Additive Local Multiplications and zero-preserving maps on $C(X)$

TL;DR

This paper characterizes topological conditions on compact Hausdorff spaces that determine when additive local multiplications on continuous function spaces are actual multiplications, and describes zero-preserving maps in terms of real and imaginary parts.

Contribution

It establishes topological criteria for when additive local multiplications are genuine multiplications and characterizes zero-preserving maps on $C(X)$.

Findings

Topological conditions equivalent to all additive local multiplications being multiplications.

Characterization of zero-preserving maps as linear combinations of real and imaginary parts.

Conditions under which additive maps on $C(X)$ are necessarily of a specific form.

Abstract

Suppose $X$ is a compact Hausdorff space. In terms of topolocical properties of $X$, we find topological conditions on $X$ that are equivalent to each of the following: 1. every additive local multiplication on $C\left( X\right) $ is a multiplication, 2. every additive local multiplication on $C_{R}\left( X\right) $ is a multiplication, and 3. every additive map on $C\left( X\right) $ that is zero-preserving (i.e., $f\left( x\right) =0$ implies $\left( Tf\right) \left( x\right) =0$) has the form $T\left( f\right) =T\left( 1\right) \operatorname{Re}f+T\left( i\right) \operatorname{Im}f$.

Equations

$$T\left(x\right)=a_{x}x\text{ (}x{a}_x\text{)}.$$

$$T\left(x\right)=ax\text{ (}xa\text{).}$$

$$a=T\left(1\right)\text{,}$$

$$T=L_{T\left(1\right)}\text{ (}{R}_{T\left(1\right)}\text{)},$$

$$f\left(x\right)=0\text{ implies }\left(Tf\right)\left(x\right)=0.$$

$$T\left(f\right)=T\left(1\right)\mathrm{Re}\left(f\right)+T\left(i\right)\mathrm{Im}\left(f\right)\text{.}$$

$$f\left(x\right)=0\Rightarrow T\left(f\right)\left(x\right)=0\text{.}$$

$$T\left(f\right)=T\left(1\right)\mathrm{Re}\left(f\right)+T\left(i\right)\mathrm{Im}\left(f\right).$$

$$T\left(r\cdot 1\right)=rT\left(1\right)=0.$$

$$f\left(x_n\right)=a-\frac{[10^{n}a]}{10^n}\text{ for every }n\in \mathbb{N}\text{, and }f\left(x\right)=0\text{.}$$

$$T\left(a-F\right)\left(x\right)=T\left(a\right)\left(x\right)-T\left(F\right)\left(x\right)$$

$$=T\left(a\right)\left(x\right)-0=T\left(a\right)\left(x\right)\ne 0\text{,}$$

$$T\left(a-F\right)\left(x\right)\ne 0\text{.}$$

$$\left(a-F\right)\left(x_k\right)=a-f\left(x_k\right)=\frac{[10^{k}a]}{10^k}=r_k\in \mathbb{Q}\text{.}$$

$$\left(a-F-r_k\right)\left(x_k\right)=0.$$

$$T\left(a-F\right)\left(x_k\right)=T\left(a-F-r_k\right)\left(x_k\right)=0\text{,}$$

$$0=T\left(g-g\left(x\right)\cdot 1\right)\left(x\right)=T\left(g\right)\left(x\right)-T\left(g\left(x\right)\cdot 1\right)=T\left(g\right)\left(x\right)\text{.}$$

$$T\left(h\right)=T\left(h_1\right)+T\left(i{h}_2\right)=T\left(i{h}_2\right)=T\left(i\right)h_2\text{.}$$

$$f\left(x_n\right)=\frac{1}{n}\cdot |\sin \left(\frac{n\pi }{2}\right)|\text{ for every }n\in \mathbb{N}\text{, and }f\left(x\right)=0\text{.}$$

$$g\left(x_n\right)=\frac{1}{n}\text{ for every }n\in \mathbb{N}\text{, and }g\left(x\right)=0\text{.}$$

$$T\left(H\right)=T\left(F+iG\right)=T\left(i\right)G=h_H\cdot \left(F+iG\right)$$

$$T\left(i\right)\left(x_n\right)G\left(x_n\right)$$

$$T\left(i\right)\left(x_n\right)\cdot \frac{1}{n}$$

$$T\left(i\right)\left(x_n\right)$$

$$T\left(i\right)\left(x\right)=i{h}_H\left(x\right)=h_H\left(x\right)+i{h}_H\left(x\right)\text{.}$$

$$T_i\left(f\right)=T\left(\tilde{f}\right){|}_{K_i}\text{ for every }f\in C\left(K_i\right)\text{,}$$

$$T\left(\widetilde{f_1}\right)-T\left(\widetilde{f_2}\right)=T\left(\widetilde{f_1}-\widetilde{f_2}\right)=h\cdot \left(\widetilde{f_1}-\widetilde{f_2}\right)$$

$$T\left(\widetilde{f_1}\right){|}_{K_i}-T\left(\widetilde{f_2}\right){|}_{K_i}$$

$$=0\text{.}$$

$$\frac{{\bar{f}}_A}{f_A}=\frac{u^2-v^2}{u^2+v^2}-i\frac{2uv}{u^2+v^2}\text{.}$$

$$h=g\cdot \left(\widetilde{h_1}-i\widetilde{h_2}\right)\text{ on }A\text{, and }h=0\text{ on }Z\left(g\right)\text{.}$$

$$k=g\cdot \left(h_r+i|h_r|\right)\text{ on }A\text{, and }k=0\text{ on }Z\left(g\right)\text{.}$$

Full text

1. Introduction

Suppose $X$ is a topological space. Let $C\left(X\right)$ and $C_{R}\left(X\right)$ be the set of all complex continuous functions and real continuous functions on $X$, respectively. This paper studies local multiplications and zero-preserving maps on the algebra $C\left(X\right)$ and $C_{R}\left(X\right)$ when $X$ is a compact Hausdorff space. We find an interesting interplay between the algebraic or linear-algebraic conditions and unusual topological properties of the space $X$.

Suppose $\mathcal{A}$ is a ring with identity. A map $T$ on $\mathcal{A}$ is a local multiplication if, for each $x\in$ $\mathcal{A}$, there is an $a_{x}\in\mathcal{A}$ such that

[TABLE]

The map $T$ is a left (right) multiplication if there is an $a\in\mathcal{A}$ such that, for every $x\in\mathcal{A}$,

[TABLE]

In this case we must have

[TABLE]

and we write

[TABLE]

i.e., left (right) multiplication by the element $T\left(1\right)$. When that algebra $\mathcal{A}$ is commutative, there is no difference between left and right, and we write $M_{T\left(1\right)}$ for $L_{T\left(1\right)}$. A map $T$ on $\mathcal{A}$ is an additive map if for every $x$ and $y$ in $\mathcal{A}$, $T\left(x+y\right)=T\left(x\right)+T\left(y\right)$.

There has been a lot of work characterizing cases in which every local multiplication of a certain type is a multiplication. In 1983 the paper [5] of D. Hadwin contains an early result on local multiplications. In 1994 the so-called ”Hadwin Lunch Bunch” [6] gave necessary conditions for local multiplications in rings with many idempotents to be multiplications.

In $1997$ D. Hadwin and J. W. Kerr [7] studied $\mathcal{R}$-linear local multiplications on an algebra $\mathcal{A}$ over commutative ring $\mathcal{R}$ with identity, and gave conditions that implied that every local multiplication on $\mathcal{A}$ is a multiplication. When $\mathcal{R}$ is the ring $\mathbb{Z}$ of integers, the $\mathcal{R}$-linear maps are simply the additive maps. When $\mathcal{A}$ is a vector space over the rational numbers $\mathbb{Q}$, the additive maps are precisely the $\mathbb{Q}$-linear maps. When $\mathcal{A}$ is a topological vector space over $\mathbb{R}$, the continuous additive maps are precisely the $\mathbb{R}$-linear ones. In [7] D. Hadwin and J. W. Kerr proved, for a large class of unital C*-algebras every additive local left (right) multiplication is a left (right) multiplication. These algebras include ones for which the set of finite-dimensional (dimension greater than $1$) separate the points of the algebra. However, for additive maps, the commutative algebras, i.e., ones for which every irreducible representation is $1$-dimensional, were not considered. A unital C*-algebra is commutative if and only if it is isomorphic to $C\left(X\right)$ for some compact Hausdorff space $X$. Note that Hadwin and Kerr’s results imply that every local left (right) multiplication on the algebra $\mathbb{M}_{2}\left(C\left(X\right)\right)$ of $2\times 2$ matrices over $C\left(X\right)$ is a left (right) multiplication.

There is a vast literature on local derivations and local automorphisms, e.g., [1], [2], [8], [9], [10], [11], [14], [16].

Another related active area of research is the study of maps that preserve a particular property. According to MATHSCINET there have been almost $200$ papers studying linear preservers and at least $48$ papers studying additive maps that preserve some property.

