On Almost Semi-Moscow Topological Groups

1. Introduction

The study of topological groups [13] connects algebraic operations with topological continuity. Over time, applications in dynamical systems, harmonic analysis, and geometric topology have shown that strict continuity often excludes naturally occurring structures where the group maps preserve neighbourhoods only in a weakened sense. This motivated the development of topological transformation groups [3,15,22], which model symmetry through continuous actions, and subsequently $\mathcal{S}$-topological transformation groups [16,17,18,19], where continuity is relaxed to semi-totally continuity. This development reflects a consistent theme in modern topology, systematically weakening continuity requirements while preserving essential algebraic structure.

Parallel to this development, many weaker forms of continuity have been introduced in general topology, and these weaker notions have also been used in the study of group structures. For instance, Singal and Singal [21] introduced almost continuous mappings, where inverse images of regularly open sets are open. Later, Munshi and Bassan [11] studied almost semi-continuous mappings, where inverse images of regularly open sets are semi-open. These ideas led to weaker forms of topological groups, such as almost topological groups [20], in which the group operations are almost continuous.

Another line of study concerns regularity, where classical regularity axioms are replaced by structurally rich alternatives. Arhangel’skii’s Moscow spaces [2] emerged as a powerful tool in topological group theory, linking $G_{\delta }$-sets to closure. Building on this foundation, Caldas et al. [4] introduced almost Moscow spaces, defined by the requirement that every $G_{\delta }$-set contains a regularly closed neighbourhood of each of its points, and semi-Moscow spaces, which separate $F_{\sigma }$-sets from points using disjoint semi-open sets. Cameron [5] introduced regularly semi-open and regularly semi-closed sets, providing a natural intermediate class between regularly open and semi-open sets. While these frameworks independently relax topological conditions and weaken the continuity of the group maps, their intersection has remained unexplored.

In this paper, we introduce almost semi-Moscow spaces. A space is called almost semi-Moscow if every $G_{\delta }$-set contains a regularly semi-closed set around each of its points. This property is weaker than the almost Moscow property and stronger than the semi-Moscow property. We prove several equivalent descriptions of this condition and give examples showing that the inclusions are proper. We then introduce almost semi-Moscow topological groups. These are groups with a topology such that multiplication and inversion are almost semi-continuous, and the space itself is almost semi-Moscow. This gives a group version of the almost semi-Moscow condition without requiring the full strength of almost continuity. We also study a stronger class called almost semi-Moscow almost topological groups. In this case the group operations are almost continuous, while the space is almost semi-Moscow. This stronger condition gives better behaviour for translations, regularly semi-open sets, regularly semi-closed sets, subgroups, quotients, and products.

A central achievement of the framework is the strict chain of inclusions established among the topological conditions, the almost semi-Moscow condition lies properly between the almost Moscow and semi-Moscow conditions. Product preservation is proved for almost semi-Moscow spaces, and two results demonstrate that no separable uncountable $T_1$ space and no first countable $T_1$ space without isolated points can be almost semi-Moscow. At the algebraic-topological level, the almost semi-Moscow almost topological groups holds closure identities, and structural results on regularly open subgroups, quotient groups, and products. Under extremally disconnectedness, the stronger class reduces to almost Moscow topological groups, yielding open Boolean subgroups and strong PT-group.

2. Preliminaries

In this section, we gather the foundational definitions, mapping classes, and properties required for our study. Throughout this paper, $(X,\tau )$ and $(Y,\sigma )$ denote topological spaces, and no separation axioms are assumed unless explicitly stated. For a subset A of X, we denote the interior, the closure, and the complement of A by $Int\left(A\right)$, $Cl\left(A\right)$, and $X\setminus A$, respectively.

Definition 1. 

Let A be a subset of a space X.

(1)

A is said to be semi-open [10] if $A\subseteq Cl(Int(A\left)\right)$. The complement of a semi-open set is called semi-closed. The semi-interior of A, denoted by $sInt\left(A\right)$, is the union of all semi-open sets contained in A. The semi-closure of A, denoted by $sCl\left(A\right)$, is the intersection of all semi-closed sets containing A.

(2)

A is said to be regularly open [8] if $A=Int(Cl(A\left)\right)$, and regularly closed if $A=Cl(Int(A\left)\right)$. The families of regularly open and regularly closed subsets of X are denoted by $RO\left(X\right)$ and $RC\left(X\right)$, respectively. The complement of a regularly open set is regularly closed, and vice versa.

(3)

A is said to be regularly semi-open [5] if there exists a regularly open set U of X such that $U\subseteq A\subseteq Cl\left(U\right)$. A subset F of X is said to be regularly semi-closed if there exists a regularly closed set C of X such that $Int\left(C\right)\subseteq F\subseteq C$. The families of all regularly semi-open and regularly semi-closed subsets of X are denoted by $RSO\left(X\right)$ and $RSC\left(X\right)$, respectively.

Definition 2. 

A mapping $f:X\to Y$ is said to be:

(1)

almost continuous [21] if $f^{-1}\left(V\right)$ is open in X for every regularly open set V of Y;

(2)

almost semi-continuous [11] if $f^{-1}\left(V\right)$ is semi-open in X for every regularly open set V of Y;

(3)

$G_{\delta }$-irresolute [4] if $f^{-1}\left(V\right)$ is a $G_{\delta }$-set in X for every $G_{\delta }$-set V of Y;

Definition 3. 

Let $f:X\to Y$ be a mapping. Then f is said to be:

(1)

pre-almost closed [9] if $f\left(F\right)$ is a regularly closed set of Y for every regularly closed set F of X;

(2)

pre-almost open [9] if $f\left(U\right)$ is a regularly open set of Y for every regularly open set U of X;

(3)

R-map [4] if $f^{-1}\left(F\right)$ is a regularly closed set of X for every regularly closed set F of Y;

(4)

$G_{\delta }$-map [4] if $f\left(U\right)$ is a $G_{\delta }$-set of Y for every $G_{\delta }$-set U of X;

(5)

almost $G_{\delta }$-continuous [4] if $f^{-1}\left(V\right)$ is a $G_{\delta }$-set of X for every regularly open set V of Y.

Definition 4. 

Let $(G,·)$ be a group endowed with a topology τ. $(G,\tau )$ is called a topological group [13] if the multiplication mapping $\phi :G\times G\to G$ and the inversion mapping $\psi :G\to G$ are continuous.

Definition 5. 

Let $(G,·)$ be a group endowed with a topology τ. $(G,\tau )$ is called an almost topological group [20] if the multiplication mapping $\phi :G\times G\to G$ and the inversion mapping $\psi :G\to G$ are almost continuous.

Definition 6. 

