H-sober spaces and Hofmann-Mislove type theorems

Introduction

The Hofmann–Mislove theorem identifies the Scott-open filters of the open-set lattice of a sober space with its compact saturated subsets [2, 3]. Its countable variants require care: countability of a filter, existence of a countable filter base, and countability of a set whose closure is irreducible impose different restrictions.

In this paper, for a $T_0$-space $X$, an open filter means a Scott-open filter on the open-set lattice $\mathcal{O}(X)$. A filter on $\mathcal{O}(X)$ is countably generated if it is generated by countably many open members of the filter. The notion of $\omega$-sobriety concerns irreducible closed sets with a countable dense subset, and does not require those closed sets themselves to be countable. In the language of irreducible subset systems, this is precisely $\mathcal{R}^{\omega}$-sobriety [5], where

Thus $\omega$-sobriety is a special case of $H$-sobriety, with the countability restriction placed on the inducing irreducible subset.

We distinguish two natural classes of proper Scott-open filters for an irreducible subset system $H$. The first consists of $\mathcal{F}_A=\{U\in\mathcal{O}(X):U\cap A\ne\varnothing\}$ for $A\in H(X)$. The second consists of $\Phi_X(\mathcal{A})=\bigcup_{G\in\mathcal{A}}\Phi_X(G)$ for $\mathcal{A}\in H(\mathcal{P}_{S}(X))$, where $\mathcal{P}_{S}(X)$ is the Smyth power space of $X$. Their representation properties characterize,

Email addresses: [email protected] (Xiaoquan Xu), [email protected] (Xinjie Huang) respectively, $H$-sobriety of $X$ and $H$-sobriety of $\mathcal{P}_{S}(X)$ (Theorems 2.10 and 2.11). These are filter formulations of the existing $H$-sobriety and super $H$-sobriety framework [4 Theorems 3.6 and 5.7], [5 Theorem 5.5]. The local property-$Q$ criterion [4 Theorem 6.4] explains when the two conditions coincide.

For an arbitrary $H$, its restriction $H^\omega$ to countable members is again an irreducible subset system, so both representation theorems apply. For $\mathcal{R}^{\omega}$ we prove property $Q$ directly, using a minimal closed set and one chosen point from each member of a countable irreducible family in $\mathcal{P}_{S}(X)$ (Lemma 3.5). This gives

Theorem 3.6 expresses this equivalence in terms of both point-induced filters and countable irreducible compact-family filters. In the latter case, countable counts the members of the family, without imposing a countable open base on the induced filter.

Representation for countably generated Scott-open filters alone is weaker than $\omega$-sobriety. Indeed, it already holds in every $\omega$-well-filtered space by the theorem of Goubault-Larrecq [1 Theorem 1 and the following remark]. We denote this open-generator representation property by $H_{\mathrm{co}}$, with “c” standing for countable and “o” for open. Goubault-Larrecq allows a broader notion of generation by descending subsets, which we distinguish in Remark 2.8. By Theorem 2.10, the filters $\mathcal{F}_{A}$ induced by countable irreducible subsets already characterize $\omega$-sobriety. Theorem 3.2 shows that one may enlarge this test class by adjoining all proper Scott-open filters with countable open bases. The resulting class $\mathcal{F}_{\omega}(X)$ is defined explicitly in Section 3. The theorem gives equivalent recovery and representation conditions for this enlarged class, whereas the countably based filters alone do not characterize $\omega$-sobriety in general.

Theorem 3.2 has a direct minimal-closed-set proof, and Theorem 4.1 gives a further equivalent formulation using traces on countable subspaces. The classical, well-filtered and $\omega$-well-filtered versions are compared through their inducing families. The examples establish that both implications $\omega$-sobriety $\Rightarrow$ $\omega$-well-filteredness $\Rightarrow$ $H_{\mathrm{co}}$ are strict.

The terminology here also differs from countably sober in [8, 9], where the defining irreducibility condition involves countable closed covers. No identification with that condition is intended.

Preliminaries and Hofmann–Mislove representations

All spaces are $T_{0}$, and “countable” includes finite. We write $\mathbb{N}=\{1,2,\ldots\}$. For a space $X$, let $\mathcal{O}(X)$ denote its lattice of open subsets, ordered by inclusion, and write $\overline{A}$ for the closure of $A\subseteq X$.

A nonempty subset $A$ of $X$ is irreducible if, whenever $A\subseteq C_{1}\cup C_{2}$ for closed subsets $C_{1},C_{2}$ of $X$, one has $A\subseteq C_{1}$ or $A\subseteq C_{2}$. Equivalently, for all $U,V\in\mathcal{O}(X)$,

A subset is irreducible if and only if its closure is irreducible. A point $x$ is a generic point of a closed set $C$ if $C=\overline{\{x\}}$. The space $X$ is sober if every irreducible closed subset has a unique generic point.

The specialization order is

For $A\subseteq X$, put

The saturation of $A$ is the intersection of all open sets containing $A$; it equals $\uparrow A$. Thus a subset is saturated if and only if it is an upper set for the specialization order. Open sets are upper sets as well. In particular, if an open or saturated set meets $\overline{\{x\}}$, then it contains $x$.