In this paper we restrict ourselves to the case where our algebra is the space $C\left(X\right)$ of complex continuous functions on a compact Hausdorff space $X$. It was shown in [5, Theorem 6] that every $\mathbb{C}$-linear map on $C\left(X\right)$ that is a local multiplication is a multiplication. We study additive or $\mathbb{R}$-linear local multiplications on $C\left(X\right)$. We also study maps $T$ that are zero-preserving, i.e., satisfying, for every $f\in C\left(X\right)$ and every $x\in X$,

[TABLE]

In Section $2$ we consider those compact Hausdorff spaces $X$ for which every additive local multiplication on $C\left(X\right)$ must be a multiplication, and we call such $X$ an $\eta$-space. If every local multiplication on $C_{R}\left(X\right)$ is a multiplication, we can $X$ a real $\eta$-space. There is a vast difference between these two concepts. We prove (Theorem 2) that if the set of points $x\in X$ that are a limit of a sequence in $X\backslash\left\{x\right\}$ is dense in $X$, then $X$ is an $\eta$-space. In particular, if $X$ is first countable, then $X$ is an $\eta$-space if and only if $X$ has no isolated points. We also prove (Theorem 3) that the closure of the union of a collection of $\eta$-subspaces of $X$ is also an $\eta$-space. Hence every compact Hausdorff space has a unique maximal compact subspace that is an $\eta$-space. It follows (Theorem 4) that the Cartesian product of an $\eta$-space with any compact Hausdorff space is an $\eta$-space. We also construct many spaces that are not $\eta$-spaces. The conjugation map on $C\left(X\right)$ is $\mathbb{R}$-linear and is never a multiplication. We prove (Theorem 6) that the conjugation map is a local multiplication on $C\left(X\right)$ if and only if $X$ is an F-space in the sense of L. Gillman and M. Jerison [4]; this is also equivalent to the set of $\mathbb{R}$-linear local multiplications on $C\left(X\right)$ being precisely the maps of the form

[TABLE]

We also prove (Theorem 7) that there is a nonzero function $g$ such that the map $T\left(f\right)=g\bar{f}$ is a local multiplication if and only if there is a nonempty open Fσ-subset (i.e., a countable union of closed sets) of $X$ that is an F-space. As a consequence, we prove (Corollary 5) that if no nonempty open Fσ-subset of $X$ is an F-space, then every $\mathbb{R}$-linear (or continuous additive) local multiplication on $C\left(X\right)$ is a multiplication. We also characterize (Proposition 1) the additive local multiplications on $\ell^{\infty}=C\left(\beta\left(\mathbb{N}\right)\right)$, where $\beta\left(\mathbb{N}\right)$ is the Stone-Čech compactification of $\mathbb{N}$.

In Section $3$ we focus on the additive maps on $C\left(X\right)$ that are zero-preserving, i.e., for every $f\in C\left(X\right)$ and for every $x\in X$,

[TABLE]

Since the conjugation map satisfies this property, we can’t expect all of these maps to be multiplications. But we can hope for them to be of the form

[TABLE]

We call a space $X$ for which every additive zero-preserving map has the above form an $\upsilon$-space. We show (Theorem 10) that this property is equivalent to every additive zero-preserving map on $C_{R}\left(X\right),$ the set of real-valued continuous functions on $X$, is a multiplication. Thus every $\upsilon$-space is a real $\eta$-space. This characterization allows us to carry over results on $\eta$-spaces to those of $\upsilon$-spaces. In particular we prove (Theorem 12) that, if the set of sequential limit points is dense in $X$ or $X$ is the closure of the union of a family of compact $\upsilon$-subspaces, or $X$ is the product of a compact Hausdorff space and an $\upsilon$-space, then $X$ is an $\upsilon$-space. This also shows that every compact Hausdorff space has a unique maximal compact subspace that is an $\upsilon$-space. We also show (Theorem 9) that every $\mathbb{R}$-linear zero-preserving map on $C_{R}\left(X\right)$ is a multiplication.

In Section 4 We introduce the notions of $\mathfrak{q}$-point and strong $\mathfrak{q}$-point, which are generalizations of a sequential limit point. We prove (Theorem 13) that if the set of strong $\mathfrak{q}$-points is dense, then $X$ is an $\eta$-space, and if the set of $\mathfrak{q}$-points is dense, then $X$ is an $\upsilon$-space. It turns out (Lemma 3) that $x$ is a $q$-point if and only if $x$ is not a P-point in the sense of L. Gillman and M. Henriksen [3], and we show that the set of $\mathfrak{q}$-points of $X$ is dense if and only if $X$ has no isolated points.

In Section 5, we present our main theorems. Our first main theorem (Theorem 14) characterizes $\upsilon$-spaces and real $\eta$-spaces: Suppose $X$ is a compact Hausdorff space. The following are equivalent:

1. (1)

$X$ is a $\upsilon$-space 2. (2)

$X$ is a real $\eta$-space 3. (3)

The set of $\mathfrak{q}$-points of $X$ is dense in $X$ 4. (4)

$X$ has no isolated points.

Our second main theorem (Theorem 15) topologically characterizes $\eta$-spaces: Suppose $X$ is a compact Hausdorff space. Then $X$ is an $\eta$-space if and only if no nonempty open Fσ set in $X$ is an F-space.

In Remark 4 we describe how to construct the maximal $\eta$-subspace and the maximal $\upsilon$-subspace of a compact Hausdorff space $X$.

We conclude with remarks about $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$, where $\beta\left(\mathbb{N}\right)$ denotes the Stone-Čech compactification of the set $\mathbb{N}$ of positive integers. We have $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ is an $\upsilon$-space since $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ has no isolated points. However, $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ is not an $\eta$-space, since it is an F-space. We also remark that W. Rudin [13] and S. Shelah (see [16]) have proved that the assertion that every point in $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ is a $q$-point is independent from the axioms of set theory (ZFC).

For topological notions we refer the reader to [15] and [4].

2. $\eta$-spaces

Suppose $Y$ is any completely regular Hausdorff space, then $\beta\left(Y\right)$ denotes the Stone-Čech Compactification of $Y$. For any real number $r$, $[r]$ is the largest integer not more than $r$.

Definition 1**.**

Suppose $X$ is a compact Hausdorff space. If every additive local multiplication on $C\left(X\right)$ is a multiplication, then we call $X$ an $\mathbf{\eta}$-space. If every additive local multiplication on $C_{R}\left(X\right)$ is a multiplication, we call $X$ a real $\eta$-space.

Example 1**.**

Suppose $X=\left\{a\right\}$ is a singleton. Thus $C\left(S\right)\cong\mathbb{C}$. Every additive map ($\mathbb{Q}$-linear map) on $\mathbb{C}$ is a local multiplication. Given a linear basis $B$ for $\mathbb{C}$ over $\mathbb{Q}$, we can get $card\left(B\right)=card\left(\mathbb{R}\right)$. Thus the cardinality of the set of all $\mathbb{Q}$-linear (i.e., additive) maps on $\mathbb{C}$ is $card\left(B\right)^{card\left(\mathbb{R}\right)}=2^{\left(2^{\aleph_{0}}\right)}$. But the cardinality of the set of all multiplications on $\mathbb{C}$ is $2^{\aleph_{0}}$. Thus no singleton is an $\eta$-space.

Definition 2**.**

Suppose $\{Y_{i}\}$ is a family of topology spaces. Let $Y$ be the disjoint union of the $Y_{i}$’s (If there are two sets intersecting, then let $Y_{i}$ be $Y_{i}\times\left\{i\right\}$). Define a subset $U$ of $Y$ to be open if and only if the intersection of $U$ and each $Y_{i}$ is open in $Y_{i}$. Thus we defined the disjoint union topology on $Y$.

Theorem 1**.**

Suppose $X$ is the disjoint union of compact Hausdorff spaces $Y$ and $Z$. Then $X$ is an $\eta$-space if and only if $Y$ and $Z$ are $\eta$-spaces.

Proof.

We have $C\left(X\right)$ is isomorphic to the direct sum of $C\left(Y\right)$ and $C\left(Z\right)$. A map $T$ is a local multiplication on $C\left(X\right)$ if and only if $T$ is the direct sum of local multiplications on $C\left(Y\right)$ and $C\left(Z\right)$. ∎

Corollary 1**.**

An $\eta$-space has no isolated points.

Proof.

If an $\eta$-space $X$ has an isolated point $x$, then $X$ is the disjoint union of $\left\{x\right\}$ and $X\backslash\left\{x\right\}$. Since $\left\{x\right\}$ is not an $\eta$-space, from Theorem 1 it follows that $X$ is not an $\eta$-space. ∎

Definition 3**.**

Suppose $X$ is a topological space and $x\in X$. If there is a sequence $\left\{x_{n}\right\}$ in $X\backslash\left\{x\right\}$ such that $x_{n}\rightarrow x$, then we call $x$ a sequential limit point of $X$.

