On $\kappa$-Pseudocompactess and Uniform Homeomorphisms of Function Spaces

Abstract

A Tychonoff space X is called $\kappa$-pseudocompact if for every continuous mapping f of X into $\mathbb {R}^\kappa$ the image f(X) is compact. This notion generalizes pseudocompactness and gives a stratification of spaces lying between pseudocompact and compact spaces. It is well known that pseudocompactness of X is determined by the uniform structure of the function space $C_p(X)$ of continuous real-valued functions on X endowed with the pointwise topology. In respect of that A.V. Arhangel’skii asked if analogous assertion is true for $\kappa$-pseudocompactness. We provide an affirmative answer to this question.

1 Introduction

In this note, by a space we mean a Tychonoff topological space. For a space X, by $C_p(X)$ we denote the space of all continuous real-valued functions on X endowed with the pointwise topology. The symbol $C_p^*(X)$ stands for the subspace of $C_p(X)$ consisting of all bounded continuous functions. Recall that X is pseudocompact if $C_p(X)=C_p^*(X)$, i.e. every real-valued continuous function on X is bounded. In 1962 J.F. Kennison [7] introduced the following generalization of psudocompactness. Let $\kappa$ be an infinite cardinal. A space X is called $\kappa$-pseudocompact if for every continuous mapping f of X into $\mathbb {R}^\kappa$ the image f(X) is compact. Clearly, $\kappa$-pseudocompactness implies $\lambda$-pseudocompactness for every infinite cardinal $\lambda \le \kappa$. Since metrizable pseudocompact spaces are compact, it is easy to see that $\omega$-pseudocompactness is precisely pseudocompactness. In particular any $\kappa$-pseudocompact space is pseudocompact. It was established by Uspenskiĭ in [12] that pseudocompactness of X is determined by the uniform structure of the function space $C_p(X)$ (see [4] for a different proof of this result). In respect of that A.V. Arhangel’skii asked in 1998 if analogous result holds for $\kappa$-pseudocompactness (see [3, Question 13] or [11, Problem 4.4.2]).

The aim of the present note is to provide an affirmative answer to this question by proving the following extension of Uspenskiĭ’s theorem:

Theorem 1.1

For any infinite cardinal $\kappa$, if $C_p(X)$ and $C_p(Y)$ are uniformly homeomorphic, then X is $\kappa$-pseudocompact if and only if Y is $\kappa$-pseudocompact.

Let us recall that that a map $\varphi :C_p(X)\rightarrow C_p(Y)$ is uniformly continuous if for each open neighborhood U of the zero function in $C_p(Y)$, there is and open neighborhood V of the zero function in $C_p(X)$ such that $(f-g)\in V$ implies $(\varphi (f)-\varphi (g))\in U$. Spaces $C_p(X)$ and $C_p(Y)$ are uniformly homeomorphic if there is a homeomorphism $\varphi$ between them such that both $\varphi$ and $\varphi ^{-1}$ are uniformly continuous.

The proof of Theorem 1.1 is inspired by author’s recent work [8] concerned with linear homeomorphisms of function spaces. The basic idea in [8] relies on the fact that certain topological properties of a space X can be conveniently characterized by the way X is positioned in its Čech-Stone compactification $\beta X$; $\kappa$-pseudocompactess is one of such properties. Indeed, Hewitt [6] gave the following description of pseudocompactness (cf. [1, Theorem 1.3.3]).

Theorem 1.2

(Hewitt). A space X is pseudocompact if and only if every nonempty $G_\delta$-subset of $\beta X$ meets X.