A subset $K\subseteq X$ is compact if every open cover of $K$ has a finite subcover; no separation assumption is included. We write $\mathcal{Q}(X)$ for the collection of nonempty compact saturated subsets of $X$.

Definition 2.1. A space $X$ is $\omega$-sober if every irreducible closed subset $A$ for which

for some countable $D \subseteq X$ is $\{x\}$ for a unique $x \in X$.

Equivalently, for every countable irreducible subset $D$ of $X$ there is a unique $x$ with $\overline{D} = \overline{\{x\}}$. This follows because a subset is irreducible if and only if its closure is irreducible. Uniqueness follows from the $T_0$ assumption. On a countable space, $\omega$-sobriety is exactly sobriety. In particular, Definition 2.1 is the $H$-sobriety condition for the countable irreducible subset system $\mathcal{R}^{\omega}$; the general definitions are given below.

A nonempty subset $D$ of a poset is directed if every pair of its members has an upper bound in $D$. For a family of open sets ordered by inclusion, this means that any two members are contained in a third member of the family.

A proper filter $\mathcal{F} \subseteq\mathcal{O}(X)$ is a nonempty upper set closed under finite intersections and not containing $\varnothing$. It is Scott-open if, for every nonempty directed family $\{U_i : i \in I\} \subseteq\mathcal{O}(X)$,

For a subset $K \subseteq X$, put

For a nonempty irreducible subset $A \subseteq X$, define

Irreducibility makes $\mathcal{F}_A$ a proper filter. It is completely prime: if an arbitrary union belongs to $\mathcal{F}_A$, some member of that union belongs to $\mathcal{F}_A$. In particular, $\mathcal{F}_A$ is Scott-open. If $A = \{x\}$, then

The set $\mathord{\uparrow} x$ is always nonempty, compact and saturated: any open set containing $x$ contains all of $\mathord{\uparrow} x$.

Proof. For a directed open cover of a compact set, a finite subcover is contained in one member of the directed family. This proves Scott openness. Saturation gives $K = \bigcap\Phi_X(K)$.

Conversely, write $K = \bigcap\mathcal{F}$. It is saturated, and properness and the displayed equality imply $K \ne\varnothing$. Given an open cover of $K$, its union belongs to $\mathcal{F}$. The finite unions of members of the cover form a directed family, so Scott openness gives a finite union in $\mathcal{F}$. That finite union contains $K$.

We recall the following minimal closed set lemma [4 Lemma 2.12], including its proof for completeness.

Proof. Consider the closed subsets of $C$ meeting every member of $\mathcal{F}$. For a nonempty descending chain $\{C_i : i \in I\}$ in this collection, suppose that $\bigcap_{i \in I} C_i$ misses some $U \in\mathcal{F}$. Then

The union belongs to $\mathcal{F}$, and the open complements form a directed family. Scott openness gives $X \setminus C_i \in\mathcal{F}$ for some $i$, a contradiction. Zorn’s lemma therefore gives a minimal member $A$.

If $A \subseteq C_1 \cup C_2$ for closed sets $C_1,C_2$ and neither contains $A$, both $A \cap C_1$ and $A \cap C_2$ are proper closed subsets of $A$. There are $U_1,U_2 \in\mathcal{F}$ with $A \cap C_i \cap U_i = \varnothing$. But then $A \cap U_1 \cap U_2 = \varnothing$, contradicting $U_1 \cap U_2 \in\mathcal{F}$.

Irreducible sets and families

The Smyth power space $P_S(X)$ is the set $Q(X)$ equipped with the upper Vietoris topology, whose basic open sets are

Its specialization order is reverse inclusion. For a nonempty family $\mathcal{A} \subseteq Q(X)$, extend the notation $\Phi_X$ by putting

The role of irreducibility is described by the following standard fact [5 Lemma 2.23 and Remark 2.24].

Proof. For each $G \in\mathcal{A}$, Lemma 2.2 shows that $\Phi_X(G)$ is Scott-open in $\mathcal{O}(X)$. Hence their union $\Phi_X(\mathcal{A})$ is a Scott-open upper set. It contains $X$ and does not contain $\varnothing$, because $\mathcal{A}$ and all its members are nonempty.

We next characterize closure under finite intersections. Suppose first that $\mathcal{A}$ is irreducible in $P_S(X)$, and let $U,V \in\Phi_X(\mathcal{A})$. Then $\mathcal{A}$ meets both $\Box U$ and $\Box V$. By irreducibility, it meets

Consequently, some $G \in\mathcal{A}$ is contained in $U \cap V$, so $U \cap V \in\Phi_X(\mathcal{A})$. Thus $\Phi_X(\mathcal{A})$ is a filter.

Conversely, suppose that $\Phi_X(\mathcal{A})$ is a filter. Let $V_1,V_2$ be open subsets of $P_S(X)$ that meet $\mathcal{A}$. Since the sets $\Box U$, $U \in\mathcal{O}(X)$, form a base for the upper Vietoris topology, there exist $U_1,U_2 \in\mathcal{O}(X)$ such that

Thus $U_1,U_2 \in\Phi_X(\mathcal{A})$, and the filter property gives $U_1 \cap U_2 \in\Phi_X(\mathcal{A})$. Hence $\mathcal{A}$ meets $\Box(U_1 \cap U_2) \subseteq V_1 \cap V_2$. This proves that $\mathcal{A}$ is irreducible in $P_S(X)$.

