$Ps$-normal and $Ps$-Tychonoff spaces

TL;DR

This paper introduces $Ps$-normal and $Ps$-Tychonoff spaces, exploring their properties and relationships with other normality concepts in topology, and establishing connections with various classes of spaces.

Contribution

It defines new classes of spaces based on bijections to normal or Tychonoff spaces that preserve homeomorphisms on pseudocompact subsets and analyzes their relations with existing normality concepts.

Findings

Established relations between $Ps$-normal, $Ps$-Tychonoff, and other normal spaces.

Connected $C$-normal, $CC$-normal, $L$-normal, $C$-Tychonoff, $CC$-Tychonoff spaces with $Ps$-normal concepts.

Provided characterizations and properties of these new classes in the context of topological normality.

Abstract

A space $X$ is called $Ps$-normal($Ps$-Tychonoff) space if there exists a normal(Tychonoff) space $Y$ and a bijection $f: X\mapsto Y$ such that $f\lvert_K:K\mapsto f(K)$ is homeomorphism for any pseudocompact subset $K$ of $X$. We establish a few relations between $C$-normal, $CC$-normal, $L$-normal, $C$-Tychonoff, $CC$-Tychonoff spaces with $Ps$-normal and $Ps$-Tychonoff spaces.

Full text

The first author thanks the NBHM, Mumbai-400 001, India, for financial support. The second author thanks to Government of West Bengal, India, for financial support.

1. Introduction

A topological space $X$ is called Pseudocompact if every real valued continuous function on $X$ is bounded.

Definition 1.1**.**

A space $X$ is called $C$-normal(resp. $CC$-normal, $L$-normal) space if there exists a normal space $Y$ and a bijection $f:X\mapsto Y$ such that $f\lvert_{K}:K\mapsto f(K)$ is homeomorphism for any compact (resp. countably compact, Lindelof) subspace $K$ of $X$.

A. V. Arhangel’skii introduced the notion of $C$-normal spaces while presenting a talk in a seminar held in Mathematics Department, King Abdulaziz University at Jeddah, Saudi Arabia in 2012. $C$-normality is a generalization of normality. Furthermore the authors in [4] initiated $CC$-normal spaces as another generalization of normal spaces. Incidentally every $CC$-normal space is $C$-normal. In the first part of this paper we offer a new generalization of normal spaces namely $Ps$-normal space. We show that each $Ps$-normal space is $CC$-normal. We give an example of a $Ps$-normal space which is not normal [Example 2.3] and a $CC$-normal space which is not $Ps$-normal [Example 2.6].

Definition 1.2**.**

A space $X$ is called $C$-Tychonoff( $L$-Tychonoff) space if there exists a Tychonoff space $Y$ and a bijection $f:X\mapsto Y$ such that $f\lvert_{K}:K\mapsto f(K)$ is homeomorphism for any compact ( Lindelof) subspace $K$ of $X$.

We define $Ps$-Tychonoff spaces in the second part of this paper. We show that $Ps$-normality and $Ps$-Tychonoffness are independent to each other. $Ps$-normality and $Ps$-Tychonoffness are both topological property and also additive. $Ps$-normality is not heriditary property but $Ps$-Tychonoff property is heriditary.

In last section we make some remarks on $C$-normal and $C$-Tychonoff spaces.

2. Ps-normal space

Theorem 2.1**.**

Every $Ps$-normal space is $CC$-normal.

Proof.

Let $X$ be $Ps$-normal space and $C$ be a countably compact subspace of $X$. $X$ is $Ps$-normal, then there exists a normal space $Y$ and a bijective map $f:X\mapsto Y$ such that $f\lvert_{K}:K\mapsto f(K)$ is homeomorphism for each pseudocompact subset $K$ of $X$. Since $C$ is countably compact then it is pseudocompact. Therefore $f\lvert_{C}:C\mapsto f(C)$ is homeomorphism. Hence $X$ is $CC$-normal space. ∎

Also every $CC$-normal space is $C$-normal. It is easy to check that every normal space is $Ps$-normal. Hence we have

Normal $\Rightarrow$ $Ps$-normal $\Rightarrow$ $CC$-normal $\Rightarrow$ $C$-normal

Theorem 2.2**.**

• a)

If $X$ is $C$-normal and any pseudocompact subspace of $X$ is contained in a compact subspace of $X$, then $X$ is $Ps$-normal.

• b)

If $X$ is $CC$-normal and any pseudocompact subspace of $X$ is contained in a countably compact subspace of $X$, then $X$ is $Ps$-compact.

Proof.

We prove $(a)$ only. $(b)$ can be prove analogously.