It was noted by Retta in [10] that the above result easily extends to $\kappa$-pseudocompactness. We need the following notation. Let $\kappa$ be an infinite cardinal. A subset A of a space Z is a $G_\kappa$-set if it is an intersection of at most $\kappa$-many open subsets of Z. The $G_{\omega }$-sets are called $G_\delta$-sets and the complement of a $G_\delta$-set is called $F_\sigma$-set. We have (see [10, Theorem 1]):

Theorem 1.3

(Retta). Let $\kappa$ be an infinite cardinal. A space X is $\kappa$-pseudocompact if and only if every nonempty $G_\kappa$-subset of $\beta X$ meets X.

The uniform structure of spaces of continuous functions was studied by many authors; the interested reader should consult the book [11]. For our purposes, the most important are some ideas developed by Gul’ko in [5].

2 Results

For a space Z and a function $f\in C^*_p(Z)$ the function ${\widetilde{f}}:\beta Z\rightarrow \mathbb {R}$ is the unique continuous extension of f over the Čech-Stone compactification $\beta Z$ of Z. Let $\varphi :C^*_p(X)\rightarrow C^*_p(Y)$ be a uniformly continuous surjection. For $y\in \beta Y$ and a subset K of $\beta X$ we define

$$\begin{aligned} a(y,K)=\sup \{|\widetilde{\varphi (f)}(y)-\widetilde{\varphi (g)}(y)|&:\;f,g\in C_p^*(X) \text{ such } \text{ that } \\&|{\widetilde{f}}(x)-{\widetilde{g}}(x)|<1 \text{ for } \text{ every } x\in K\} \end{aligned}$$

Note that $a(y,\emptyset )=\infty$ since $\varphi$ is onto.

For $y\in \beta Y$ define the family

$$\begin{aligned} \mathscr {A}(y)=\{K\subseteq \beta X: K \text{ is } \text{ compact } \text{ and } a(y,K)<\infty \}. \end{aligned}$$

Similarly, for $y\in \beta Y$ and $n\in \mathbb {N}$ let

$$\begin{aligned} \mathscr {A}_n(y)=\{K\subseteq \beta X: K \text{ is } \text{ compact } \text{ and } a(y,K)\le n\}. \end{aligned}$$

It may happen that for some n the family $\mathscr {A}_n(y)$ is empty. However, we have the following:

Proposition 2.1

For every $y\in Y$, there exists n for which $\mathscr {A}_n(y)$ contains a nonempty finite subset of X. In particular, for this n the family $\mathscr {A}_n(y)$ is nonempty.

Proof

By uniform continuity of $\varphi$, there is $n\in \mathbb {N}$ and a finite subset F of X such that

$$\begin{aligned} \begin{aligned}&\text{ if } |f(x)-g(x)|<1/n \text{ for } \text{ every } x\in F,\\ {}&\text{ then } |\varphi (f)(y)-\varphi (g)(y)|<1. \end{aligned} \end{aligned}$$

(1)

We claim that $F\in \mathscr {A}_n(y)$. To see this, take arbitrary functions $f,g\in C_p^*(X)$ such that $|f(x)-g(x)|<1$, for every $x\in F$. Put $f_k=f+\tfrac{k}{n}(g-f)$, for $k=0,1,\ldots , n$. Then $f_0=f$, $f_n=g$ and $|f_k(x)-f_{k+1}(x)|<1/n$ for $x\in F$. Hence, by (1) we get

$$\begin{aligned}&|\varphi (f)(y)-\varphi (g)(y)|\\&\quad \le |\varphi (f_0)(y)-\varphi (f_1)(y)|+\cdots + |\varphi (f_{n-1})(y)-\varphi (f_n)(y)|<n, \end{aligned}$$

as required. $\square$

Clearly, for every $y\in \beta Y$ we have $\mathscr {A}(y)=\bigcup _{n\in \mathbb {N}}\mathscr {A}_n(y)$. In particular, for $y\in Y$ the family $\mathscr {A}(y)$ is always nonempty.