Using $G = \bigcap\Phi_X(G)$ for each saturated $G$, we obtain

Finally, for every $U \in\mathcal{O}(X)$, the open set $\Box U$ meets $\mathcal{A}$ if and only if it meets $\overline{\mathcal{A}}^{P_S(X)}$. By the definition of $\Phi_X$, this gives the second identity in (5).

The following combines the classical Hofmann–Mislove theorem with its irreducible-set and Smyth-power formulations; see [2, 3], [4 Theorems 3.6 and 5.7], and [7 Corollaries 4.3 and 4.4].

(3) For every irreducible subset $A \subseteq X$, there exists $K \in\mathcal{Q}(X)$ such that $\mathcal{F}_A = \Phi_X(K)$.

(4) For every irreducible subset $A \subseteq P_S(X)$, there exists $K \in\mathcal{Q}(X)$ such that $\Phi_X(A) = \Phi_X(K)$.

In (3) and (4), one may equivalently restrict the respective test sets to irreducible closed sets. Each representation may equivalently be written as recovery from the intersection of the filter; the representing set is unique.

Proof. The equivalence of (1) and (2) is the classical Hofmann–Mislove theorem. Condition (2) implies (3) and, by Lemma 2.4, (4). Conversely, suppose (3), fix an irreducible $A$, and write $\mathcal{F}_A = \Phi_X(K)$. The set $K$ meets $C = \overline{A}$, since otherwise the open set $X \setminus C$ would belong to $\mathcal{F}_A$. Choose $x \in K \cap C$. Every open neighbourhood of every $y \in C$ meets $A$ and hence contains $K$, so it contains $x$. Thus $C \subseteq\{x\}$; the reverse inclusion follows from $x \in C$. This proves (1).

For (4)$\Rightarrow$(3), the map $\eta_X : X \to P_S(X)$, $x \mapsto\uparrow x$, is continuous because $\eta_X^{-1}(\Box U) = U$. Consequently $\eta_X(A)$ is irreducible and

Finally, $\mathcal{F}_A = \overline{\mathcal{F}_A}$ and the second identity in (5) justify the closed-set restrictions. The assertions about recovery and uniqueness follow from Lemma 2.2.

Filtered compact families and countable open bases

A nonempty family $\mathcal{K} \subseteq\mathcal{Q}(X)$ is filtered if for $G,H \in\mathcal{K}$ some $L \in\mathcal{K}$ satisfies $L \subseteq G \cap H$. Such a family is irreducible in $P_S(X)$ and therefore induces the proper Scott-open filter in (2.3). Recall that $X$ is well-filtered if, for every such family and every $U \in\mathcal{O}(X)$,

Restricting to countable filtered families gives $\omega$-well-filteredness [6].

Proposition 2.6 (The well-filtered versions). Let $X$ be a $T_0$-space.

(1) $X$ is well-filtered if and only if, for every nonempty filtered family $\mathcal{K} \subseteq\mathcal{Q}(X)$, $\Phi_X(\mathcal{K}) = \Phi_X(\bigcap\mathcal{K})$.

(2) $X$ is $\omega$-well-filtered if and only if the same equality holds for every nonempty countable filtered family $\mathcal{K} \subseteq\mathcal{Q}(X)$.

In either assertion, the equality implies that $\bigcap\mathcal{K}$ is nonempty, compact and saturated. Equivalently, one may require the existence of $K \in\mathcal{Q}(X)$ with $\Phi_X(\mathcal{K}) = \Phi_X(K)$. In (2), it suffices to test descending sequences in $\mathcal{Q}(X)$.

Proof. For a filtered $\mathcal{K}$, equation (6) says precisely that an open set containing $\bigcap\mathcal{K}$ belongs to $\Phi_X(\mathcal{K})$. The reverse inclusion is automatic. Apply Lemmas 2.2 and 2.4 to obtain all the representation assertions. This is the filter form of well-filteredness in [4 Theorem 3.11], with the same argument for its countable version.

If $\mathcal{K} = \{G_n : n \in\mathbb{N}\}$ is filtered, choose $K_1 = G_1$ and recursively choose $K_{n+1} \in\mathcal{K}$ with $K_{n+1} \subseteq K_n \cap G_{n+1}$. Then $\{K_n\}$ is a decreasing cofinal subset. It has the same intersection and induces the same filter as $\mathcal{K}$. This also proves the final assertion; compare [6 Definition 3.9 and Proposition 3.10].

Definition 2.7. For a nonempty filtered family $\mathcal{U} \subseteq\mathcal{O}(X) \setminus\{\varnothing\}$, put

The space $X$ satisfies $(H_{\mathrm{co}})$ if, for every such countable family $\mathcal{U}$ for which $\Psi_X(\mathcal{U})$ is Scott-open, there exists $K \in\mathcal{Q}(X)$ such that $\Psi_X(\mathcal{U}) = \Phi_X(K)$. Here “c” stands for countable and “o” for open.

