Topological spaces with the Freese--Nation property II
Judyta B\k{a}k, Andrzej Kucharski

TL;DR
This paper investigates topological spaces with the FNS and π-FNS properties, establishing their connections to openly generated and skeletally generated spaces, respectively, through properties of bases and generation methods.
Contribution
It demonstrates that compact spaces with the FNS property are openly generated, and those with the π-FNS property are skeletally generated, linking these properties to space generation concepts.
Findings
Compact FNS spaces are openly generated.
π-FNS spaces are skeletally generated.
Connections between FNS properties and space generation methods.
Abstract
The aim of this paper is to study the class of spaces with the FNS property and property. We shown that compact spaces with the FNS property for some base consisting of cozero-sets are openly generated spaces and spaces with the property are skeletally generated spaces.
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Topological spaces with the Freese–Nation property II
Judyta Bąk
Judyta Bąk
Institute of Mathematics, Jan Kochanowski University, Świętokrzyska 15, 25-406 Kielce and Institute of Mathematics, University of Silesia
Bankowa 14, 40-007 Katowice
Poland
and
Andrzej Kucharski
Andrzej Kucharski
Institute of Mathematics, University of Silesia in Katowice
Bankowa 14, 40-007 Katowice
Poland
Abstract.
The aim of this paper is to study the class of spaces with the FNS property and property. We shown that compact spaces with the FNS property for some base consisting of cozero-sets are openly generated spaces and spaces with the property are skeletally generated spaces.
Key words and phrases:
FNS property, inverse limit, openly generated space, skeletally generated space
2000 Mathematics Subject Classification:
Primary: 54G20, 91A44; Secondary: 54F99
1. Introduction
This paper is a continuation of the Freese-Nation property research introduced in the paper [1] for topological spaces. The Freese-Nation property was introduced by R. Freese and J.B. Nation [5]. In [1] it is shown that spaces with the FNS property satisfy ccc and any product of such spaces also satisfies ccc. All metrizable spaces have the FN property. L. Heindorf and L.B. Shapiro [9] showed that a family of all clopen sets of 0-dimensional compact space has the FNS property if and only if is openly generated. The concept of openly generated spaces was introduced by E.V. Shchepin in [20] and developed in [21] and [22]. The FNS property and some versions of it for compact spaces were studied in [9], [6], [7], [16], [17].
We say that a family of open sets has * property* if there exists a map such that if are disjoint then there are disjoint sets such that .
We say that a topological space has * property* if there exists a base which has property. We say that a space has * property* if there exists a -base which has property.
A topological space has the property if there exists a base such that for every there are two finite sets and such that if , then .
A regular space with countable weight has the property and a regular space with countable -weight has the property. We estabilish in Section 2 that every compact Hausdorff space with the property for some base consisting of cozero-sets is openly generated. We prove that Player I in open-open game has winning strategy in every space with the property. It follows from the result [11] that every compact Hausdorff space with the property is skeletally generated. In Section 3 we show that developable spaces have the FN property. We indicate examples of base without property. Section 4 contains the proof that a space which is co-absolute to a space with the property has the property.
2. Openly generated spaced and skeletally generared spaces
We say that a compact space is openly generated if it is homeomorphic to , where
- (1)
is a compact metrizable space for , 2. (2)
is an open surjection, 3. (3)
is –complete, i.e. for any chain there exists , 4. (4)
the inverse system is continuous, i.e. .
We say that a map is skeletal [15] if for every nonempty open set .
Similarly as openly generated space we define skeletally generated space, by replacing open surjections with skeletal surjections. Skeletally generated spaces were introduced in [24].
Let and be a topological spaces. We say that a function is d-open [23] if for every open set .
Lemma 1** ([23]).**
Let and for every set , where is a base of regular space , a map is closed, then is open.
Lemma 2** ([12]).**
For any map the following conditions are equivalent:
- (1)
f is d-open,
- (2)
there exists a base of a space such that a family satisfies the following condition: for every family and a point , there is an open neighborhood of such that ,
Let be a family of open subsets of a space . We introduce an equivalence relation on a space in the following way
[TABLE]
Let be a set of all equivalence classes with the quotient topology and let denote a quotient map (see [4]).