A space X is said to be:

(1)

almost Moscow [4] if for every $G_{\delta }$-set V of X and each $x\in V$, there exists a regularly closed set F of X such that $x\in F\subseteq V$;

(2)

semi-Moscow [4] if for each $F_{\sigma }$-set D of X and each point $x\in X\setminus D$, there exist disjoint semi-open sets U and V in X such that $x\in U$ and $D\subseteq V$;

(3)

weakly $P_{\Sigma }$ [9] if for any regularly open set V of X and any $x\in V$, there exists a regularly closed set F of X such that $x\in F\subseteq V$.

Definition 7. 

A space X is said to be:

(1)

S-closed [12] if every cover of X by regularly closed sets admits a finite subcover. A subset $A\subseteq X$ is said to be S-closed relative to X if every cover of A by regularly closed sets of X admits a finite subcover;

(2)

locally countably S-closed [7] if each point $x\in X$ has an open U that is countably S-closed relative to X;

(3)

km-perfect [6] if for every regularly open set $U\subseteq X$ and each $x\in U$, there exists a sequence ${\left\{G_n\right\}}_{n\in \omega }$ of open sets such that $U\subseteq {\bigcup }_{n\in \omega }{G}_n$ and $x\notin {\bigcup }_{n\in \omega }Cl\left(G_n\right)$;

(4)

extremally disconnected [1] if the closure of every open subset of X is open;

Definition 8. 

[23] For a group G, the Boolean subgroup is $B\left(G\right)=\{x\in G:x^2=e\}$, where e denotes the identity element.

Definition 9. 

[1] A topological group G is a PT-group if the group operations on G extend uniquely to its Dieudonné completion $\mu G$, making $\mu G$ a topological group containing G as a dense topological subgroup. G is a strong PT-group if G is C-embedded in its Raĭkov completion $\varrho G$. Every strong PT-group is a PT-group.

Lemma 1. 

[20] Let G be an almost topological group. For every $g\in G$ and every regularly open set $U\in RO\left(G\right),gU,Ug,U^{-1}\in RO\left(G\right),$ and $Cl\left(gU\right)=gCl\left(U\right),Cl\left(Ug\right)=Cl\left(U\right)g,Cl\left(U^{-1}\right)=Cl{\left(U\right)}^{-1}.$ Similarly, regularly closed sets are preserved by left translation, right translation, and inversion.

Theorem 1. 

[14] Every extremally disconnected almost Moscow topological group contains an open Boolean subgroup.

Theorem 2. 

[14] Let G be an almost Moscow topological group. If G is locally countably S-closed, km-perfect, and semi-regular, then G is a topological group that is extremally disconnected. Consequently, G is a strong PT-group and hence a PT-group.

2.1. Almost Semi-Moscow Spaces

Definition 10. 

A space $(X,\tau )$ is called almost semi-Moscow if for every $G_{\delta }$-set Vof X and each point $x\in V$, there exists a regularly semi-closed set F of X such that $x\in F\subseteq V.$

Remark 1. 

A space $(X,\tau )$ is almost semi-Moscow if and only if for every $F_{\sigma }$-set D of X and each point $x\in X\setminus D$, there exists a regularly semi-closed set F of X such that $x\in F$ and $D\cap F=\varnothing .$

Example 1. 

Every discrete space and every indiscrete space is almost semi-Moscow.

Example 2. 

Let $X=\{a,b,c\}$ and let $\tau =\{\varnothing ,\{a\},\{b\},\{a,b\},X\}.$ The regularly closed subsets of X are $\varnothing ,X,\{a,c\},\{b,c\}.$ The sets $\left\{a\right\}$ and $\left\{b\right\}$ are regularly semi-closed. Indeed, $\{a,c\}$ is regularly closed and $Int\left(\right\{a,c\left\}\right)=\left\{a\right\}\subseteq \left\{a\right\}\subseteq \{a,c\},$ so $\left\{a\right\}$ is regularly semi-closed. Similarly, $\left\{b\right\}$ is regularly semi-closed. Since X is finite, every $G_{\delta }$-set is open. Hence the $G_{\delta }$-sets of X are $\varnothing ,\left\{a\right\},\left\{b\right\},\{a,b\},X.$ For the $G_{\delta }$-set $\left\{a\right\}$, the point a is contained in the regularly semi-closed set $\left\{a\right\}$. For the $G_{\delta }$-set $\left\{b\right\}$, the point b is contained in the regularly semi-closed set $\left\{b\right\}$. For the $G_{\delta }$-set $\{a,b\}$, the point a is contained in $\left\{a\right\}$ and the point b is contained in $\left\{b\right\}$. For X, take X itself. Therefore X is almost semi-Moscow.

Example 3. 

Let D be an uncountable set and let $p\notin D$. Put $X=D\cup \left\{p\right\}$. Define a topology on X by taking a subset $U\subseteq X$ to be open if either $p\notin U$, or $p\in U$ and $D\setminus U$ is finite. Then X is almost Moscow, and hence almost semi-Moscow by Theorem 4(i).

Theorem 3. 

If $X,Y$ are almost semi-Moscow spaces, then $X\times Y$ is almost semi-Moscow.

Proof. 

Let V be a $G_{\delta }$-set of $X\times Y$ and $(x,y)\in V$. Let $V={\bigcap }_n{W}_n$, $W_n$ open in $X\times Y$. For each n, choose open $A_n$ of x, $B_n$ of y with $A_n\times B_n\subseteq W_n$. Put $V_1={\bigcap }_n{A}_n$, $V_2={\bigcap }_n{B}_n$, these are $G_{\delta }$-sets of $X,Y$ with $V_1\times V_2\subseteq V$ and $(x,y)\in V_1\times V_2$. Choose regularly semi-closed $F_1$ of x, $F_1\subseteq V_1$ and $F_2$ of y, $F_2\subseteq V_2$, the regularly closed $C_1,C_2$ with $Int\left(C_i\right)\subseteq F_i\subseteq C_i$, $C_1\times C_2$ is regularly closed in $X\times Y$, and $Int(C_1\times C_2)=Int\left(C_1\right)\times Int\left(C_2\right)\subseteq F_1\times F_2\subseteq C_1\times C_2$, so $F_1\times F_2$ is regularly semi-closed in $X\times Y$, with $(x,y)\in F_1\times F_2\subseteq V$. □

Theorem 4. 

For a topological space $(X,\tau )$, the following implications hold:

(i)

Almost Moscow ⇒ almost semi-Moscow.

(ii)

Almost semi-Moscow ⇒ semi-Moscow.

Consequently, every almost Moscow space is a semi-Moscow space.

Proof. (i) Let X be an almost Moscow space. Let V be a $G_{\delta }$-set of X and let $x\in V$. By the definition of an almost Moscow space, there exists a regularly closed set F of X such that $x\in F\subseteq V.$ Every regularly closed set is regularly semi-closed. Hence F is regularly semi-closed. Therefore X is almost semi-Moscow.

