#### Abstract

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 Szymański [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 .*

#### Acknowledgment

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