1. Introduction

We consider the homeomorphism of the spaces of continuous functions on generalized ordered spaces. A space \( X \) is a generalized ordered space if \( X \) is a subspace of some linearly ordered topological space \( Y \) (see [1]). In fact, these are the linearly ordered spaces in which a neighborhood base of \( x\in X \) is given by intervals \( (y,z) \), with \( x\in(y,z) \), or by half-intervals \( [x,y) \) or \( (y,x] \) or by the singleton \( \{x\} \). Among the examples of these spaces are the Sorgenfrey line, the Michael line, the Hattori spaces, etc.

In generalized ordered spaces, the tightness \( t(X) \) of a space \( X \) coincides with the functional tightness of \( (X) \). Therefore, if \( t(X)\neq t(Y) \) for generalized ordered spaces \( X \) and \( Y \) then the spaces \( C_{p}(X) \) and \( C_{p}(Y) \) are nonhomeomorphic [2]. In the article, considering generalized ordered spaces \( X \) and \( Y \) of the same tightness but having the nonequipollent sets of the points of maximum tightness, we find the sufficient conditions for \( X \) and \( Y \) to be not \( t \)-equivalent, which means that \( C_{p}(Y) \) and \( C_{p}(X) \) are nonhomeomorphic.

As a consequence, using the homeomorphism theorems for \( C_{p}[0,\alpha] \), we obtain the topological classification of \( C_{p}(S_{\alpha}) \) on the “long Sorgenfrey lines \( S_{\alpha} \).” Note that the topological and linear classifications of \( C_{p}[0,\alpha] \) depend on the \( t \)-equivalence of the ordinal segments \( [0,\alpha] \). The theorems in this article imply that the topological classification of the spaces \( C_{p}(S_{\alpha}) \) coincides with the linear homeomorphic classification of \( C_{p}(S_{\alpha}) \) which was obtained in [3].

2. The Main Notations and Definitions

Throughout the sequel, \( {𝕉} \) and \( {𝕅} \) are the sets of reals and naturals respectively; \( \aleph_{0} \) is a countable cardinal, \( \aleph_{1} \) is the first uncountable cardinal; and \( |A| \) is the cardinality of a set \( A \). Given a generalized linear ordered space \( X \), as usual, we put

$$ (x,\rightarrow)=\{y\in X:y>x\}\quad\text{and}\quad(\leftarrow,x)=\{y\in X:y<x\}. $$

The spaces \( [x,\rightarrow) \), \( (\leftarrow,x] \), \( [y,x) \), and \( (x,y] \) are defined likewise. Observe that every linear ordered topological space \( X \) is normal (see [4, 1.7.5]) and so is every subspace \( Y\subset X \) (see [4, 2.7.5]). Therefore, all generalized ordered spaces are normal.

Definition 1

Let \( X \) be a generalized ordered space. If \( x \) is a limit point of \( (\leftarrow,x) \) then the cofinality of \( x \) is the cardinal

$$ \operatorname{cf}x=\min\{|A|:A\text{ is a cofinal subset of }(\leftarrow,x)\}. $$

If \( x \) is an isolated point in \( (\leftarrow,x]\subset X \) then we put \( \operatorname{cf}x=0 \). The coinitiality of \( x\in X \) is defined similarly: \( \operatorname{cn}x=0 \) if \( x \) is an isolated point in \( [x,\rightarrow) \) and

$$ \operatorname{cn}x=\min\{|A|:A\text{ is a coinitial subset of }(x,\rightarrow)\} $$

if \( x \) is a limit point of \( (x,\rightarrow) \).

Denote by \( t(X) \) the tightness of a space \( X \) and designate as \( t(x,X) \) the tightness of \( X \) at a point \( x \). If \( X \) is a generalized ordered space, then clearly \( t(x,X)=\max\{\operatorname{cf}x,\operatorname{cn}x\} \). Let \( t(X)=\tau \). Consider the set

$$ T(X)=\{x\in X:t(x,X)=\tau\} $$

and its subsets

$$ T_{l}(X)=\{x\in T(X):\operatorname{cf}x=\tau,\operatorname{cn}x<\tau\}; $$
$$ T_{r}(X)=\{x\in T(X):\operatorname{cn}x=\tau,\operatorname{cf}x<\tau\}; $$
$$ T_{lr}(X)=\{x\in T(X):\operatorname{cn}x=\operatorname{cf}x=\tau\}. $$

Clearly,

$$ T(X)=T_{l}(X)\sqcup T_{r}(X)\sqcup T_{lr}(X). $$

Definition 2

Let \( Z\subset X \) and let \( \lambda \) be a cardinal. A point \( x\in Z \) is \( \lambda \)-inaccessible in \( Z \) if from \( x\notin A\subset Z \) and \( |A|\leq\lambda \) it follows that \( x\notin\overline{A} \).

