Clearly, every subset of a topological space $X$ can be obtained from open (or closed) subsets of $X$ by the operations of intersection and union, and many classes of topological spaces have been defined in such a way. Among them are $s$-spaces (see [1-3]) and Lindelöf $\Sigma$-spaces (see [10]). Fix a space $Z$ and an arbitrary family $\mathcal{S}$ of open (or closed) subsets of $Z$, and let $\mathcal{S}_{\delta}$ be the family of all sets that can be represented as the intersection of some subfamily of $\mathcal{S}$. We will say that the family $\mathcal{S}$ is an open (or closed) source of a subspace $X$ in $Z$ if $X$ is the union of some subfamily of $\mathcal{S}_{\delta}$. A space $X$ is called an $s$-space if there exists a countable open source for $X$ in some (every) compactification $bX$ of $X$. A space $X$ is called a Lindelöf $\Sigma$-space if there exists a countable closed source for $X$ in some (every) compactification $bX$ of $X$.

The following statement establishes the relationship between $s$-spaces and Lindelöf $\Sigma$-spaces.

Theorem 1.1 [2] Suppose that $X$ is a nowhere locally compact space with a remainder $Y$. Then $X$ is a Lindelöf $\Sigma$-space if and only if $Y$ is an $s$-space.

The next result about remainders is due to M. Henriksen and J.R. Isbell [9].

Theorem 1.2 A Tychonoff space $X$ is of countable type if and only if the remainder in any (or some) Hausdorff compactification of $X$ is Lindelöf.

Recall thata space $X$ is of countable type if every compact subset $P$ of $X$ is contained in a compact subset $F\subset X$ that has a countable base in $X$. Every $p$-space, as well as each metrizable space, is of countable type. From Theorem 1.1 and Theorem 1.2 above we can see that every $s$-space is of countable type.

Recall that a paratopological group $G$ is a group $G$ with a topology such that the multiplication is jointly continuous. A semitopological group $G$ is a group $G$ with a topology such that the multiplication is separately continuous. A paratopological group with the inversion being continuous is called a topological group. Clearly, every topological group is a paratopological group, and a paratopological group is a semitopological group. It is well known that each semitopological group is homogeneous.

In this paper, we investigate addition theorems for $s$-spaces, and obtain several sufficient conditions that the union of some family of $s$-spaces is also an $s$-space. A sufficient condition for an $s$-space to be sequential is established. We also study the remainders about $s$-spaces, and some results about topological groups with a remainder being an $s$-space are obtained.

Throughout this paper, a space always means a Tychonoff topological space. By a remainder of a Tychonoff space $X$, we mean the subspace $bX\setminus X$ of a Hausdorff compactification $bX$ of $X$. $\overline{A}^{X}$ stands for the closure of $A$ in $X$.

In general, we follow [6] in terminology and notation.

In [2], Arhangel'skii studied the addition theorem for $s$-spaces, and established the following statement.

Theorem 2.1 [2] If a space $X$ is the union of a countable family $\eta$ of dense subspaces of $X$ such that each $Z \in \eta$ is an $s$-space, then $X$ is also an $s$-space.

We complement Arhangel'skii's result above as follows.

Lemma 2.2 The sum space of a countable family of $s$-spaces is an $s$-space.

