ is open.

Proof

Let V be an open subset of $Y(\infty _{H})$. To show that $\rho (V)$ is open in $\rho (Y(\infty _{H}))$, pick $z\in \rho (V)$. There exists $t\in V$ such that $z=\rho (t)$.

Case 1. If $t\in \text {Coz}(H)$, then $z=\pi (t)=\{t\}$. Note that $t\notin \mathfrak {Z}(H)$ and $\mathfrak {Z}(H)$ is a compact subset of $\beta X$. Hence, there exists an open neighborhood W of t in $\beta X$ such that $W\cap \mathfrak {Z}(H)=\emptyset$ and $W\cap X\subseteq V$. Therefore, ${\widetilde{W}}:=\{\pi (t): t\in W\}$ is open in $\beta X/\mathfrak {Z}(H)$. Now consider

$$\begin{aligned} \mathcal {O}:= {\widetilde{W}}\cap \rho \left( Y(\infty _{H})\right) . \end{aligned}$$

The set $\mathcal {O}$ is an open neighborhood of z in $\rho \left( Y(\infty _{H})\right)$. Clearly $\mathcal {O}$ is a subset of $\rho (V)$. Thus, z is an interior point of $\rho (V)$.

Case 2. If $z=\mathfrak {Z}(H)$, then $\infty _{H}\in V$. There exists $h\in H$ such that

$$\begin{aligned} \left( \text {Coz}(H){\setminus } |h|^{-1}\left( [1,+\infty )\right) \right) \cup \{\infty _{H}\}\subseteq V. \end{aligned}$$

The mapping $h^{\beta }$ vanishes on $\mathfrak {Z}(H)$. Therefore, there exists $g\in C_{B}\left( \beta X/\mathfrak {Z}(H)\right)$ such that $h^{\beta }=g\circ \pi$. The subset $|g|^{-1}\left( [1,+\infty )\right)$ is closed in $\beta X/\mathfrak {Z}(H)$. Let $\mathcal {W}$ be the complement of $|g|^{-1}\left( [1,+\infty )\right)$ in $\beta X/\mathfrak {Z}(H)$. We claim that

$$\begin{aligned} \mathcal {W}\cap \rho \left( Y(\infty _{H})\right) \subseteq \rho (V). \end{aligned}$$

Suppose that x is in $\text {Coz}(H)$ such that $w=\rho (x)$ belongs to $\mathcal {W}\cap \rho \left( Y(\infty _{H})\right)$. Note that $|g(w)|<1$. Since $w=\pi (x)$, thus

$$\begin{aligned} g(w)=g\circ \pi (x)=h^{\beta }(x)=h(x). \end{aligned}$$

Therefore, $x\notin |h|^{-1}\left( [1,+\infty )\right)$ and hence $x\in V$. This implies that $w\in \rho (V)$ and hence w is an interior point of $\rho (V)$.

$\square$

Corollary 7.2

Let X be a Tychonoff space, and let Y be a Tychonoff one-point extension of X. Then, there exists a non-vanishing closed ideal H of $C_{B}(X)$ such that the Čech–Stone compactification of Y is equal to $\beta X/\mathfrak {Z}(H)$.

As an application of Theorem 7.1, some topological properties of special closed ideals of $C_{B}(X)$ are characterized in Proposition 7.4. Recall that a closed ideal I in a commutative unital Banach algebra $\left( {\textbf{A}},\Vert \cdot \Vert \right)$ has a countable topological generator if there exists a sequence $(a_{n}: n\in \mathbb {N})$ in I such that $\text {cl}_{\Vert \cdot \Vert }\left( \langle a_{n}: n\in \mathbb {N}\rangle \right) =I$, where $\langle a_{n}: n\in \mathbb {N}\rangle$ stands for the ideal generated by the sequence $(a_{n}: n\in \mathbb {N})$. If there exists an element $b\in I$ such that $\text {cl}_{\Vert \cdot \Vert }\left( \langle b\rangle \right) =I$, then I is called a topological principal ideal. In that case b is said to be a topological generator for I.

We will need to use the following result (see, e.g., [18]).