Definition 3

A function \( f:X\rightarrow{𝕉} \) is \( \lambda \)-continuous if, for every \( A\subset X \), with \( |A|\leq\lambda \), the function \( f|_{A} \) is continuous on \( A \). If for every set \( A\subset X \) such that \( |A|\leq\lambda \), there exists a continuous function \( h:X\rightarrow{𝕉} \) such that \( f|_{A}=h|_{A} \) then \( f \) is strictly \( \lambda \)-continuous.

Note that for a normal space \( X \), every \( \lambda \)-continuous function is strictly \( \lambda \)-continuous (see [5]).

It is not hard to see that the characteristic function \( \chi_{\{x\}}:X\rightarrow{𝕉} \) is \( \lambda \)-continuous if and only if \( x \) is \( \lambda \)-accessible in \( X \). As usual, \( C_{p}(X) \) means the space of all real-valued continuous functions on \( X \) which is endowed with the topology of pointwise convergence, while \( C_{\omega}(X) \) is the space of all \( \omega \)-continuous functions on \( X \) which is also endowed with the topology of pointwise convergence. The notation \( C_{p}(X)\sim C_{p}(Y) \) means that \( C_{p}(X) \) and \( C_{p}(Y) \) are homeomorphic.

Given a Tychonoff space \( X \), denote the Hewitt completion of \( X \) by \( \nu X \). The following are well known:

Theorem 1 [4]

If \( X \) and \( Y \) are Tychonoff spaces and \( \varphi:X\rightarrow Y \) is a homeomorphism then there exists a homeomorphism \( \tilde{\varphi}:\nu X\rightarrow\nu Y \) such that \( \tilde{\varphi}|_{X}=\varphi \).  ☐

Theorem 2 [5]

If \( X \) is a normal space then

$$ \nu C_{p}(X)=\{f\in{𝕉}^{X}:f\text{ is }\omega\text{-continuous on }X\}.\quad\text{☐} $$

3. The Main Results

Lemma 1

Let \( X \) be a generalized ordered space, \( t(X)>\aleph_{0} \), and let \( P_{l}\subset T_{l}(X) \) be a right discrete space in \( T_{l}(X) \); i.e., every \( x\in P_{l} \) is isolated in \( T_{l}(X)\cap[x,\rightarrow) \). Then there exists \( A(P_{l})\subset C_{\omega}(X)\setminus C_{p}(X) \), with \( A(P_{l})=\{f_{x}:x\in P_{l}\} \), such that for every \( x\in P_{l} \) the function \( f_{x} \) is continuous on \( X\setminus\{x\} \), right continuous at \( x\in X \), and

$$ f^{-1}_{x^{{}^{\prime}}}({𝕉}\setminus\{0\})\cap f^{-1}_{x^{\prime\prime}}({𝕉}\setminus\{0\})=\varnothing $$

for all \( x^{\prime},x^{\prime\prime}\in P_{l} \), \( x^{\prime}\neq x^{\prime\prime} \).

Proof

Since \( P_{l} \) is a right discrete set in \( T_{l}(X) \), for every \( x\in P_{l} \) there exists a neighborhood \( U(x) \) such that \( U(x)\cap[x,\rightarrow)\cap T_{l}(X)=\{x\} \). Putting \( f_{x}(x)=1 \) and \( f_{x}(t)=0 \) if \( t<x \) or \( t\in[x,\rightarrow)\setminus U(x) \), we extend \( f_{x} \) by continuity to \( [x,\rightarrow) \). It is not hard to see that \( f_{x} \) is left discontinuous at \( x \), continuous on \( [x,\rightarrow) \), and \( \omega \)-continuous since \( \operatorname{cf}x=\tau>\aleph_{0} \). The set \( A(P_{l})=\{f_{x}:x\in P_{l}\} \) satisfies the hypotheses of the lemma.  ☐

Obviously, if \( T_{r}(X) \) contains a left discrete set \( P_{r} \); then, as in Lemma 1, we can construct \( A(P_{r})\subset C_{\omega}(X)\setminus C(X) \).