(ii) Let X be an almost semi-Moscow space. Let D be an $F_{\sigma }$-set of X and let $x\in X\setminus D$. By Remark 1, there exists a regularly semi-closed set F of X such that $x\in F$ and $D\cap F=\varnothing .$ We show that F is both semi-open and semi-closed. Since F is regularly semi-closed, there exists a regularly closed set C of X such that $Int\left(C\right)\subseteq F\subseteq C.$ Since C is regularly closed, $C=Cl(Int(C\left)\right).$ Because $Int\left(C\right)\subseteq F$, we have $Int\left(C\right)\subseteq Int\left(F\right)$. Hence $C=Cl(Int(C\left)\right)\subseteq Cl(Int(F\left)\right).$ Since $F\subseteq C$, it follows that $F\subseteq Cl(Int(F\left)\right),$ so F is semi-open. On the other hand, since $F\subseteq C$ and C is closed, $Cl\left(F\right)\subseteq C.$ Therefore $Int(Cl(F\left)\right)\subseteq Int\left(C\right)\subseteq F,$ which shows that F is semi-closed. Now put $U=F$ and $W=X\setminus F.$ Since F is semi-open, U is a semi-open set containing x. Since F is semi-closed, W is semi-open. Moreover, $U\cap W=\varnothing$ and $D\subseteq X\setminus F=W.$ Thus X is semi-Moscow. □

Remark 2. 

The converse of Theorem 4(i) does not hold in general. This is given in the following example.

Example 4. 

In Example 2, X is almost semi-Moscow. However, X is not almost Moscow. The set $\left\{a\right\}$ is a $G_{\delta }$-set, but there is no regularly closed set F such that $a\in F\subseteq \left\{a\right\}.$ The regularly closed sets containing a are $\{a,c\}$ and X, and neither is contained in $\left\{a\right\}$. Hence X fails to be almost Moscow.

Remark 3. 

By Theorem 4 and Example 4, we have

$$\mathrm{Almost}\mathrm{Moscow}⊊\mathrm{Almost}\mathrm{semi}-\mathrm{Moscow}\subseteq \mathrm{Semi}-\mathrm{Moscow}.$$

Definition 11. 

A subset A of a space X is called semi-θ-open if A is a union of regularly semi-closed subsets of X. Equivalently, A is semi-θ-open if for each $x\in A$, there exists a regularly semi-closed set F of X such that $x\in F\subseteq A.$ The complement of a semi-θ-open set is called semi-θ-closed.

Theorem 5. 

For a topological space X, the following conditions are equivalent:

(1)

X is almost semi-Moscow.

(2)

For every $F_{\sigma }$-set D of X and each point $x\in X\setminus D$, there exists a regularly semi-closed set F of X such that $x\in F$ and $D\cap F=\varnothing$.

(3)

Every $G_{\delta }$-set in X is a union of regularly semi-closed sets.

(4)

Every $F_{\sigma }$-set in X is an intersection of regularly semi-open sets.

(5)

Every $F_{\sigma }$-set in X is semi-θ-closed.

Proof. (1) ⇒ (2): Let D be an $F_{\sigma }$-set of X and let $x\in X\setminus D$. Then $V=X\setminus D$ is a $G_{\delta }$-set containing x. Since X is almost semi-Moscow, there exists a regularly semi-closed set F such that $x\in F\subseteq V.$ Thus $D\cap F=\varnothing$.

(2) ⇒ (3): Let V be a $G_{\delta }$-set of X. Then $D=X\setminus V$ is an $F_{\sigma }$-set. For each $x\in V$, condition (2) gives a regularly semi-closed set $F_x$ such that $x\in F_x$ and $D\cap F_x=\varnothing .$ Hence $F_x\subseteq X\setminus D=V$. Therefore $V={\bigcup }_{x\in V}{F}_x,$ so V is a union of regularly semi-closed sets.

(3) ⇒ (4): Let D be an $F_{\sigma }$-set of X. Then $X\setminus D$ is a $G_{\delta }$-set. By (3), $X\setminus D={\bigcup }_{\alpha \in \Lambda }{F}_{\alpha },$ where each $F_{\alpha }$ is regularly semi-closed. Taking complements, $D={\bigcap }_{\alpha \in \Lambda }(X\setminus F_{\alpha }).$ The complement of a regularly semi-closed set is regularly semi-open. Hence D is an intersection of regularly semi-open sets.

(4) ⇒ (5): Let D be an $F_{\sigma }$-set of X. By (4), $D={\bigcap }_{\alpha \in \Lambda }{U}_{\alpha },$ where each $U_{\alpha }$ is regularly semi-open. Hence $X\setminus D={\bigcup }_{\alpha \in \Lambda }(X\setminus U_{\alpha }).$ The complement of a regularly semi-open set is regularly semi-closed. Thus $X\setminus D$ is semi-$\theta$-open, and therefore D is semi-$\theta$-closed.

(5) ⇒ (1): Let V be a $G_{\delta }$-set of X. Then $X\setminus V$ is an $F_{\sigma }$-set. By (5), $X\setminus V$ is semi-$\theta$-closed, so V is semi-$\theta$-open. Hence, for each $x\in V$, there exists a regularly semi-closed set F such that $x\in F\subseteq V.$ Therefore X is almost semi-Moscow. □

Corollary 1. 

A space X is almost semi-Moscow if and only if every $G_{\delta }$-set of X is semi-θ-open.

Proof. 

This is exactly the equivalence of conditions (1) and (3) in Theorem 5, together with Definition 11. □

Definition 12. 

Let $f:X\to Y$ be a mapping. Then f is said to be:

(1)

pre-almost semi-closed if $f\left(F\right)$ is regularly semi-closed in Y for every regularly semi-closed set F of X;

(2)

pre-almost semi-open if $f\left(U\right)$ is regularly semi-open in Y for every regularly semi-open set U of X;

(3)

an R-semi-map if $f^{-1}\left(F\right)$ is regularly semi-closed in X for every regularly semi-closed set F of Y;

(4)

R-semi-$G_{\delta }$-open if $f\left(U\right)$ is regularly semi-open in Y for every $G_{\delta }$-set U in X;

(5)

almost semi-$G_{\delta }$-continuous if $f^{-1}\left(V\right)$ is a $G_{\delta }$-set in X for every regularly semi-open set V of Y.

Definition 13. 

A space X is called weakly semi-$P_{\Sigma }$ if for every regularly semi-open set V of X and each point $x\in V$, there exists a regularly semi-closed set F of X such that $x\in F\subseteq V.$

Theorem 6. 