Let $Y$ be a normal space and $f:X\mapsto Y$ be bijective mapping such that $f\lvert_{K}:K\mapsto f(K)$ is a homeomorphism for every compact subspace $K$ of $X$.

Let $K$ be a pseudocompact subspace of $X$. By hypothesis there exists a compact subspace $A$ of $X$ such that $K\subseteq A$. Since $X$ is $C$-normal, then $f\lvert_{A}:A\mapsto f(A)$ is homeomorphism and hence $f\lvert_{K}:K\mapsto f(K)$ is homeomorphism. Therefore $X$ is $Ps$-compact. ∎

The following example gives us a $Ps$-normal space which is not normal.

Example 2.3**.**

The square of Sorgenfrey line $\mathbb{R}_{l}\times\mathbb{R}_{l}$ is $C$-normal from Theorem 1.2 in [2] but it is not normal. We want to show $\mathbb{R}_{l}\times\mathbb{R}_{l}$ is $Ps$-normal. Let $S$ be a pseudocompact subset of $\mathbb{R}_{l}\times\mathbb{R}_{l}$ and $S_{1},S_{2}$ be the first and second projections of $S$ respectively. Then $S_{1},S_{2}$ are pseudocompact. $\mathbb{R}_{l}$ is hereditarily normal. Therefore $S_{1},S_{2}$ are normal, $T_{1}$ and pseudocompact. This implies that $S_{1},S_{2}$ are countably compact. Again $\mathbb{R}_{l}$ is hereditarily Lindel$\ddot{o}$f. Thus $S_{1},S_{2}$ are countably compact and Lindel$\ddot{o}$f. Therefore $S_{1},S_{2}$ are compact in $\mathbb{R}_{l}\times\mathbb{R}_{l}$. Hence $S_{1}\times S_{2}$ is compact in $\mathbb{R}_{l}\times\mathbb{R}_{l}$. Also $S\subseteq S_{1}\times S_{2}$. Hence by Theorem 2.2(a), $\mathbb{R}_{l}\times\mathbb{R}_{l}$ is $Ps$-normal.

Theorem 2.4**.**

$Ps$-normality is a topological property.

Proof.

Let $X$ be a $Ps$-normal and $X$ be homeomorphic to $Z$. Let $Y$ be a normal space and $f:X\mapsto Y$ be bijective mapping such that $f\lvert_{S}:S\mapsto f(S)$ is homeomorphism for any pseudocompact subspace $K$ of $X$. Let $g:X\mapsto Z$ be the homeomorphism. Then $f\circ g^{-1}:Z\mapsto Y$ is the required map. ∎

Theorem 2.5**.**

If $X$ is pseudocompact but not normal, then $X$ is not $Ps$-normal.

Proof.

Let $X$ be pseudocompact but not normal. If possible let there exist a normal space $X$ and a bijective mapping $f:X\mapsto Y$ such that $f\lvert_{K}:K\mapsto f(K)$ is a homeomorphism for any pseudocompact subspace $K$ of $X$. In particular $f:X\mapsto Y$ is a homeomorphism. This is a contradiction as $Y$ is normal and $X$ is not normal. Hence $X$ is not $Ps$-normal. ∎

We have seen that every $Ps$-normal space is $CC$-normal. The following example shows that the converse is not true.

Example 2.6**.**

Consider $(\mathbb{R},\tau_{c})$, where $\tau_{c}$ is the co-countable topology on set of all real numbers $\mathbb{R}$. In Example 2.6, [4] it is shown that $(\mathbb{R},\tau_{c})$ is $CC$-compact. $(\mathbb{R},\tau_{c})$ is pseudocompact as each real valued continuous function on it is constant but it is not normal. Hence by Theorem 2.5, $(\mathbb{R},\tau_{c})$ is not $Ps$-normal.

More generally every irreducible space (i.e a space which has no pair of disjoint nonempty open sets) which is not normal is not $Ps$-normal.

Theorem 2.7**.**

Let $X=\oplus_{\alpha\in\Lambda}X_{\alpha}$. $C\subseteq X$ is psedocompact if and only if $\Lambda_{\circ}=\{\alpha\in\Lambda:C\cap X_{\alpha}\neq\phi\}$ is finite and $C\cap X_{\alpha}$ is pseudocomapct for each $\alpha\in\Lambda$.

Proof.

We shall show that if $C\subseteq X$ is pseudocompact then $\Lambda_{\circ}$ is finite. Others parts are trivial.