The Scott-open requirement in Definition 2.7 is an additional hypothesis. In contrast, for filtered compact saturated families the induced filter is automatically Scott-open. The same cofinal-sequence argument as above allows $\mathcal{U}$ to be replaced by a decreasing sequence of nonempty open sets. Hence $\mathrm{H}_{\mathrm{co}}$ is exactly representation for proper Scott-open filters with countable open bases.

Remark 2.8 (The scope of Goubault-Larrecq’s theorem). Goubault-Larrecq [1 Theorem 1 and subsequent remark] permits descending nonempty subsets $B_1 \supseteq B_2 \supseteq\cdots$, not necessarily open, as generators:

Whenever $\mathcal{G}$ is Scott-open, his result represents it under $\omega$-well-filteredness. Taking the $B_n$ open gives $\omega$-well-filteredness $\Rightarrow\mathrm{H}_{\mathrm{co}}$.

Requiring representation for all such Scott-open filters with arbitrary subset generators is in fact equivalent to $\omega$-well-filteredness: the converse follows by taking $B_n=K_n\in\mathcal{Q}(X)$ and applying Proposition 2.6. Thus $\mathrm{H}_{\mathrm{co}}$ is specifically the open-generator case of the result.

Irreducible subset systems and the two $H$-versions

We use the irreducible subset systems of [5]; their Hofmann–Mislove formulation is developed in [4]. Here $\mathbf{Top}_0$ is the category of $T_0$-spaces and continuous maps, and $\mathbf{Set}$ is the category of sets and maps.

Definition 2.9. An irreducible subset system is a functor $H:\mathbf{Top}_0\to\mathbf{Set}$ such that, for each space $X$,

and, for every continuous map $f:X\to Y$, the map $H(f)$ is given by $H(f)(A)=f(A)$. A member of $H(X)$ is called an $H$-set. Put

A closed $H$-set means a member of $H_c(X)$. The space $X$ is $H$-sober if every $A\in H(X)$ has $\overline{A}=\{\overline{x}\}$ for a unique $x\in X$. It is super $H$-sober if $P_S(X)$ is $H$-sober.

The use of $H_c(X)$ matters: it consists of closures of $H$-sets, rather than just those members of $H(X)$ that happen to be closed. The two filter classes below consist of proper Scott-open filters, by (1) and Lemma 2.4. We first state the point-induced formulation of [4 Theorems 3.6 and 5.7].

Theorem 2.10 (The point-induced $H$-version). Let $H$ be an irreducible subset system. For a $T_0$-space $X$, the following conditions are equivalent.

1. $X$ is $H$-sober.

2. For every $A\in H(X)$, $\mathcal{F}_A=\Phi_X(\bigcap\mathcal{F}_A)$.

3. For every $A\in H(X)$, the set $\bigcap\mathcal{F}_A$ belongs to $\mathcal{Q}(X)$ and $\mathcal{F}_A=\Phi_X(\bigcap\mathcal{F}_A)$.

4. For every $A\in H(X)$, there exists $K\in\mathcal{Q}(X)$ such that $\mathcal{F}_A=\Phi_X(K)$.

The representing set is unique and equals $\bigcap\mathcal{F}_A=\mathord{\uparrow}x$, where $\overline{A}=\{\overline{x}\}$. The same equivalences hold with the test sets taken from $H_c(X)$.

Proof. If $X$ is $H$-sober, equation (2) gives (4). Conversely, suppose (4), fix $A\in H(X)$, and write $\mathcal{F}_A=\Phi_X(K)$. Since $X\setminus\overline{A}\notin\mathcal{F}_A$, there is $x\in K\cap\overline{A}$. Every open neighbourhood of any $y\in\overline{A}$ meets $A$, belongs to $\mathcal{F}_A$, and therefore contains $x$. Thus $\overline{A}\subseteq\{\overline{x}\}$. The reverse inclusion follows from $x\in\overline{A}$, and uniqueness follows from $T_0$. This proves (1).

Lemma 2.2 gives the equivalence of (2), (3) and (4), as well as uniqueness of the representing set. Equation (2) identifies it as $\mathord{\uparrow}x$. Finally, $\mathcal{F}_A=\mathcal{F}_{\overline{A}}$ gives the formulation using $H_c(X)$. ∎

The second formulation concerns $H$-sets of the Smyth power space. It is the filter representation form of [5 Theorem 5.5].

Proof. For $A \in H(P_{S}(X))$, $K \in Q(X)$ and $U \in\mathcal{O}(X)$,

Since the sets $\Box U$ form a base, it follows that

This proves (1)$\Longleftrightarrow$(4). The filter $\Phi_{X}(A)$ is proper and Scott-open, and $\bigcap\Phi_{X}(A) = \bigcap A$, by Lemma 2.4. Lemma 2.2 therefore gives the remaining equivalences and uniqueness. The second identity in (5) gives the formulation using $H_{c}(P_{S}(X))$.

These two classes are related by the continuous map

Functoriality gives $\eta_{X}(A) \in H(P_{S}(X))$ for $A \in H(X)$, and $\Phi_{X}(\eta_{X}(A)) = \mathcal{F}_{A}$. Consequently,

In particular, super $H$-sobriety always implies $H$-sobriety. The class on the right also contains every $\Phi_{X}(K)$ with $K \in Q(X)$, by taking $A = \{K\}$.