Let $f:X\to Y$ be a $G_{\delta }$-irresolute, pre-almost semi-closed surjection. If X is almost semi-Moscow, then Y is almost semi-Moscow.

Proof. 

Let V be a $G_{\delta }$-set of Y and let $y\in V$. Since f is surjective, choose $x\in X$ such that $f\left(x\right)=y$. Since f is $G_{\delta }$-irresolute, $f^{-1}\left(V\right)$ is a $G_{\delta }$-set of X, and $x\in f^{-1}\left(V\right).$ Since X is almost semi-Moscow, there exists a regularly semi-closed set F of X such that $x\in F\subseteq f^{-1}\left(V\right).$ Because f is pre-almost semi-closed, $f\left(F\right)$ is regularly semi-closed in Y. Also, $y=f\left(x\right)\in f\left(F\right)$ and $f\left(F\right)\subseteq f\left(f^{-1}\left(V\right)\right)=V.$ Therefore Y is almost semi-Moscow. □

Theorem 7. 

Let $f:X\to Y$ be an almost semi-$G_{\delta }$-continuous, pre-almost semi-closed surjection. If X is almost semi-Moscow, then Y is weakly semi-$P_{\Sigma }$.

Proof. 

Let V be a regularly semi-open set of Y and let $y\in V$. Since f is surjective, choose $x\in X$ such that $f\left(x\right)=y$. Since f is almost semi-$G_{\delta }$-continuous, $f^{-1}\left(V\right)$ is a $G_{\delta }$-set of X, and $x\in f^{-1}\left(V\right).$ Since X is almost semi-Moscow, there exists a regularly semi-closed set F of X such that $x\in F\subseteq f^{-1}\left(V\right).$ Because f is pre-almost semi-closed, $f\left(F\right)$ is regularly semi-closed in Y. Also, $y\in f\left(F\right)\subseteq V.$ Thus Y is weakly semi-$P_{\Sigma }$. □

Theorem 8. 

Let $f:X\to Y$ be an injective mapping which is both an R-semi-map and a $G_{\delta }$-map. If Y is almost semi-Moscow, then X is almost semi-Moscow.

Proof. 

Let U be a $G_{\delta }$-set of X and let $x\in U$. Since f is a $G_{\delta }$-map, $f\left(U\right)$ is a $G_{\delta }$-set of Y, and $f\left(x\right)\in f\left(U\right)$. Since Y is almost semi-Moscow, there exists a regularly semi-closed set F of Y such that $f\left(x\right)\in F\subseteq f\left(U\right).$ Since f is an R-semi-map, $f^{-1}\left(F\right)$ is regularly semi-closed in X. By injectivity, $x\in f^{-1}\left(F\right)\subseteq f^{-1}\left(f\left(U\right)\right)=U.$ Therefore X is almost semi-Moscow. □

Theorem 9. 

Let $f:X\to Y$ be an injective R-semi-map.

(i)

If f is R-semi-$G_{\delta }$-open and Y is weakly semi-$P_{\Sigma }$, then X is almost semi-Moscow.

(ii)

If f is pre-almost semi-open and Y is weakly semi-$P_{\Sigma }$, then X is weakly semi-$P_{\Sigma }$.

Proof. (i) Let U be a $G_{\delta }$-set of X and let $x\in U$. Since f is R-semi-$G_{\delta }$-open, $f\left(U\right)$ is regularly semi-open in Y. Since Y is weakly semi-$P_{\Sigma }$, there exists a regularly semi-closed set F of Y such that $f\left(x\right)\in F\subseteq f\left(U\right).$ Since f is an R-semi-map, $f^{-1}\left(F\right)$ is regularly semi-closed in X. By injectivity, $x\in f^{-1}\left(F\right)\subseteq f^{-1}\left(f\left(U\right)\right)=U.$ Hence X is almost semi-Moscow.

(ii) Let U be a regularly semi-open set of X and let $x\in U$. Since f is pre-almost semi-open, $f\left(U\right)$ is regularly semi-open in Y. Since Y is weakly semi-$P_{\Sigma }$, there exists a regularly semi-closed set F of Y such that $f\left(x\right)\in F\subseteq f\left(U\right).$ Since f is an R-semi-map, $f^{-1}\left(F\right)$ is regularly semi-closed in X. Again, injectivity gives $x\in f^{-1}\left(F\right)\subseteq f^{-1}\left(f\left(U\right)\right)=U.$ Therefore X is weakly semi-$P_{\Sigma }$. □

Lemma 2. 

Let $(X,\tau )$ be a $T_1$ space with no isolated points. If there exists a point $x\in X$ such that the singleton $\left\{x\right\}$ is a $G_{\delta }$-set, then X is not almost semi-Moscow.

Proof. 

Suppose, for contradiction, that X is almost semi-Moscow. Since $\left\{x\right\}$ is a $G_{\delta }$-set, by almost semi-Moscow there is a regularly semi-closed set F of X such that $x\in F\subseteq \left\{x\right\}.$ Hence $F=\left\{x\right\}$. By definition, there exists a regularly closed set C of X such that $Int\left(C\right)\subseteq \left\{x\right\}\subseteq C.$ Since $Int\left(C\right)$ is open and $Int\left(C\right)\subseteq \left\{x\right\}$, there are two cases. If $Int\left(C\right)=\left\{x\right\}$, then $\left\{x\right\}$ is open, making x an isolated point. This contradicts the hypothesis. If $Int\left(C\right)=\varnothing$, then since C is regularly closed, $C=Cl(Int(C\left)\right)=Cl(\varnothing )=\varnothing ,$ which contradicts $\left\{x\right\}\subseteq C$. Both cases gives a contradiction. Therefore X is not almost semi-Moscow. □

Remark 4. 

Lemma 2 shows that many familiar spaces are not almost semi-Moscow. For instance, $\mathbb{R}$ with the usual topology is $T_1$, has no isolated points, and every singleton is a $G_{\delta }$-set. Therefore $\mathbb{R}$ is not almost semi-Moscow. The same reasoning applies to ${\mathbb{R}}^n$ with the usual topology for every $n\ge 1$.

Lemma 3. 

Let $(X,\tau )$ be a $T_1$ space with a countable dense subset D with $D⊊X$. Then X is not almost semi-Moscow.

Proof. 

