# On Open-Open Games of Uncountable Length

Authors: Andrzej Kucharski
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

The aim of this paper is to investigate the open-open
game of uncountable length. We introduce a cardinal numberμ(X), which says how long the Player I has to play to ensure a victory. It is proved that c(X)≤μ(X)≤c(X)+. We also introduce the class 𝒞κ of topological spaces that can be represented as the inverse
limit of κ-complete system {Xσ,πρσ,Σ} with w(Xσ)≤κ and
skeletal bonding maps. It is shown that product of spaces which
belong to 𝒞κ also belongs to this class and μ(X)≤κ whenever X∈𝒞κ.


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## Body

## 1. Introduction

The following game is due to Daniels et al. [1]: two players take turns playing on a topological space X; a round consists of Player I choosing a nonempty open set U⊆X and Player II choosing a nonempty open set V⊆U; 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 X. This game is called the open-open game.In this paper, we consider what happens if one drops restrictions on the length of games. Ifκ is an infinite cardinal and rounds are played for every ordinal number less than κ, then this modification is called the open-open game of lengthκ. The examination of such games is a continuation of [2–4]. A cardinal number μ(X) is introduced such that c(X)≤μ(X)≤c(X)+. Topological spaces, which can be represented as an inverse limit of κ-complete system {Xσ,πϱσ,Σ} with w(Xσ)≤κ and each Xσ is T0 space and skeletal bonding map πϱσ, are listed as the class 𝒞κ. If μ(X)=ω, then X∈Cω. There exists a space X with X∉Cμ(X). The class 𝒞κ is closed under any Cartesian product. In particular, the cellularity number of XI is equal κ whenever X∈𝒞κ. This implies Theorem of Kurepa that c(XI)≤2κ, whenever c(X)≤κ. Undefined notions and symbols are used in accordance with books [5–7]. For example, if κ is a cardinal number, then κ+ denotes the first cardinal greater than κ.


## 2. When Games Favor Player I

LetX be a topological space. Denote by 𝒯 the family of all nonempty open sets of X. For an ordinal number α, let 𝒯α denote the set of all sequences of the length α consisting of elements of 𝒯. The space X is called κ-favorable whenever there exists a function
(2.1)s:⋃{(T)α:α<κ}⟶T,
such that for each sequence {Bα+1:α<κ}⊆𝒯 with B1⊆s(∅) and Bα+1⊆s({Bγ+1:γ<α}), for each α<κ, the union ⋃‍{Bα+1:α<κ} is dense in X. We may also say that the function s is witness to κ-favorability of X. In fact, s is a winning strategy for Player I. For abbreviation we say that s is κ-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 familyℬ of open non-empty subset is called a π-base for X if every non-empty open subset U⊆X contains a member of ℬ. The smallest cardinal number |ℬ|, where ℬ is a π-base for X, is denoted by π(X).Proposition 2.1.
Any topological spaceX is π(X)-favorable.Proof.
Let{Uα:α<π(X)} be a π-base. Put s(f)=Uα for any sequence f∈𝒯α. Each family {Bγ:Bγ⊆Uγandγ<π(X)} of open non-empty sets is again a π-base for X. So, its union is dense in X.According to [6, p. 86] the cellularity of X is denoted by c(X). Let sat(X) be the smallest cardinal number κ such that every family of pairwise disjoint open sets of X has cardinality <κ, compare [8]. Clearly, if sat(X) is a limit cardinal, then sat(X)=c(X). In all other cases, sat(X)=c(X)+. Hence, c(X)≤sat(X)≤c(X)+. Let
(2.2)μ(X)=min⁡{κ:Xisaκ-favorableandκisacardinalnumber}.
Proposition 2.1 implies μ(X)≤π(X). The next proposition gives two natural strategies and gives more accurate estimation than c(X)≤μ(X)≤c+(X).Proposition 2.2.
c
(
X
)
≤
μ
(
X
)
≤
sat
(
X
).Proof.
Supposec(X)>μ(X). Fix a family {Uξ:ξ<μ(X)+} of pairwise disjoint open sets. If Player II always chooses an open set, which meets at most one Uξ, then he will not lose the open-open game of the length μ(X), a contradiction.
Suppose sets{Bγ+1:γ<α} are chosen by Player II. If the set
(2.3)X∖cl⁡⋃{Bγ+1:γ<α}
is non-empty, then Player I choses it. Player I wins the open-open game of the length sat(X), when he will use this rule. This gives μ(X)≤sat(X).Note that,ω0=c({0,1}κ)=μ({0,1}κ)≤sat({0,1}κ)=ω1, where {0,1}κ is the Cantor cube of weight κ. There exists a separable space X which is not ω0-favorable, see Szymański [9] or [1, p.207-208]. Hence we get
(2.4)ω0=c(X)<μ(X)=sat⁡(X)=ω1.