Property $Q$ and countable restrictions

Property $Q$ was introduced in [5 Definition 5.10]. We use its local form from [4 Definition 6.2].

The following exact comparison is [4 Theorem 6.4]. We include a proof to exhibit how the two filter representations enter the argument.

Proof. Suppose that $P_S(X)$ is $H$-sober. Equation (9) and the two representation theorems show that $X$ is $H$-sober. Let $\mathcal{A}$ and $C$ be as in Definition 2.12. Theorem 2.11 gives $K = \bigcap\mathcal{A} \in Q(X)$ and $\Phi_X(\mathcal{A}) = \Phi_X(K)$. If $C \cap K = \varnothing$, then $X \setminus C \in\Phi_X(\mathcal{A})$, so some $G \in\mathcal{A}$ misses $C$, a contradiction. Choose $x \in C \cap K$. The closed set $B = \{x\} \in H_c(X)$ is contained in $C$ and meets every $G \in\mathcal{A}$, proving property $Q$ for $X$.

Conversely, suppose that $X$ is $H$-sober and $H$ has property $Q$ for $X$. Let $\mathcal{A} \in H(P_S(X))$ and suppose that $\bigcap\mathcal{A} \subseteq U$ for an open set $U$. If no $G \in\mathcal{A}$ is contained in $U$, the closed set $C = X \setminus U$ meets every $G \in\mathcal{A}$. Property $Q$ gives $B \in H_c(X)$, contained in $C$, that still meets every $G$. By $H$-sobriety, $B = \{x\}$ for some $x \in X$. Each saturated $G$ meeting $\{x\}$ contains $x$. Hence $x \in C \cap\bigcap\mathcal{A}$, a contradiction. We have proved $\Phi_X(\bigcap\mathcal{A}) \subseteq\Phi_X(\mathcal{A})$; the opposite inclusion is automatic. Apply Theorem 2.11.

Definition 2.14 (Countable restriction). For an irreducible subset system $H$, define

This is again an irreducible subset system, since direct images preserve countability. Its closed-set family is

In particular, writing $\mathcal{R}(X)$ for all nonempty irreducible subsets of $X$, its countable restriction is

Thus $(H^\omega)_c(X)$ consists of closures of countable $H$-sets; its members need not themselves be countable. In particular,

Applying Theorems 2.10 and 2.11 to $H^\omega$ gives the two countable $H$-versions: the point-induced filters characterize $H^\omega$-sobriety of $X$, and the Smyth-induced filters characterize $H^\omega$-sobriety of $P_S(X)$. Theorem 2.13 compares them when $H^\omega$ has property $Q$ for $X$. For $\mathcal{R}^\omega$ this property will be proved in Lemma 3.5.

For a countable family $\mathcal{A} \subseteq P_S(X)$, the countability condition concerns the number of compact saturated members of $\mathcal{A}$. Each such member may be uncountable, and the induced Scott-open filter need not have a countable open base.

An enlarged class of Scott-open filters

Definition 3.1. A proper filter $\mathcal{F} \subseteq\mathcal{O}(X)$ is countably generated, or countably based, if there are $U_n \in\mathcal{F}$, $n \in\mathbb{N}$, such that

The base may equivalently be required to be a countable filtered family of nonempty open sets. Indeed, finite intersections give a decreasing base; conversely, from a countable filtered open base one can choose a decreasing cofinal sequence of its members. Thus we may assume $U_{n+1} \subseteq U_n$. These sets form a base of the filter, not necessarily a base of the topology of $X$.

Let $\operatorname{OFilt}_{\mathrm{cb}}(X)$ denote the family of proper Scott-open filters on $\mathcal{O}(X)$ having a countable open base, where the subscript “cb” stands for “countably based”. We define the enlarged class

Every member of $\mathcal{F}_\omega(X)$ is a proper Scott-open filter. Members of the second subfamily need not have countable open bases.

By Theorem 2.10, applied to $H = \mathbb{R}^{\omega}$, $\omega$-sobriety is already equivalent to compact saturated representation of all filters $\mathcal{F}_{A}$ induced by countable irreducible subsets $A$ of $X$. The next theorem shows that adjoining $\operatorname{OFilt}_{\mathrm{cb}}(X)$ to this test class preserves the equivalence. Its additional assertion is that $\omega$-sobriety also represents every proper Scott-open filter having a countable open base. In general, the filters $\mathcal{F}_{A}$ cannot be omitted: representation for $\operatorname{OFilt}_{\mathrm{cb}}(X)$ alone is strictly weaker, as Remark 3.4 and Example 5.3 explain.

Proof. We first prove (1)$\Rightarrow$(4). Suppose that $X$ is $\omega$-sober and $\mathcal{F} \in\operatorname{OFilt}_{\mathrm{cb}}(X)$, with a countable open base $\{U_{n}: n \in\mathbb{N}\}$. Let $W \in\mathcal{O}(X) \setminus\mathcal{F}$. Every member of $\mathcal{F}$ meets $C = X \setminus W$: otherwise it would be contained in $W$, forcing $W \in\mathcal{F}$. Apply Lemma 2.3 to obtain a minimal irreducible closed set $A \subseteq C$ meeting every member of $\mathcal{F}$.