Following [10] (see also [2]) let us call a family of open subsets of a spaces *completely regular *if for each there exist two sequences and in such that
[TABLE]
J. Kerstan has proved the following characterization of complete regularity.
Theorem** ([10],(see also [2])).**
A -space is completely regular if and only if it has a completely regular base.
We say that a family of open subsets of a spaces *weakly completely regular * if for each and any there exist two sequences such that
[TABLE]
If a space has a completely regular family then not so hard to see that closed under the finite intersection and unions of elements of is completely regular family. Then is weakly completely regular family. Next theorem is similar to Theorem 2.16. [14] but we assume only that family is a countable and weakly completely regular.
Theorem 1**.**
If is a countable and weakly completely regular family of then is metrizable.
Proof.
We will show that a map is continuous. First we check that is a base of the space . To show that take any and . We are not losing generality assuming that Since is weakly completely regular there is a sequence such that Hence we get . This proves that . Now we shall prove that
[TABLE]
Take any Clearly . Take an arbitrary , then there is such that . By the definition of the relation we get The same reasoning shows that so we proved that . Since is the weakly completely regular family there is a sequence such that Therefore there exists such that . This proves that the family is the base of the space .
The map is continuous, because for all . To show that the space is Hausdorff space, take There exists such that and . Since is the weakly completely regular family there are sequences which satisfy condition for the set There is such that and . Therefore there are neighborhoods of the point and of the point . Since we have and .
We shall prove that is regular. Take any closed set and a point . Since is a base of the space there is such that and By the property of there are sequences which satisfy condition for the set So there is such that and Since the space is regular. By the Urysohn metrization theorem is metrizable. ∎
We get a similar characterization of complete regularity by the property obtained by J. Kerstan [10] ( see also [2]).
Corollary 1**.**
A -space is completely regular if and only if it has a weakly completely regular base.
Proof.
Clearly the family of all cozero-set of a space is weakly comletely regular. Let be a weakly completely regular base for a space . To show that is completely regular take any . One can assume that there is a countable subfamily such that is weakly completely regular and . By Theorem 1 is metrizable. Since for all and is an open set in the metrizable space , is a cozero-set, this completes the proof. ∎
Theorem 2**.**
Every compact Hausdorff space with the property for some base consisting of cozero-sets is openly generated.
Proof.
Let be a base consisting of cozero-sets of a space which has the property. For every countable family there exists a countable weakly completely regular family such that and for all Indeed, if , then and is set. Therefore there exists a continuous function such that
[TABLE]
[TABLE]
Let , then for . Since the space is regular, for every there exists such that . Set is compact for , hence . Let Similarly for we get such that , then . For every we have , hence .
By Theorem 1 the space is metrizable. We shall show that is an open map. The map is closed map as a continuous map from a compact space to a Hausdorff space. According to the lemma 1 it is sufficient to show that a map is d-open. By the lemma 2 a map is d-open if and only if a family satisfies the condition (2) of this lemma. Take a family and There exists such that and . Put Since the base has the FNS property there are sets such that , and for every . Therefore . Since is the weakly completely regular family there is a sequence such that Since there is and such that . Hence this proves that the map is d-open.
Put
[TABLE]
The family ordered by inclusion is directed by the first part of the proof. If then a map is an open surjection onto the compact metrizable space If , where , then one can define a map such that . Clearly is open. If is an increasing chain in then the space , is homeomorphic to where . Thus is a -complete inverse system, all spaces are compact and metrizable and all bonding maps are open. A map given by the formula
[TABLE]
is the homeomorphism. ∎
Theorem 3**.**
A family of all regular open sets in regular infinite space does not have the property.
Proof.
Suppose that a regular infinite space has a map witnessing the property for Consider the Stone space of the Boole’a algebra with a topology generated by a base
[TABLE]
where
[TABLE]
Let be given by the formula
[TABLE]
Since for all the family has the property.
is the compact Hausdorff space with the property. By Theorem 2 is openly generated and where is a -complete inverse system, all spaces are compact and metrizable and all bonding maps are open. Since the space is extremally disconnected and each is an open map, every is extremally disconnected. The space is extremally disconnected and compact metric, therefore it has to be finite.
Since is the infinite regular space, For each there exists such that and is a chain. Since is the -complete inverse system where a contradiction. ∎
Corollary 2**.**
A topology of regular infinite space does not have the property.
Proof.