For $n\in \mathbb {N}$ we set

$$\begin{aligned} Y_n=\{y\in \beta Y:\mathscr {A}_n(y) \text{ is } \text{ nonempty }\}. \end{aligned}$$

Note that $y\in Y_n$ if and only if $\beta X\in \mathscr {A}_n(y)$. Using this observation it is easy to show the following:

Lemma 2.2

For every $n\in \mathbb {N}$ the set $Y_n$ is closed in $\beta Y$; hence compact.

Proof

Pick $y\in \beta Y\setminus Y_n$. Since $\beta X\notin \mathscr {A}_n(y)$, there are functions $f,g\in C_p^*(X)$ satisfying $|{\widetilde{f}}(x)-{\widetilde{g}}(x)|<1$ for every $x\in \beta X$, and $|\widetilde{\varphi (f)}(y)-\widetilde{\varphi (g)}(y)|>n$. The set

$$\begin{aligned} U=\{z\in \beta Y:|\widetilde{\varphi (f)}(z)-\widetilde{\varphi (g)}(z)|>n\} \end{aligned}$$

is an open neighborhood of y in $\beta Y$. Moreover if $z\in U$, then f and g witness that $\beta X\notin \mathscr {A}_n(z)$; thus $U\cap Y_n=\emptyset$. $\square$

For a space X and a positive integer m, we denote by $[X]^{\le m}$ the space of all nonempty at most m-element subsets of X endowed with the Vietoris topology, i.e. basic open sets in $[X]^{\le m}$ are of the form

$$\begin{aligned} \langle U_1,\ldots ,U_k\rangle =\left\{ F\in [X]^{\le m}:\forall i\le k\quad F\cap U_i\ne \emptyset \; \text {and}\;F\subseteq \bigcup _{i=1}^k U_i\right\} , \end{aligned}$$

where $\{U_1,\ldots ,U_k\}$ is a finite collection of open subset of X.

For any positive integers n, m we define

$$\begin{aligned} Y_{n,m}=\{y\in \beta Y:\mathscr {A}_n(y) \cap [\beta X]^{\le m}\ne \emptyset \} \end{aligned}$$

Note that $Y_{n,m}\subseteq Y_n$ and by Proposition 2.1 we have

$$\begin{aligned} Y\subseteq \bigcup _{n,m}Y_{n,m} \end{aligned}$$

(2)

We claim that $Y_{n,m}$ is closed in $Y_n$ and hence it is compact:

Lemma 2.3

The set $Y_{n,m}$ is closed in $Y_n$, hence it is compact.

Proof

Consider the following subset Z of the product $Y_n\times [\beta X]^{\le m}$

$$\begin{aligned} Z=\{(y,F)\in Y_n\times [\beta X]^{\le m}:F\in \mathscr {A}_n(y)\}. \end{aligned}$$

We show that Z is closed. Pick $(y,F)\in (Y_n\times [\beta X]^{\le m}){\setminus } Z$. Then $F\notin \mathscr {A}_n(y)$ and thus there are $f,g\in C_p^*(X)$ satisfying $|{\widetilde{f}}(x)-{\widetilde{g}}(x)|<1$ for every $x\in F$, and $|\widetilde{\varphi (f)}(y)-\widetilde{\varphi (g)}(y)|>n$. Let $U=\{x\in \beta X:|{\widetilde{f}}(x)-{\widetilde{g}}(x)|<1\}$ and $V=\{z\in Y_n:|\widetilde{\varphi (f)}(z)-\widetilde{\varphi (g)}(z)|>n\}$. The set $V\times \langle U \rangle$ is an open neighborhood of (y, F) in $Y_n\times [\beta X]^{\le m}$ disjoint from Z.

The set Z, being is closed in the compact space $Y_n\times [\beta X]^{\le m}$, is compact. Since the set $Y_{n,m}$ is the image of Z under the projection map it must be compact. $\square$

Corollary 2.4

Suppose that Y is pseudocompact and let $\varphi :C_p^*(X)\rightarrow C_p^*(Y)$ be a uniformly continuous surjection. For every $y\in \beta Y$, there exist n and m such that $y\in Y_{n,m}$.