Proof Assume that $X=\bigoplus_\limits{i\in \omega}X_{i}$ and each $X_{i}$ is an $s$-space. Fix a compactification $bX$ of $X$, and let $bX_{i}$ be the closure of $X_{i}$ in $bX$ for $i\in \omega$. Then each $bX_{i}$ is a compactification of $X_{i}$. Since $X_{i}$ is an $s$-space, there exists a countable open source $\mathcal{O}_{i}$ in $bX_{i}$. Observing that each $X_{i}$ is open in $X$, we can fix an open subset $U_{i}$ of $bX$ such that $U_{i}\cap X=X_{i}$. Clearly, $X_{i}$ is dense in $U_{i}$. Hence $\overline{U_{i}}^{bX}=\overline{X_{i}}^{bX}=bX_{i}$, which follows that $U_{i}$ is contained in $bX_{i}$. Put $\mathcal{S}_{i}=\{O\cap U_{i}:O\in \mathcal{O}_{i}\}$ for each $i\in \omega$. It is easy to see that $\mathcal{S}_{i}$ is a countable open source of $X_{i}$ in $bX$. Therefore, $\bigcup_\limits{i\in\omega}\mathcal{S}_{i}$ is a countable family of open subsets of $bX$. It remains to show that $\bigcup_\limits{i\in\omega}\mathcal{S}_{i}$ is a source of $X$ in $bX$. Take any distinct points $x, y$ such that $x\in X, y\in bX\setminus X$. There exists $X_{i}$ such that $x\in X_{i}$. Since $\mathcal{S}_{i}$ is a source of $X_{i}$ in $bX$, $\bigcap\{S:x\in S\in\mathcal{S}_{i}\}$ is contained in $X_{i}$. Hence, we can take $S\in\mathcal{S}_{i}\subset\bigcup_\limits{i\in\omega}\mathcal{S}_{i}$ such that $x\in S\subset bX\setminus\{y\}$. Therefore, $X$ is an $s$-space.

Theorem 2.3 Let $X$ be the union of a countable family $\eta$ of closed subspaces such that each $Z\in \eta$ is an $s$-space. If $\eta$ is locally finite in $X$, then $X$ is an $s$-space.

Proof Let $Y$ be the sum space of $\eta$, i.e., $Y=\bigoplus\eta$. By Lemma 2.2, $Y$ is an $s$-space. Let $f: Y\rightarrow X$ be the canonical mapping that restricts to the identity on each $Z\in \eta$. Since $\eta$ is a family of closed subsets of $X$ and locally finite in $X$, it follows that $f$ is a perfect mapping. By Theorem 2.13 in [2], the image of an $s$-space under a perfect mapping is an $s$-space. Therefore, $X$ is an $s$-space.

Corollary 2.4 If $X$ be the union of a finite family $\eta$ of closed subspaces and each $Z\in \eta$ is an $s$-space, then $X$ is an $s$-space.

The following example shows that the assumption in Theorem 2.3 that $\eta$ is locally finite cannot be dropped.

Example 1 The union of a countable family of closed $s$-subspaces need not be an $s$-space.

Proof Fix a $\sigma$-compact $X$ such that $X$ is not a $p$-space (for instance, the $\sigma$-product of $\omega_{1}$ copies of two-elements topological group). Since each compact space is an $s$-space, $X$ is the union of a countable family of closed $s$-spaces. We claim that $X$ is not an $s$-space. Assume the contrary. Let $bX$ be a compactification of $X$. By Theorem 1.1, $bX\setminus X$ is a Lindelöf $\Sigma$-space. Notice that $X$ is also a Lindelöf $\Sigma$-space. By Corollary 6.3 in [2], a Lindelöf $\Sigma$-space $Y$ is a $p$-space provided that $Y$ is a subspace of a Lindelöf $p$-space $Z$ and $Z\setminus Y$ is also a Lindelöf $\Sigma$-space. Hence, $X$ is a $p$-space since $X$ is a Lindelöf $\Sigma$-space. This is a contradiction. Therefore, $X$ is not an $s$-space.

For open $s$-subspaces the circumstances is different, which can be seen from the following result whose proof is similar with Lemma 2.2.

Theorem 2.5 If $X$ is the union of a countable family $\eta$ of open subspaces such that each $Z\in \eta$ is an $s$-space, then $X$ is an $s$-space.

Theorem 2.6 Let $X$ be the union of a countable family $\eta$ of metrizable subspaces. If $X$ is an $s$-space, then $X$ is a sequential space.