Let $C\subseteq X$ be pseudocompact. If possible $\Lambda_{\circ}$ be infinite. Let $\{\alpha_{n}:n\in\mathbb{N}\}$ be a countably infinite subset of $\Lambda_{\circ}$. Define a function $f:C\mapsto\mathbb{R}$ by $f(x)=n$ for $x\in X_{\alpha_{n}}$ and zero otherwise. Then $f\lvert_{C}:C\mapsto\mathbb{R}$ is continuous and unbounded function on $C$. This implies that $C$ is not pseudocompact, a contradiction. ∎

Theorem 2.8**.**

$Ps$-normality is an additive property.

Proof.

Let $X=\oplus_{\alpha\in\Lambda}X_{\alpha}$ and each $X_{\alpha}$ is pseudocompact. Then for each $\alpha\in\Lambda$ there exists a normal space $Y_{\alpha}$ and a bijective mapping $f_{\alpha}:X_{\alpha}\mapsto Y_{\alpha}$ such that $f_{\alpha}\lvert_{C_{\alpha}}:C_{\alpha}\mapsto f_{\alpha}(C_{\alpha})$ is homeomorphism for each pseudocompact subspace $C_{\alpha}$ of $X_{\alpha}$. Then $Y=\oplus_{\alpha\in\Lambda}Y_{\alpha}$ is a normal space. Consider the function $f=\oplus_{\alpha\in\Lambda}f_{\alpha}:X\mapsto Y$. Let $C\subseteq X$ be pseudocompact subspace of $X$. Then by Theorem 2.7, $\Lambda_{\circ}=\{\alpha\in\Lambda:C\cap X_{\alpha}\neq\phi\}$ is finite and $C\cap X_{\alpha}$ is pseudocomapct for each $\alpha\in\Lambda$. Then $f\lvert_{C}:C\mapsto f(C)$ is a homeomorphism. Hence $X$ is $Ps$-normal. ∎

A space $X$ is called Frechet if for any subset $B$ of $X$ and $x\in\overline{B}$ there exist a sequence $\{b_{n}\}$ of points of $B$ such that $b_{n}\rightarrow x$.

Theorem 2.9**.**

If $X$ is $Ps$-normal and Frechet, $f:X\mapsto Y$ is witness of $Ps$-normality, then $f$ is a continuous mapping.

Proof.

It is sufficient to show that for any $A\subseteq X,f(\overline{A})\subseteq\overline{f(A)}$.

Let $A\subseteq X$ and $y\in f(\overline{A})$. Let $x$ be the unique point in $\overline{A}$ such that $f(x)=y$. Since $X$ is Frechet, then there exists a sequence $\{a_{n}\}$ in $A$ such that $x$ is a limit of $\{a_{n}\}$. Let $K=\{a_{n}:n\in\mathbb{N}\}\cup\{x\}$. Then $K$ is compact and therefore pseudocompact. Then $f\lvert_{K}:K\mapsto f(K)$ is homeomorphism as $X$ is $Ps$-normal.

Let $U$ be any open set in $Y$. Then $U\cap f(K)$ is open set in $f(K)$ containing $y$. Now $f(\{a_{n}:n\in\mathbb{N})\subseteq f(K)\cap f(A)$ and $U\cap f(K)\neq\phi$. Then $U\cap f(A)\neq\phi$. Therefore $y\in\overline{f(A)}$. Hence $f(\overline{A})\subseteq\overline{f(A)}$. ∎

3. $Ps$-Tychonoff spaces

Definition 3.1**.**

A space $X$ is called $Ps$-Tychonoff($C$-Tychonoff) space if there exists a Tychonoff space $Y$ and a bijective mapping $f:X\mapsto Y$ such that $f\lvert_{C}:C\mapsto f(C)$ is homeomorphism for every Pseudocompact(compact) subset $C$ of $X$.

On using the arguments of Theorems 2.1, 2.5, 2.9, 2.2, 2.4, 2.8 closely, we get the following properties respectively.

Every Tychonoff space is $Ps$-Tychonoff space.

Theorem 3.2**.**

Every $Ps$-Tychonoff space is $C$-Tychonoff.

Theorem 3.3**.**

If $X$ is Pseudocompact but not Tychonoff, then $X$ is not $Ps$-normal.

Therefore $\mathbb{R}$ with co-finite topology is not $Ps$-normal.

Theorem 3.4**.**

If $X$ is $Ps$-Tychonoff and Frechet, $f:X\mapsto Y$ is witness of $Ps$-Tychonoffness, then $f$ is continuous.

Theorem 3.5**.**

Let $X$ be an $L$-Tychonoff space. If each Pseudocompact subset in $X$ is contained in a Lindeloff subset, then $X$ is $Ps$-Tychonoff.

Theorem 3.6**.**

$Ps$-Tychonoffness is a topological property.