## 3. On Inverse Systems with Skeletal Bonding Maps

Recall that, a continuous surjection isskeletal if for any non-empty open sets U⊆X the closure of f[U] has non-empty interior. If X is a compact space and Y is a Hausdorff space, then a continuous surjection f:X→Y is skeletal if and only if Intf[U]≠∅, for every non-empty and open U⊆X, see Mioduszewski and Rudolf [10].Lemma 3.1.
A skeletal image ofκ-favorable space is a κ-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 Σ is said to be κ-complete if any chain of length ≤κ consisting of its elements has the least upper bound in Σ. An inverse system {Xσ,πϱσ,Σ} is said to be a κ-complete, whenever Σ is κ-complete and for every chain A⊆Σ, where |A|≤κ, such that σ=supA∈Σ we get
(3.1)Xσ=lim⁡←⁡{Xα,παβ,A}.
In addition, we assume that bonding maps are surjections.Forω-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.
IfY⊆X is dense, then X is κ-favorable if and only if Y is κ-favorable.Proof.
Let a functionσX be a witness to κ-favorability of X. Put
(3.2)σY(∅)=σX(∅)∩Y.
If Player II chooses open set V1∩Y⊆σY(∅), then put
(3.3)V1′=V1∩σX(∅)⊆σX(∅).
We get V1′∩Y=V1∩Y⊆σY(∅), since V1∩Y⊂σX(∅)∩Y. Then we put
(3.4)σY(V1∩Y)=σX(V1′)∩Y.
Suppose we have already defined
(3.5)σY({Vα+1∩Y:α<γ})=σX({Vα+1′:α<γ})∩Y,
for γ<β<κ. If Player II chooses open set Vβ+1∩Y⊆σY({Vα+1∩Y:α<β}), then put
(3.6)Vβ+1′=Vβ+1∩σX({Vα+1′:α<β})⊆σX({Vα+1′:α<β}).
Finally, put
(3.7)σY({Vα+1∩Y:α≤β})=σX({Vα+1′:α≤β})∩Y
and check that σY is witness to κ-favorability of Y.
Assume thatσY is a witness to κ-favorability of Y. If σY(∅)=U0∩Y and U0⊆X is open, then put σX(∅)=U0. If Player II chooses open set V1⊆σX(∅), then V1∩Y⊆σY(∅). Put σX(V1)=U1, where σY(V1∩Y)=U1∩Y and U1⊆X is open. Suppose
(3.8)σY({Vα+1∩Y:α<γ})=Uγ∩Y,σX({Vα+1:α<γ})=Uγ
have been already defined for γ<β<κ. If II Player chooses open set Vβ+1⊆σX({Vα+1:α<β}), then put σX({Vα+1:α<β+1})=Uβ+1, where open set Uβ+1⊆XX is determined by σY({Vα+1∩Y:α<β+1})=Uβ+1∩Y.The next theorem is similar to [12, Theorem 2]. We replace a continuous inverse system with indexing set being a cardinal, by κ-complete inverse system, and also c(X) is replaced by μ(X). Let κ be a fixed cardinal number.Theorem 3.3.
LetX be a dense subset of the inverse limit of the κ-complete system {Xσ,πϱσ,Σ}, where κ=sup{μ(Xσ):σ∈Σ}. If all bonding maps are skeletal, then μ(X)=κ.Proof.
By Lemma3.2, one can assume that X=lim←{Xσ,πϱσ,Σ}. Fix functions sσ:𝒯σ<κ→Tσ, each one is a witness to μ(Xσ)-favorability of Xσ. This does not reduce the generality, because μ(Xσ)≤κ for every σ∈Σ. In order to explain the induction, fix a bijection f:κ→κ×κ such that  (1)
iff(α)=(β,ζ), then β,ζ≤α;(2)
f-1(β,γ)<f-1(β,ζ) if and only if γ<ζ;(3)
f-1(γ,β)<f-1(ζ,β) if and only if γ<ζ.
One can take asf an isomorphism between κ and κ×κ, with canonical well-ordering, see [7]. The function f will indicate the strategy and sets that we have taken in the following induction.
We construct a functions:𝒯<κ→𝒯 which will provide κ-favorability of X. The first step is defined for f(0)=(0,0). Take an arbitrary σ1∈Σ and put
(3.9)s(∅)=πσ1-1(sσ1(∅)).
Assume that Player II chooses non-empty open set B1=πσ2-1(V1)⊆s(∅), where V1⊆Xσ2 is open. Let
(3.10)s({B1})=πσ1-1(sσ1({Intcl⁡πσ1(B1)∩sσ1(∅)}))
and denote D00=Intclπσ1(B1)∩sσ1(∅). So, after the first round and the next respond of Player I, we know: indexes σ1 and σ2, the open set B1⊆X and the open set D00⊆Xσ1.
Suppose that sequences of open sets{Bα+1⊆X:α<γ}, indexes {σα+1:α<γ}, and sets {Dζφ:f-1(φ,ζ)<γ} have been already defined such that.
Ifα<γ and f(α)=(φ,η), then
(3.11)Bα+1=πσα+2-1(Vα+1)⊆s({Bξ+1:ξ<α})=πσφ+1-1(sσφ+1({Dνφ:ν<η})),
where Dνφ=Intclπσφ+1(Bf-1((φ,ν))+1)∩sσφ+1({Dζφ:ζ<ν}) and Vα+1⊆Xσα+2 are open.
Iff(γ)=(θ,λ) and β<λ, then take
(3.12)Dβθ=Intcl⁡πσθ+1(Bf-1((θ,β))+1)∩sσθ+1({Dζθ:ζ<β})
and put
(3.13)s({Bα+1:α<γ})=πσθ+1-1(sσθ+1({Dαθ:α<λ})).
Since Σ is κ-complete, one can assume that the sequence {σα+1:α<κ} is increasing and σ=sup{σξ+1:ξ<κ}∈Σ.
We will prove that⋃α<κ‍Bα+1 is dense in X. Since πσ-1(πσ(Bα+1))=Bα+1 for each α<κ and πσ is skeletal map, it is sufficient to show that ⋃α<κ‍πσ(Bα+1) is dense in Xσ. Fix arbitrary open set (πσξ+1σ)-1(W) where W is an open set of Xξ+1. Since sσξ+1 is winning strategy on Xσξ+1, there exists Dαξ such that Dαξ∩W≠∅, and Dαξ⊆Intclπσξ+1(Bf-1((ξ,α))+1). Therefore we get
(3.14)(πσξ+1σ)-1(W)∩πσ(Bδ+1)≠∅,
where δ=f-1((ξ,α)). Indeed, suppose that (πσξ+1σ)-1(W)∩πσ(Bδ+1)=∅. Then
(3.15)∅=πσξ+1σ[(πσξ+1σ)-1(W)∩πσ(Bδ+1)]=W∩πσξ+1σ[πσ(Bδ+1)]=W∩πσξ+1(Bδ+1).
Hence we haveW∩Intclπσξ+1(Bδ+1)=∅, a contradiction.Corollary 3.4.
IfX is dense subset of an inverse limit of μ(X)-complete system {Xσ,πϱσ,Σ}, where all bonding map are skeletal, then
(3.16)c(X)=sup⁡{c(Xσ):σ∈Σ}.Proof.
LetX=lim←{Xσ,πϱσ,Σ}. Since c(X)≥c(Xσ), for every σ∈Σ, we will show that
(3.17)c(X)≤sup⁡{c(Xσ):σ∈Σ}.
Suppose thatsup{c(Xσ):σ∈Σ}=τ<c(X). Using Proposition 2.2 and Theorem 3.3, check that
(3.18)μ(X)=sup⁡{μ(Xσ):σ∈Σ}≤sup⁡{c(Xσ)+:σ∈Σ}≤τ+≤c(X).
So, we get μ(X)=c(X)=τ+. Therefore, there exists a family ℛ, of size τ+, which consists of pairwise disjoint open subset of X. We can assume that
(3.19)R⊆{πσ-1(U):UisanopensubsetofXσ,σ∈Σ}.
Since {Xσ,πϱσ,Σ} is μ(X)-complete inverse system and |ℛ|=μ(X), there exists β∈Σ such that
(3.20)R⊆{πβ-1(U):UisanopensubsetofXβ},
a contradiction with c(Xβ)<τ+.The above corollary is similar to [12, Theorem 1], but we replaced a continuous inverse system, whose indexing set is a cardinal number by κ-complete inverse system.