Note that if $X$ is a $T_{1}$ space, then $x\in X$ is a sequential limit point if and only if there is a sequence$\left\{x_{n}\right\}$ in $X\backslash\left\{x\right\}$ whose terms are different from each other such that $x_{n}\rightarrow x$.

Theorem 2**.**

Suppose $X$ is a compact Hausdorff space, and let $A$ be the set of all sequential limit points of $X$. If $X=\bar{A}$, then $X$ is an $\eta$-space.

Proof.

Suppose $T$ is an additive local multiplication on $C\left(X\right)$, $T$ is $\mathbb{Q}$-linear. Since the set of local multiplications on $C\left(X\right)$ is closed under linear combinations and compositions, $T$ is a (local) multiplication if and only if** $T-M_{T\left(1\right)}$** is a (local) multiplication. We may suppose $T\left(1\right)=0$, and prove $T=0$.

First we prove $T\left(a\cdot 1\right)=T\left(a\right)=0$ for every $a\in\mathbb{R}$. Since $T$ is $\mathbb{Q}$-linear, we know, for every $r\in\mathbb{Q}$, that

[TABLE]

Suppose $a\in\mathbb{R}$. Assume, via contradiction that $T\left(a\right)\left(y\right)\neq 0$ for some $y\in X$. Since $A$ is dense in $X$, there is an $x\in A$ such that $T\left(a\right)\left(x\right)\neq 0$. Since $x\in A$, there is a sequence $\left\{x_{n}\right\}$ in $X\backslash\left\{x\right\}$ whose terms are different from each other such that $x_{n}\rightarrow x$. Let $K=\left\{x_{n}:n\in\mathbb{N}\right\}\cup\left\{x\right\}$. Clearly, $K$ is a closed subset of $X$. Define a function $f:K\rightarrow[0,1]$ by

[TABLE]

Clearly, $f$ is continuous on $K$. Since $X$ is compact and Hausdorff, $X$ is normal. By the Tietze extension theorem, there is a continuous function $F$ from $X$ to $[0,1]$ such that $F|_{K}=f$. Since $F\left(x\right)=0$ and $T$ is a local multiplication, we know that $T\left(F\right)\left(x\right)=0.$ Thus

[TABLE]

[TABLE]

i.e.,

[TABLE]

Since $T\left(a-F\right)\in C\left(X\right)$, and $x_{n}\rightarrow x$, there exists a $k\in\mathbb{N}$ such that $T\left(a-F\right)\left(x_{k}\right)\neq 0$. But

[TABLE]

Thus

[TABLE]

[TABLE]

since $T\left(r_{k}\right)=r_{k}\cdot T\left(1\right)=0$. This is a contradiction. Thus for every $y\in X$, $T\left(a\right)\left(y\right)=0$, whence $T\left(a\right)=0$.

For every $g\in C_{R}\left(X\right)$ and every $x\in X$, $\left(g-g\left(x\right)\cdot 1\right)\left(x\right)=0,$ and since $T$ is a local multiplication,

[TABLE]

Thus $T\left(g\right)=0$.

Now suppose $h\in C\left(X\right)$, and let $h=h_{1}+ih_{2}$, where $h_{1}$, $h_{2}\in C_{R}\left(X\right)$. We have

[TABLE]

(Set $L\left(f\right)=T\left(if\right)$ for every $f\in C\left(X\right)$. Thus $L$ is an additive local multiplication on $C\left(X\right)$. From above, if $L\left(1\right)=0$, then $L\left(g\right)=0$ for every $g\in C_{R}\left(X\right)$. Thus $L\left(g\right)=L\left(1\right)g$, i.e., $T\left(ig\right)=T\left(i\right)g$ for any $g\in C_{R}\left(X\right)$.) To prove $T\left(h\right)=0$, it’s enough to show $T\left(i\right)=0$.

For every $x\in A$, there is a sequence$\left\{x_{n}\right\}$ in $X\backslash\left\{x\right\}$ whose terms are different from each other such that $x_{n}\rightarrow x$. Let $K=\left\{x_{n}:n\in\mathbb{N}\right\}\cup\left\{x\right\}$. Then $K$ is a closed subset of $X$. Define a function $f:K\rightarrow[0,1]$ by

[TABLE]

Thus $f$ is continuous on $K$. By the Tietze extension theorem, there is a continuous function $F$ from $X$ to $[0,1]$ such that $F|_{K}=f$. Similarly, define $g:K\rightarrow[0,1]$ by

[TABLE]

We get a continuous function $G$ from $X$ to $[0,1]$ such that $G|_{K}=g$. Set $H=F+iG\in C\left(X\right)$. Thus

[TABLE]

for some $h_{H}\in C\left(X\right)$. Thus

[TABLE]

for every $n\in\mathbb{N}$.

Clearly, we have $T\left(i\right)\left(x_{n}\right)=ih_{H}\left(x_{n}\right)$ when $n$ is even, and $T\left(i\right)\left(x_{n}\right)=h_{H}\left(x_{n}\right)+ih_{H}\left(x_{n}\right)$ when $n$ is odd. Since $T\left(i\right)\left(x_{n}\right)\rightarrow T\left(i\right)\left(x\right)$ as $n\rightarrow\infty$, we have

[TABLE]

Thus $h_{H}\left(x\right)=0$. Thus $T\left(i\right)\left(x\right)=0$ for every $x\in A$. Since $X=\bar{A}$, $T\left(i\right)=0$. Thus $T\left(h\right)=0$ for every $h\in C\left(X\right)$, i.e., $T=0$. Thus every additive local multiplication on $C\left(X\right)$ is a multiplication, i.e., $X$ is an $\eta$-space**.** ∎

Remark 1**.**

The preceding theorem tells us whether $X$ is an $\eta$-space has nothing to do with the connectedness of $X$. From the theorem, we know the Cantor set and the closed interval $[0,1]$ are $\eta$-spaces but the former is totally disconnected and the latter is connected. Note that in this theorem $X=\bar{A}$ implies has no isolated points.

Corollary 2**.**

Suppose $X$ is a compact Hausdorff space. If $X$ is first countable, then $X$ is an $\eta$-space if and only if $X$ has no isolated points.

Corollary 3**.**

Suppose $Y$ is a completely regular Hausdorff space, and let $A$ be the set of all sequential limit points of $Y$. If $A$ is dense in $Y$, then $\beta\left(Y\right)$ is an $\eta$-space.

Next theorem is another version of [7, Theorem 6].

Theorem 3**.**

Suppose $X$ is a compact Hausdorff space, and $\left\{K_{i}\right\}$ is a collection of closed subset of $X$. If each $K_{i}$ is an $\eta$-space, then the closure of the union of $K_{i}$’s is an $\eta$-space.

Proof.

Let $K$ be the closure of the union of $K_{i}$’s. Suppose $T$ is an additive local multiplication on $C\left(K\right)$ and $T\left(1\right)=0$. For each $i$ in $I$, define $T_{i}$ on $C\left(K_{i}\right)$ by

[TABLE]

where $\widetilde{f}\in C\left(K\right)$ is a Tietze extension of $f$. The definition is well-defined. In fact, suppose $\widetilde{f_{1}}$, $\widetilde{f_{2}}$ $\in C\left(K\right)$ and $\widetilde{f_{1}}|_{K_{i}}=$ $\widetilde{f_{2}}|_{K_{i}}$. Thus

[TABLE]

for some $h\in C\left(K\right)$. Then

[TABLE]

Thus $T\left(\widetilde{f_{1}}\right)|_{K_{i}}=T\left(\widetilde{f_{2}}\right)|_{K_{i}}$. Clearly, $T_{i}$ is an additive local multiplication on $C\left(K_{i}\right)$. Since $K_{i}$ is an $\eta$-space and $T_{i}\left(1\right)=0$, we have $T_{i}=0$. Thus for every $g\in C\left(K\right)$, $T\left(g\right)|_{K_{i}}=0$ for each $i$ in $I$. Since $K$ is the closure of the union of $K_{i}$’s, $T\left(g\right)=0$. Since $g$ is arbitrary, we have $T=0$. Thus $K$ is an $\eta$-space. ∎

Theorem 4**.**

Suppose $Y$ is a compact Hausdorff space, and $K$ is an $\eta$-space. Then $Y\times K$ is an $\eta$-space.

Proof.

For any $y\in Y$, $\left\{y\right\}\times K$ is a closed $\eta$-subspace of $Y\times K$. Thus from the last theorem, $Y\times K=\cup_{y\in Y}\left\{y\right\}\times K$ is an $\eta$-space. ∎

Corollary 4**.**

Every compact Hausdorff space is homeomorphic to a subspace of an $\eta$-space.

Theorem 5**.**

Every compact Hausdorff space has a unique maximal $\eta$-subspace.

Proof.