Proof

By Lemma 2.3, the set $\bigcup _{n,m} Y_{n,m}$ is $F_\sigma$ in $\beta Y$ and contains Y, by (2). It follows from Theorem 1.2 that $\bigcup _{n,m}Y_{n,m}=\beta Y$. $\square$

For $y\in \bigcup _{n,m} Y_{n,m}$ we define

$$\begin{aligned} K(y)=\bigcap \mathscr {A}(y). \end{aligned}$$

Remark

For $y\in Y$ the set K(y) is the support introduced by Gul’ko in [5] (see also [2, 4, 9]).

Lemma 2.5

For every $y\in \bigcup _{n,m} Y_{n,m}$ the set K(y) is a nonempty finite subset of $\beta X$. Moreover, $K(y)\in \mathscr {A}(y)$. If $y\in Y$, then K(y) is a subset of X.

Proof

We show that the family $\mathscr {A}(y)$ is closed under finite intersections. Pick $K_1,K_2\in \mathscr {A}(y)$ and let $f,g\in C_p^*(X)$ be such that $|{\widetilde{f}}(x)-{\widetilde{g}}(x)|<1$ for every $x\in K_1\cap K_2$. Let

$$\begin{aligned} U=\{x\in \beta X: |{\widetilde{f}}(x)-{\widetilde{g}}(x)|<1\}. \end{aligned}$$

The set U is open in $\beta X$ and $K_1\cap K_2\subseteq U$.

Since $K_1$ and $K_2{\setminus } U$ are disjoint closed subsets of the compact space $\beta X$, by Urysohn’s lemma there is a continuous function $u:\beta X\rightarrow [0,1]$ such that

$$\begin{aligned} u(x)= \left\{ \begin{aligned}&1{} & {} \text{ for } x\in K_1\\&0 \quad{} & {} \text{ for } x\in K_2\setminus U \end{aligned} \right. \end{aligned}$$

Let

$$\begin{aligned} {\widetilde{h}}=u\cdot ({\widetilde{f}}-{\widetilde{g}})+{\widetilde{g}} \end{aligned}$$

and let $h\in C^*_p(X)$ be the restriction of ${\widetilde{h}}$ to X. We have:

• ${\widetilde{h}}(x)={\widetilde{f}}(x)$ for $x\in K_1$,

• ${\widetilde{h}}(x)={\widetilde{g}}(x)$ for $x\in K_2{\setminus } U$ and

• if $x\in U$, then $|{\widetilde{h}}(x)-{\widetilde{g}}(x)|=|u(x)|\cdot |{\widetilde{f}}(x)-{\widetilde{g}}(x)|<1$, by definition of U and the fact that u maps into [0, 1].

In particular, since $K_1\cap K_2\subseteq U$, we get

• $|{\widetilde{h}}(x)-{\widetilde{g}}(x)|<1$ for $x\in K_2$.

Since $K_1\in \mathscr {A}(y)$ and ${\widetilde{h}}(x)={\widetilde{f}}(x)$ for $x\in K_1$, we get $|\widetilde{\varphi (f)}(y)-\widetilde{\varphi (h)}(y)|\le a(y,K_1)<\infty$. Similarly, since $|{\widetilde{h}}(x)-{\widetilde{g}}(x)|<1$ for $x\in K_2$ and $K_2\in \mathscr {A}(y)$, we have $|\widetilde{\varphi (g)}(y)-\widetilde{\varphi (h)}(y)|\le a(y,K_2)<\infty$. Hence,

$$\begin{aligned} |\widetilde{\varphi (f)}(y)-\widetilde{\varphi (g)}(y)|{} & {} \le |\widetilde{\varphi (f)}(y)-\widetilde{\varphi (h)}(y)| + |\widetilde{\varphi (h)}(y)-\widetilde{\varphi (g)}(y)|\\{} & {} \le a(y,K_1)+a(y,K_2). \end{aligned}$$

So $a(y,K_1\cap K_2)\le a(y,K_1)+a(y,K_2)$ and thus $K_1\cap K_2\in \mathscr {A}(y)$. By induction the result follows for any finite intersection.