Assume that the topology of regular infinite space has the property. From the Proposition 2.2 [1] it follows that the family of all regular open sets has the property, a contardiction with Proposition 3. ∎
From the above proposition follows that has the property and does not have the property.
Example 1**.**
Suppose that there exists a base of with the property. Then is a base consisting of clopen sets. It’s easy to see that has the property. By Theorem 2 there exists a -complete inverse system such that all spaces are compact and metrizable and all maps are open and a contradiction because is extremally disconnected (compare the proof of Theorem 3). Since has a countable base closed under the complement it easy to define operator which is witness on the property (compare [1, Proposition 5]).
Corollary 3**.**
If is a regular infinite space, then the base does not have the property.
Now we shall prove that if is a regular infinite space, then the bases and the topology of do not have the FN property.
Consider a cardinal number with the discrete topology. It easy to see that a base has the property. One can check that for all have the desired properties. Next theorem gvies a negative answer to the question: Does the base ( topology) have the property, whenever there exists some base with property?
Theorem 4**.**
A family of all regular open sets in regular infinite space does not have the property.
Proof.
Let be a regular infinite space. Suppose that operators are witnesses on the property for a family of all regular open sets Since the space is regular and infinite there is an infinite maximal family of pairwise disjoint set. We may define by a straight forward recursion a sequence and a sequence of finite sets such that
- (1)
and for all 2. (2)
for all 3. (3)
or for all and 4. (4)
is finite for all 5. (5)
If then , for each and .
Suppose that we have just defined and with the property for . For each there is such that and Therefore there is a finite family such that and , whenever Finaly we get and which satisfy properties Let and For every we get
[TABLE]
Hence for every by property , a contradiction. ∎
Corollary 4**.**
A topology of regular infinite space does not have the property.
Proof.
Assume that the topology of regular infinite space has the property. Suppose that operators are witnesses on the property for a family of all open sets. Put
[TABLE]
for every The operators are witnesses on the property for a family of all regular open sets, a contradition with Theorem 4. ∎
In the paper [1] we have proved that if there exists a base closed under the finite intersections with the FNS property, then one can enlarge the base to a base with FNS that contains both families and . Now we can strengthen the mentioned result.
Theorem 5**.**
If a 0-dimensional compact space has the property for some base consisting of cozero-sets, then has the property.
Proof.
By Theorem 2 is openly genereted space, i.e. is a -complete inverse system, all spaces are compact and metrizable and all bonding maps are open and . We may assume without loss of generality that are 0-dimensional. Applying Heindorf and Shapiro’s result [9, Theorem 2.2.3] ( see also [1, Theorem 21]) we get an operator which witnesses the FNS property for the family . ∎
The open-open game was introduced by P. Daniels, K. Kunen and H. Zhou [3] for an arbitrary topological space . Two players take countably many turns. A round consists of player I choosing a nonempty set and player II choosing a nonempty with . Player I wins if the union of II’s open sets is dense in , otherwise player II wins. Denote this game by . Consider the following game on a topological space . At the -th round Player I chooses a finite family of open non-empty open sets. Then Player II chooses a finite family of non-empty open subsets of such that for each there exists with . Player I wins if the union of II’s open sets is dense in , otherwise player II wins. It’s well known that the game is equvalent to the game (see [3]). We say that Player I has a winning strategy in the game whenever there exists a function
[TABLE]
where all families and are finite and consists of non-empty open sets such that for each game
[TABLE]
the union is dense in . Assume that is an base, then I Player has a winning strategy in the game if and only if there is a winning strategy defined on base ( see [3]).
Theorem 6**.**
In every compact Hausdorff space with the property I Player has a winning strategy in open-open game.
Proof.
Assume that is a compact Hausdorff space there are base and which witnessing the property. We shall define a winning strategy for I player in the game . Take any and put . Assume that we have just defined
[TABLE]
where all families and are finite and consists of non-empty open sets such for each there exists with Let For each finite family with the non-empty intersection we choose such that Put
[TABLE]
Suppose that is not dense in . There is a set such that . For each there is such that and There is such that
[TABLE]
Hence there exists such that , a contradiction with ∎
Similar theorem to the Theorem 2 is true for spaces with property and skeletaly generated spaces.
Theorem 7**.**
Every compact Hausdorff space with the property is skeletally generated.
Proof.