Since X is $T_1$, every singleton is closed. Because D is countable, $D={\bigcup }_{d\in D}\left\{d\right\}$ is an $F_{\sigma }$-set. Hence $V=X\setminus D$ is a $G_{\delta }$-set. Since $D⊊X$, the set V is non-empty. Since D is dense in X, $Int\left(V\right)=Int(X\setminus D)=X\setminus Cl\left(D\right)=X\setminus X=\varnothing .$ Let $x\in V$ and suppose there exists a regularly semi-closed set F of X with $x\in F\subseteq V$. By definition, there exists a regularly closed set C of X such that $Int\left(C\right)\subseteq F\subseteq C.$ Since $F\subseteq V$, it follows that $Int\left(C\right)\subseteq V$. Because $Int\left(C\right)$ is open and is contained in V, the definition of interior gives $Int\left(C\right)\subseteq Int\left(V\right)=\varnothing .$ Since C is regularly closed, $C=Cl(Int(C\left)\right)=Cl(\varnothing )=\varnothing .$ This gives $x\in F\subseteq C=\varnothing$, a contradiction. Therefore X is not almost semi-Moscow. □

Remark 5. 

Lemma 3 shows that no separable $T_1$ space without isolated points is almost semi-Moscow. This includes every separable metrizable space without isolated points, such as ${\mathbb{R}}^n$ with the usual topology, and the irrational numbers with the subspace topology.

Theorem 10. 

Let $(X,\tau )$ be an extremally disconnected space. Then $RSC\left(X\right)=RC\left(X\right).$ Consequently, X is almost semi-Moscow if and only if X is almost Moscow.

Proof. 

The inclusion $RC\left(X\right)\subseteq RSC\left(X\right)$ holds in every topological space, since every regularly closed set is regularly semi-closed. For the reverse inclusion, let $F\in RSC\left(X\right)$. By definition, there exists a regularly closed set C of X such that $Int\left(C\right)\subseteq F\subseteq C.$ Since C is regularly closed, $C=Cl(Int(C\left)\right).$ The set $Int\left(C\right)$ is open. Since X is extremally disconnected, the closure of every open set is open. Therefore $Cl(Int(C\left)\right)$ is open, and hence C is open. Since C is both open and closed, C is clopen. Because C is open,$C\subseteq Int\left(C\right).$ The reverse inclusion $Int\left(C\right)\subseteq C$ always holds. Therefore $Int\left(C\right)=C$. Then $C\subseteq F\subseteq C,$ which gives $F=C$. Hence $F\in RC\left(X\right)$, proving $RSC\left(X\right)\subseteq RC\left(X\right)$. Therefore, in an extremally disconnected space, every regularly semi-closed set is regularly closed, and hence X is almost semi-Moscow if and only if X is almost Moscow. □

2.2. Almost Semi-Moscow Topological Groups

Definition 14. 

Let $(G,·)$ be a group endowed with a topology τ. The pair $(G,\tau )$ is called an almost semi-Moscow topological group if the following two conditions hold:

(i)

the multiplication mapping $\phi :G\times G\to G$, defined by $\phi (x,y)=x·y$, and the inversion mapping $\psi :G\to G$, defined by $\psi \left(x\right)=x^{-1}$, are almost semi-continuous, where $G\times G$ carries the product topology;

(ii)

G is an almost semi-Moscow space.

Example 5. 

Any discrete or indiscrete group is an almost semi-Moscow topological group.

Example 6. 

Let $(G,\tau )$ be any Hausdorff topological group and ${\tau }_{\delta }$ is the topology generated by taking every τ-$G_{\delta }$ set as a base. Then, $(G,{\tau }_{\delta })$ is a topological group and a Hausdorff P-space, therefore almost Moscow[4], hence almost semi-Moscow. Since the group operations are continuous, they are almost continuous, and hence almost semi-continuous. Therefore $(G,{\tau }_{\delta })$ forms an almost semi-Moscow topological group.

Definition 15. 

A group $(G,\tau )$ is called an almost $\mathcal{S}$-topological group if both the multiplication mapping $\phi :G\times G\to G$ and the inversion mapping $\psi :G\to G$ are almost semi-continuous.

Example 7. 

The additive group $(\mathbb{R},+)$ with the usual topology is an almost $\mathcal{S}$-topological group but not an almost semi-Moscow topological group.

Theorem 11. 

Let $(G,\tau )$ be a group endowed with a topology. The following implications hold:

(i)

Almost Moscow topological group ⇒ almost semi-Moscow topological group.

(ii)

Almost semi-Moscow topological group ⇒ almost $\mathcal{S}$-topological group.

(iii)

Almost topological group ⇒ almost $\mathcal{S}$-topological group.

Proof. (i) If $(G,\tau )$ is an almost Moscow topological group, the multiplication and inversion mappings are almost continuous. Since every almost continuous mapping is almost semi-continuous, both mappings are almost semi-continuous. Every almost Moscow space is almost semi-Moscow by Theorem 4(i), so the underlying space G is almost semi-Moscow. Therefore $(G,\tau )$ is an almost semi-Moscow topological group.

(ii) This follows directly from Definition 14, which requires the group operations to be almost semi-continuous.

(iii) If $(G,\tau )$ is an almost topological group, the multiplication and inversion mappings are almost continuous, and therefore almost semi-continuous. Thus $(G,\tau )$ is an almost $\mathcal{S}$-topological group. □

Theorem 12. 

Let G and H be almost $\mathcal{S}$-topological groups, and let $f:G\to H$ be a surjective group homomorphism. If f is $G_{\delta }$-irresolute and pre-almost semi-closed, and if G is an almost semi-Moscow topological group, then H is an almost semi-Moscow topological group.

Proof. 

Since H is an almost $\mathcal{S}$-topological group, its multiplication and inversion mappings are almost semi-continuous. It remains to prove that H is almost semi-Moscow. Let V be a $G_{\delta }$-set in H and let $y\in V$. Choose $x\in G$ with $f\left(x\right)=y$. Since f is $G_{\delta }$-irresolute, $f^{-1}\left(V\right)$ is a $G_{\delta }$-set in G, and $x\in f^{-1}\left(V\right)$. Because G is almost semi-Moscow, there exists a regularly semi-closed set F of G such that $x\in F\subseteq f^{-1}\left(V\right)$. Since f is pre-almost semi-closed, $f\left(F\right)$ is regularly semi-closed in H. Moreover, $y=f\left(x\right)\in f\left(F\right)$ and $f\left(F\right)\subseteq f\left(f^{-1}\left(V\right)\right)=V$. Thus H is almost semi-Moscow. □

Theorem 13. 

Let G and H be almost $\mathcal{S}$-topological groups, and let $f:G\to H$ be a surjective group homomorphism. If f is almost semi-$G_{\delta }$-continuous and pre-almost semi-closed, and if G is an almost semi-Moscow topological group, then H is a weakly semi-$P_{\Sigma }$ almost $\mathcal{S}$-topological group.

Proof. 