Suppose $X$ is a compact Hausdorff space, and let $K$ be the closure of the union of all $\eta$-subspaces of $X$. Thus $K$ is an $\eta$-space. Any $\eta$-subspace of $X$ is a subset of $K$. Thus $K$ is the unique maximal $\eta$-subspace of $X$. ∎

Definition 4**.**

Suppose $X$ is a completely regular Hausdorff space. For every $f\in C\left(X\right)$, let $Z\left(f\right)=\left\{x\in X:f\left(x\right)=0\right\}$ be the* zero-set of $f$.** Any set that is a zero-set of some function in $C\left(X\right)$ is called a zero-set in $X$. We call the complement of a zero-set a cozero-set. We say a subspace $S$ of $X$ is $R$-embedded if every bounded function in $C_{R}\left(S\right)$ can be extended to a bounded function in $C_{R}\left(X\right)$. If every cozero-set in $X$ is $R$-embedded, we call $X$ an F-space [4, Theorem 14.25(6)].*

If $g\in C\left(X\right)$ is nonzero, then the map $T\left(f\right)=g\bar{f}$ is not a multiplication. We now characterize the spaces $X$ for which $T$ is a local multiplication.

For any $a$, $b\in\mathbb{R}$, the symbol $a\vee b$ denotes $\sup\left\{a\text{, }b\right\}$. Likewise, $a\wedge b$ denotes $\inf\left\{a\text{, }b\right\}$. For any $f$, $g\in C_{R}\left(X\right)$, define $\left(f\vee g\right)\left(x\right)=f\left(x\right)\vee g\left(x\right)$, for every $x\in X$. Thus $f\vee g\in C_{R}\left(X\right)$. Dually, we defined $f\wedge g\in C_{R}\left(X\right)$. In the proof of the next theorem, we use an equivalent condition for $X$ to be an F-space [4, Theorem 14.25(5)], i.e., given $f\in C_{R}\left(X\right)$, there exists $k\in C_{R}\left(X\right)$ such that $f=k\cdot\left|f\right|$.

Lemma 1**.**

Suppose $X$ is a compact Hausdorff space, and $0\neq g\in C\left(X\right)$. Let $A=X\backslash Z\left(g\right)$. Define $T\left(f\right)=g\cdot\bar{f}$ for every $f\in C\left(X\right)$. Then $T$ is a local multiplication if and only if $A$ is an F-space.

Proof.

$\Leftarrow$: Suppose $A$ is an F-space. For every $f\in C\left(X\right)$, $f|_{A}\in C\left(A\right)$. Denote $f|_{A}$ by $f_{A}$. Set $f_{A}=u+iv$, where $u$, $v\in C_{R}\left(A\right)$. Thus on $A\backslash Z\left(f_{A}\right)$, we have

[TABLE]

Let $h_{1}=\frac{u^{2}-v^{2}}{u^{2}+v^{2}}$ and $h_{2}=\frac{2uv}{u^{2}+v^{2}}$. Clearly, $h_{1}$, $h_{2}$ are bounded real continuous functions on $A\backslash Z\left(f_{A}\right)=A\backslash Z\left(u^{2}+v^{2}\right)$, where $u^{2}+v^{2}\in C_{R}\left(A\right)$. Since $A$ is an F-space, $h_{1}$, $h_{2}$ have bounded continuous extensions on $A$, say $\widetilde{h_{1}}$, $\widetilde{h_{2}}$. Define

[TABLE]

Thus $h\in C\left(X\right)$ and $h\cdot f=g\cdot\bar{f}$.

Thus $T\left(f\right)=g\cdot\bar{f}$ is a local multiplication.

$\Longrightarrow$: Suppose $T$ is a local multiplication. For each $h\in C_{R}\left(A\right)$, if $h$ is not bounded, choose an $r>0$ and let $h_{r}=\left(-r\vee h\right)\wedge r$, then $h_{r}$ is a bounded real continuous function on $A$. Define

[TABLE]

Thus $k\in C\left(X\right)$. Since $T$ is a local multiplication, there is an $h_{k}\in C\left(X\right)$ such that $T\left(k\right)=g\cdot\bar{k}=h_{k}\cdot k$. Thus on $A\backslash Z\left(h_{r}\right)$, we have

[TABLE]

i.e., $h_{k}=-i\bar{g}\cdot\frac{h_{r}}{|h_{r}|}$. Since $\bar{g}\neq 0$ on $A$, we have

[TABLE]

on $A\backslash Z\left(h_{r}\right)$. Set $\frac{ih_{k}}{\bar{g}}=f_{1}+if_{2}$, where $f_{1}$, $f_{2}\in C_{R}\left(A\right)$. From above, $f_{1}=\frac{h_{r}}{|h_{r}|}$, and $f_{2}=0$ on $A\backslash Z\left(h_{r}\right)$. Thus $h_{r}=f_{1}\cdot|h_{r}|$ on $A$.

We say $h=f_{1}\cdot|h|$ on $A$. Since if $h\left(x\right)<-r$ for some $x\in A$, we have $h_{r}\left(x\right)=-r$. Then $f_{1}\left(x\right)=-1$ and $h\left(x\right)=-|h\left(x\right)|=f_{1}\left(x\right)\cdot|h\left(x\right)|$. For $h\left(x\right)>r$, we get the same result. Thus we proved that for every $h\in C_{R}\left(A\right)$, there exists an $f_{1}\in C_{R}\left(A\right)$ such that $h=f_{1}\cdot|h|$, i.e., $A$ is an F-space. ∎

Theorem 6**.**

Suppose $X$ is a compact Hausdorff space. The following are equivalent:

1. (1)

The conjugation map is a local multiplication on $C\left(X\right)$. 2. (2)

$X$* is an F-space.* 3. (3)

The set of $\mathbb{R}$-linear local multiplications on $C\left(X\right)$ consists of all maps of the form

[TABLE]

Proof.

The equivalent of $\left(1\right)$ and $\left(2\right)$ is immediately derived from Lemma 1.

$\left(2\right)\Leftrightarrow\left(3\right)$: Suppose $T$ is an $\mathbb{R}$-linear local multiplication on $C\left(X\right)$. It is clear $T$ has the form $T\left(f\right)=T\left(1\right)\operatorname{Re}\left(f\right)+T\left(i\right)\operatorname{Im}\left(f\right)$ for any $f\in C\left(X\right)$. Conversely, if $T$ is a map on $C\left(X\right)$ which has the form $T\left(f\right)=T\left(1\right)\operatorname{Re}\left(f\right)+T\left(i\right)\operatorname{Im}\left(f\right)$, then of course it is $\mathbb{R}$-linear. Also we have

[TABLE]

Thus $T$ is a local multiplication if and only if $\left[\frac{T\left(1\right)+iT\left(i\right)}{2}\right]\cdot\bar{f}$ is a local multiplication. If $X$ is an F-space, then every cozero-set in $X$ is an F-space [4, 14.26]. From Lemma 1, we have $\left[\frac{T\left(1\right)+iT\left(i\right)}{2}\right]\cdot\bar{f}$ is a local multiplication, i.e., $T$ is a local multiplication. Thus we prove $\left(2\right)\Rightarrow\left(3\right)$.

If every $T$ of the form $T\left(f\right)=T\left(1\right)\operatorname{Re}\left(f\right)+T\left(i\right)\operatorname{Im}\left(f\right)$ is a local multiplication, then $\left[\frac{T\left(1\right)+iT\left(i\right)}{2}\right]\cdot\bar{f}$ is a local multiplication. From Lemma 1, $X\backslash Z\left(\frac{T\left(1\right)+iT\left(i\right)}{2}\right)$ is an F-space. Choose $T\left(1\right)$ and $T\left(i\right)$ such that $Z\left(\frac{T\left(1\right)+iT\left(i\right)}{2}\right)$ is empty. Then $X$ is an F-space. ∎

The following theorem is an immediate consequence of Lemma 1 and the fact that $A\subseteq X$ is an open Fσ if and only if there is a $g\in C\left(X\right)$ such that $A=X\backslash Z\left(g\right)$ (See [4].)

Theorem 7**.**

Suppose $X$ is a compact Hausdorff space. The following are equivalent:

1. (1)

There is a nonzero function $g$ $\in C\left(X\right)$ such that the map $T\left(f\right)=g\bar{f}$ is a local multiplication on $C\left(X\right)$. 2. (2)

There is a nonempty open Fσ subset of $X$ that is an F-space.

Corollary 5**.**

No nonempty open Fσ-subset of $X$ is an $F$-space if and only if every $\mathbb{R}$-linear local multiplication on $X$ is a multiplication.

Proof.

The sufficiency follows from Theorem 7, since a map $T\left(f\right)=g\bar{f}$ with $g\neq 0$ is never a multiplication. Conversely, suppose no nonempty open Fσ-subset of $X$ is an $F$-space. Suppose $T$ is a real linear local multiplication on $C\left(X\right)$. We may suppose $T\left(1\right)=0$. Thus $T\left(u\right)=0$ for every $u\in C_{R}\left(X\right)$. Thus $T\left(f\right)=T\left(i\right)\operatorname{Im}f$ for any $f$, $g\in C_{R}\left(X\right)$. If $T\left(i\right)\neq 0$, we have

[TABLE]

is a local multiplication. It follows from Theorem 7 that $X\backslash Z\left(T\left(i\right)\right)$ is a nonempty F-space, which is a contradiction.