Proposition 7.3

[18, Proposition 7.140] Let X be a compact Hausdorff space. Then, the maximal ideal $M_{X}(x_{0})$ in $C_{B}(X)$ has a topological generator if and only if $x_{0}$ is a $G_{\delta }$-point of X.

In the first step, we intend to improve Proposition 7.3 and extend it to Tychonoff spaces.

Proposition 7.4

Let X be a Tychonoff space. For $p\in \beta X$, the following statements are equivalent.

• (1) The point p is a $G_{\delta }$-point of $\beta X$ and hence $p\in X$, (see e.g., [6], Corollary 9.6).

• (2) The ideal $M_{X}(p)$ has a topological generator.

• (3) The ideal $M_{X}(p)$ has a countable topological generator.

Proof

Consider the isometric $*$-isomorphism $\beta$ from $C_{B}(X)$ onto $C(\beta X)$ such that $\beta (f)=f^{\beta }$, for each $f\in C_{B}(X)$ (Sect. 2.1). Note that

$$\begin{aligned}\beta \left( M_{X}(p)\right) =M_{\beta X}(p).\end{aligned}$$

Now Proposition 7.3 implies that the statements (1) and (2) are equivalent.

(3) implies (2). Let $(f_{n}:n\in \mathbb {N})$ be a sequence in $M_{X}(p)$ such that

$$\begin{aligned} \text {cl}_{\Vert \cdot \Vert }\left( \langle f_{n}: n\in \mathbb {N}\rangle \right) =M_{X}(p). \end{aligned}$$

The mapping $|f_{n}|\vee {\textbf{1}}$ is a unit of $C_{B}(X)$, for every $n\in \mathbb {N}$. Define $g_{n}\in C_{B}(X)$, for each $n\in \mathbb {N}$ as follows.

$$\begin{aligned} g_{n}:={\left\{ \begin{array}{ll} \frac{f_{n}}{|f_{n}|\vee {\textbf{1}}} &{} \quad \text {if } \Vert f_{n}\Vert _{X}>1,\\ f_{n} &{}\quad \text {if } \Vert f_{n}\Vert _{X}\le 1. \end{array}\right. } \end{aligned}$$

Since

$$\begin{aligned} \langle f_{n}: n\in \mathbb {N}\rangle =\langle g_{n}: n\in \mathbb {N}\rangle , \end{aligned}$$

the sequence $(g_{n}:n\in \mathbb {N})$ is also a topological generator for $M_{X}(p)$. Note that $\Vert g_{n}\Vert _{X}\le 1$, for each $n\in \mathbb {N}$. Define

$$\begin{aligned} g:=\sum _{n=1}^{+\infty }\frac{\sqrt{|g_{n}|}}{2^{n}}:X\rightarrow \mathbb {C}. \end{aligned}$$

Note that each partial sum $\sum _{k=1}^{n}\frac{\sqrt{|g_{k}|}}{2^{k}}$ belongs to $M_{X}(p)$, for each $n\in \mathbb {N}$. Since $M_{X}(p)$ is a closed ideal, we infer that $g\in M_{X}(p)$. Note that since $|g_{n}|\le 2^{2n}g^{2}$, for each $n\in \mathbb {N}$, the mapping

$$\begin{aligned} k_{n}(x):={\left\{ \begin{array}{ll} \frac{g_{n}(x)}{g(x)} &{}\quad \text {if }\ x\in \text {coz}(g)\\ 0 &{}\quad \text {if }\ x\in Z(g)\end{array}\right. } \end{aligned}$$

belongs to $C_{B}(X)$ and $k_{n}g=g_{n}$, for each $n\in \mathbb {N}$. Therefore

$$\begin{aligned} M_{X}(p)=\text {cl}_{\Vert \cdot \Vert }\left( \langle g_{n}: n\in \mathbb {N}\rangle \right) \subseteq \text {cl}_{\Vert \cdot \Vert }\left( \langle g\rangle \right) . \end{aligned}$$

Since $g\in M_{X}(p)$, we infer that $\text {cl}_{\Vert \cdot \Vert }\left( \langle g\rangle \right)$ is a subset of $M_{X}(p)$ and hence $M_{X}(p)$ has a topological generator. $\square$

The results below give a complete description of a closed ideal of $C_{B}(X)$ with countable topological generator.