Since H is an almost $\mathcal{S}$-topological group, it is enough to show that H is weakly semi-$P_{\Sigma }$. Let W be a regularly semi-open subset of H and let $y\in W$. Choose $x\in G$ with $f\left(x\right)=y$. Since f is almost semi-$G_{\delta }$-continuous, $f^{-1}\left(W\right)$ is a $G_{\delta }$-set in G, and $x\in f^{-1}\left(W\right)$. Since G is almost semi-Moscow, there exists a regularly semi-closed set F of G such that $x\in F\subseteq f^{-1}\left(W\right)$. Because f is pre-almost semi-closed, $f\left(F\right)$ is regularly semi-closed in H. Also, $y=f\left(x\right)\in f\left(F\right)\subseteq f\left(f^{-1}\left(W\right)\right)=W$. Thus H is weakly semi-$P_{\Sigma }$. □

Theorem 14. 

Let G and H be almost $\mathcal{S}$-topological groups, and let $f:G\to H$ be an injective group homomorphism. If f is an R-semi-map and a $G_{\delta }$-map, and if H is an almost semi-Moscow topological group, then G is an almost semi-Moscow topological group.

Proof. 

Since G is an almost $\mathcal{S}$-topological group, it is enough to prove that G is almost semi-Moscow. Let U be a $G_{\delta }$-set in G and let $x\in U$. Since f is a $G_{\delta }$-map, $f\left(U\right)$ is a $G_{\delta }$-set in H, and $f\left(x\right)\in f\left(U\right)$. Since H is almost semi-Moscow, there exists a regularly semi-closed set F of H such that $f\left(x\right)\in F\subseteq f\left(U\right)$. Since f is an R-semi-map, $f^{-1}\left(F\right)$ is regularly semi-closed in G. By injectivity, $x\in f^{-1}\left(F\right)\subseteq f^{-1}\left(f\left(U\right)\right)=U$. Therefore G is almost semi-Moscow. □

Corollary 2. 

No non-discrete, first countable, Hausdorff topological group is an almost semi-Moscow topological group.

Proof. 

By Definition 14(ii), the space of an almost semi-Moscow topological group is almost semi-Moscow. Let G be a non-discrete, first countable, Hausdorff topological group. Since G is Hausdorff, it is $T_1$. Since G is a non-discrete topological group, translations are homeomorphisms, and hence G is homogeneous. If one point were isolated, every point would be isolated, G discrete. Therefore G has no isolated points. Since G is first countable, the singleton $\left\{e\right\}$ is a $G_{\delta }$-set. By Lemma 2, G is not almost semi-Moscow. Therefore G is not an almost semi-Moscow topological group. □

Corollary 3. 

No separable, uncountable, Hausdorff topological group is an almost semi-Moscow topological group.

Proof. 

Let G be a separable, uncountable, Hausdorff topological group. Since G is separable, there exists a countable dense subset D of G. Since G is uncountable, $D⊊G$. Since G is Hausdorff, it is $T_1$. By Lemma 3, G is not almost semi-Moscow. Therefore G is not an almost semi-Moscow topological group. □

2.3. Almost Semi-Moscow Almost Topological Groups

Definition 16. 

Let $(G,·)$ be a group endowed with a topology τ. The pair $(G,\tau )$ is called an almost semi-Moscow almost topological group if the following conditions hold:

(i)

the multiplication mapping $\phi :G\times G\to G$, defined by $\phi (x,y)=x·y$, and the inversion mapping $\psi :G\to G$, defined by $\psi \left(x\right)=x^{-1}$, are almost continuous, where $G\times G$ carries the product topology;

(ii)

G is an almost semi-Moscow space.

Remark 6. 

Equivalently, an almost semi-Moscow almost topological group is an almost topological group whose topological space is almost semi-Moscow.

Example 8. 

Let G be a group and N a proper non-trivial normal subgroup. Equip G with the topology of all unions of cosets of N. Then G is a topological group in which every coset of N is clopen. Every $G_{\delta }$-set is a union of cosets of N. If x belongs to a $G_{\delta }$-set V, then the coset $xN$ is a clopen, hence regularly closed, subset of V containing x. Thus the space is almost Moscow, and therefore almost semi-Moscow. Since the operations are continuous, G is an almost semi-Moscow almost topological group.

Example 9. 

Let $G=\mathbb{Z}$ and $N=2\mathbb{Z}$. Equip $\mathbb{Z}$ with the topology $\tau =\{\varnothing ,2\mathbb{Z},1+2\mathbb{Z},\mathbb{Z}\}$, whose open sets are precisely the unions of cosets of $2\mathbb{Z}$. Both cosets $2\mathbb{Z}$ and $1+2\mathbb{Z}$ are clopen. The group operations are continuous with respect to τ, so $(\mathbb{Z},\tau )$ is a topological group. The only $G_{\delta }$-sets are ∅, $2\mathbb{Z}$, $1+2\mathbb{Z}$, and $\mathbb{Z}$. For any point x belonging to a non-empty $G_{\delta }$-set V, the coset $x+2\mathbb{Z}$ is a clopen, hence regularly closed, subset of V containing x. Therefore space is almost Moscow, and consequently almost semi-Moscow. Since the operations are continuous, they are almost continuous. Hence $(\mathbb{Z},\tau )$ is an almost semi-Moscow almost topological group.

Example 10. 

Let $(G,\tau )$ be any Hausdorff topological group and ${\tau }_{\delta }$ is the topology generated by taking every τ-$G_{\delta }$ set as a base. Then, by Example 6$(G,{\tau }_{\delta })$ forms an almost semi-Moscow almost topological group.

Proposition 1. 

The following implications hold:

(i)

Every almost Moscow topological group is an almost semi-Moscow almost topological group.

(ii)

Every almost semi-Moscow almost topological group is an almost semi-Moscow topological group.

(iii)

Every almost semi-Moscow almost topological group is an almost topological group.

(iv)

Every topological group whose topological space is almost semi-Moscow is an almost semi-Moscow almost topological group.

Proof. (i) Since an almost Moscow topological group has almost continuous, and every almost Moscow space is almost semi-Moscow. Therefore every almost Moscow topological group is an almost semi-Moscow almost topological group.

(ii) Almost continuity implies almost semi-continuity, so condition (i) of Definition 14 is satisfied, and the space is almost semi-Moscow. So every almost semi-Moscow almost topological group is an almost semi-Moscow topological group.

(iii) This is immediate from Definition 16.

(iv) In a topological group the operations are continuous, hence almost continuous, and if the space is almost semi-Moscow, the group is an almost semi-Moscow almost topological group. □

Remark 7. 

The converse of the Proposition 1 (iii) is not true, which is given in the following example.

Example 11. 

Let $G={\mathbb{Z}}_2=\{0,1\}$ with the topology $\tau =\{\varnothing ,\{0\},G\}.$ The only regularly open sets are ∅ and G. Hence the multiplication and inversion mappings are almost continuous, and $(G,\tau )$ is an almost topological group. However, the singleton $\left\{0\right\}$ is a $G_{\delta }$-set. The only regularly semi-closed subsets of G are ∅ and G. Therefore $\left\{0\right\}$ contains no regularly semi-closed set of 0. Thus the space is not almost semi-Moscow. Consequently, $(G,\tau )$ is not an almost semi-Moscow almost topological group.