Lemma 2

Suppose that \( Y \) is a generalized ordered space, \( y_{0}\in Y \), \( \operatorname{cf}y_{0}>\aleph_{0} \), and there exists a cofinal countably compact set \( A\subset(\leftarrow,y_{0}) \). Then for every function \( f\in C_{\omega}(Y) \) there exists \( y_{f}\in Y \) such that \( y_{f}<y_{0} \) and \( f(y)=f(y_{f}) \) for every \( y\in[y_{f},y_{0}) \). The same holds for \( y_{0}\in Y \) such that \( \operatorname{cn}y_{0}>\aleph_{0} \).

Proof

Show that

(\( \ast \)) for every \( \varepsilon>0 \) there is \( y_{\varepsilon}<y_{0} \) such that \( |f(y^{\prime\prime})-f(y^{\prime})|<\varepsilon \) if \( y^{\prime},y^{\prime\prime}\in(y_{\varepsilon},y_{0}) \).

Indeed, if (\( \ast \)) fails then there is \( \varepsilon_{0}>0 \) such that for every \( y<y_{0} \) we have points \( y^{\prime},y^{\prime\prime}>y \) for which \( |f(y^{\prime})-f(y^{\prime\prime})|\geq\varepsilon_{0} \). By induction, we can choose some increasing sequence of points

$$ y_{1}<y^{\prime}_{1}<y^{\prime\prime}_{1}<\dots<y_{n}<y^{\prime}_{n}<y^{\prime\prime}_{n}<\cdots $$

such that \( \{y_{n}\}^{\infty}_{n=1}\subset A \) and \( |f(y^{\prime\prime}_{n})-f(y^{\prime}_{n})|\geq\varepsilon_{0} \) for each \( n\in{𝕅} \). Since \( A \) is countably compact, there exists \( a=\sup\{y_{n}:n\in{𝕅}\}\in A \). Clearly, \( a \) is also a limit point of \( \{y^{\prime}_{n}:n\in{𝕅}\} \) and \( \{y^{\prime\prime}_{n}:n\in{𝕅}\} \). We come to a contradiction to the continuity of \( f \) on the countable set \( \{y^{\prime}_{n}:n\in{𝕅}\}\cup\{y^{\prime\prime}_{n}:n\in{𝕅}\}\cup\{a\} \). Consequently, (\( \ast \)) holds.

Applying (\( \ast \)) to \( \varepsilon=\frac{1}{n} \), we obtain some sequence \( y_{1}<y_{2}<\cdots \) such that \( y_{n}\in A \) and \( |f(y^{\prime\prime})-f(y^{\prime})|<\frac{1}{n} \) for \( y^{\prime},y^{\prime\prime}\in(y_{n},y_{0}) \). Then for \( y_{f}=\sup\{y_{n}:n\in{𝕅}\} \) we have \( f(y)=f(y_{f}) \) for all \( y\in[y_{f},y_{0}) \).  ☐

Lemma 3

Suppose that \( X \) and \( Y \) be generalized ordered sets, \( t(X)=t(Y)=\tau>\aleph_{0} \), and \( \Phi:C_{\omega}(X)\rightarrow C_{\omega}(Y) \) is a homeomorphism such that \( \Phi(C_{p}(X))=C_{p}(Y) \). If \( f_{x}\in A(P_{l}) \) (or \( f_{x}\in A(P_{r})) \) then \( \Phi(f_{x}) \) is \( \lambda \)-continuous for all \( \lambda<\tau \).