The family $\mathscr {A}(y)$, consisting of closed subsets of $\beta X$, is closed under finite intersections and $\emptyset \notin \mathscr {A}(y)$ so by compactness the intersection $\bigcap \mathscr {A}(y)$ must be nonempty. It is finite because $y\in Y_{n,m}$ guarantees that the family $\mathscr {A}(y)$ contains a subset of $\beta X$ which is at most m-element.

Since $\mathscr {A}(y)$ contains a finite subset F of $\beta X$, the set $K(y)\subseteq F$ is an intersection of finitely many elements of $\mathscr {A}(y)$ so the first part of the proof implies that $K(y)\in \mathscr {A}(y)$.

Finally, if $y\in Y$ then $y\in \bigcup _{n,m}Y_{n,m}$, by (2). So K(y) is well defined. The inclusion $K(y)\subseteq X$ follows from Proposition 2.1. $\square$

For $y\in \bigcup _{n,m} Y_{n,m}$ we define

$$\begin{aligned} a(y)=a(y,K(y)) \end{aligned}$$

By Lemma 2.5, $K(y)\in \mathscr {A}(y)$ so $a(y)<\infty$. For a subset A of $\beta X$ we set

$$\begin{aligned} K^{-1}(A)=\left\{ y\in \bigcup _{n,m} Y_{n,m}:K(y)\cap A\ne \emptyset \right\} . \end{aligned}$$

Combining Corollary 2.4 and Lemma 2.5 we get

Proposition 2.6

Suppose that $\varphi :C^*_p(X)\rightarrow C^*_p(Y)$ is a uniformly continuous surjection. If Y is pseudocompact then, for every $y\in \beta Y$, the set K(y) is a well-defined nonempty finite subset of $\beta X$ that belongs to the family $\mathscr {A}(y)$. Also, a(y) is a well-defined number, for every $y\in \beta Y$.

The proof of the next lemma is analogous to the proof of [2, Lemma 1.3].

Lemma 2.7

Suppose that $U\subseteq \beta X$ is open and let n be a positive integer. For every $y\in K^{-1}(U)\cap Y_n$ there exists an open neighborhood V of y in $\beta Y$ such that for every $z\in V\cap Y_n$ and every $A\in \mathscr {A}_n(z)$ we have $A\cap U\ne \emptyset$.

Proof

Fix $x_0\in K(y)\cap U$ witnessing $y\in K^{-1}(U)$. Since K(y) is finite, shrinking U if necessary we can assume that $U\cap K(y)=\{x_0\}$. Note that $\beta X{\setminus } U\notin \mathscr {A}(y)$ for otherwise K(y) would be a subset of $\beta X\setminus U$ and this is not the case because $x_0\in K(y)\cap U$. It follows that there are $f,g\in C^*_p(X)$ such that

$$\begin{aligned}&|{\widetilde{f}}(x)-{\widetilde{g}}(x)|<1 \text{ for } \text{ every } x\in \beta X\setminus U\quad \text{ and } \end{aligned}$$

(3)

$$\begin{aligned}&|\widetilde{\varphi (f)}(y)-\widetilde{\varphi (g)}(y)|>a(y)+n \end{aligned}$$

(4)

Let ${\widetilde{h}}\in C_p(\beta X)$ be a function satisfying

$$\begin{aligned} {\widetilde{h}}(x)={\widetilde{f}}(x) \text{ for } \text{ every } x\in \beta X\setminus U,\quad \text{ and } \quad {\widetilde{h}}(x_0)={\widetilde{g}}(x_0), \end{aligned}$$

(5)

and let $h\in C^*_p(X)$ be the restriction of ${\widetilde{h}}$.