Thus $T\left(i\right)=0$ and $T=0$, i.e., $T$ is a multiplication. ∎

Example 2**.**

If $X$ is an F-space, then the maximal $\eta$-subspace is the empty set. If $X$ is the union of a compact $\eta$-space $A$ and a compact F-space $Y$, then $A$ is the maximal $\eta$-subspace of $X$. To see this, suppose $A\subseteq K$ and $A\neq K$ and $K$ is compact. Choose $b\in K\backslash A$ and choose a continuous function $g:X\rightarrow\left[0,1\right]$ such that $g|_{A}=0$ and $g\left(b\right)=1$. Suppose $f\in C\left(X\right)$. Since $Y$ is an $F$-space, we know from Theorem 6 and the Tietze extension theorem that there is an $h\in C\left(X\right)$ such that $\bar{f}=hf$ on $Y.$ Since $g|A=0,$ we see that

[TABLE]

on $X.$ Thus $T\left(f\right)=g\bar{f}$ is a local multiplication that is not a multiplication. Thus $T\left(f|K\right)=\left(g|_{K}\right)\left(\bar{f}|_{K}\right)$ defines a local multiplication on $C\left(K\right)$ that is not a multiplication.

If $Y$ is a completely regular Hausdorff space, let $C_{b}\left(Y\right)$ denote the bounded continuous functions on $Y$. Then $C_{b}\left(Y\right)\cong C\left(\beta\left(Y\right)\right)$. We can say $Y$ is an $\eta$-space when $\beta\left(Y\right)$ is an $\eta$-space.

Theorem 8**.**

Suppose $I$ is an infinite set. Suppose $Y_{i}$ is a nonempty compact $\eta$-space for each $i$ in $I$, and $Y$ is the disjoint union of $Y_{i}$’s. Let $K=\beta\left(Y\right)\backslash Y$. Then $K$ is an F-space if and only if $I$ is countable.

Proof.

A function $f$ in $C_{b}\left(Y\right)$ corresponds to a uniformly bounded family $\left\{f_{i}\right\}$ with $f_{i}$ in $C\left(Y_{i}\right)$. The function $f$ is [math] on $K$ if and only if for every $r>0$, there is a finite subset $D$ of $I$ so that $|f_{i}|<r$ whenever $i$ is not in $D$. Let $J$ be the set of all functions in $C_{b}\left(Y\right)$ which vanish on $K$. Thus $C\left(K\right)\cong C_{b}\left(Y\right)/J$.

$\Longrightarrow$: Suppose $K$ is an F-space, thus the conjugation map is a local multiplication on $C\left(K\right)\cong C_{b}\left(Y\right)/J$. For every $f\in C_{b}\left(Y\right)$, we have $\bar{f}-h_{f}\cdot f\in J$ for some $h_{f}\in C_{b}\left(Y\right)$. Thus for every $n\in\mathbb{N}$, there is a finite subset $D_{\left(f,n\right)}$ of $I$ so that $|\bar{f}_{i}-h_{f_{i}}\cdot f_{i}|<\frac{1}{n}$ whenever $i$ is not in $D_{\left(f,n\right)}$. Set $D_{f}={\textstyle\bigcup}D_{\left(f,n\right)}$. Thus $D_{f}$ is countable.

Since each $Y_{i}$ is a compact $\eta$-space, the conjugation map is not a local multiplication on $C\left(Y_{i}\right)$. Thus there is a $g_{i}\in C\left(Y_{i}\right)$ such that $\bar{g}_{i}-h\cdot g_{i}\neq 0$ for every $h\in C\left(Y_{i}\right)$. Let $g={\textstyle\bigcup}g_{i}\in C\left(Y\right)$. For every $n\in\mathbb{N}$, set

[TABLE]

Thus $g\left(n\right)\in C_{b}\left(Y\right)$ for every $n\in\mathbb{N}$. Set $D={\textstyle\bigcup}D_{g\left(n\right)}$ which $D_{g\left(n\right)}$ is defined similarly with $D_{f}$ for every $n\in\mathbb{N}$. Since each $D_{g\left(n\right)}$ is countable, we have $D$ is countable. Also $D\subseteq I$.

If there exists an $m\in I$ and $m\notin D$, we have $g_{m}\in C\left(Y_{m}\right)=C_{b}\left(Y_{m}\right)$. There is a $k\in\mathbb{N}$ so that $g_{m}=g\left(k\right)_{m}$. Since $g\left(k\right)\in C_{b}\left(Y\right)$, we have $\overline{g\left(k\right)}-h\cdot g\left(k\right)\in J$ for some $h\in C_{b}\left(Y\right)$. Thus for every $n\in\mathbb{N}$, we have $|\overline{g\left(k\right)}_{m}-h_{m}\cdot g\left(k\right)_{m}|<\frac{1}{n}$, since $m\notin D\supseteq D_{g\left(k\right)}$. Thus $|\overline{g\left(k\right)}_{m}-h_{m}\cdot g\left(k\right)_{m}|=0$. Since $g_{m}=g\left(k\right)_{m}$, we have $\bar{g}_{m}=h_{m}\cdot g_{m}$ where $h_{m}\in C\left(Y_{m}\right)$. This contradicts our choice of $g_{m}$. Thus $I=D$, i.e., we get $I$ is countable.

$\Leftarrow$: Suppose $I$ is countable. Let $I=\mathbb{N}$. For every $f\in C_{b}\left(Y\right)$, $f_{n}\in C\left(Y_{n}\right)$ for every $n\in\mathbb{N}$. Let $F_{n}=\left\{x\in Y_{n}\text{: }|f_{n}\left(x\right)|\geq\frac{1}{n}\right\}$, and $F_{n}$ is a closed subset of $Y_{n}$. Thus $\frac{\bar{f}_{n}}{f_{n}}$ is continuous on $F_{n}$ and $|\frac{\bar{f}_{n}}{f_{n}}|=1$. By the Tietze extension theorem, there is a continuous function $h_{n}$ from $Y_{n}$ to $\mathbb{C}$ such that. $h_{n}|_{F_{n}}=\frac{\bar{f}_{n}}{f_{n}}$ and $|h_{n}|\leq 2$. Thus on $Y_{n}\backslash F_{n}$, we have

[TABLE]

On $F_{n}$, we have $|\bar{f}_{n}-h_{n}\cdot f_{n}|=0$. Thus $|\bar{f}_{n}-h_{n}\cdot f_{n}|<\frac{3}{n}$ on $Y_{n}$.

Set $h={\textstyle\bigcup}h_{n}$. Since $|h_{n}|<2$ for every $n\in\mathbb{N}$, we have $h\in C_{b}\left(Y\right)$. Thus $\bar{f}-h\cdot f\in J$. Thus the conjugation map is a local multiplication on $C_{b}\left(Y\right)/J\cong C\left(K\right)$, i.e., we get $K$ is an F-space. ∎

Remark 2**.**

Suppose $X$ is a compact $\eta$-space. We know that every clopen subset of $X$ is an $\eta$-space. From Theorem 8, we get a closed subset $K=\beta\left(Y\right)\backslash Y$ of a compact $\eta$-space $\beta\left(Y\right)$ which is not an $\eta$-space, in fact, an F-space.

Corollary 6**.**

Suppose $I$ is uncountable and $Y$ is the disjoint union of nonempty compact Hausdorff spaces $Y_{i}$’s with $i$ in $I$. Let $K=\beta\left(Y\right)\backslash Y$. Then $K$ is an F-space if and only if all but countable many $Y_{i}$’s are F-spaces.

Let $\mathbb{N}$ be all positive integers with discrete topology. Since the conjugation map is a local multiplication on $C_{b}\left(\mathbb{N}\right)\cong C\left(\beta\left(\mathbb{N}\right)\right)$, we get that $\beta\left(\mathbb{N}\right)$ is an F-space. We have characterized all real linear local multiplications on $C\left(\beta\left(\mathbb{N}\right)\right)$. Next we will see how an additive local multiplication on $C\left(\beta\left(\mathbb{N}\right)\right)$ relates to a real linear local multiplication.

Proposition 1**.**

Suppose $T$ is an additive local multiplication on $C\left(\beta\left(\mathbb{N}\right)\right)\cong C_{b}\left(\mathbb{N}\right)$. Then $T\left(f\right)=T\left(1\right)\cdot f$ for every $f\in C_{R}\left(\beta\left(\mathbb{N}\right)\right)$ except finite many points of $\mathbb{N}$.

Proof.