Proof

Suppose that there exists a subset \( Y_{0}\subset Y \), with \( |Y_{0}|<\tau \), and \( \Phi(f_{x})|_{Y_{0}} \) is discontinuous at some point \( y_{0}\in Y_{0} \). Given \( y\in Y_{0} \) and \( n\in N \), consider the standard neighborhoods of \( \Phi(f_{x}) \); i.e.,

$$ \begin{gathered}\displaystyle U(y,n)=U\left(\Phi(f_{x}),y,y_{0},\frac{1}{n}\right)\\ \displaystyle=\left\{g\in C_{\omega}(Y):|g(y)-\Phi(f_{x})(y)|<\frac{1}{n},\ |g(y_{0})-\Phi(f_{x})(y_{0})|<\frac{1}{n}\right\}.\end{gathered} $$

Then for every \( g\in\bigcap\{U(y,n):y\in Y_{0} \), \( n\in{𝕅}\} \), we have \( g|_{Y_{0}}=\Phi(f_{x})|_{Y_{0}} \). Hence,

$$ \bigl{(}\bigcap\{U(y,n):y\in Y_{0},\ n\in{𝕅}\}\bigr{)}\cap C(Y)=\varnothing. $$

Since \( \Phi(C_{p}(X))=C_{p}(Y) \); therefore,

$$ \Bigl{(}\bigcap\{\Phi^{-1}U(y,n):y\in Y_{0},\ n\in{𝕅}\}\Bigr{)}\cap C_{p}(X)=\varnothing. $$
(1)

On the other hand, for every neighborhood \( \Phi^{-1}U(y,n) \) of \( f_{x} \), there exists a neighborhood

$$ V\left(f_{x},F(y,n),\frac{1}{k(y,n)}\right)\subset\Phi^{-1}U(y,n), $$

where \( F(y,n)\subset X \) is a finite subset and \( k(y,n)\in 𝕅 \). Obviously, \( |\bigcup\{(F(y,n):y\in Y_{0} \), \( n\in 𝕅\}|<\tau \) and hence \( F=\bigcup\{F(y,n):y\in Y_{0} \), \( n\in 𝕅\} \) is not cofinal to \( x \). Since \( f_{x}\in A(P_{l}) \) is \( \lambda \)-continuous for all \( \lambda<\tau \) and hence strictly \( \lambda \)-continuous (see [5]), there exists a continuous function \( h\in C_{p}(X) \) such that \( f_{x}|_{F}=h|_{F} \). Clearly,

$$ h\in\bigcap\left\{V\left(f_{x},F(y,n),\frac{1}{k(y,n)}\right):y\in Y_{0},\ n\in{𝕅}\right\}\subset\bigcap\{\Phi^{-1}U(y,n):y\in Y_{0},\ n\in{𝕅}\}, $$

which contradicts (1).  ☐

Lemma 4

Suppose that \( X \) and \( Y \) are generalized ordered spaces, \( t(X)=t(Y)=\tau>\aleph_{0} \), and \( \Phi:C_{\omega}(X)\rightarrow C_{\omega}(Y) \) is a homeomorphism such that \( \Phi(C_{p}(X))=C_{p}(Y) \). If \( x\in T_{lr}(X) \) then \( \Phi(\chi_{\{x\}}) \) is \( \lambda \)-continuous for all \( \lambda<\tau \).

Proof

Observe that the function \( \chi_{\{x\}} \) is \( \lambda \)-continuous for \( \lambda<\tau \) provided that \( x\in T_{lr}(X) \). The rest of the proof is finished as in Lemma 3.  ☐

Theorem 3

Suppose that \( X \) and \( Y \) are generalized ordered spaces, \( t(X)=t(Y)=\tau>\aleph_{0} \), \( |T(X)|>|T(Y)|\geq\aleph_{0} \), and the following are fulfilled:

\( (1) \) Either \( |T_{lr}(X)|>|T(Y)| \) or there exists a right discrete subset \( P_{l}\subset T_{l}(X) \) such that \( |P_{l}|>|T(Y)| \) or there exists a left discrete set \( P_{r}\subset T_{r}(X) \) such that \( |P_{r}|>|T(Y)| \).

\( (2) \) If \( y\in Y \) and \( \operatorname{cf}y=\tau \) then there exists a cofinal countably compact subset in \( (\leftarrow,y) \). If \( \operatorname{cn}y=\tau \) then there exists a coinitial countably compact subset in \( (y,\rightarrow) \).

Then \( C_{p}(Y) \) and \( C_{p}(X) \) are not homeomorphic.