Theorem 3.7**.**

$Ps$-Tychonoffness is additive property.

Theorem 3.8**.**

If $X$ is $T_{1}$ space and only Pseudocompact subset of $X$ is finite, then $X$ is $Ps$-Tychonoff.

Proof.

Let $Y=X$ and $Y$ be discrete topological space. Let $I:X\mapsto X$ be the identity mapping. Let $K$ be pseudocompact subset of $X$. Then by the hypothesis $K$ is finite. Therefore $K$ is finite and $T_{1}$. therefore $K$ is a discrete topological space. Thus $I:K\mapsto I(K)$ is homeomorphism. Henece $X$ is $Ps$-Tychonoff. ∎

Example 3.9**.**

Let $X=\mathbb{R}\cup\{i,j\}$, where $i\notin\mathbb{R}$ and $j\notin\mathbb{R}$. Define a topology on $X$ as follows: Each point of $\mathbb{R}$ is isolated. A basic open neighbourhood of $i$ is a subset $U$ of X containing $i$ and $X\setminus U$ is countable. Also a basic neighbourhood of $j$ is a subset $V$ of $X$ containing $j$ and $X\setminus V$ is countable. Clearly $X$ is $T_{1}$ space. But $X$ is not $T_{2}$. Therefore $X$ is not Tychonoff.

We want to show that the Pseudocompact subsets of $X$ are finite. Let $A$ be infinite subset of $X$. If $i,j\notin A$, then $X$ is infinite discrete and so it is not Pseudocompact.

If $i,j\in A$. Let $B=\{a_{n}:n\in\mathbf{N}\}$ be countably infinite subset of $A$ and $i,j\notin B$. Define $f:A\mapsto\mathbb{R}$ as follows: $f(a_{n})=n$ for each $n\in\mathbb{N}$ and zero otherwise. Each point of $A\setminus\{i,j\}$ is isolated and therefore $f$ is continuous on $A\setminus\{i,j\}$. $f(i)=0$. Let $U=A\setminus B$. Then $U$ is an open neighbourhood of $i$ and $f(U)=\{0\}$. Therefore $f$ is continuous at $i$. Similarly $f$ is continuous at $j$. Therefore $f$ is an unbounded continuous function on $A$. Therefore $A$ is not Pseudocompact.

Similarly we can show that if either $i\in A$ or $j\in A$, then $A$ is not Pseudocompact.

Therefore $X$ is $T_{1}$ and each Pseudocompact subset is finite. Therefore $X$ is $Ps$-Tychonoff.

Theorem 3.10**.**

Every $Ps$-Tychonoff Frechet Lindeloff space is $Ps$-normal.

Proof.

Let $X$ be $Ps$-Tychonoff Frechet Lindeloff space. Then there exists a Tychonoff space $Y$ and a bijective mapping $f:X\mapsto Y$ such that $f\lvert_{K}:K\mapsto f(K)$ is homeomorphism for any Pseudocompact subset $K$ of $X$. By Theorem 3.4, $f$ is continuous. Since continuous image of Lindeloff space is Lindeloff, therefore $Y$ is Lindeloff. Thus $Y$ is regular Lindeloff, this implies that $Y$ is normal. Hence $X$ is $Ps$-normal. ∎

We now show that $Ps$-normality and $Ps$-Tychonoffness are independent properties. We reproduce the following two examples from [1] for our purpose.

Example 3.11**.**

Let $\tau=\{\emptyset,\mathbb{R}\}\cup\{(x,\infty):a\in\mathbb{R}\}$ be the topology on $\mathbb{R}$. Then $(\mathbb{R},\tau)$ is a normal space and therefore it is $Ps$-normal space. From Example 4,[1], $(\mathbb{R},\tau)$ is not a $C$-Tychonoff space. Then by Theorem 3.2 it is not a $Ps$-Tychonoff space.

Example 3.12**.**

Let $G=\prod_{\alpha\in\omega_{1}}D_{\alpha}$, where $D_{\alpha}=\{0,1\}$ for each $\alpha$ and $\omega_{1}$ is the first uncountable ordinal. Let $H$ be the set of all points of $G$ with atmost contably many non-zero co-ordinates. Let $M=G\times H$. By Example 4 in [1] $M$ is a Tychonoff space and not $C$-normal. Therefore $M$ is a $Ps$-Tychonoff space. But every $Ps$-normal space is $C$-normal and $M$ is not $C$-normal, we have that $M$ is not $Ps$-normal.

Theorem 3.13**.**

Each subset of $Ps$-Tychonoff space is $Ps$-Tychonoff..

Proof.