Theorem 7.5

Let X be a Tychonoff space, and let H be a proper closed ideal of $C_{B}(X)$. The following statements are equivalent.

• (1) The ideal H has a topological generator.

• (2) The ideal H has a countable topological generator.

• (3) The point $\infty _{H}$ is a $G_{\delta }$-point of $\beta \left( Y\left( \infty _{H}\right) \right)$, where $Y=\text {Coz}(H)$.

• (4) The spectrum $\mathfrak {s}\mathfrak {p}(H)$ is $\sigma$-compact.

• (5) The spectrum $\mathfrak {s}\mathfrak {p}(H)$ is Lindelöf.

• (6) There exists $h\in H$ such that $\mathfrak {s}\mathfrak {p}(H)=\text {coz}(h^{\beta })$.

• (7) The set-ideal $\mathfrak {I}_{H}$ has a countable generator.

Proof

If H is generated by an idempotent, conditions $(1)-(7)$ are all satisfied. Thus, we only focus on the case that H is not generated by an idempotent. Theorem 6.4 implies that there exists an isometric $*$-isomorphism ${\bar{\Theta }}$ from $H+\mathbb {C}$ onto $C_{0}^{\mathfrak {I}_{H}}(Y)+\mathbb {C}$ such that ${\bar{\Theta }}\left( H\right) =M_{Y}(\infty _{H})$, where $Y=\text {Coz}(H)$. Following Proposition 7.4, the statements (1), (2) and (3) are equivalent. Moreover, since $\mathfrak {s}\mathfrak {p}(H)$ is locally compact, by [5, Exercise 3.8.C(b)], the space is Lindelöf if and only if it is $\sigma$-compact. Therefore, the statements (4) and (5) are equivalent, too.

(3) implies (4). By Theorem 7.1, there exists an embedding $\rho$ from the space $Y(\infty _{H})$ onto the subspace $\left( X\cup \mathfrak {Z}(H)\right) /\mathfrak {Z}(H)$ of $\beta X/\mathfrak {Z}(H)$ of $\beta X/\mathfrak {Z}(H)$ such that $\rho (\infty _{H})=\mathfrak {Z}(H)$ and the Čech–Stone compactification of $Y(\infty _{H})$ is equal to the quotient space $\beta X/\mathfrak {Z}(H)$. Thus by our hypothesis, since $\infty _{H}$ is a $G_{\delta }$-point of $\beta \left( Y\left( \infty _{H}\right) \right)$, we infer that $\mathfrak {Z}(H)$ is a $G_{\delta }$-point of $\beta X/\mathfrak {Z}(H)$. If $\pi : \beta X\rightarrow \beta X/\mathfrak {Z}(H)$ is the corresponding quotient mapping of the quotient space $\beta X/\mathfrak {Z}(H)$, then $\beta X/\mathfrak {Z}(H)$ is the one-point compactification of $\pi \left( \mathfrak {s}\mathfrak {p}(H)\right)$. Hence, $\pi \left( \mathfrak {s}\mathfrak {p}(H)\right)$ is $\sigma$-compact. Note that the quotient mapping $\pi$ is perfect and injective on $\mathfrak {s}\mathfrak {p}(H)$. Therefore, $\mathfrak {s}\mathfrak {p}(H)$ is $\sigma$-compact as well.

(4) implies (3). Note that since $\beta X/\mathfrak {Z}(H)$ is homeomorphic to the one-point compactification of the locally compact space $\mathfrak {s}\mathfrak {p}(H)$ and since $\mathfrak {s}\mathfrak {p}(H)$ is $\sigma$-compact, the point $\mathfrak {Z}(H)$ is a $G_{\delta }$-point of $\beta X/\mathfrak {Z}(H)$. Now since $\rho$ maps $Y(\infty _{H})$ onto the subspace $\left( X\cup \mathfrak {Z}(H)\right) /\mathfrak {Z}(H)$ with $\rho (\infty _{H})=\mathfrak {Z}(H)$, the point $\infty _{H}$ is a $G_{\delta }$-point of $\beta \left( Y\left( \infty _{H}\right) \right)$.