Choose $x_{n} \in A \cap U_{n}$ and put $D = \{x_{n}: n \in\mathbb{N}\}$. The closed set $\overline{D} \subseteq A$ still meets every member of $\mathcal{F}$, because every member contains some $U_{n}$. Minimality gives $A = \overline{D}$. Thus $A$ is an irreducible closed set with a countable dense subset. By $\omega$-sobriety, $A = \{x\}$ for some $x \in X$. Every $U \in\mathcal{F}$ meets $A$, so it contains $x$. Since $x \in A \subseteq C$, $x \in\bigcap\mathcal{F} \setminus W$. Consequently an open set contains $\bigcap\mathcal{F}$ if and only if it belongs to $\mathcal{F}$. Lemma 2.2 supplies the required nonempty compact saturated representing set.

For the other subfamily in (3.2), let $A$ be countable and irreducible. By $\omega$-sobriety, $\overline{A} = \{x\}$ for some $x \in X$. Equation (2) gives $\mathcal{F}_{A} = \Phi_{X}(\uparrow x)$, proving (4) for every member of $\mathcal{F}_{\omega}(X)$.

The implication (4)$\Rightarrow$(2) follows from $K = \bigcap\Phi_{X}(K)$ for saturated $K$. The implication (2)$\Rightarrow$(3) is Lemma 2.2, and (3)$\Rightarrow$(4) is immediate.

Finally, (2) applied to the filters $\mathcal{F}_{A}$ induced by countable irreducible subsets is condition (2) of Theorem 2.10 for $H = \mathbb{R}^{\omega}$. That theorem gives $\omega$-sobriety and hence (1). Uniqueness of the representing set follows from saturation, and (2.2) gives its stated form for $\mathcal{F}_{A}$. □

A subset $K$ has countable outer character if its open-neighbourhood filter has a countable open base. Let $Q_{\mathrm{co}}(X)$ consist of the members of $Q(X)$ having this property.

Proof. Theorem 3.2 gives the bijection and its inverse. For saturated $K,L$, the equivalence $K \supseteq L \Longleftrightarrow\Phi_{X}(K) \subseteq\Phi_{X}(L)$ follows by intersecting their open neighbourhoods. □

Remark 3.4. The restriction to $Q_{\mathrm{co}}(X)$ is necessary: a compact saturated set need not have a countable open-neighbourhood base.

The enlargement in (12) is necessary in general for the converse in Theorem 3.2. On an uncountable set $X$ with the cofinite topology, every proper countably generated filter has the form

and $X \setminus E$ is nonempty, compact and saturated. Consequently, all of conditions (2)–(4) in Theorem 3.2 hold when their test class is restricted to $\operatorname{OFilt}_{\mathrm{cb}}(X)$. Nevertheless, $X$ is not $\omega$-sober. For every countably infinite $D \subseteq X$, the filter $\mathcal{F}_D = \mathcal{O}(X) \setminus\{\varnothing\}$ has empty intersection and no countable open base. It belongs to $\mathcal{F}_{\omega}(X) \setminus\operatorname{OFilt}_{\mathrm{cb}}(X)$ and fails the representation condition. The details are given in Example 5.3.

If $X$ is first-countable, however, then $\mathcal{F}_{\omega}(X) = \operatorname{OFilt}_{\mathrm{cb}}(X)$. Indeed, for a countable irreducible subset $A \subseteq X$, choose a countable base $\mathcal{B}_a$ of open neighbourhoods at each $a \in A$. The countable family $\bigcup_{a\in A}\mathcal{B}_a$ is an open base of $\mathcal{F}_A$. Thus, within the class of first-countable $T_0$-spaces, representation for $\operatorname{OFilt}_{\mathrm{cb}}(X)$ alone does characterize $\omega$-sobriety.

We retain the usual meaning of countably generated; the whole class $\mathcal{F}_{\omega}(X)$ need not consist of countably generated filters.

The countable irreducible subset system

We now connect the enlarged-filter theorem with both general $H$-versions. The following countable selection argument is the essential step.

Proof. Let $X$ be a $T_0$-space, $A = \{K_n : n \in\mathbb{N}\} \in\mathcal{R}^{\omega}(\mathcal{P}_S(X))$, and let $C$ be a closed set meeting every $K_n$. For a finite family, we allow repetitions in this enumeration. By Lemma 2.4,

is a proper Scott-open filter. A closed set $B$ meets every member of $\mathcal{F}$ if and only if it meets every $K_n$. Indeed, meeting each $K_n$ suffices, since each member of $\mathcal{F}$ contains some $K_n$. Conversely, if $B$ misses $K_n$, then $X \setminus B \in\Phi_X(K_n) \subseteq\mathcal{F}$ witnesses failure to meet every member of $\mathcal{F}$.

