On some functional generalizations of the regularity of topological spaces

TL;DR

This paper explores generalized notions of regularity in topological spaces, linking these concepts to continuity properties and characterizing regularity in first-countable Hausdorff spaces via the absence of a specific topological structure.

Contribution

It introduces new generalizations of regular spaces motivated by continuity considerations and characterizes regularity in first-countable Hausdorff spaces through the Gutik hedgehog.

Findings

First-countable Hausdorff space is regular iff it lacks a Gutik hedgehog

New generalizations of regularity are proposed based on continuity properties

Characterization of regularity via topological substructures

Abstract

We introduce and study some generalizations of regular spaces, which were motivated by studying continuity properties of functions between (regular) topological spaces. In particular, we prove that a first-countable Hausdorff topological space is regular if and only if it does not contain a topological copy of the Gutik hedgehog.

Equations

$$\mbox{$\theta$-weakly discontinuous $\Rightarrow$ weakly discontinuous $\Rightarrow$ scatteredly continuous}.$$

$$f(O_z\cap A_{\gamma })=f(O_z\cap O_x\cap A_{\gamma })\subset f(O_z)\cap f(O_x\cap A_{\gamma })\subset (f(W)\setminus {\bar{O}}_{f(x)})\cap {\bar{O}}_{f(x)}=\varnothing$$

$$\{\mathbb{Q}\}\cup \{(-\infty ,a),(a,+\infty ):a\in \mathbb{R}\}.$$

$$\chi(x)=\begin{cases}1&\mbox{if $x\in\mathbb{Q}$};\\ 0&\mbox{if $x\in\mathbb{R}\setminus\mathbb{Q}$;}\end{cases}$$

$$f(x)=\begin{cases}1&\mbox{if $x\in\mathbb{Q}$};\\ 0&\mbox{if $x\notin\mathbb{Q}$}\end{cases}$$

$${\tau }_E\cup \{W_n:n\in \omega \}\text{mbox}where{W}_n=\mathbb{R}\setminus \left\{\frac{1}{2^k{3}^m}:m\in \omega ,k\ge n\right\}.$$

$$\{(U\cap \mathbb{Q})\cup \{x\}:x\in U\in {\tau }_E\}.$$

$$\{\{x\}:x\in {\mathbb{N}}^2\}\cup \{U_n:n\in \mathbb{N}\}\cup \{U_{n,m}:n,m\in \mathbb{N}\}$$

$$U_n=\{\varnothing \}\cup \{(i,j)\in {\mathbb{N}}^2:i\ge n\}\text{mbox}and{U}_{n,m}=\{(n)\}\cup \{(n,j):j\ge m\}\subset {\mathbb{N}}^1\cup {\mathbb{N}}^2$$

$$x_n\in {\overline{U}}_{k_n}\subset {\overline{U}}_{k_i}\subset X\setminus V_i\subset X\setminus \{x_i\},$$

$$sw$$

$$\begin{gathered}\mbox{weakly regular $\Rightarrow$ $sw$-regular},\\ \mbox{$\theta$-weakly regular $\Rightarrow$ locally regular},\\ \mbox{locally regular $\Rightarrow$ $w\theta$-regular}\end{gathered}$$

Full text

In this paper we introduce and study some generalizations of regular spaces, which were motivated by continuity properties of functions between (regular) topological spaces. First we introduce the necessary definitions.

A subset $U$ of a topological space $X$ is called $\theta$-open if each point $x\in U$ has a neighborhood $O_{x}\subset X$ such that $\bar{O}_{x}\subset U$. It is clear that each $\theta$-open set is open. Moreover, a topological space is regular if and only if each open subset of $X$ is $\theta$-open.

For a function $f:X\to Y$ between topological spaces by $C(f)$ we denote the set of continuity points of $f$.

• •

scatteredly continuous if for any non-empty subset $A\subset X$ the set $C(f|A)$ is not empty;

• •

weakly discontinuous if if for any non-empty subset $A\subset X$ the set $C(f|A)$ has non-empty interior in $A$;

• •

$\theta$-weakly discontinuous if if for any non-empty subset $A\subset X$ the set $C(f|A)$ contains a non-empty $\theta$-open subset of $A$.

So, we have the implications:

[TABLE]

The first and last implications can be reversed for functions with regular domain and range, respectively.

A proof the Theorem 4 can be found in [1], [8]. More information on various sorts of generalized continuity can be found in [2]–[12].

Motivated by Theorems 3 and 4, let us introduce the following definition.