Clearly, $C_{b}\left(\mathbb{N}\right)$ is the set of all bounded complex sequences. Let $C_{0}\left(\mathbb{N}\right)$ be the set of all sequences which converge to [math]. We may suppose $T\left(1\right)=0$. Thus $T\left(r\cdot 1\right)=T\left(r\right)=0$ for every $r\in\mathbb{Q}$. Also for every bounded rational sequences $\left\{r_{n}\right\}$, $T\left(\left\{r_{n}\right\}\right)=0$. Since for every $n\in\mathbb{N}$, $T\left(\left\{r_{n}\right\}\right)\left(n\right)=T\left(\left\{r_{n}\right\}-r_{n}\cdot 1\right)\left(n\right)=h\left(n\right)\cdot\left(r_{n}-r_{n}\right)=0$ for some $h\in C_{b}\left(\mathbb{N}\right)$. For every $a\in\mathbb{R}$, there is a sequence $\left\{a_{n}\right\}$ in $\mathbb{Q}$ such that. $a_{n}\rightarrow a$. Thus

[TABLE]

for some $h\in C_{b}\left(\mathbb{N}\right)$. Thus for every $n\in\mathbb{N}$, $-T\left(a\right)\left(n\right)=h\left(n\right)\cdot\left(a_{n}-a\right)$. Let $n\rightarrow\infty$,

[TABLE]

since $h$ is a bounded sequence. Thus $T\left(a\right)\in C_{0}\left(\mathbb{N}\right)$ for every $a\in\mathbb{R}$.

Let $f$ be any bounded real sequence. It has a convergent subsequence $\left\{f\left(n_{k}\right)\right\}$ and $f\left(n_{k}\right)\rightarrow b$ as $k\rightarrow\infty$ for some $b\in\mathbb{R}$. Thus

[TABLE]

for some $h_{b}\in C_{b}\left(\mathbb{N}\right)$. For every $k\in\mathbb{N}$,

[TABLE]

Let $k\rightarrow\infty$, we get $T\left(f\right)\left(n_{k}\right)\rightarrow 0$.

Thus every subsequence $\left\{T\left(f\right)\left(n_{k}\right)\right\}$ of $T\left(f\right)$ has a subsequence $\left\{T\left(f\right)\left(n_{k_{j}}\right)\right\}$ that converges to [math]. Then $T\left(f\right)\in C_{0}\left(\mathbb{N}\right)$ for every bounded $f\in C_{R}\left(\mathbb{N}\right)$. Thus $T\left(f\right)$ is [math] on $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ for every $f\in C_{R}\left(\beta\left(\mathbb{N}\right)\right)$.

Suppose $B$ is a linear basis for $\mathbb{R}$ over $\mathbb{Q}$ and $1\in B$. For every $n\in\mathbb{N}$, let $I_{n}$ be the sequence with $1$ in the $n$-th place and [math] other places. For every $n\in\mathbb{N}$, define $T_{n}:B\rightarrow\mathbb{C}$ by

[TABLE]

If $k\neq n$, then $T\left(b\cdot I_{n}\right)\left(k\right)=h_{b}\left(k\right)\cdot b\cdot I_{n}\left(k\right)=0$. Thus we have $T_{n}\left(b\right)\cdot I_{n}=T\left(b\cdot I_{n}\right)$.

We have known $T_{n}\left(1\right)=0$ for every $n\in\mathbb{N}$. Next we prove there is a $n_{T}\in\mathbb{N}$ such that. $T_{n}=0$ for every $n\geq n_{T}$.

Else, we can get a subsequence $\left\{b_{n_{k}}\right\}\subseteq B$ such that. $T_{n_{k}}\left(b_{n_{k}}\right)\neq 0$ for every $k\in\mathbb{N}$. For every $k\in\mathbb{N}$, choose an $q_{k}\in\mathbb{Q}$ such that. $q_{k}>\frac{1}{|T_{n_{k}}\left(b_{n_{k}}\right)|}$. We know $q_{k}\cdot b_{n_{k}}\in\mathbb{R}$ and choose an $r_{k}\in\mathbb{Q}$ such that. $|q_{k}\cdot b_{n_{k}}-r_{k}|<1$. Define a sequence

[TABLE]

for every $n\in\mathbb{N}$. Thus $\left\{a\left(n\right)\right\}$ is a real bounded sequence.

For every $k\in\mathbb{N}$,

[TABLE]

since $q_{k}$, $r_{k}\in\mathbb{Q}$. But $|q_{k}\cdot T_{n_{k}}\left(b_{n_{k}}\right)|>1$ for every $k\in\mathbb{N}$. Thus $T\left(a\right)\notin C_{0}\left(\mathbb{N}\right)$. This is a contradiction.

Thus there is a $n_{T}\in\mathbb{N}$ such that. $T_{n}=0$ for every $n\geq n_{T}$. Now for any bounded $f\in C_{R}\left(\mathbb{N}\right)$,

[TABLE]

for every $n\geq n_{T}$. Let $D=\left\{n\in\mathbb{N}\text{: }1\leq n\leq n_{T}\right\}$. Thus $T\left(f\right)=0$ on $\beta\left(\mathbb{N}\right)\backslash D$ for every $f\in C_{R}\left(\beta\left(\mathbb{N}\right)\right)$.

Then $T\left(f\right)=T\left(1\right)\cdot f$ on $\beta\left(\mathbb{N}\right)\backslash D$ for every $f\in C_{R}\left(\beta\left(\mathbb{N}\right)\right)$. ∎

Remark 3**.**

Note that the $n_{T}$ is in terms of $T$, so we cannot say that there is a finite subset $D$ of $\mathbb{N}$ such that for every additive local multiplication $T$ on $C\left(\beta\left(\mathbb{N}\right)\right)$, $T\left(f\right)=T\left(1\right)\cdot f$ on $\beta\left(\mathbb{N}\right)\backslash D$ for every $f\in C_{R}\left(\beta\left(\mathbb{N}\right)\right)$. In fact, we can get that for every additive local multiplication $T$ on $C\left(\beta\left(\mathbb{N}\right)\right)$, $T\left(f\right)=T\left(1\right)\cdot f$ on $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ for every $f\in C_{R}\left(\beta\left(\mathbb{N}\right)\right)$.

3. $\upsilon$-Spaces

Definition 5**.**

Suppose $X$ is a compact Hausdorff space. Call a map $T$ on $C\left(X\right)$ ($C_{R}\left(X\right)$) zero-preserving if for every $f\in C\left(X\right)$($C_{R}\left(X\right)$) and every $x\in X$, $f\left(x\right)=0$ implies $T\left(f\right)\left(x\right)=0$.

The zero-preserving property is weaker than being a local multiplication. Every local multiplication has this property. However, in general, not every zero-preserving additive map is a local multiplication. For example, any map $T$ of the form

[TABLE]

For every additive $T$ on $C\left(X\right)$ there corresponds four additive maps on $C_{R}\left(X\right)$, namely,

[TABLE]

and it turns out that $T$ is zero-preserving if and only if each of these four maps is zero-preserving and $T$ has the form

[TABLE]

Definition 6**.**

We call a compact Hausdorff space $X$ a $\mathbf{\upsilon}$-space if every zero-preserving additive map on $C_{R}\left(X\right)$ is a multiplication. Equivalently $X$ is a $\upsilon$-space if and only if every zero-preserving map on $C\left(X\right)$ has the form

[TABLE]

The relationship between local multiplications and zero-preserving maps is given in the following lemma.

Lemma 2**.**

Suppose $T$ is any map on $C\left(X\right)$. Then $T$ is a local multiplication if and only if $T$ leaves invariant every ideal of $C\left(X\right)$ and $T$ is zero-preserving if and only if $T$ leaves invariant every closed ideal of $C\left(X\right)$.

Proof.

Suppose $T$ is a local multiplication and $I$ is an ideal of $C\left(X\right)$. Then for every $f\in I$, $T\left(f\right)=h_{f}\cdot f\in I$. Now suppose $T$ leaves invariant every ideal of $C\left(X\right)$. Suppose $f\in C\left(X\right)$ and $I_{f}$ is an ideal generated by $f$. Since $T\left(f\right)\in I_{f}$, $T\left(f\right)=h\cdot f$ for some $h\in C\left(X\right)$. $T$ is a local multiplication.

Suppose $T$ is zero-preserving and $I$ is a closed ideal of $C\left(X\right)$. Then there is a closed subset $K$ of $X$ such that $I$ is the set of all functions in $C\left(X\right)$ which vanish on $K$. For every $f\in I$, $f=0$ on $K$. Thus $T\left(f\right)=0$ on $K$, i.e., $T\left(f\right)\in I$. Now suppose $T$ leaves invariant every closed ideal of $C\left(X\right)$. For every $f\in C\left(X\right)$ and every $x\in X$, if $f\left(x\right)=0$, let $I_{x}$ be the ideal of all functions in $C\left(X\right)$ which vanish on $\left\{x\right\}$. Thus $I_{x}$ is closed. Thus $T\left(f\right)\in I_{x}$, $T\left(f\right)\left(x\right)=0$, i.e., $T$ is zero-preserving. ∎

Here we prove a purely algebraic result that relates to zero-preserving maps, since the evaluation maps at points of $X$ are algebra homomorphisms of $C\left(X\right)$ to $\mathbb{C}$ and $C_{R}\left(X\right)$ into $\mathbb{R}$, respectively.