Proof Since $X$ is an $s$-space, $X$ is a $k$-space by Corollary 2.12 in [2]. In fact, it follows from the fact that every space of point-countable type is a $k$-space. Fix any non-closed subset $A$ of $X$. Then there is a compact subset $K$ of $X$ such that $A\cap K$ is not closed in $K$. By [11], every compact space that is the union of a countable family of metrizable subspaces is sequential. Since $K$ is the union of a countable metrizable subspaces, it follows that $K$ is sequential. Hence, there is a sequence $\{x_{n}:n\in\omega\}$ of $A\cap K$ converging to a point $x\in K\setminus A\subset X\setminus A$. Therefore, $X$ is a sequential space.

Corollary 2.7 Let $X$ be a topological group that is an $s$-space. If $X$ is the union of a countable family $\eta$ of metrizable subspaces, then $X$ is metrizable.

Proof By the assumption and Theorem 2.6, $X$ is sequential. Hence, $X$ has countable tightness. Since $X$ is an $s$-space, it is of countable type. Fix a compact subset $K$ of $X$ such that $K$ has a countable base in $X$. Since $K$ is a compact space with countable tightness, $K$ has countable $\pi$-character by [12]. Then it follows from the fact $K$ having a countable base in $X$ that $X$ has countable $\pi$-character at each point of $K$. Since $X$ is homogeneous, $X$ has countable $\pi$-character. Hence, $X$ is first countable since it is a topological group (see Proposition 5.2.6 in [4]). Therefore, $X$ is metrizable (see Theorem 3.3.12 in [4]).

In [2], Arhangel'skii proved that $s$-spaces are preserved by a perfect mapping in both directions. It is also known that the image of a Lindelöf $\Sigma$-space under any continuous mapping is also a Lindelöf $\Sigma$-space [10]. However, the image of an $s$-space under a continuous closed mapping need not be an $s$-space.

Theorem 2.8 The image of an $s$-space under a continuous closed mapping need not be an $s$-space.

Proof Let $X=\bigoplus\limits_{ i\in \omega}I_{i}$ be the sum space of $\omega$ copies of closed unit interval $I$, where each $I_{i}$ is homeomorphic to $I$. Let $0_{i}$ be the zero of $I_{i}$, and identify all $0_{i}$s to be one point $0$. Then we obtain a quotient space $Y$ of $X$ with respect to the canonical mapping $f:X\rightarrow Y$ defined by $f(0_{i})=0$ for each $i\in \omega$, and $f(x)=x$ for each $x\in X\setminus\{0_{i}: i\in \omega\}$. Clearly, $f$ is a continuous closed mapping. Since $X$ is separable and metrizable, $X$ is an $s$-space.

Claim $Y$ is not an $s$-space.

Assume the contrary. By Theorem 7.1 in [2], $w(Z)=nw(Z)$ provided that $Z$ is an $s$-space, where $w(Z)$ and $nw(Z)$ denotes the weight and network weight of $Z$ respectively. Since $Y$ has a countable network, it follows that $Y$ has a countable base, which contradicts with the fact that $Y$ is not first countable.

The following results complement Theorem 1.1.

Theorem 2.9 If $B$ is a compact space and a subspace $X$ of $ B$ is a Lindelöf $\Sigma$-space, then the subspace $B\setminus X$ of $B$ is an $s$-space.

Proof Let $Y$ be the closure of $X$ in $B$. Then $Y$ is a compactification of $X$, and there exists a countable closed source $\mathcal{F}$ of $X$ in $Y$. Clearly, $\mathcal{O}=\{B\setminus F:F\in \mathcal{F}\}$ is a countable family of open subsets of $B$. Let $Z$ be the closure of $B\setminus X$ in $B$, and $\mathcal{S}=\{O\cap Z: O\in\mathcal{O}\}$. Obviously, $Z$ is a compactification of $B\setminus X$ and $\mathcal{S}$ is a countable open source of $B\setminus X$ in $Z$. Therefore, $B\setminus X$ is an $s$-space.