Note that by (3) and (5), $|{\widetilde{h}}(x)-{\widetilde{g}}(x)|<1$ for every $x\in K(y)$. Therefore,

$$\begin{aligned} |\widetilde{\varphi (h)}(y)-\widetilde{\varphi (g)}(y)|\le a(y). \end{aligned}$$

(6)

According to (4) and (6) we have

$$\begin{aligned} |\widetilde{\varphi (f)}(y)-\widetilde{\varphi (h)}(y)| \ge |\widetilde{\varphi (f)}(y)-\widetilde{\varphi (g)}(y)|- |\widetilde{\varphi (h)}(y)-\widetilde{\varphi (g)}(y)|>n. \nonumber \\ \end{aligned}$$

(7)

Let

$$\begin{aligned} V=\{z\in \beta Y:|\widetilde{\varphi (f)}(z)-\widetilde{\varphi (h)}(z)|>n\}. \end{aligned}$$

The set V is open and $y\in V$, by (7). We show that V is as required.

Take $z\in V\cap Y_n$ and let $A\in \mathscr {A}_n(z)$. If $A\cap U=\emptyset$ then ${\widetilde{h}}(x)={\widetilde{f}}(x)$ for every $x\in A$, by (5). So $|\widetilde{\varphi (f)}(z)-\widetilde{\varphi (h)}(z)|\le n$, contradicting $z\in V$. $\square$

Proposition 2.8

Suppose that Y is pseudocompact. If $U\subseteq \beta X$ is open, then the set $K^{-1}(U)$ is a $G_\delta$-subset of $\beta Y$.

Proof

By Proposition 2.6, for every $y\in \beta Y$, the set K(y) is a nonempty finite subset of $\beta X$. For $n=1,2,\ldots$, let

$$\begin{aligned} L_n=K^{-1}(U)\cap Y_n. \end{aligned}$$

For $y\in L_n$ let $V_n^y$ be an open neighborhood of y in $\beta Y$ provided by Lemma 2.7, i.e.

$$\begin{aligned} \text{ if } z\in V_n^y\cap Y_n \text{ and } A\in \mathscr {A}_n(z), \text{ then } A\cap U\ne \emptyset . \end{aligned}$$

(8)

Let

$$\begin{aligned} V_n=\bigcup \{V_n^y:y\in L_n\}. \end{aligned}$$

We claim that

$$\begin{aligned} K^{-1}(U)=\bigcap _{m=1}^\infty \bigcup _{n=m}^\infty V_n. \end{aligned}$$

Indeed, pick $y\in K^{-1}(U)$ and fix an arbitrary $m\ge 1$. Since $\beta Y=\bigcup _{n=1}^\infty Y_n$ (cf. Corollary 2.4), there is i such that $y\in Y_i$. Since $Y_n\subseteq Y_{n+1}$, we can assume that $i>m$. We have $y\in L_i$ whence $y\in V^y_i\subseteq V_i\subseteq \bigcup _{n=m}^\infty V_n$, because $i>m$.

To prove the opposite inclusion, take $z\in \bigcap _{m=1}^\infty \bigcup _{n=m}^\infty V_n$. Again, there is i such that $z\in Y_i$. Let j be a positive integer satisfying $j>\max \{a(z),i\}$. By our assumption, $z\in \bigcup _{n=j}^\infty V_n$, so there is $k\ge j$ such that $z\in V_k$. Clearly, $z\in Y_k$ and since $k>a(z)$ we have

$$\begin{aligned} K(z)\in \mathscr {A}_k(z). \end{aligned}$$

(9)

By definition of $V_k$, there is $y\in L_k$ such that $z\in V^y_k$. Now, from (8) and (9) we get $z\in K^{-1}(U)$. $\square$

Remark 2.9

If $\varphi :C^*_p(X)\rightarrow C^*_p(Y)$ is a uniform homeomorphism, we may consider the inverse map $\varphi ^{-1}:C^*_p(Y)\rightarrow C^*_p(X)$ and apply all of the above results to $\varphi ^{-1}$. In particular, if X is pseudocompact, then for every $x\in \beta X$ we can define the set $K(x)\subseteq \beta Y$ and the real number a(x) simply by interchanging the roles of X and Y above.