Proof

Suppose that \( P_{l}\subset T_{l}(X) \) is right discrete in \( T_{l}(X) \) and \( |P_{l}|>|T(Y)| \). Consider

$$ A(P_{l})\subset C_{\omega}(X)\setminus C_{p}(X) $$

and suppose that there exists a homeomorphism \( \Phi:C_{p}(X)\rightarrow C_{p}(Y) \). Without loss of generality, we may assume that \( \Phi(0)=0 \). Using Theorems 1 and 2, we can extend \( \Phi \) to a homeomorphism \( \Phi:C_{\omega}(X)\rightarrow C_{\omega}(Y) \). Since the function \( f\equiv 0 \) is a limit point of the nonstationary sequence \( \{f_{x_{n}}\}_{n=1}^{\infty}\subset A(P_{l}) \), we see that

$$ |\{x\in P_{l}:\Phi(f_{x})(y)\neq 0\}|\leq\aleph_{0}\quad\text{for all }y\in Y. $$

Since \( |T(Y)|<|P_{l}| \), there exists \( P_{l1}\subset P_{l} \) such that \( |P_{l1}|=|P_{l}| \) and \( \Phi(f_{x})(y)=0 \) for all \( x\in P_{l} \) and \( y\in T(Y) \).

Since all \( \Phi(f_{x}) \) do not belong to \( C_{p}(Y) \) and are \( \lambda \)-continuous for \( \lambda<\tau \) by Lemma 3, for every \( x\in P_{l1} \) there exists \( y\in T(Y) \) at which \( \Phi(f_{x}) \) is left discontinuous if \( y\in T_{l}(Y) \) and right discontinuous if \( y\in T_{r}(Y) \).

Let \( y\in T(Y) \) and

$$ P_{ly}=\{x\in P_{l1}:\Phi(f_{x})\text{ is discontinuous at }y\}. $$

Since \( \bigcup\{P_{ly}:y\in T(Y)\}=P_{l1} \) and \( |P_{l1}|=|P_{l}|>T(Y) \), there exists \( y_{0}\in T(Y) \) such that \( |P_{ly_{0}}|>\aleph_{0} \). Without loss of generality, we may assume that all functions \( \{\Phi(f_{x}):x\in P_{ly_{0}}\} \) are left discontinuous at \( y_{0} \). By Lemma 2, for every \( \Phi(f_{x}) \) there exists a point \( y_{x}\in Y \) such that \( \Phi(f_{x})(y)=\Phi(f_{x})(y_{x}) \) for all \( y\in[y_{x},y_{0}) \); moreover, \( \Phi(f_{x})(y_{x})\neq 0 \) because \( \Phi(f_{x}) \) is discontinuous at \( y_{0} \) and \( \Phi(f_{x})(y_{0})=0 \). Since \( |P_{ly_{0}}|>\aleph_{0} \), for some \( n\in{𝕅} \) there is an uncountable set \( P_{ly_{0}n}=\bigl{\{}x\in P_{ly_{0}}:\Phi(f_{x})(y_{x})\geq\frac{1}{n}\bigr{\}} \).

Consider an arbitrary countable subset \( B\subset P_{ly_{0}n} \). Since \( \{y_{x}:x\in B\} \) is not cofinal to \( (\leftarrow,y_{0}) \), there exists a point \( y_{1} \) such that \( y_{x}<y_{1}<y_{0} \) for every \( x\in B \); i.e., the identically zero function is not limit point of \( \{\Phi(f_{x}):x\in B\} \). This contradicts the fact that the function \( f\equiv 0 \) is a limit point of \( \{f_{x}:x\in B\} \).

For the case of the existence of a left discrete subset \( P_{r} \) in \( T_{r}(X) \) such that \( |P_{r}|>|T(Y)| \), the proof is similar. If \( |T_{lr}(X)|>|T(Y)| \); then, instead of \( A(P_{l}) \), we must consider the set \( \{\chi_{\{x\}}:x\in T_{lr}(x)\} \) and use Lemma 4.  ☐