Theorem 15. 

Let G be an almost semi-Moscow almost topological group. If $U\in RSO\left(G\right)$, then for every $g\in G,gU\in RSO\left(G\right),Ug\in RSO\left(G\right),U^{-1}\in RSO\left(G\right).$ Consequently, if $F\in RSC\left(G\right)$, then $gF\in RSC\left(G\right),Fg\in RSC\left(G\right),F^{-1}\in RSC\left(G\right).$

Proof. 

Let $U\in RSO\left(G\right)$. By definition, there exists a regularly open set $O\in RO\left(G\right)$ such that $O\subseteq U\subseteq Cl\left(O\right).$ By Lemma 1, $gO$ is regularly open and $Cl\left(gO\right)=gCl\left(O\right).$ Multiplying the inclusion $O\subseteq U\subseteq Cl\left(O\right)$ by g gives $gO\subseteq gU\subseteq gCl\left(O\right)=Cl\left(gO\right).$ Thus $gU$ is regularly semi-open. The proofs for $Ug$ and $U^{-1}$ are identical.

If $F\in RSC\left(G\right)$, then $G\setminus F\in RSO\left(G\right)$. Since translations and inversion preserve regularly semi-open sets, the sets $g(G\setminus F),(G\setminus F)g,{(G\setminus F)}^{-1}$ are regularly semi-open. Using $g(G\setminus F)=G\setminus gF,(G\setminus F)g=G\setminus Fg,{(G\setminus F)}^{-1}=G\setminus F^{-1},$ it follows that $gF$, $Fg$, and $F^{-1}$ are regularly semi-closed. □

Theorem 16. 

Let G be an almost semi-Moscow almost topological group. If $U\in RSO\left(G\right)$, then for every $g\in G,Cl\left(gU\right)=gCl\left(U\right),Cl\left(Ug\right)=Cl\left(U\right)g,Cl\left(U^{-1}\right)=Cl{\left(U\right)}^{-1}.$

Proof. 

Let $U\in RSO\left(G\right)$. Choose $O\in RO\left(G\right)$ such that $O\subseteq U\subseteq Cl\left(O\right).$ Then $Cl\left(U\right)=Cl\left(O\right).$ By Lemma 1,$Cl\left(gO\right)=gCl\left(O\right).$ Since $gO\subseteq gU\subseteq gCl\left(O\right)=Cl\left(gO\right),$ it follows that $Cl\left(gU\right)=Cl\left(gO\right)=gCl\left(O\right)=gCl\left(U\right).$ The right translation and inversion identities follow in the same way. □

Corollary 4. 

Let G be an almost semi-Moscow almost topological group. If $F\in RSC\left(G\right)$, then for every $g\in G,sInt\left(gF\right)=gsInt\left(F\right),sInt\left(Fg\right)=sInt\left(F\right)g,sInt\left(F^{-1}\right)=sInt{\left(F\right)}^{-1}.$

Proof. 

Every regularly semi-closed set is semi-open. Hence $sInt\left(F\right)=F$. By Theorem 15, the sets $gF$, $Fg$, and $F^{-1}$ are regularly semi-closed, and therefore semi-open. Thus $sInt\left(gF\right)=gF=gsInt\left(F\right),$ and the remaining identities follow similarly. □

Theorem 17. 

Let G and H be almost semi-Moscow almost topological groups, and let $f:G\to H$ be a group homomorphism. Suppose that for every regularly open V of $e_H$ in H, there exists a regularly open U of $e_G$ in G such that $f\left(U\right)\subseteq V.$ Then f is almost continuous.

Proof. 

Let $x\in G$ and let W be a regularly open set of $f\left(x\right)$ in H. Since H is an almost topological group, left translation by $f{\left(x\right)}^{-1}$ preserves regularly open sets. Hence $V=f{\left(x\right)}^{-1}W$ is a regularly open of $e_H$. By hypothesis, there exists a regularly open U of $e_G$ such that $f\left(U\right)\subseteq V.$ Since G is an almost topological group, left translation by x preserves regularly open sets. Therefore $xU$ is a regularly open, hence open set of x. Moreover, $f\left(xU\right)=f\left(x\right)f\left(U\right)\subseteq f\left(x\right)V=W.$ Thus for every $x\in f^{-1}\left(W\right)$, there exists an open set $xU$ of x contained in $f^{-1}\left(W\right)$. Hence $f^{-1}\left(W\right)$ is open. Since W is an arbitrary regularly open set in H, the mapping f is almost continuous. □

Lemma 4. 

Let Y be a clopen subspace of a space X. If $F\subseteq Y$ is regularly semi-closed in X, then F is regularly semi-closed in Y.

Proof. 

Since F is regularly semi-closed in X, there exists a regularly closed set C of X such that ${Int}_X\left(C\right)\subseteq F\subseteq C.$ Put $C_Y=C\cap Y$. Because Y is clopen, $C_Y$ is regularly closed in Y, and ${Int}_Y\left(C_Y\right)={Int}_X\left(C\right)\cap Y.$ Since $F\subseteq Y$, we have ${Int}_Y\left(C_Y\right)={Int}_X\left(C\right)\cap Y\subseteq F.$ Also $F\subseteq C\cap Y=C_Y$. Hence F is regularly semi-closed in Y. □

Theorem 18. 

Let G be an almost semi-Moscow almost topological group, and let H be a regularly open subgroup of G. Then H is clopen in G, and H, with the subspace topology, is an almost semi-Moscow almost topological group.

Proof. 

Since H is regularly open, $H\in RO\left(G\right)$. By Lemma 1, each coset $gH$ is regularly open. The complement of H is the union of all cosets distinct from H, and therefore is open. Hence H is closed. Thus H is clopen. We next show that H is an almost topological group. Let $W\in RO\left(H\right)$. Since H is clopen, $W\in RO\left(G\right)$. The multiplication mapping ${\phi }_H:H\times H\to H$ satisfies ${\phi }_H^{-1}\left(W\right)={\phi }_G^{-1}\left(W\right)\cap (H\times H).$ Because ${\phi }_G$ is almost continuous and $W\in RO\left(G\right)$, the set ${\phi }_G^{-1}\left(W\right)$ is open in $G\times G$. Hence ${\phi }_H^{-1}\left(W\right)$ is open in $H\times H$. Thus ${\phi }_H$ is almost continuous. The same argument applies to inversion. Finally, we prove that the space H is almost semi-Moscow. Let V be a $G_{\delta }$-set in H and let $x\in V$. Since H is open in G, V is also a $G_{\delta }$-set in G. Since G is almost semi-Moscow, there exists a regularly semi-closed set F of G such that $x\in F\subseteq V.$ Because $V\subseteq H$, we have $F\subseteq H$. By Lemma 4, F is regularly semi-closed in H. Hence H is almost semi-Moscow. Therefore H is an almost semi-Moscow almost topological group. □

Lemma 5. 