Lemma 2.10

Suppose that both X and Y are pseudocompact spaces. Let $\varphi :C^*_p(X)\rightarrow C^*_p(Y)$ be a uniform homeomorphism. For any $x\in \beta X$ there is $y\in K(x)$ such that $x\in K(y)$.

Proof

Let $x\in \beta X$. Applying Proposition 2.6, first to $\varphi ^{-1}$ and then to $\varphi$, we infer that the set $F=\bigcup \{K(y):y\in K(x)\}\subseteq \beta X$ is finite being a finite union of finite sets. Let M be a positive integer such that

$$\begin{aligned} M>\max \{a(y):y\in K(x)\}. \end{aligned}$$

Striving for a contradiction, suppose that $x\notin F$. Let ${\widetilde{f}},{\widetilde{g}}\in C_p(\beta X)$ be functions satisfying

$$\begin{aligned} {\widetilde{f}}(z)={\widetilde{g}}(z) \text{ for } \text{ every } z\in F \text{ and } |{\widetilde{f}}(x)-{\widetilde{g}}(x)|> M\cdot a(x). \end{aligned}$$

(10)

Let $f\in C^*_p(X)$ and $g\in C^*_p(X)$ be the restrictions of ${\widetilde{f}}$ and ${\widetilde{g}}$, respectively. Since for every $y\in K(x)$ the functions ${\widetilde{f}}$ and ${\widetilde{g}}$ agree on $K(y)\subseteq F$, we have

$$\begin{aligned} |\widetilde{\varphi (f)}(y)-\widetilde{\varphi (g)}(y)|\le a(y)<M, \text{ for } \text{ every } y\in K(x). \end{aligned}$$

(11)

For $k\in \{0,1,\ldots , M\}$ define a function $\widetilde{h_k}\in C_p(\beta Y)$ by the formula

$$\begin{aligned} \widetilde{h_k}=\widetilde{\varphi (f)}+\tfrac{k}{M}\left( \widetilde{\varphi (g)}-\widetilde{\varphi (f)}\right) . \end{aligned}$$

Obviously, $\widetilde{h_0}=\widetilde{\varphi (f)}$ and $\widetilde{h_M}=\widetilde{\varphi (g)}$. Moreover, by (11), we have

$$\begin{aligned} |\widetilde{h_{k+1}}(y)-\widetilde{h_{k}}(y)|=\tfrac{1}{M}|\widetilde{\varphi (g)}(y)-\widetilde{\varphi (f)}(y)|<1, \text{ for } \text{ every } y\in K(x). \end{aligned}$$

(12)

For $k\in \{0,1,\ldots , M\}$ let $h_k\in C^*_p(Y)$ be the restriction of $\widetilde{h_k}$. Using (12) we get:

$$\begin{aligned} |{\widetilde{f}}(x)-{\widetilde{g}}(x)|&=|\widetilde{\varphi ^{-1}(\varphi (f))}(x)-\widetilde{\varphi ^{-1}(\varphi (g))}(x)|= |\widetilde{\varphi ^{-1}(h_0)}(x)-\widetilde{\varphi ^{-1}(h_M)}(x)|\\&\le |\widetilde{\varphi ^{-1}(h_0)}(x)-\widetilde{\varphi ^{-1}(h_1)}(x)|+\cdots \\&\quad + |\widetilde{\varphi ^{-1}(h_{M-1})}(x)-\widetilde{\varphi ^{-1}(h_M)}(x)|\le M\cdot a(x) \end{aligned}$$

This however contradicts (10). $\square$

Now we are ready to prove of our main result.