Assume that is compact Hausdorff space with the property. By Theorem 6 I Player has a winning strategy. Using [11, Theorem 12] we get is skeletally generated. ∎
3. Developable spaces have the FN property
Let be a cover of a space and . Recall [4] that a set is called a star of the point with respect to the cover . A sequence of covers of a space is called a development whenever for every point and an open set such that there exists such that . A space which has a development is called developable.
A family of sets is called a point-finite if for every the set is finite. We say that a cover is a refinement of a cover if for every there exists such that .
Proposition 8**.**
A topological space having a development consisting of point-finite covers has property.
Proof.
Let be a a development consisting of point-finite covers of . Without loss of generality it can be assumed that there exists a development consisting of point-finite covers such that the cover is a refinement of the cover .
For a family of sets we shall denote by subfamily of consisting of all maximal elements, i.e. of set such that if then . Since each is point-finite cover is well defined and it is the point-finite cover. Two distinct elements in are incomparable by the inclusion. The sequence is the development. Indeed, take a point and a neighborhood of . Since is the development there is such that If then Therefore The family is a base for the space . For each put
[TABLE]
If , then and for or . If , then for there is such that . Since elements of are incomparable by the inclusion, we have . If , then . ∎
Since every metrizable space has a development consisting of point-finite covers [4, Theorem 5.4.8] we get the following
Corollary 5** ([1]).**
Metric spaces satisfy FN property.
4. property and absolutes
We will need a notion of a small image and some facts connected with it.
Let be a map from a space to a space . A small image of a set under a map is called a set .
Lemma 3**.**
Let be a map from to , then small image of a set under the map is given by .
Proof.
We will show that for every set .
Let , then , hence .
Let and , then . Hence f\big{(}f^{-1}(y)\cap(X\setminus U)\big{)}=\{y\}\cap f(X\setminus U)=\emptyset and . ∎
Lemma 4**.**
Let be an irreducible map and be open sets, then if and only if .
Proof.
Let and . Suppose that , then there exists such that , this is a contradiction.
Let . Suppose that , then we get \emptyset\neq X\setminus f\big{(}Z\setminus(U\cap V)\big{)}=f^{\#}(U\cap V)\subseteq f^{\#}(U)\cap f^{\#}(V) and this contradiction ends the proof. ∎
Lemma 5**.**
Let be a map from onto . For every set we have .
Proof.
Let us take any set and , then . Hence there exists and . ∎
We say that a continuous, closed map from onto is irreducible if is not the image of any proper closed subset of . The absolute of a space is the space which is mapped and irreducibly onto , and is such that any irreducible inverse image of the space is homeomorphic to it. We say that topological spaces are coabsolute if their absolutes are homeomorphic.
Lemma 6**.**
If is a –base for and is an irreducible and closed map, then is a –base for .
Proof.
We will show that for every open set there exists a set such that Let us take any nonempty, open set , then and it is closed. Hence and it is open, then there exists an open set such that . We get . Hence , and . ∎
Theorem 9**.**
Every space which is co-absolute to a space with the property has the property.
Proof.
Let a space be an absolute of the space and , be irreducible, onto maps. Let be a -base for which has property and witnesses this property.
We will show that the family is a -base for . Let be an open set in . By Lemma 6 there exists a set such that . Hence by Lemma 5 .
We will show that the family has property. We define a map by the formula
[TABLE]
Let for some . By Lemma 4 it has to be , then there exist disjoint sets . We have , and . By Lemma 4 we have . This ends the proof. ∎
Dugundji spaces were introduced in [18]. Skeletally Dugundji spaces are skeletal analogue of Dugundji spaces, see [13]. We say that a compact space is Dugundji (skeletally Dugundji) whenever can be represented as the limit space of a continuous inverse system such that is a compact metrizable with open bonding (skeletal) maps and each map has a metrizable kernel, see [8]. L. Shapiro [19] proved that every skeletally Dugundji spaces are co-absolute to a 0-dimensional Dugundji spaces. By Theorem 2.2.3 [9] ( see also Theorem 21 [1]) and Theorem 9 every 0-dimensional skeletally Dugundji space has the property.
Question 1**.**
Do openly generated spaces have the property?
Question 2**.**
Do skeletally generated spaces have the property?
The reference list from the paper itself. Each links out to its DOI / PubMed record.
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- 4[4] R. Engelking, General topology , Heldermann Verlag, Berlin (1989).
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