Apply Lemma 2.3 to obtain a minimal irreducible closed set $B \subseteq C$ meeting every $K_n$. Choose $x_n \in B \cap K_n$ and put $D = \{x_n : n \in\mathbb{N}\}$. The closed set $\overline{D} \subseteq B$ still meets every $K_n$, so minimality gives $B = \overline{D}$. Since $B$ is irreducible, $D$ is irreducible. Thus $D \in\mathcal{R}^{\omega}(X)$ and $B \in(\mathcal{R}^{\omega})_c(X)$, as required by Definition 2.12. $\square$

The lemma requires the family $A$ to be countable and irreducible in $\mathcal{P}_S(X)$; it need not be filtered under inclusion. It proves that the closed witness has a countable dense irreducible subset, without requiring the closed witness itself to be countable.

These conditions are also equivalent to the conditions of Theorem 3.2. All representing compact saturated sets are unique: in (2) the set is $\hat{x}$, where $\hat{A} = \{\hat{x}\}$, and in (6) it is $\bigcap A$.

In (2), one may instead test irreducible closed subsets of $X$ having a countable dense subset. In (4)–(6), one may instead test irreducible closed subsets of $\mathcal{P}_S(X)$ having a countable dense subfamily.

Proof. Equation (10) and Theorem 2.10 give (1)$\Leftrightarrow$(2). Lemma 3.5 and Theorem 2.13, applied to $H = \mathcal{R}^{\omega}$, give (1)$\Leftrightarrow$(3). Theorem 2.11 gives the equivalence of (3)–(6) and identifies the representing set in (6). Theorem 2.10 identifies the set in (2). The closed-set formulations follow from the descriptions of $(\mathcal{R}^{\omega})_c(X)$ and $(\mathcal{R}^{\omega})_c(\mathcal{P}_S(X))$. $\square$

Traces on countable subspaces

A further equivalent formulation uses filters pulled back from countable subspaces. For $D \subseteq X$ let

where $D$ carries the subspace topology. This map preserves finite intersections and arbitrary unions. Hence, if $\mathcal{G}$ is a proper Scott-open filter on $\mathcal{O}(D)$, then

is a proper Scott-open filter on $\mathcal{O}(X)$. Its membership is determined by the traces of open sets on $D$. For irreducible $D$, the completely prime filter $\mathcal{F}_D$ from (1) is the pullback of $\mathcal{O}(D)\setminus\{\varnothing\}$.

Proof. Assume (1), and fix a countable $D$, a proper Scott-open $\mathcal{G}$, and $\mathcal{F}=\rho_D^{-1}(\mathcal{G})$. Let $C$ be a closed set meeting every member of $\mathcal{F}$. By Lemma 2.3, it contains a minimal irreducible closed set $A$ with that property.

We claim that $A\cap D$ still meets every member of $\mathcal{F}$. Otherwise, for some $U\in\mathcal{F}$ we would have

The left side belongs to $\mathcal{G}$, so upperness of $\mathcal{G}$ would give $X\setminus A\in\mathcal{F}$, a contradiction. Thus $\overline{A\cap D}\subseteq A$ is a closed set meeting every member of $\mathcal{F}$. By minimality, $A=\overline{A\cap D}$. The set $A\cap D$ is countable. Consequently $A=\{x\}$ for some $x\in X$, and $x\in C\cap\bigcap\mathcal{F}$.

We have proved the closed-set intersection property for $\mathcal{F}$ and every closed $C$. Its contrapositive, applied to complements of open sets, gives $\mathcal{F}=\Phi_X(\bigcap\mathcal{F})$. Lemma 2.2 proves (2).

The implication (2)$\Rightarrow$(3) follows by saturation. For (3)$\Rightarrow$(4), take a countable irreducible subspace $D$ and $\mathcal{G}=\mathcal{O}(D)\setminus\{\varnothing\}$. This is a proper completely prime filter, and its inverse image is $\mathcal{F}_D$.

Assume (4). The open set $X\setminus\overline{D}$ does not belong to $\mathcal{F}_D$, so it cannot contain $\bigcap\mathcal{F}_D$. This proves (5).

Finally, assume (5) and choose $x\in\overline{D}\cap\bigcap\mathcal{F}_D$. Every open set meeting $D$ contains $x$. If $y\in\overline{D}$, each open neighbourhood of $y$ meets $D$ and therefore contains $x$. Hence $y\in\{x\}$, proving $\overline{D}\subseteq\{x\}$. The reverse inclusion follows from $x\in\overline{D}$. Uniqueness follows from $T_0$. Moreover, $\mathcal{F}_D=\{U:x\in U\}=\Phi_X(\uparrow x)$.

Remark 4.2

The filters in Theorem 4.1 are not required to have countable open bases. For example,

is a countable union of point-neighbourhood filters when $D$ is countable. This does not make it a countably generated filter on $\mathcal{O}(X)$. The filter of all nonempty opens in Example 5.3 is an explicit instance. The completely prime formulation (4) is the $H=\mathbb{R}^{\omega}$ case of the general approach in [4], Theorems 3.6 and 5.7.

$\omega$-well-filteredness and strictness

We now compare $\omega$-sobriety with the filtered-family and open-generator representation properties from § 2. Write $\mathcal{D}(X)$ for the nonempty directed subsets of $X$ in the specialization order, and let $\mathcal{D}^{\omega}$ be its countable restriction. Both are irreducible subset systems. The $\mathcal{D}$-sobriety condition is the usual $d$-space property. Its countable version, the $\omega$-$d$-space property, says that the closure of every countable directed set is a point closure [5, 6].