4. On the \( t \)-Equivalence of the Spaces \( X_{\alpha} \)

Let \( X \) be a separable generalized ordered spaces and let \( \alpha \) be an ordinal. Endow the product \( [0,\alpha)\times X \) with the order relation \( (\gamma_{1},x_{1})\leq(\gamma_{2},x_{2}) \) if \( \gamma_{1}<\gamma_{2} \) or \( \gamma_{1}=\gamma_{2} \) and \( x_{1}\leq x_{2} \). Let \( B(x) \) be a neighborhood base of \( x\in X \) and let \( x_{0} \) be the first element in \( X \) (if existent). Define the neighborhood base of \( (\gamma,x)\in[0,\alpha)\times X \) as follows:

$$ {\mathcal{B}}(\gamma,x)=\{\{\gamma\}\times O(x):O(x)\in{\mathcal{B}}(x)\} $$

if \( x\neq x_{0} \)

$$ {\mathcal{B}}(\gamma,x_{0})=\{(\beta,\gamma)\times X\sqcup\{\gamma\}\times O(x_{0}):\beta<\gamma,\ O(x_{0})\in{\mathcal{B}}(x_{0})\}, $$

if \( \gamma \) is a limit ordinal and

$$ {\mathcal{B}}(\gamma,x_{0})=\{\{\gamma-1\}\times(x,\rightarrow)\sqcup\{\gamma\}\times O(x_{0}):x\in X,\ O(x_{0})\in{\mathcal{B}}(x_{0})\} $$

if \( \gamma \) is a nonlimit ordinal.

Denote the space \( [0,\alpha)\times X \) with the topology base \( \{B(\gamma,x):\gamma\in[0,\alpha),\ x\in X\} \) by \( X_{\alpha} \).

Considering the segment \( [0,\alpha] \), we obtain the space \( X_{\alpha+1} \). In particular, if \( X=[0,1)\subset{𝕉} \) and \( \alpha=\omega_{1} \) then we get a “long line,” and if \( S \) is a Sorgenfrey line with neighborhood base \( {\mathcal{B}}(x)=\{(a,x]:a<x\} \) and \( X=[0,1)\subset S \) then \( [0,\alpha)\times X=S_{\alpha} \) is a “long Sorgenfrey line.” As \( X \), we can take the “two arrows” or \( [0,1)\subset H(A) \), where \( H(A) \) is the Hattori space (see [6]). Note that the existence of the first element \( x_{0}\in X \) makes it possible to define some mapping \( \varphi(\gamma)=(\gamma,x_{0}) \) that is a homeomorphic embedding of the ordinal interval \( [0,\alpha) \) onto the closed subspace \( \{(\gamma,x_{0}):\gamma\in[0,\alpha)\}\subset[0,\alpha)\times X \). If \( \tau \) and \( \sigma \) are initial ordinals, \( \omega\leq\sigma\leq\tau \), and \( \tau \) is a regular ordinal; then \( t(X_{\tau\sigma+1})=\tau \) and \( |T(X_{\tau\sigma+1})|=|\sigma| \). It is not hard to see that \( T(X_{\tau\sigma+1}) \) has the form

$$ \{(\tau(\gamma+1),x_{0}):1\leq\gamma<\sigma\}, $$

is right discrete, and \( T(X_{\tau\sigma+1})=T_{l} \). (In case \( \sigma=\tau \), we must add the point \( (\tau\cdot\tau,x_{0}) \).)

The set \( \{(\tau\cdot\gamma+\beta,x_{0}):1\leq\beta<\tau\} \) is homeomorphic to the ordinal interval \( [1,\tau) \) and, hence, it is countably compact. Moreover, this is a cofinal subset of \( (\leftarrow,(\tau(\gamma+1)),x_{0}) \). Thus, \( X_{\tau\sigma+1} \) satisfy conditions (1) and (2) of Theorem 3. Consequently, we have the following

Theorem 4

Suppose that \( X \) is a separable generalized ordered space with the first element; \( \tau \), \( \lambda \), and \( \sigma \) are initial ordinals, and \( \tau \) is a regular ordinal. If \( \omega\leq\sigma<\lambda\leq\tau \) then \( C_{p}(X_{\tau\sigma+1}) \) and \( C_{p}(X_{\tau\lambda+1}) \) are nonhomeomorphic.  ☐

Theorem 5

Let \( \alpha \) and \( \beta \) be infinite ordinals and let \( X \) be a separable generalized ordered space with the first element. The space \( C_{p}(X_{\alpha+1}) \) is homeomorphic to \( C_{p}(X_{\beta+1}) \) if and only if \( C_{p}[0,\alpha] \) is homeomorphic to \( C_{p}[0,\beta] \).