• •

$sw$-regular if any scatteredly continuous function $f:Z\to X$ defined on a topological space $Z$ is weakly discontinuous;

• •

$w\theta$-regular if any weakly discontinuous function $f:X\to Y$ to any topological space $Y$ is $\theta$-weakly discontinuous.

Theorems 3 and 4 imply that each regular space is $sw$-regular and $w\theta$-regular.

The following theorem characterizes $w\theta$-regular spaces.

Now we prove the “only if” part. Assume that the space $X$ is $w\theta$-regular. Given any subset $A\subset X$ and a non-empty open subset $U\subset A$, consider the closures $\bar{A}$ and $\overline{A\setminus U}$ of the sets $A$ and $A\setminus U$ in $X$. Observe that $\tilde{U}:=\bar{A}\setminus\overline{A\setminus U}$ is an open set in $\bar{A}$ with $\tilde{U}\cap A=U$ and $\tilde{U}\subset\overline{U}$. Consider the topological sum $Y=\tilde{U}\oplus(X\setminus\tilde{U})$ and observe that the identity map $f:X\to Y$ is weakly discontinuous. The $w\theta$-regularity of the space $X$ ensures that $f$ is $\theta$-weakly discontinuous. Consequently, the closure $\bar{U}$ of $U$ in $\bar{A}$ contains a non-empty $\theta$-open subset $V\subset\bar{U}$ such that $f|V$ is continuous. The continuity of $f|V$ ensures that $V\subset\tilde{U}$. We claim that $V$ is $\theta$-open in $\bar{A}$. Since $V$ is $\theta$-open in $\bar{U}$, for any $x\in V$ there exists a neighborhood $O_{x}$ of $x$ such that $O_{x}$ is open in $\bar{U}$ and $O_{x}\subset\overline{O}_{x}\subset V\subset\tilde{U}$. So, $O_{x}$ is open in $\tilde{U}$ and hence is open in $\bar{A}$.

Taking into account that $V$ is a non-empty $\theta$-open subset of $\bar{A}$, we conclude that $V\cap A\subset\tilde{U}\cap A=U$ is a non-empty $\theta$-open subset of $A$, contained in the set $U$. ∎

We shall prove that $sw$-regular and $w\theta$-regular spaces are preserved by $\theta$-weak homeomorphisms.

We shall need the following proposition describing the continuity properties of compositions of scatteredly continuous, weakly discontinuous and $\theta$-weakly discontinuous functions.

1. (1)

If $f,g$ are weakly discontinuous, then $g\circ f$ is weakly discontinuous. 2. (2)

If $f,g$ are $\theta$-weakly discontinuous, then $g\circ f$ is $\theta$-weakly discontinuous. 3. (3)

If $f$ is weakly discontinuous and $g$ is scatteredly continuous, then $g\circ f$ is scatteredly continuous. 4. (4)

If $f$ is scatteredly continuous and $g$ is $\theta$-weakly discontinuous, then $g\circ f$ is scatteredly continuous.

1. Assume that $f,g$ are weakly discontinuous. To prove that $g\circ f$ is weakly discontinuous, we need to show that for any non-empty subset $A\subset X$ the set $C(g\circ f|A)$ has non-empty interior in $A$. By the weak discontinuity of $f$, the set $C(f|A)$ contains a non-empty open subset $U\subset A$. By the weak discontinuity of $g$, the set $C(g|f(U))$ contains a non-empty open set $V\subset f(U)$. By the continuity of $f|U$, the set $W=(f|U)^{-1}(V)$ is open in $U$ and hence open in $A$. Since $f(W)\subset V$, the continuity of the restrictions $f|W$ and $g|V$ implies the continuity of the restriction $g\circ f|W$. So, $W\subset C(g\circ f|A)$.

2. Assume that $f,g$ are $\theta$-weakly discontinuous. To prove that $g\circ f$ is $\theta$-weakly discontinuous, we need to show that for any non-empty subset $A\subset X$ the set $C(g\circ f|A)$ contains a non-empty $\theta$-open subset $W\subset A$. By the $\theta$-weak discontinuity of $f$, the set $C(f|A)$ contains a non-empty $\theta$-open subset $U\subset A$. By the $\theta$-weak discontinuity of $g$, the set $C(g|f(U))$ contains a non-empty $\theta$-open set $V\subset f(U)$. By the continuity of $f|U$, the set $W=(f|U)^{-1}(V)$ is $\theta$-open in $U$ and hence $\theta$-open in $A$, by Lemma 1. Since $f(W)\subset V$, the continuity of the restrictions $f|W$ and $g|V$ implies the continuity of the restriction $g\circ f|W$. Now we see that the set $C(g\circ f|A)$ contains the non-empty $\theta$-open subset $W$ of $A$, witnessing that $g\circ f$ is $\theta$-weakly discontinuous.