Proof of Theorem 1.1

Let $\kappa$ be an infinite cardinal and let $\varphi :C_p(X)\rightarrow C_p(Y)$ be a uniform homeomorphism. By symmetry it is enough to show that if Y is $\kappa$-pseudocompact then so is X. So let us assume that Y is $\kappa$-pseudocompact. Then, in particular Y is pseudocompact and hence by Uspenskiĭ’s theorem [12, Corollary] (cf. [11, V.136]), so is X. Hence, $C_p(Y)=C^*_p(Y)$ and $C_p(X)=C^*_p(X)$. In order to prove that X is $\kappa$-pseudocompact we will employ Theorem 1.3. For this purpose fix a nonempty $G_\kappa$-subset G of $\beta X$. It suffices to prove that $G\cap X\ne \emptyset$.

Claim 1

The set $K^{-1}(G)=\{y\in \beta Y:K(y)\cap G\ne \emptyset \}$ is nonempty.

Proof

The set G is nonempty so let us fix $x\in G$. According to Lemma 2.10 there is $y\in K(x)$ such that $x\in K(y)$. In particular, $y\in K^{-1}(G)$. $\square$

Claim 2

The set $K^{-1}(G)=\{y\in \beta Y:K(y)\cap G \ne \emptyset \}$ is a $G_\kappa$-set in $\beta Y$.

Proof

Write $G=\bigcap \{U_\alpha :\alpha <\kappa \}$, where each $U_\alpha$ is an open subset of $\beta X$. We can also assume that the family $\{U_\alpha :\alpha <\kappa \}$ is closed under finite intersections. It follows from Proposition 2.8 that for each $\alpha <\kappa$, the set $K^{-1}(U_\alpha )$ is $G_\delta$ in $\beta Y$. Thus, it is enough to show that

$$\begin{aligned} K^{-1}(G)=\bigcap _{\alpha < \kappa }K^{-1}(U_\alpha ). \end{aligned}$$

To this end, take $y\in \bigcap _{\alpha < \kappa }K^{-1}(U_\alpha )$. According to Proposition 2.6, the set K(y) is a nonempty finite subset of $\beta X$. Enumerate $K(y)=\{x_1,\ldots ,x_k\}$, where k is a positive integer. If $y\notin K^{-1}(G)$, then for every $i\le k$ there is $\alpha _i<\kappa$ such that

$$\begin{aligned} x_i\notin U_{\alpha _i}. \end{aligned}$$

(13)

The family $\{U_\alpha :\alpha <\kappa \}$ is closed under finite intersections, so there is $\gamma <\kappa$ with $U_\gamma =U_{\alpha _1}\cap \cdots \cap U_{\alpha _k}$. But $y\in \bigcap _{\alpha < \kappa }K^{-1}(U_\alpha )\subseteq K^{-1}(U_\gamma )$. Hence, there is $j\le k$ such that $x_j\in U_\gamma \subseteq U_{\alpha _j}$, which is a contradiction with (13). Therefore, we must have $y\in K^{-1}(G)$. This provides the inclusion $K^{-1}(G)\supseteq \bigcap _{\alpha < \kappa }K^{-1}(U_\alpha )$. The opposite inclusion is immediate. $\square$

It follows from Claims 1 and 2 that the $K^{-1}(G)$ is a nonempty $G_\kappa$-subset of $\beta Y$. Hence, by Theorem 1.3, there exists $p\in K^{-1}(G)\cap Y$. We have $K(p)\cap G\ne \emptyset$ and since $p\in Y$, we infer from Lemma 2.5 that K(p) is a nonempty finite subset of X. Therefore, $\emptyset \ne K(p)\cap G\subseteq X\cap G$. $\square$