Example 2 There exists a compact space $B$ and its subspace $X$ which is an $s$-space, the subspace $B\setminus X$ of $B$ need not be a Lindelöf $\Sigma$-space.

Proof Let $B=C_{1}\cup C_{2}$ be the Alexandroff double of the circle, where $C_{i}=\{(x, y):x^{2}+y^{2}=i\}, i=1, 2$ (see Example 3.1.26 in [6]). It is known that $B$ is a compact space, $C_{1}$ is a compact subspace of $B$, and hence $C_{1}$ is an $s$-space. Since $C_{2}$ is an open discrete subspace of $B$ with cardinality $2^{\omega}$, it follows that $C_{2}$ is not a Lindelöf $\Sigma$-space.

Theorem 2.10 Suppose that $B$ is a compact space, and $X$ is a subspace of $B$ such that $X$ is dense in some open subspace $U$ of $B$. If $X$ is an $s$-space, then $B\setminus X$ is a Lindelöf $\Sigma$-space.

Proof Let $Y$ be the closure of $X$ in $B$. Then $Y$ is a compactification of $X$, and there exists a countable open source $\mathcal{O}$ of $X$ in $Y$. Notice that $U$ is an open subspace of $Y$. It is easy to see that $\mathcal{V}=\{O\cap U: O\in\mathcal{O}\}$ is a family of open subsets of $B$ and a source of $X$ in $Y$. Let $Z$ be the closure of $B\setminus X$ in $B$. Then the family $\mathcal{F}=\{(B\setminus V)\cap Z: V\in\mathcal{V}\}$ is a countable closed source of $B\setminus X$ in $Z$. Therefore, $B\setminus X$ is a Lindelöf $\Sigma$-space.

In the end, we study some spaces with a compactification such that the remainder is (locally) an $s$-space.

Theorem 2.11 Let $X$ be a non-locally compact homogeneous space with a compactification $bX$ such that the remainder $Y=bX\setminus X$ is locally an $s$-space. Then $X$ is a Lindelöf $\Sigma$-space and $Y$ is an $s$-space.

Proof Since $Y$ is locally an $s$-space and every closed subspace of an $s$-space is also an $s$-space, we can fix an open subspace $U$ of $Y$ such that the closure of $U$ in $Y$, denoted by $F$, is an $s$-space. Since $X$ be a non-locally compact homogeneous space, it is nowhere locally compact. Therefore, $Y$ is dense in $bX$. Let $Z$ be the closure of $F$ in $bX$. Then $Z\setminus F$ is a Lindelöf $\Sigma$-space and contained in $X$. Clearly, $Z\setminus F$ is a closed subspace of $X$ and has non-empty interior in $X$. It follows that $X$ is locally a Lindelöf $\Sigma$-space. Since every $s$-space is of countable type, $Y$ is of locally countable type. By [13], every space of locally countable type is of countable type. Hence, $Y$ is of countable type. By Theorem 1.2, $X$ is a Lindelöf space. It follows that $X$ is covered by a countable family of its Lindelöf $\Sigma$-subspaces. Therefore, $X$ is a Lindelöf $\Sigma$-space (see Proposition 5.3.8 in [4]). Hence, $Y$ is an $s$-space by Theorem 1.1.

Corollary 2.12 If a first-countable paratopological group $G$ has a compactification $bG$ such that the remainder $Y=bG\setminus G$ is locally an $s$-space, then $G$ is metrizable.

Proof If $G$ is locally compact, then $G$ is a topological group by [5]. A first-countable topological group is metrizable by Theorem 3.3.12 in [4]. Therefore, $G$ is metrizable.

If $G$ is non-locally compact, then $G$ is a Lindelöf $\Sigma$-space by Theorem 2.11. Since a semitopological group with countable $\pi$-character has a $G_{\delta}$-diagonal (see Corollary 5.7.5 in [4]), $G$ has a $G_{\delta}$-diagonal. Hence, $G$ has a countable network, since every $\Sigma$-space with a $G_{\delta}$-diagonal is a $\sigma$-space [8] and every Lindelöf $\sigma$-space has a countable network. By Proposition 5.7.14 in [4], a first-countable paratopological group with a countable network has a countable base, so has $G$. Therefore, $G$ is metrizable.