(5) implies (6). Since $\mathfrak {s}\mathfrak {p}(H)$ is Lindelöf, there exists a sequence $(h_{n}: n\in \mathbb {N})$ in H with $\Vert h_{n}\Vert _{X}\le 1$, for each $n\in \mathbb {N}$ such that

$$\begin{aligned} \mathfrak {s}\mathfrak {p}(H)=\bigcup _{n\in \mathbb {N}}\text {coz}(h_{n}^{\beta }). \end{aligned}$$

The mapping

$$\begin{aligned} h:=\sum _{n=1}^{+\infty }\frac{|h_{n}|}{2^{n}}:X\rightarrow \mathbb {C} \end{aligned}$$

is well-defined and continuous by the Weierstrass M-test. Also, since H is closed, we infer that $h\in H$. Furthermore, note that

$$\begin{aligned} \bigcup _{n\in \mathbb {N}}\text {coz}(h_{n}^{\beta })=\text {coz}(h^{\beta }). \end{aligned}$$

(6) implies (7). Suppose that $g\in H$. Then, the set $\text {cl}_{\beta X}|g|^{-1}\left( [1,+\infty )\right)$ is a subset of $\mathfrak {s}\mathfrak {p}(H)$ and since it is compact, there exists $m\in \mathbb {N}$ such that

$$\begin{aligned} \text {cl}_{\beta X}|g|^{-1}\left( [1,+\infty )\right) \subseteq |mh^{\beta }|^{-1}\left( [1,+\infty )\right) . \end{aligned}$$

Hence, the set-ideal $\mathfrak {I}_{H}$ is generated by the following family:

$$\begin{aligned} \left\{ |mh|^{-1}\left( [1,+\infty )\right) : m\in \mathbb {N}\right\} . \end{aligned}$$

Therefore, the set-ideal $\mathfrak {I}_{H}$ has a countable generator.

(7) implies (4). Following Corollary 2.4,

$$\begin{aligned}\mathfrak {s}\mathfrak {p}(H)=\bigcup _{h\in H}\text {cl}_{\beta X} |h|^{-1}\left( [1,+\infty )\right) . \end{aligned}$$

Since the set-ideal $\mathfrak {I}_{H}$ has a countable generator, thus there exists a sequence $(h_{n}: n\in \mathbb {N})$ of H such that

$$\begin{aligned} \mathfrak {I}_{H}:=\langle \left\{ |h_{n}|^{-1}\left( [1,+\infty )\right) : n\in \mathbb {N}\right\} \rangle . \end{aligned}$$

Therefore, we infer that

$$\begin{aligned} \mathfrak {s}\mathfrak {p}(H)=\bigcup _{n\in \mathbb {N}}\text {cl}_{\beta X} |h_{n}|^{-1}\left( [1,+\infty )\right) , \end{aligned}$$

which means that $\mathfrak {s}\mathfrak {p}(H)$ is $\sigma$-compact. $\square$

8 Multiplier Algebras of Closed Ideals of $C_{B}(X)$

In Sect. 6, for a Tychonoff space X, we established a connection between the minimal unitization of closed ideals of $C_B(X)$ which do not have an idempotent generator and Tychonoff one-point extensions of special open subsets of X. Note that the minimal unitization of a closed ideal H of $C_{B}(X)$, in some sense, can be regarded as a minimal $C^{*}$-algebra containing H as an ideal. In this section, we consider the problem of determining the multipliers of any closed ideal of the Banach algebra $C_{B}(X)$. We recall that the multiplier algebra of a $C^{*}$-algebra A is a maximal unitization of A, in the category of $C^{*}$-algebras, in which A is an essential ideal (see, e.g., [2]). We need some preliminary results about the multipliers of a commutative semi-simple Banach algebra.

We recall that for a $C^{*}$-algebra A, the linear operator $T:A\rightarrow A$ is called a multiplier of A if $aT(b)=T(a)b$ holds for all $a,b\in A$. M(A) denotes the collection of all multipliers of A. It is well known that all elements of M(A) are bounded linear operators and hence M(A) equipped with the operator-norm forms a $C^{*}$-algebra.

Let X be a locally compact Hausdorff space. It is well known that the multiplier algebra of the closed ideal $C_{0}(X)$ is equal to $C_{B}(X)$ (see, e.g., [1]). As a more subtle result, for an arbitrary Tychonoff space, the characterization of the multiplier algebra of an arbitrary closed ideal of $C_{B}(X)$ is given.