Let $X$ be a $Ps$-Tychonoff space and $A$ be a subspace of $X$. Now there exists a Tychonoff space $Y$ and a bijective map $f:X\mapsto Y$ such that for every Pseudocompact subset $C$ of $X$, $f\lvert_{C}:C\mapsto f(C)$ is homeomorphism.

Let $K$ be a Pseudocompact subset of $A$. Then $K$ is also a Pseudocompact subset of $X$. Then $f\lvert_{K}:K\mapsto f(K)$ is homeomorphism. Therefore $f\lvert_{A}$ is the witness of $Ps$-Tychonoffness of $A$. ∎

Let $X$ be any topological space. Lat $X^{\prime}=X\times\{1\}$. Late $A(X)=X\cup X^{\prime}$. For $x\in X$ we denote $<x,1>$ as $x^{\prime}$ and for $B\subseteq X$ let $B^{\prime}=\{x^{\prime}:x\in B\}=B\times\{1\}$. Let $\{\beta(x):x\in X\}\cup\{\beta(x^{\prime}):x^{\prime}\in X^{\prime}\}$ be neighbourhood system for some topology $\tau$ on $A(X)$, where for $x\in X,\beta(x)=\{U\cup(U^{\prime}\setminus\{x^{\prime}\}):U$ is an open set in $X$ containing $x\}$ and for $x^{\prime}\in X^{\prime},\beta(x^{\prime})=\{\{x^{\prime}\}\}$. $(A(X),\tau)$ is called Alexandroff Duplicate of $X$.

The following problems are open.

Question 3.14**.**

Is $A(X)$ $Ps$-normal, whwn $X$ is $Ps$-normal?

Question 3.15**.**

Is $A(X)$ $Ps$-Tychonoff, when $X$ is $Ps$-Tychonoff?

4. Some special results

Theorem 4.1**.**

If $X$ is $C$-Tychonoff and Frechet, then for any compact set $A$ in $X$ and $x\notin A$, there exists a continuous function $h:X\mapsto[0,1]$ such that $h(x)=0$ and $h(A)=\{1\}$.

Proof.

Let $f:X\mapsto Y$ be witness of $C$-Tychonoff. Since $X$ is $C$-compact and Frechet, then by Theorem 5[1], $f$ is continuous. Since $f$ is continuous and $A$ is compact, then f(A) is compact in $Y$. $Y$ is Tychonoff and so it is $T_{2}$. Therefore $f(A)$ is closed in $Y$ and also $f(x)\in f(A)$. Since $Y$ is Tychonoff, therefore there exists a continuous function $g:Y\mapsto[0,1]$ such that $g(f(x))=0$ and $g(f(A))=\{1\}$. Then $h=g\circ f:X\mapsto[0,1]$ is a continuous function such that $h(x)=0$ and $h(A)=\{1\}$. This completes the proof. ∎

We give an alternative proof of the following result.

Corollary 4.2** (Corollary 2[1]).**

Every $C$-Tychonoff Frechet space is Urysonh.

Proof.

Let $X$ be $C$-Tychonoff Frechet space and contains at least two points. Let $a,b$ be two distinct points of $X$. Then $\{b\}$ is a compact set and $a\notin\{b\}$. Then there exists a continuous functions$h:X\mapsto[0,1]$ such that $f(a)=0$ and $f(b)=1$. Let $U=\{x\in X:f(x)\leq\frac{1}{3}\}$ and $V=\{x\in X:f(x)\geq\frac{1}{2}\}$. Then $U,V$ are disjoint closed neighbourhoods of $a,b$ respectively. Hence $X$ is Urysonh. ∎

Theorem 4.3**.**

Let $X$ be a $C$-normal Frechet space. Let $f:X\mapsto Y$ be witness of $C$-normality of $X$. In addition if $Y$ is $T_{2}$ space, then for any two disjoint compact sets $H,K$ in $X$, there exists a continuous function $h:X\mapsto[0,1]$ such that $h(H)=\{0\}$ and $h(K)=\{1\}$.

Proof.

$A$ is $C$-normal and Frechet, then $f$ is continuous. Then $f(H),f(K)$ are disjoint compact sets in the $T_{2}$ space $Y$. By normality of $Y$, there exists a continuous function $g:Y\mapsto[0,1]$ such that $g(f(H))=\{0\}$ and $g(f(K))=\{1\}$. Then $h=g\circ f:X\mapsto[0,1]$ is a continuous function such that $h(H)=\{0\}$ and $h(K)=\{1\}$. This completes the proof. ∎

Acknowledgments: The authors would like to thank to Professor Sudip Kumar Acharyya and Professor Asit Baran Raha for their valuable suggestions.