Proof

Consider the closed subset \( A=[0,\alpha]\times\{x_{0}\}\subset X_{\alpha+1} \) and show that there exists a continuous linear extension operator \( \Psi:C_{p}(A)\overset{{in}}{\rightarrow}C_{p}(X_{\alpha+1}) \). On every segment \( I_{\gamma}=[(\gamma,x_{0}),(\gamma+1,x_{0})] \), \( 0\leq\gamma<\alpha \), fix \( x_{1}\in X \), with \( x_{1}>x_{0} \). By Urysohn’s Lemma, there is a linear function \( g_{0}:I_{\gamma}\rightarrow[0,1] \) such that \( g_{0}([(\gamma,x_{0}),(\gamma,x_{1})])\subset\{1\} \) and \( g_{0}(\gamma+1,x_{0})=0 \). In a similar fashion, define the function \( g_{1}:I_{\gamma}\rightarrow[0,1] \), \( g_{1}([(\gamma,x_{1}),(\gamma+1,x_{0})])\subset\{1\} \), and \( g_{1}(\gamma,x_{0})=0 \). Putting \( f_{0}=\frac{g_{0}}{g_{0}+g_{1}} \) and \( f_{1}=\frac{g_{1}}{g_{0}+g_{1}} \), we obtain the partition of unity \( \{f_{0},f_{1}\} \). Consider the operator \( \Psi:C_{p}(A)\rightarrow C_{p}(X_{\alpha+1}) \) defined by the formula

$$ \Psi(f)(\gamma,x)=f(\gamma,x_{0})f_{0}(\gamma,x)+f(\gamma+1,x_{0})f_{1}(\gamma,x) $$

if \( 0\leq\gamma<\alpha \) and \( \Psi(f)(\alpha,x)=f(\alpha,x_{0}) \). The function \( \Psi(f)|_{A} \) is equal to \( f \), whereas \( \Psi(f) \) takes values between \( f(\gamma,x_{0}) \) and \( f(\gamma+1,x_{0}) \) on \( I_{\gamma} \) and hence \( \Psi(f) \) is continuous on \( X_{\alpha+1} \). It is easy to check that \( \Psi \) is linear and continuous. In this event (see [5, 1.5]) \( C_{p}(X_{\alpha+1}) \) is linearly homeomorphic to \( C_{p}(A)\times C_{p}^{0}(X_{\alpha+1},A) \), where \( C_{p}^{0}(X_{\alpha+1},A)=\{f\in C_{p}(X_{\alpha+1}):f(A)\subset\{0\}\} \).

By the compactness of the ordinal segment \( [0,\alpha] \), the set \( \{\gamma:\sup\nolimits_{x\in X}|f(\gamma,x)|\geq\varepsilon\} \) is finite for all \( f\in C_{p}^{0}(X_{\alpha+1},A) \) and \( \varepsilon>0 \). Therefore, \( C_{p}^{0}(X_{\alpha+1},A) \) is linearly homeomorphic to the space \( \big{(}\prod\nolimits_{0\leq\gamma\leq\alpha}C_{p}^{0}(I_{\gamma})\big{)}_{c_{0}} \) defined as follows:

$$ \begin{gathered}\displaystyle\bigg{(}\prod\limits_{0\leq\gamma\leq\alpha}C_{p}^{0}(I_{\gamma})\bigg{)}_{c_{0}}\\ \displaystyle=\bigg{\{}f=\{f_{\gamma}\}_{\gamma\leq\alpha}\in\prod\limits_{0\leq\gamma\leq\alpha}C_{p}^{0}(I_{\gamma}):\{\gamma:\sup\limits_{(\gamma,x)\in I_{\gamma}}|f_{\gamma}(\gamma,x)|\geq\varepsilon\}\text{ is finite for any }\varepsilon>0\bigg{\}},\end{gathered} $$

where

$$ C_{p}^{0}(I_{\gamma})=\{f\in C_{p}(I_{\gamma}):f(\gamma,x_{0})=f(\gamma+1,x_{0})=0\} $$

if \( 0\leq\gamma<\alpha \) and

$$ C_{p}^{0}(I_{\alpha})=\{f\in C_{p}(I_{\alpha}):f(\alpha,x_{0})=0\}. $$

Since all \( I_{\gamma} \)’s, with \( 0\leq\gamma<\alpha \), are homeomorphic to \( I_{0} \) and \( I_{\alpha} \) is homeomorphic to \( X \), the space