Theorem 9**.**

Suppose $\mathcal{A}$ is an algebra with identity $1$ over a field $\mathbb{F}$. Suppose also that, whenever $x\in\mathcal{A}$ and $x\neq 0$, there is an algebra homomorphism $h$ from $\mathcal{A}$ to $\mathbb{F}$ such that $h\left(x\right)\neq 0$. Let $S$ be the set of unital algebra homomorphism from $\mathcal{A}$ to $\mathbb{F}$. Suppose $T$ is a linear map from $\mathcal{A}$ to $\mathcal{A}$ such that, for every $a$ in $\mathcal{A}$ and every $s$ in $S$, $s\left(a\right)=0$ implies $s\left(T\left(a\right)\right)=0$. Then $T$ is left multiplication by $T\left(1\right)$.

Proof.

If there exists an $a\in\mathcal{A}$ such that. $T\left(a\right)\neq T\left(1\right)a$, i.e., $T\left(a\right)-T\left(1\right)a\neq 0$. From the assumption, there is an algebra homomorphism $h$ from $\mathcal{A}$ to $\mathbb{F}$ such that. $h\left(T\left(a\right)-T\left(1\right)a\right)\neq 0$. But

[TABLE]

Since $h\in S$ and

[TABLE]

we have

[TABLE]

This is a contradiction. ∎

Corollary 7**.**

Suppose $X$ is a compact Hausdorff space. Then any linear map on $C\left(X\right)$ ($\mathbb{R}$-linear map on $C_{R}\left(X\right)$) that is zero-preserving is a multiplication.

Corollary 8**.**

Suppose $X$ is a compact Hausdorff space. Then $T$ is an $\mathbb{R}$-linear zero-preserving map on $C\left(X\right)$

if and only if, $T$ has the form

[TABLE]

The equivalence of $\left(1\right)$ and $\left(5\right)$ in the following theorem, shows that if $X$ is an $\upsilon$-space, then $X$ is like a ”real” $\eta$-space.

Theorem 10**.**

Suppose $X$ is a compact Hausdorff space. The following are equivalent:

1. (1)

$X$* is a $\upsilon$-space.* 2. (2)

Every additive zero-preserving map on $C\left(X\right)$ is $\mathbb{R}$-linear. 3. (3)

Every additive zero-preserving map on $C\left(X\right)$ is continuous. 4. (4)

Every additive zero-preserving map on $C\left(X\right)$ has the form

[TABLE] 5. (5)

Every additive zero-preserving map on $C_{R}\left(X\right)$ is a multiplication.

Proof.

$\left(5\right)\Rightarrow\left(4\right)$: We have already known that for every additive $T$ on $C\left(X\right)$ there corresponds four additive maps on $C_{R}\left(X\right)$, namely,

[TABLE]

and it turns out that $T$ is zero-preserving if and only if each of these four maps is zero-preserving and $T$ has the form

[TABLE]

Since every additive zero-preserving map on $C_{R}\left(X\right)$ is a multiplication. We have

[TABLE]

$\left(4\right)\Rightarrow\left(3\right)$: Suppose $\left\{f_{n}\right\}$ is a sequence in $C\left(X\right)$ and $f_{n}\rightarrow f$. Then $T\left(f_{n}\right)=T\left(1\right)\operatorname{Re}\left(f_{n}\right)+T\left(i\right)\operatorname{Im}\left(f_{n}\right)\rightarrow T\left(1\right)\operatorname{Re}\left(f\right)+T\left(i\right)\operatorname{Im}\left(f\right)=T\left(f\right)$. Thus $T$ is continuous.

$\left(3\right)\Rightarrow\left(2\right)$: Suppose $a\in\mathbb{R}$ and $\left\{r_{n}\right\}$ is a rational sequence which converges to $a$. Then $r_{n}\cdot f\rightarrow a\cdot f$. Thus

[TABLE]

Thus $T$ is $\mathbb{R}$-linear.

$\left(2\right)\Rightarrow\left(1\right)$: Suppose $T$ is an additive zero-preserving map on $C_{R}\left(X\right)$. If $T$ is not $\mathbb{R}$-linear. Define $T_{0}$ on $C\left(X\right)$ by $T_{0}\left(f\right)=T\left(\operatorname{Re}\left(f\right)\right)+iT\left(\operatorname{Im}\left(f\right)\right)$. Thus $T_{0}$ is an additive zero-preserving map on $C\left(X\right)$ which is not $\mathbb{R}$-linear. This is a contradiction. Then $T$ is $\mathbb{R}$-linear. From Corollary 7, $T$ is a multiplication. Thus $X$ is a $\upsilon$-space.

$\left(1\right)\Rightarrow\left(5\right)$: This follows immediately from the definition of $\upsilon$-space. ∎

Example 3**.**

Suppose $X=\left\{a\right\}$. Then $C_{R}\left(X\right)$ is isomorphic to $\mathbb{R}$. Every additive map on $\mathbb{R}$ is zero-preserving. The number of additive (i.e., $\mathbb{Q}$-linear) maps on $\mathcal{\mathbb{R}}$ is $2^{2^{\aleph_{0}}}$ but the number of multiplications is $2^{\aleph_{0}}$, so $X$ is not a $\upsilon$-space.

As in the $\eta$-space case, we get the following for free.

Theorem 11**.**

Suppose $X$ is the disjoint union of compact Hausdorff spaces $Y$ and $Z$. Then $X$ is a $\upsilon$-space if and only if $Y$ and $Z$ are $\upsilon$-spaces.

Corollary 9**.**

A $\upsilon$-space has no isolated points.

If we examine the proofs of Theorems in the preceding section, we immediately obtain the following results.

Theorem 12**.**

Suppose $X$ is a compact Hausdorff space. Then

1. (1)

If the set of sequential limit points is dense in $X$, then $X$ is a $\upsilon$-space. 2. (2)

If $X$ is first countable, then $X$ is a $\upsilon$-space if and only if $X$ has no isolated points. 3. (3)

If $\left\{K_{i}:i\in I\right\}$ is a collection of closed subsets of $X$ and if each $K_{i}$ is a $\upsilon$-space, then the closure of the union of $K_{i}$’s is a $\upsilon$-space. 4. (4)

$X$* has a unique maximal compact $\nu$-subspace.* 5. (5)

If $Y$ is a compact Hausdorff $\upsilon$-space, then $X\times Y$ is a $\upsilon$-space.

4. P-points and $\mathfrak{q}$-points

We now want to generalize Theorem 2. We call a point $x\in X$ a $q$-point if and only if there is a disjoint sequence $\left\{K_{n}\right\}$ of compact subsets of $X$ such that

1. (1)

Each $K_{n}$ is disjoint from the closure of $\cup_{m\in\mathbb{N},m\neq n}K_{m}$; 2. (2)

$x\in\left(\cup_{n\in\mathbb{N}}K_{n}\right)^{-}\backslash\left(\cup_{n\in\mathbb{N}}K_{n}\right)$.

We say that $x$ is a strong $\mathfrak{q}$-point if there is a disjoint sequence $\left\{K_{n}\right\}$ of compact sets satisfying

$x\in\left(\cup_{n\in\mathbb{N}}K_{2n}\right)^{-}\cap\left(\cup_{n\in\mathbb{N}}K_{2n-1}\right)^{-}$.

It is clear that these conditions on the sequence $\left\{K_{n}\right\}$ is precisely what is needed to ensure that if $\left\{z_{n}\right\}$ is a sequence of complex numbers converging to $z$, then the function $f$ on $\cup_{n=1}^{\infty}K_{n}$ defined by $f|_{K_{n}}=z_{n}$ extends to $\left(\cup_{n\in\mathbb{N}}K_{2n}\right)^{-}$, which by the Tietze extension theorem extends to a continuous function on $X$. By examining the proof of Theorem 2, we easily obtain the following.

Theorem 13**.**

Suppose $X$ is a compact Hausdorff space.

1. (1)

If the set of $\mathfrak{q}$-points is dense in $X$, then every additive local multiplication $T$ on $X$ has the form $T\left(f\right)=T\left(1\right)\operatorname{Re}\left(f\right)+T\left(i\right)\operatorname{Im}\left(f\right)$ for some $T\left(1\right)$,$T\left(i\right)\in C\left(X\right)$. 2. (2)

If the set of strong $\mathfrak{q}$-points is dense in $X$, then $X$ is an $\eta$-space.

The notion of a $\mathfrak{q}$-point is very closely related to the classical notion of a P-point defined by L. Gillman and M. Henriksen [3], which can be defined as follows:

$x\in X$ is a P-point in $X$ if and only if every continuous function in $C\left(X\right)$ is constant on some open set containing $x$.

Lemma 3**.**

Suppose $X$ is a compact Hausdorff space. Then

1. (1)

If $x\in X$, then $x$ is a $\mathfrak{q}$-point if and only if $x$ is not a P-point. 2. (2)

If $K\subseteq X$ is compact and every point of $K$ is a P-point of $X$, then $K$ is finite. 3. (3)

If $A$ is the set of all $\mathfrak{q}$-points of $X$, then $X\backslash\bar{A}$ is the set of isolated points of $X$.

