The aim of this paper is to investigate the open-open game of uncountable length. We introduce a cardinal number , which says how long the Player I has to play to ensure a victory. It is proved that . We also introduce the class of topological spaces that can be represented as the inverse limit of -complete system with and skeletal bonding maps. It is shown that product of spaces which belong to also belongs to this class and whenever .

1. Introduction

The following game is due to Daniels et al. [1]: two players take turns playing on a topological space ; a round consists of Player I choosing a nonempty open set and Player II choosing a nonempty open set ; a round is played for each natural number. Player I wins the game if the union of open sets which have been chosen by Player II is dense in . This game is called the open-open game.

In this paper, we consider what happens if one drops restrictions on the length of games. If is an infinite cardinal and rounds are played for every ordinal number less than , then this modification is called the open-open game of length . The examination of such games is a continuation of [2–4]. A cardinal number is introduced such that . Topological spaces, which can be represented as an inverse limit of -complete system with and each is space and skeletal bonding map , are listed as the class . If , then . There exists a space with . The class is closed under any Cartesian product. In particular, the cellularity number of is equal whenever . This implies Theorem of Kurepa that , whenever . Undefined notions and symbols are used in accordance with books [5–7]. For example, if is a cardinal number, then denotes the first cardinal greater than .

2. When Games Favor Player I

Let be a topological space. Denote by the family of all nonempty open sets of . For an ordinal number , let denote the set of all sequences of the length consisting of elements of . The space is called -favorable whenever there exists a function such that for each sequence with and , for each , the union is dense in . We may also say that the function is witness to -favorability of . In fact, is a winning strategy for Player I. For abbreviation we say that is -winning strategy. Sometimes we do not precisely define a strategy. Just give hints how a player should play. Note that, any winning strategy can be arbitrary on steps for limit ordinals.

A family of open non-empty subset is called a -base for if every non-empty open subset contains a member of . The smallest cardinal number , where is a -base for , is denoted by .

Proposition 2.1. Any topological space is -favorable.

Proof. Let be a -base. Put for any sequence . Each family of open non-empty sets is again a -base for . So, its union is dense in .

According to [6, p. 86] the cellularity of is denoted by . Let be the smallest cardinal number such that every family of pairwise disjoint open sets of has cardinality , compare [8]. Clearly, if is a limit cardinal, then . In all other cases, . Hence, . Let Proposition 2.1 implies . The next proposition gives two natural strategies and gives more accurate estimation than .

Proposition 2.2. .

Proof. Suppose . Fix a family of pairwise disjoint open sets. If Player II always chooses an open set, which meets at most one , then he will not lose the open-open game of the length , a contradiction.
Suppose sets are chosen by Player II. If the set is non-empty, then Player I choses it. Player I wins the open-open game of the length , when he will use this rule. This gives .

Note that, , where is the Cantor cube of weight . There exists a separable space which is not -favorable, see SzymaΕ„ski [9] or [1, p.207-208]. Hence we get

3. On Inverse Systems with Skeletal Bonding Maps

Recall that, a continuous surjection is skeletal if for any non-empty open sets the closure of has non-empty interior. If is a compact space and is a Hausdorff space, then a continuous surjection is skeletal if and only if , for every non-empty and open , see Mioduszewski and Rudolf [10].

Lemma 3.1. A skeletal image of -favorable space is a -favorable space.

Proof. A proof follows by the same method as in [11, Theorem 4.1]. In fact, repeat and generalize the proof given in [4, Lemma 1].

According to [5], a directed set is said to be -complete if any chain of length consisting of its elements has the least upper bound in . An inverse system is said to be a -complete, whenever is -complete and for every chain , where , such that we get In addition, we assume that bonding maps are surjections.

For -favorability, the following lemma is given without proof in [1, Corollary 1.4]. We give a proof to convince the reader that additional assumptions on topology are unnecessary.

Lemma 3.2. If is dense, then is -favorable if and only if is -favorable.

Proof. Let a function be a witness to -favorability of . Put If Player II chooses open set , then put We get , since . Then we put Suppose we have already defined for . If Player II chooses open set , then put Finally, put and check that is witness to -favorability of .
Assume that is a witness to -favorability of . If and is open, then put . If Player II chooses open set , then . Put , where and is open. Suppose have been already defined for . If II Player chooses open set , then put , where open set is determined by .

The next theorem is similar to [12, Theorem 2]. We replace a continuous inverse system with indexing set being a cardinal, by -complete inverse system, and also is replaced by . Let be a fixed cardinal number.

Theorem 3.3. Let be a dense subset of the inverse limit of the -complete system , where . If all bonding maps are skeletal, then .

Proof. By Lemma 3.2, one can assume that . Fix functions , each one is a witness to -favorability of . This does not reduce the generality, because for every . In order to explain the induction, fix a bijection such that (1)if , then ; (2) if and only if ; (3) if and only if .
One can take as an isomorphism between and , with canonical well-ordering, see [7]. The function will indicate the strategy and sets that we have taken in the following induction.
We construct a function which will provide -favorability of . The first step is defined for . Take an arbitrary and put Assume that Player II chooses non-empty open set , where is open. Let and denote . So, after the first round and the next respond of Player I, we know: indexes and , the open set and the open set .
Suppose that sequences of open sets , indexes , and sets have been already defined such that.
If and , then where and are open.
If and , then take and put Since is -complete, one can assume that the sequence is increasing and .
We will prove that is dense in . Since for each and is skeletal map, it is sufficient to show that is dense in . Fix arbitrary open set where is an open set of . Since is winning strategy on , there exists such that , and . Therefore we get where . Indeed, suppose that . Then
Hence we have , a contradiction.