$$ \bigg{(}\prod\limits_{0\leq\gamma\leq\alpha}C_{p}^{0}(I_{\gamma})\bigg{)}_{c_{0}} $$

is linearly homeomorphic to

$$ \bigg{(}\prod\limits_{0\leq\gamma<\alpha}C_{p}^{0}(I_{0})\bigg{)}_{c_{0}}\times C_{p}^{0}(I_{\alpha})\sim\bigg{(}\prod\limits_{|\alpha|}C_{p}^{0}(I_{0})\bigg{)}_{c_{0}}\times C_{p}^{0}(X). $$

Suppose that \( C_{p}[0,\alpha] \) is homeomorphic to \( C_{p}[0,\beta] \). Clearly, in this case \( |\alpha|=|\beta| \). Since \( A \) is homeomorphic to \( [0,\alpha] \), we obtain

$$ \begin{gathered}\displaystyle C_{p}(X_{\alpha+1})\sim C_{p}(A)\times C_{p}^{0}(X_{\alpha+1},A)\sim C_{p}[0,\alpha]\times\bigg{(}\prod\limits_{|\alpha|}C_{p}^{0}(I_{0})\bigg{)}_{c_{0}}\times C_{p}^{0}(X)\\ \displaystyle\sim C_{p}[0,\beta]\times\bigg{(}\prod\limits_{|\beta|}C_{p}^{0}(I_{0})\bigg{)}_{c_{0}}\times C_{p}^{0}(X)\sim C_{p}(X_{\beta+1}).\end{gathered} $$

If \( C_{p}[0,\alpha] \) is not homeomorphic to \( C_{p}[0,\beta] \) then this means that (see [7, 8]) either

(a) \( |\alpha|\neq|\beta| \)

or

(b) \( |\alpha|=|\beta|=|\tau| \), where \( \tau \) is an initial regular ordinal and there exist initial ordinals \( \sigma,\lambda,\sigma<\lambda\leq\tau \) such that \( \tau\sigma\leq\alpha<\tau\sigma^{+} \) and \( \tau\lambda\leq\beta<\tau\lambda^{+} \).

In case (a), granted the separability of \( X \), we obtain

$$ d(X_{\alpha+1})\neq d(X_{\beta+1}), $$

and so \( C_{p}(X_{\alpha+1}) \) and \( C_{p}(X_{\beta+1}) \) are nonhomeomorphic.

In case (b), \( C_{p}[0,\alpha]\sim C_{p}[0,\tau\sigma] \) and \( C_{p}[0,\beta]\sim C_{p}[0,\tau\lambda] \); therefore, by the above,

$$ C_{p}(X_{\alpha+1})\sim C_{p}(X_{\tau\sigma+1}), $$

and, respectively, \( C_{p}(X_{\beta+1})\sim C_{p}(X_{\tau\lambda+1}) \). By Theorem 4, we conclude that \( C_{p}(X_{\alpha+1}) \) and \( C_{p}(X_{\beta+1}) \) are nonhomeomorphic.  ☐

Remark

If \( m,n\in{𝕅} \) and \( m\neq n \); then, essentially repeating the proof in [9], we can prove that \( C_{p}(X_{\tau n+1}) \) and \( C_{p}(X_{\tau m+1}) \) are nonhomeomorphic.

If \( \sigma=n\in{𝕅} \) and \( \omega\leq\lambda<\tau \) then \( C_{p}(X_{\tau\sigma+1}) \) is nonhomeomorphic to its square, whereas \( C_{p}(X_{\tau\lambda+1}) \) is homeomorphic to its square by Theorem 5. Therefore, \( C_{p}(X_{\tau\sigma+1}) \) and \( C_{p}(X_{\tau\lambda+1}) \) are nonhomeomorphic.

Corollary 6

Let \( \alpha \) and \( \beta \) be infinite ordinals and let \( S_{\alpha+1} \) and \( S_{\beta+1} \) be “long Sorgenfrey lines.” The spaces \( C_{p}(S_{\alpha+1}) \) and \( C_{p}(S_{\beta+1}) \) are homeomorphic if and only if so are \( C_{p}[0,\alpha] \) and \( C_{p}[0,\beta] \).  ☐