3. Assume that $f$ is weakly discontinuous and $g$ is scatteredly continuous. To prove that $g\circ f$ is scatteredly continuous, we need to show that for any non-empty subset $A\subset X$ the function $g\circ f|A$ has a continuity point. By the weak discontinuity of $f$, the set $C(f|A)$ contains a non-empty open subset $U\subset A$. By the scattered continuity of $g$, the function $g|f(U)$ has a continuity point $y$. Then any point $x\in U\cap f^{-1}(y)$ is a continuity point of the restriction $g\circ f|A$.

4. Assume that $f$ is scatteredly continuous and $g$ is $\theta$-weakly discontinuous. Given a non-empty subset $A\subset X$, we need to show that the restriction $g\circ f|A$ has a continuity point. Let $A_{0}:=A$ and $A_{\alpha}:=\bigcap_{\beta<\alpha}A_{\beta}\setminus C(f|A_{\beta})$ for any non-zero ordinal $\alpha$. In particular, $A_{\alpha+1}=A_{\alpha}\setminus C(f|A_{\alpha})$ for any ordinal $\alpha$.

Let $\delta$ be the smallest ordinal such that $A_{\delta}$ is not dense in $A$ and let $W=A\setminus\overline{A}_{\delta}$. It follows that $W=\bigcup_{\alpha<\delta}W\cap C(f|A_{\alpha})$ and each set $W\cap C(f|A_{\alpha})$ is dense in $W$ (by the scattered continuity of $f$).

Since the function $g$ is $\theta$-weakly discontinuous, the set $C(g|f(W))$ contains a non-empty $\theta$-open subset $V\subset f(W)$. Since $W=\bigcup_{\alpha<\delta}W\cap C(f|A_{\alpha})$, we can choose the smallest ordinal $\gamma<\delta$ such that $W\cap C(f|A_{\gamma})\cap f^{-1}(V)\neq\emptyset$. Choose a point $x\in W\cap C(f|A_{\gamma})\cap f^{-1}(V)$. Since the set $V$ is $\theta$-open in $f(W)$, the point $f(x)\in V$ has a closed neighborhood $\bar{O}_{f(x)}\subset f(W)$ such that $\bar{O}_{f(x)}\subset V$. By the continuity of the map $f|A_{\gamma}$ at $x$, there exists an open neighborhood $O_{x}\subset W$ of $x$ such that $f(O_{x}\cap A_{\gamma})\subset\bar{O}_{f(x)}\subset V$.

We claim that $\gamma=0$. To derive a contradiction, assume that $\gamma>0$. In this case $W\cap C(f|A_{0})\cap f^{-1}(V)=\emptyset$ and hence $x\notin C(f|A_{0})=C(f|A)$. By the density of $C(f|A)$ in $A$, there exists a point $z\in O_{x}\cap C(f|A)$. It follows that $f(z)\in W\setminus V\subset W\setminus\bar{O}_{f(x)}$. By the continuity of $f|W$ at $z$, there exists an open neighborhood $O_{z}\subset O_{x}$ such that $f(O_{z})\subset f(W)\setminus\bar{O}_{f(x)}$. Then

[TABLE]

and hence $O_{z}\cap A_{\gamma}=\emptyset$, which contradicts the density of $A_{\gamma}$ in $A$. This contradiction shows that $\gamma=0$ and hence $x\in C(f|A_{\gamma})=C(f|A)$ is a continuity point of $f|A$ with $f(O_{x})\subset V$. The continuity of the restriction $g|V$ implies that $g\circ f|A$ is continuous at $x$. So, $g\circ f|A$ has a continuity point. ∎

[TABLE]

It can be shown that the topological space $X=(\mathbb{R},\tau)$ is weakly regular. The identity map $\mathbb{R}\to X$ is scatteredly continuous but not weakly discontinuous, which implies that the space $X$ is not $sw$-regular. On the other hand, the function $\chi:X\to\{0,1\}\subset\mathbb{R}$ defined by

[TABLE]

is weakly discontinuous but not $\theta$-weakly discontinuous, witnessing that the space $X$ is not $w\theta$-regular. Theorem 15 implies that the space $X$ is not $\theta$-weakly regular.