Corollary 3.4. If is dense subset of an inverse limit of -complete system , where all bonding map are skeletal, then

Proof. Let . Since , for every , we will show that
Suppose that . Using Proposition 2.2 and Theorem 3.3, check that So, we get . Therefore, there exists a family , of size , which consists of pairwise disjoint open subset of . We can assume that Since is -complete inverse system and , there exists such that a contradiction with .

The above corollary is similar to [12, Theorem 1], but we replaced a continuous inverse system, whose indexing set is a cardinal number by -complete inverse system.

4. Classes

Let be an infinite cardinal number. Consider inverse limits of -complete system with . Let be a class of such inverse limits with skeletal bonding maps and being -space. Now, we show that the class is stable under Cartesian products.

Theorem 4.1. The Cartesian product of spaces from belongs to .

Proof. Let where each . For each , let be a -complete inverse system with skeletal bonding map such that each -space has the weight . Consider the union Introduce a partial order on as follows: where is the partial order on . The set with the relation is upward directed and -complete.
If , then denotes the Cartesian product If , then put where is the projection of onto and is the Cartesian product of the bonding maps . We get the inverse system which is -complete, bonding maps are skeletal and . So, we can take .
Now, define a map by the following formula: where and and . By the property the map is well defined and it is injection.
The map is surjection. Indeed, let . For each and each we fix such that and . Let be a projection for each .
For each let define , where . We will prove that an element is a thread of the space . Indeed, if and , then take functions and . For abbreviation, denote and . Define a function in the following way: The function is element of and . Note that and . Since we get It is clear that .
We shall prove that the map is continuous. Take an open subset such that where is open subset. A map is projection from the inverse limit to . It is sufficient to show that where and and is the projection and . We have
Since the map is bijection and for any subbase subset , the map is open.

In the case we have well-known results that product of I-favorable space is I-favorable space (see [1] or [2]).

Corollary 4.2. Every I-favorable space is stable under any product.

If is a set and is cardinal number then we denote by .

The following result probably is known but we give a proof for the sake of completeness.

Theorem 4.3. Let be an infinite cardinal and let be a set such that . If and for all then there exists a set such that and and for every and every , where

Proof. Assume that is regular cardinal. Let and let for . Let . Assume that we have defined for such that . Put Calculate the size of the set :
Let , so we get . Fix a sequence and . Since there exists such that and for some .
In the second case , we proceed the above induction up to . Let , so we get and . Similarly to the first case we get that is closed under all function , .

Theorem 4.4. If belongs to the class then .

Proof. If then by Theorems 3.3 and Proposition 2.2 we get .

We apply some facts from the paper [3]. Let be a family of open subset of topological space and . We say that if and only if for every . The family of all sets we denote by . Define a map as follows . The set is equipped with topology generated by all images where .

Recall Lemma from paper [3]: if is a family of open set of and is closed under finite intersection then the mapping is continuous. Moreover if then the family is a base for the topology .

Notice that if has a property then and by [3, Lemma 3] the topology is Hausdorff. Moreover if is closed under finite intersection then by [3, Lemma 4] the topology is regular. Theorem 5 and Lemma 9 [3] yeild.

Theorem 4.5. If is a set of open subset of topological space such that (1)is closed under -winning strategy, finite union and intersection, (2)has property , then with topology is completely regular space and is skeletal.

If a topological space has the cardinal number then , but for equals for instance we get only .

Theorem 4.6. Each Tichonov space with can be dense embedded into inverse limit of a system , where all bonding map are skeletal, indexing set is complete each is Tichonov space with and

Proof. Let be a -base for topological space consisting of cozero sets and be a -winning strategy. We can define a function of finite intersection property and finite union property as follows: and . For each cozero set fix a continuous function such that . Put and . By Theorem 4.3 for each and all functions there is subset such that (1) , where (2) , (3) is closed under -winning strategy , function of finite intersection property and finite union property, (4) is closed under , , hence holds property .
Therefore by Theorem 4.5 we get skeletal mapping . Let be a set of families which satisfies above condition (1), (2), (3) and the (4). If is directed by inclusion. It is easy to check that is -complete. Similar to [3, Theorem 11] we define a function as follows , where and . If and , then . Thus is a thread, that is, . It easy to see that is homeomorphism onto its image and is dense in , compare [3, proof of Theorem 11].

Theorem 4.6 suggests question.

Does each space belong to ?

Fleissner [13] proved that there exists a space such that and . Hence, we get , by Theorem 3.3 and Corollary 4.2. Suppose that then , by Theorem 4.4, a contradiction.

Corollary 4.7. If is topological space with then and

Proof. By Theorem 4.3 we get . Hence by Theorems 4.4 and 4.1 we have .

By above Corollary we get the following.

Corollary 4.8 (see [14, Kurepa]). If is a family of topological spaces and for each , then .


The author thanks the referee for careful reading and valuable suggestions.