Proposition 8.1

Let X be a Tychonoff space, and let H be a closed ideal of $C_{B}(X)$. Then, the multiplier algebra M(H) is isometrically $*$-isomorphic with the Banach algebra $C_{B}\left( \text {Coz}(H)\right)$.

Proof

Let $f\in C_{B}\left( \text {Coz}(H)\right)$. Define the mapping $T_{f}(h):H\rightarrow H$ as follows.

$$\begin{aligned} T_{f}(h)(x):={\left\{ \begin{array}{ll} f(x)h(x) &{}\quad \text {if } x\in \text {Coz}(H),\\ 0 &{}\quad \text {if } x\in X{\setminus } \text {Coz}(H). \end{array}\right. } \end{aligned}$$

There exists $K>0$ such that $\Vert f\Vert _{\text {Coz}(H)}\le K$. It is clear that $T_{f}(h)\in C_{B}(X)$. We claim that $T_{f}(h)\in H$. For an arbitrary positive number $\epsilon >0$,

$$\begin{aligned} |T_{f}(h)|^{-1}\left( [\epsilon ,+\infty )\right) \subseteq |h|^{-1}\left( [\frac{\epsilon }{M}, +\infty )\right) =|\frac{M}{\epsilon } h|^{-1}\left( [1,+\infty )\right) . \end{aligned}$$

Note that H is an ideal of $C_{B}(X)$ and hence $|T_{f}(h)|^{-1}\left( [\epsilon ,+\infty )\right) \in \mathfrak {I}_{H}$. Therefore $T_{f}(h)\in C_{0}^{\mathfrak {I}_{H}}(X)=H$.

Clearly $T_{f}$ is a multiplier on H and $\Vert T_{f}\Vert \le \Vert f\Vert _{\text {Coz}(H)}$. Conversely, let T be an arbitrary multiplier of H. For every $x\in \text {Coz}(H)$, there exists $h\in H$ such that $h(x)\ne 0$. For any two $h_{1}, h_{2}\in H$ with $h_{1}(x)\ne 0$ and $h_{2}(x)\ne 0$, we have

$$\begin{aligned} \frac{T(h_{1})(x)}{h_{1}(x)}=\frac{T(h_{2})(x)}{h_{2}(x)}, \end{aligned}$$

since $h_{2}T(h_{1})=T(h_{2})h_{1}$. Thus we define the mapping $f:\text {Coz}(H)\rightarrow \mathbb {C}$ by the rule $f(x)=\frac{T(g)(x)}{g(x)}$, for some $g\in H$ such that $g(x)\ne 0$. Note that since g is nonzero on a neighborhood of x, the mapping f is continuous at each $x\in \text {Coz}(H)$. Moreover, it is clear that

$$\begin{aligned} T(g)(x)=T_{f}(g)(x), \qquad \left( \text {for all} x\in X \text {and all} g\in H \right) . \end{aligned}$$

Now, let $x\in \text {Coz}(H)$. There exists $h\in H$ such that $h(x)=1$. Define $h_{1}=\frac{h}{|h|\vee {\textbf{1}}}$. Clearly $h_{1}\in C_{B}(X)$ and $\Vert h_{1}\Vert _{X}=1$. Therefore

$$\begin{aligned} |f(x)|=|T(h_{1})(x)|\le \Vert T(h_{1})\Vert _{X}\le \Vert T\Vert \Vert h_{1}\Vert _{X}=\Vert T\Vert . \end{aligned}$$

It follows that the mapping $f\mapsto T_{f}$ provides an isometric $*$-isomorphism between $C_{B}\left( \text {Coz}(H)\right)$ and M(H). $\square$

Evidently for every non-vanishing closed ideal H of $C_{B}(X)$,$\text {Coz}(H)$ is equal to X. For instance, the subsequent result is an immediate consequence.

Corollary 8.2

Let X be a Tychonoff space. For every non-vanishing closed ideal H of $C_{B}(X)$, the multiplier algebra M(H) is isometrically $*$-isomorphic with the Banach algebra $C_{B}(X)$.