Since the specialization order of $P_S(X)$ is reverse inclusion, its directed subsets are exactly the filtered families in $Q(X)$. Thus Proposition 2.6 is the specialization of Theorem 2.11 to $\mathcal{D}$ and $\mathcal{D}^{\omega}$, respectively. Together with Theorems 2.5 and 3.6, this gives Table 1.

$H$	Point-induced version ( $X$ is $H$-sober)	Smyth version ( $P_S(X)$ is $H$-sober)
$\mathcal{R}$	Sobriety	Sobriety
$\mathcal{D}$	$d$-space property	Well-filteredness
$\mathcal{D}^{\omega}$	$\omega$-$d$-space property	$\omega$-well-filteredness
$\mathcal{R}^{\omega}$	$\omega$-sobriety	$\omega$-sobriety

The distinction between $\omega$-well-filteredness and $\omega$-sobriety is therefore the distinction between testing countable filtered compact families and testing all countable irreducible subsets of $P_S(X)$.

Proposition 5.1. Every $\omega$-sober space is $\omega$-well-filtered.

Proof. Every nonempty countable filtered family in $Q(X)$ is a countable irreducible subset of $P_S(X)$. If $X$ is $\omega$-sober, Theorem 3.6(4) gives the recovery equality for each such family. Proposition 2.6(2) now yields $\omega$-well-filteredness.

With $H_{\mathrm{co}}$ as in Definition 2.7, Theorem 3.2 and Proposition 5.1, together with the known representation theorem under $\omega$-well-filteredness [1 Theorem 1 and subsequent remark], give

By Theorem 3.2, the first property is equivalent to the representation property for all filters in $\mathcal{F}_{\omega}(X)$. Both implications in (14) are strict, as the following two examples show.

Example 5.2 (Well-filtered but not $\omega$-sober). Let $\mathcal{U}$ be a free ultrafilter on $\mathbb{N}$ and equip $X = \mathbb{N}$ with

This is a topology. Every cofinite set belongs to $\mathcal{U}$, so $X$ is $T_1$. Two nonempty open sets have nonempty intersection, so $X$ is an irreducible closed set. It is countable and has no generic point, because point closures are singletons. Thus $X$ is not $\omega$-sober.

We prove that its compact subsets are exactly the finite subsets. Suppose $K \subseteq\mathbb{N}$ is infinite. Partition $K$ into two infinite pieces; at least one piece, say $B$, does not belong to $\mathcal{U}$. Enumerate $B = \{b_1,b_2,\ldots\}$ and put

Each $V_n$ is open, since $\mathbb{N} \setminus B \in\mathcal{U}$. These sets increase to $X$, but no finite subfamily covers $K$. Hence $K$ is not compact.

Since $X$ is $T_1$, its nonempty compact saturated sets are exactly the nonempty finite sets. In any filtered family of such sets, choose a member $K_0$ of minimal cardinality. A common refinement of $K_0$ and any other member must equal $K_0$. Thus $K_0$ is contained in every member, and the intersection of the family is $K_0$. This proves well-filteredness, and hence $\omega$-well-filteredness. The example uses the usual existence of a free ultrafilter in ZFC.

Remark 5.4 (Strict weakening of sobriety). For completeness, $\omega$-sobriety is itself weaker than sobriety. On an uncountable set with the countable topology, each countable subspace is closed and discrete. Consequently, every countable irreducible subset is a singleton, so the space is $\omega$-sober. However, the whole space is irreducible and has no generic point. It is therefore not sober.

Scope of the countable version

The two natural classes of filters induced by an irreducible subset system $H$ encode $H$-sobriety of $X$ and of $P_S(X)$. Their relation is given by the local property-$Q$ criterion [4 Theorem 6.4]. The point-induced formulation belongs to the framework of [4 Theorems 3.6 and 5.7], and the Smyth formulation is the filter version of [5 Theorem 5.5]. Theorem 3.6 is the countable specialization obtained by proving that $\mathcal{R}^{\omega}$ has property $Q$. In particular, the $\omega$-sober spaces considered here are exactly the $\mathcal{R}^{\omega}$-sober spaces.

Theorem 3.2 incorporates the proper Scott-open filters with countable open bases into an exact representation characterization, by adjoining the filters induced by countable irreducible subsets. Theorem 4.1 supplies the equivalent countable-trace formulation. The Smyth formulation tests countable irreducible families of compact saturated sets; restricting those families to filtered ones gives $\omega$-well-filteredness.

The examples distinguish these conditions from the open-generator property $H_{\mathrm{co}}$. The implication $\omega$-well-filteredness $\Rightarrow H_{\mathrm{co}}$ is the open-generator case of [1], and both implications in (14) are strict. These distinctions keep the different roles of countability explicit throughout the representation theorems.

Declaration of generative AI and AI-assisted technologies in the manuscript preparation process

During the revision of this manuscript, ChatGPT (OpenAI) was used to assist with checking mathematical arguments, clarifying definitions and proofs, organizing the exposition, and language editing. The authors take responsibility for the content of the manuscript.