Let G be an almost topological group, and let $\pi :G\to G/N$ be an open continuous surjective homomorphism. Then $G/N$ is an almost topological group.

Proof. 

Let $W\in RO(G/N)$. Since $\pi$ is open and continuous, for every subset $A\subseteq G/N$, ${\pi }^{-1}(Cl\left(A\right))=Cl\left({\pi }^{-1}\left(A\right)\right)$ and ${\pi }^{-1}(Int\left(A\right))=Int\left({\pi }^{-1}\left(A\right)\right).$ Therefore ${\pi }^{-1}\left(W\right)\in RO\left(G\right)$. Let $\overline{\phi }:(G/N)\times (G/N)\to G/N$ be the multiplication mapping on $G/N$. Then ${(\pi \times \pi )}^{-1}\left({\overline{\phi }}^{-1}\left(W\right)\right)={\phi }_G^{-1}\left({\pi }^{-1}\left(W\right)\right).$ Since ${\pi }^{-1}\left(W\right)\in RO\left(G\right)$ and ${\phi }_G$ is almost continuous, the right-hand side is open in $G\times G$. Because $\pi \times \pi$ is an open surjection, it is a quotient mapping. Hence ${\overline{\phi }}^{-1}\left(W\right)$ is open in $(G/N)\times (G/N)$. Thus $\overline{\phi }$ is almost continuous. The proof for inversion is identical. Hence $G/N$ is an almost topological group. □

Theorem 19. 

Let G be an almost semi-Moscow almost topological group, and let N be a normal subgroup of G. Let $\pi :G\to G/N$ be the canonical projection. Suppose that π is open, continuous, and pre-almost semi-closed.Then $G/N$ is an almost semi-Moscow almost topological group.

Proof. 

By Lemma 5, $G/N$ is an almost topological group. It remains to prove that the space $G/N$ is almost semi-Moscow. Let V be a $G_{\delta }$-set in $G/N$ and let $\overline{x}\in V$. Choose $x\in G$ with $\pi \left(x\right)=\overline{x}$. Since $\pi$ is continuous, ${\pi }^{-1}\left(V\right)$ is a $G_{\delta }$-set in G, and $x\in {\pi }^{-1}\left(V\right)$. Since G is almost semi-Moscow, there exists a regularly semi-closed set F of G such that $x\in F\subseteq {\pi }^{-1}\left(V\right).$ Since $\pi$ is pre-almost semi-closed, $\pi \left(F\right)$ is regularly semi-closed in $G/N$. Moreover, $\overline{x}=\pi \left(x\right)\in \pi \left(F\right)$ and $\pi \left(F\right)\subseteq \pi \left({\pi }^{-1}\left(V\right)\right)=V.$ Thus $G/N$ is almost semi-Moscow. Therefore $G/N$ is an almost semi-Moscow almost topological group. □

Theorem 20. 

Let G be an almost semi-Moscow almost topological group. If G is semi-regular, then G is a topological group.

Proof. 

Since G is semi-regular, the family $RO\left(G\right)$ is a base for the topology of G. Let W be an open subset of G. Then $W={\bigcup }_{\alpha \in \Lambda }{U}_{\alpha }$ for some family ${\left\{U_{\alpha }\right\}}_{\alpha \in \Lambda }\subseteq RO\left(G\right)$. Since the multiplication mapping $\phi$ is almost continuous, ${\phi }^{-1}\left(U_{\alpha }\right)$ is open in $G\times G$ for each $\alpha$. Hence ${\phi }^{-1}\left(W\right)={\bigcup }_{\alpha \in \Lambda }{\phi }^{-1}\left(U_{\alpha }\right)$ is open in $G\times G$. Thus $\phi$ is continuous. The same argument applies to the inversion mapping $\psi$. Therefore G is a topological group. □

Theorem 21. 

Let $G,H$ be semi-regular almost semi-Moscow almost topological groups. Then $G\times H$, with the product topology, is an almost semi-Moscow almost topological group; in fact $G\times H$ is a topological group.

Proof. 

By Theorem 20, G and H are topological groups. The direct product of topological groups is a topological group, the multiplication and inversion maps of $G\times H$ are continuous. In particular they are almost continuous. By Theorem 3, $G\times H$ is almost semi-Moscow as a space. Since $G\times H$ is a topological group whose space is almost semi-Moscow, Proposition 1(iv) gives that $G\times H$ is an almost semi-Moscow almost topological group. □

Theorem 22. 

If G is an almost semi-Moscow almost topological group whose space is extremally disconnected, then G is an almost Moscow topological group.

Proof. 

By Definition 16, $\phi ,\psi$ are almost continuous, and the space is almost semi-Moscow; since the space is extremally disconnected, by Theorem 10, it is almost Moscow. Therefore G is an almost Moscow topological group. □

Corollary 5. 

Every extremally disconnected almost semi-Moscow almost topological group contains an open Boolean subgroup.

Proof. 

By Theorem 22, G is an extremally disconnected almost Moscow topological group, therefore by Theorem 1, G contains an open Boolean subgroup. □

Corollary 6. 

If G is an almost semi-Moscow almost topological group whose space is extremally disconnected, locally countably S-closed, km-perfect, and semi-regular, then G is a topological group and a strong PT-group.

Proof. 

By Theorem 22, G is an almost Moscow topological group, and by Theorem 2, G is a topological group and a strong PT-group. □

3. Conclusions

The introduction of almost semi-Moscow topological groups establishes a framework combining almost semi-continuous group operations with a $G_{\delta }$-regularity on regularly semi-closed sets. The almost Moscow spaces form a proper subclass of almost semi-Moscow spaces, which in turn form a subclass of semi-Moscow spaces. Product stability for spaces and characterizations via semi-$\theta$-open sets are proved. The stronger class of almost semi-Moscow almost topological groups, requiring almost continuous operations, holds closure identities, results on regularly open subgroups, quotient groups, and products under semi-regularity. Under extremally disconnectedness, the class reduces to almost Moscow topological groups, yielding open Boolean subgroups and strong PT-group structure under suitable conditions. These results confirm that $G_{\delta }$-regularity through regularly semi-closed sets compensates for the failure of almost semi-continuous mappings to preserve exact closures, enabling homomorphic and quotient stability without full continuity. Future work may characterize pre-almost semi-closed quotient maps, and extend the theory to transformation groups.