Proof.

$\left(1\right)$. Suppose $x$ is a $\mathfrak{q}$-point. We can choose a disjoint collection $\left\{K_{1},K_{2},\ldots\right\}$ of compact sets such that each is disjoint from the union of the others, $x\in\left(\cup_{n=1}^{\infty}K_{n}\right)^{-}\backslash\left(\cup_{n=1}^{\infty}K_{n}\right)$. We can define a continuous function $f:\left(\cup_{n=1}^{\infty}K_{n}\right)^{-}\rightarrow\left[0,1\right]$ so that $f\left(a\right)=1/n$ when $a\in K_{n}$ and $f\left(a\right)=0$ on $\left(\cup_{n=1}^{\infty}K_{n}\right)^{-}\backslash\left(\cup_{n=1}^{\infty}K_{n}\right)$. By the Tietze extension theorem, we can assume $f\in C\left(X\right)$. It is clear that $f\left(x\right)=0$ but there is no neighborhood of $x$ on which $f$ is [math]. Thus $x$ is not a P-point.

Conversely, suppose $x$ is not a P-point and suppose $g\in C\left(X\right)$ and $g\left(x\right)=0$ but $g$ is not [math] on a neighborhood of $x$. Thus

[TABLE]

By replacing $g$ with $\left|g\right|/\left(1+\left|g\right|\right)$ we can assume that $0\leq g\leq 1$. For each $n\in\mathbb{N}$, let $E_{n}=\left\{x\in X:\frac{1}{n+1}\leq g\left(x\right)\leq\frac{1}{n}\right\}$. We know that $x$ is in the closure of the union of the $E_{n}$’s. Thus $x$ is either in the union of the closure of $\cup_{n\in\mathbb{N}}E_{2n}$ or the closure of $\cup_{n\in\mathbb{N}}E_{2n-1}$. In the former case we let $K_{n}=E_{2n}$ for each $n$, and in the latter case we let $K_{n}=E_{2n-1}$ for each $n$. In either case we see that $x$ is a $\mathfrak{q}$-point.

$\left(2\right).$ Suppose $K\subseteq X$ is compact and every point of $K$ is a P-point of $X.$ It follows from the Tietze extension theorem that every point of $K$ is a P-point of $K$. It follows from Proposition 4.1 in [12] that $K$ is finite.

$\left(3\right)$. Let $E$ be the closure of the set of all $\mathfrak{q}$-points of $X$. Suppose $x_{0}\in X\backslash E$. By Urysohn’s lemma we can find a continuous $h:X\rightarrow\left[0,1\right]$ such that $h\left(x_{0}\right)=1$ and $h|_{E}=0$. Thus $\left\{x\in X:h\left(x\right)\geq 1/2\right\}$ is a compact subset of $X$ for which every point is a P-point of $X$, which means it is finite. Thus $\left\{x\in X:g\left(x\right)>1/2\right\}$ is a finite open set containing $x_{0}$. Hence $x_{0}$ is an isolated point of $X$. Also no isolated point is in $E$, so $X\backslash E$ is the set of isolated points of $X$. ∎

5. Main Results

We can now give complete characterizations of $\eta$-spaces, real $\eta$-spaces and $\upsilon$-spaces. Here is our first main theorem, which shows that being a real $\eta$-space and being a $\upsilon$-space are the same as having no isolated points.

Theorem 14**.**

Suppose $X$ is a compact Hausdorff space. Then the following are equivalent:

1. (1)

$X$* is a $\upsilon$-space.* 2. (2)

$X$* is a real $\eta$-space.* 3. (3)

The set of $\mathfrak{q}$-points of $X$ is dense in $X$. 4. (4)

$X$* has no isolated points.*

Proof.

We already proved that $\left(3\right)\Rightarrow\left(1\right)\Rightarrow\left(2\right)\Rightarrow\left(4\right)$. The proof of $\left(4\right)\Rightarrow\left(3\right)$ follows from part $\left(3\right)$ of Lemma 3. ∎

The following result characterizes $\eta$-spaces.

Theorem 15**.**

Suppose $X$ is a compact Hausdorff space. The the following are equivalent:

1. (1)

$X$* is an $\eta$-space.* 2. (2)

No nonempty open Fσ set in $X$ is an F-space. 3. (3)

$X$* has no isolated points and, for every $0\neq g\in C\left(X\right)$, the map $T\left(f\right)=g\bar{f}$ is not a local multiplication.*

Proof.

$\left(1\right)\Rightarrow\left(2\right)$: If $X$ is an $\eta$-space, it follows from Theorem 7 that no nonempty open Fσ set in $X$ is an F-space.

$\left(2\right)\Rightarrow\left(3\right)$: Suppose that no nonempty open Fσ set in $X$ is an F-space. It follows that $X$ has no isolated points. Also, by Lemma 1, for every $0\neq g\in C\left(X\right)$, the map $T\left(f\right)=g\bar{f}$ is not a local multiplication.

$\left(3\right)\Rightarrow\left(1\right)$: By Theorem 14, $X$ is an $\upsilon$-space. Suppose $T$ is an additive local multiplication on $X.$ Since $X$ is an $\upsilon$-space, $T$ must have the form

[TABLE]

Thus $T$ is $\mathbb{R}$-linear. It follows from Corollary 5 that $T$ is a multiplication. ∎

We now describe how the maximal subsets of a compact Hausdorff space $X$ that are $\eta$-spaces or $\upsilon$-spaces can be constructed.

Remark 4**.**

One might guess that the maximal $\upsilon$-subspace of $X$ is the closure of the set of $\mathfrak{q}$-points of $X$. However, if $X=\left\{0,1,1/2,1/3,1/4,\ldots\right\}$ then the closure of the set of $\mathfrak{q}$-points of $X$ is precisely $\left\{0\right\}$. However, this is not a $\mathfrak{u}$-space. We have to use a transfinite construction argument. If $K$ is a compact Hausdorff space we define

[TABLE]

[TABLE]

We let $E_{0}=X$, and suppose $\alpha>0$ is an ordinal such that, for all $\beta>\alpha$, $E_{\beta}$ is defined. We define $E_{\alpha}$ by

[TABLE]

Since $\left\{E_{\beta}\backslash E_{\beta+1}:\beta\text{ is an ordinal}\right\}$ is a disjoint collection of subsets of $X$, there is a smallest ordinal $\alpha$ such that $E_{\alpha}=E_{\alpha+1}$. It is clear that if $K$ is a compact subset of $X$ having no isolated points, then $K\subset E_{\beta}$ for every ordinal $\beta$; in particular, $K\subset E_{\alpha}$. Since $E_{\alpha}=E_{\alpha+1}=\mathfrak{n}\left(E_{\alpha}\right)$, $E_{\alpha}$ has no isolated points. Thus $E_{\alpha}$ is the maximal compact subset of $X$ that has no isolated points, i.e., the maximal compact subset of $X$ that is a $\upsilon$-space.

The maximal $\eta$-subspace is a little more complicated. First suppose $Y$ is a compact subset of $X$ and $Y$ is an $\eta$-space. Suppose also that $V$ is an open $F_{\sigma}$ subset of $X$ and that $V$ is an F-space. It follows from Lemma 1 that there is a $g\in C\left(X\right)$ such that $X\backslash Z\left(g\right)=V$ and the map $T\left(f\right)=g\bar{f}$ is a local multiplication on $C\left(X\right)$. If $g_{Y}=g|_{Y}\neq 0$, then the map $S\left(f\right)=g_{Y}\bar{f}$ is a local multiplication on $C\left(Y\right)$ that is not a multiplication. Since $Y$ is an $\eta$-space, $g|_{Y}=0,$ or $Y\cap V=\varnothing$.

We now imitate the process above. If $K$ is a compact Hausdorff space, define

[TABLE]

We define $F_{0}=X$, and for $\alpha>0$ define

[TABLE]

Then we can choose $\alpha$ to be the smallest ordinal such that $F_{\alpha}=F_{\alpha+1}$. Then $F_{\alpha}$ is the maximal compact $\eta$-subspace of $X$.

We conclude with remarks on the space $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$.

Remark 5**.**

Since $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ is an F-space, we know that $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ is not an $\eta$-space. However, $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ has no isolated points. Thus, by Theorem 14, $X$ is an $\upsilon$-space. Thus every local multiplication on $C_{R}\left(X\right)$ is a multiplication, but the same is not true for $C\left(X\right)$.

Remark 6**.**

Another interesting fact is that the question of whether every point in $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ is a $q$-point is independent from the axioms of set theory (ZFC). W. Rudin [13] proved that if you assume the continuum hypothesis, then $\beta\left(\mathbb{N}\right)\backslash\mathbb{N}$ contains a P-point. Later S. Shelah (see [16]) proved that there is a model of set theory in which every point of $\beta\left(\mathbb{N}\right)/\mathbb{N}$ is a $\mathfrak{q}$-point.