Theorem 2.13 Let $G$ be a non-locally compact topological group with a compactification $bG$ such that the remainder $Y=bG\setminus G$ is an $s$-space and is the union of a countable family $\eta$ of metrizable subspaces. Then $G$ is separable and metrizable.

Proof Since $Y$ is an $s$-space, it is of countable type. Take an arbitrary point $y\in Y$ and a compact subset $K\subset Y$ such that $y\in K$ and $K$ has a countable base in $Y$. From the proof of Corollary 2.7 we can see that $Y$ has countable $\pi$-character at $y$. It follows that $Y$ has countable $\pi$-character. Since $Y$ is dense in $bG$, it follows that $bG$ has countable $\pi$-character at each point of $Y$.

By [7], every countably compact space that is the union of a countable family of $D$-spaces is compact. Since every metrizable space is a $D$-space and $Y$ is non-compact, it follows that $Y$ is not countably compact. Then there is a countable closed subset $A\subset Y$ which is discrete in $Y$. Since $bG$ is compact, there exists a point $c\in G$ such that $c$ is a accumulation point of $A$.

For every $a\in A$, we take a countable $\pi$-base $\eta_{a}$ of $bG$ at $a$. Then the family $\bigcup_\limits{a\in A}\eta_{a}$ is a countable $\pi$-base of $bG$ at $c$. Put $\mathcal{O}=\{O\cap G:O\in \bigcup_\limits{a\in A}\eta_{a}\}$. Since $G$ is dense in $bG$, it follows that $\mathcal{O}$ is a countable $\pi$-base of $G$ at $c$. Hence, $G$ has countable $\pi$-character since it is homogeneous. Therefore, it follows from $G$ being a topological group that $G$ is metrizable. Clearly, $G$ is Lindelöf, since it is a Lindelöf $\sigma$-space. Therefore, $G$ is separable and metrizable.

Theorem 2.14 Let $G$ be a non-locally compact semitopological group with a compactification $bG$ such that the remainder $Y=bG\setminus G$ is an $s$-space and is the union of a countable family $\eta$ of metrizable subspaces. Then $G$ has a countable network.

Proof From the proof of Corollary 2.7 we can see that $G$ has countable $\pi$-character. Since $G$ is a paratopological group, it follows that $G$ has a $G_{\delta}$-diagonal. Clearly, $G$ is a Lindelöf $\sigma$-space. Therefore, $G$ has a countable network.

Theorem 2.15 Let $G$ be a non-locally compact topological group with a compactification $bG$ such that the remainder $Y=bG\setminus G$ is hereditarily an $s$-space. Then either $G$ is separable and metrizable, or $G$ is $\sigma$-compact.

Proof Clearly, both $G$ and $Y$ are dense in $bG$. Then $Y$ is nowhere locally compact, which implies that $Y$ is dense-in-itself. By Theorem 7.11 in [2], if a dense-in-itself space is hereditarily an $s$-space, then it is first-countable. Hence, $Y$ is first-countable.

If $Y$ is not countably compact, then from the proof of Theorem 2.13 we can see that $G$ is separable and metrizable. If $Y$ is countably compact, then $Y$ is Čech-complete by Theorem 3.6 in [2]. It follows that $G$ is $\sigma$-compact.

Corollary 2.16 Let $G$ be a non-locally compact topological group with a compactification $bG$ such that the remainder $Y=bG\setminus G$ is hereditarily an $s$-space. If $G$ has the Baire property, then $G$ is separable and metrizable.

Proof Suppose to the contrary that $G$ is not separable and metrizable. Then $G$ is $\sigma$-compact by Theorem 2.15. Hence, it follows from $G$ having the Baire property that $G$ is locally compact. This is a contradiction.