Theorem 3, 4 and Corollary 12 imply:

Given any subspace $A\subset X$, we need to find a non-empty $\theta$-open regular subspace $W\subset A$. Since the map $h$ is $\theta$-weakly discontinuous, there exists a non-empty $\theta$-open subset $U\subset A$ such that $h|U$ is continuous. Since $h^{-1}$ is $\theta$-weakly discontinuous, the non-empty subspace $h(U)$ of $Y$ contains a non-empty $\theta$-open subspace $V$ such that $h^{-1}|V$ is continuous. The continuity of the map $h|U$ implies that the set $W:=(h|U)^{-1}(V)$ is $\theta$-open in $U$ and hence $\theta$-open in $A$ (by Lemma 1). The continuity of maps $h|W$ and $h^{-1}|h(W)$ implies that $h|W:W\to h(W)$ is a homeomorphism. The regularity of the topological space $Y$ implies the regularity of its subspace $h(W)$ and the regularity of the topological copy $W$ of $h(W)$. Therefore, $W$ is a required non-empty $\theta$-open regular subspace of $A$.

Now assume that each non-empty closed subspace $A\subset X$ contains a non-empty $\theta$-open regular subspace. Let $A^{\theta}$ be the union of all $\theta$-open regular subspaces of $A$. It is clear that the subspace $A^{\theta}$ is $\theta$-open in $A$ and regular. Let $X_{0}:=X$ and $X_{\alpha}=\bigcap_{\beta<\alpha}X_{\beta}\setminus X_{\beta}^{\theta}$ for each ordinal $\alpha$. It follows that for any ordinal $\alpha$ with $X_{\alpha}\neq\emptyset$ the set $X_{\alpha+1}=X_{\alpha}\setminus X_{\alpha}^{\theta}$ is closed in $X_{\alpha}$ and has non-empty complement $X_{\alpha+1}\setminus X_{\alpha}=X_{\alpha}^{\theta}$. Consequently, $X_{\gamma}=\emptyset$ for some $\gamma$ and hence $X=\bigcup_{\alpha<\gamma}X_{\gamma}^{\theta}$.

Let $Y:=\bigoplus_{\alpha<\gamma}X_{\alpha}^{\theta}$ be the topological sum of the regular spaces $X_{\alpha}^{\theta}$ for $\alpha<\gamma$. It is clear that the space $Y$ is regular and the identity map $i:Y\to X$ is continuous. We claim that the identity map $i^{-1}:X\to Y$ is $\theta$-weakly discontinuous. Given any non-empty subset $A\subset X$ find the smallest ordinal $\beta\leq\gamma$ such that $A\not\subset X_{\beta}$. Then $A\subset X_{\alpha}$ for all $\alpha<\beta$, which implies that $\beta$ is a successor ordinal. Write $\beta=\alpha+1$ for some $\alpha$ and observe that $U=A\cap X_{\alpha}^{\theta}=A\cap(X_{\alpha}\setminus X_{\alpha+1})$ is a non-empty $\theta$-open subspace of $A$ such that $i^{-1}|U$ is continuous. This means that $i^{-1}$ is $\theta$-weakly discontinuous and $i:X\to Y$ is a $\theta$-weak homeomorphism of $X$ onto the regular space $Y$. ∎

By analogy we can prove a characterization of weakly regular spaces.

A topological space $X$ is called

• •

quasi-regular if each non-empty open subset of $X$ contains the closure of some non-empty open set in $X$;

• •

hereditarily quesi-regular if each subspace of $X$ is quesi-regular.

Theorem 6 implies

Theorems 16 and 15 imply:

The $T_{1}$-requirement in Corollary 19 is essential as shown by the following example.

[TABLE]

is scatteredly continuous but not weakly discontinuous as $C(f)=\mathbb{Q}$ has empty interior in $\mathbb{R}$. Consequently, $D$ is not $sw$-regular and hence not $\theta$-weakly regular.

The identity map $i:D\to\{0,1\}$ to the discrete doubleton is weakly discontinuous but not $\theta$-weakly discontinuous. This means that $D$ is not $w\theta$-regular.

Theorem 17 implies that each locally regular space is weakly regular.

By the scattered continuity of $f$, the map $f|A$ has a continuity point $a\in A$. By our assumption, the point $f(a)$ is contained in an open regular subspace $U\subset Y$. By the continuity of $f$ at $a$, there exists an open neighborhood $O_{a}\subset A$ of $a$ such that $f(O_{a})\subset U$. Since $U$ is regular, the set $C(f|O_{a})$ has non-empty interior in $O_{a}$ and then the set $C(f)\supset C(f|O_{a})$ has non-empty interior in $A$. ∎

[TABLE]

It can be shown that the space $X=(\mathbb{R},\tau)$ is $\theta$-weakly regular but not locally regular.

A topological space $X$ is called regular at a point $x\in X$ if any neighborhoodof $x$ in $X$ contains a closed neighborhood of $x$ in $X$. A topological space $X$ is called nowhere regular if $X$ is not regular at each point $x\in X$.

[TABLE]

The space $(\mathbb{R},\tau)$ is locally regular and hence $sw$-regular. On the other hand, it is nowhere regular, not quasi-regular and not $w\theta$-regular.

Now, we describe the smallest non-regular first-countable Hausdorff space, which is called the Gutik hedgehog. The Gutik hedgehog is the space $\mathbb{N}^{\leq 2}=\mathbb{N}^{0}\cup\mathbb{N}^{1}\cup\mathbb{N}^{2}$ endowed with the topology generated by the base

[TABLE]

where

[TABLE]

for $n,m\in\omega$. Here $\emptyset$ is the unique element of the set $\mathbb{N}^{0}$. For the first time, the Gutik hedgehog has appeared in the paper [9] of Gutik and Pavlyk.

The following properties of the Gutik hedgehog can be derived from its definition.

Moreover, the following theorem shows that the Gutik hedgehog is the smallest space among non-regular first-countable spaces.

To prove the “only if” part, assume that a first-countable Hausdorff space $X$ is not regular at some point $x$. Then we can find a neighborhood $U_{0}\subset X$ of $x$ that does not contain the closure of any neighborhood $V$ of $x$. Fix a neighborhood base $\{U_{n}\}_{n\in\mathbb{N}}$ at $x$ such that $U_{n}\subset U_{n-1}$ for all $n\in\mathbb{N}$. Let $k_{1}=0$, choose any point $x_{1}\in\overline{U}_{k_{1}}\setminus U_{0}$, and using the Hausdorff property of $X$, find a neighborhood $V_{1}$ of $x_{1}$ such that $V_{1}\cap U_{k_{2}}=\emptyset$ for some number $k_{2}>k_{1}$.

Proceeding by induction, we can choose an increasing number sequence $(k_{n})_{n\in\omega}$ and a sequence $(x_{n})_{n\in\mathbb{N}}$ of points in $X$ such that for every $n\in\mathbb{N}$, the point $x_{n}$ belongs to $\overline{U}_{k_{n}}\setminus U_{0}$ and has an open neighborhood $V_{n}$, disjoint with the neighborhood $U_{k_{n+1}}$ of $x$. Observe that for every $i<n$, we have

[TABLE]

which implies that $x_{n}\notin\{x_{i}\}_{i<n}$. Replacing $V_{n}$ by a smaller neighborhood of $x_{n}$, we can assume that its closure $\overline{V}_{n}$ does not contain the points $x_{1},\dots,x_{n-1}$.

Since $X$ is first-countable, for every $n\in\mathbb{N}$ we can choose a sequence $\{x_{n,i}\}_{i\in\mathbb{N}}$ of pairwise distinct points in $V_{n}\cap U_{k_{n}}$ that converge to $x_{n}$. Observe that for any $n<m$ the sets $\overline{U}_{k_{n+1}}\supset\overline{U}_{k_{m}}\supset\{x_{m,i}\}_{i\in\mathbb{N}}$ and $V_{n}\supset\{x_{n,i}\}_{i\in\mathbb{N}}$ are disjoint, which implies that the points $x_{n,i}$, $n,i\in\mathbb{N}$, are pairwise disjoint. Consider the subspace $\tilde{H}:=\{x\}\cup\{x_{n}:n\in\mathbb{N}\}\cup\{x_{n,i}:n,i\in\mathbb{N}\}$ and observe that the map $h:H\to\tilde{H}$, defined by $h(\emptyset)=x$, $h(n)=x_{n}$ and $h(n,m)=x_{n,m}$ for $n,m\in\mathbb{N}$, is a homeomorphism. ∎

Finally let us draw a diagram of all provable implications between various regularity properties.

[TABLE]

Examples 14, 23 and 24 show that none of the implications

[TABLE]

holds in general.

Acknowledgements. The authors express their sincere thanks to Alex Ravsky for careful reading the paper and many valuable suggestions improving the presentation.