## 4. Classes𝒞κ

Letκ be an infinite cardinal number. Consider inverse limits of κ-complete system {Xσ,πϱσ,Σ} with w(Xσ)≤κ. Let 𝒞κ be a class of such inverse limits with skeletal bonding maps and Xσ being T0-space. Now, we show that the class 𝒞κ is stable under Cartesian products.Theorem 4.1.
The Cartesian product of spaces from𝒞κ belongs to 𝒞κ.Proof.
LetX=∏‍{Xs:s∈S} where each Xs∈𝒞κ. For each s∈S, let Xs=lim←{Xσ,sρσ,Σs} be a κ-complete inverse system with skeletal bonding map such that each T0-space Xσ has the weight ≤κ. Consider the union
(4.1)Γ=⋃‍{∏s∈AΣs:A∈[S]κ}.
Introduce a partial order on Γ as follows:
(4.2)f⪯g⟺dom⁡(f)⊆dom⁡(g),∀a∈dom⁡(f)f(a)≤ag(a),
where ≤a is the partial order on Σa. The set Γ with the relation ⪯ is upward directed and κ-complete.
Iff∈Γ, then Yf denotes the Cartesian product
(4.3)∏‍{Xf(a):a∈dom⁡(f)}.
If f⪯g, then put
(4.4)pfg=(∏a∈dom⁡(f)af(a)g(a))∘πdom⁡(f)dom⁡(g),
where πdom(f)dom(g) is the projection of ∏‍{Xg(a):a∈dom(g)} onto ∏‍{Xg(a):a∈dom(f)} and ∏a∈dom(f)‍af(a)g(a) is the Cartesian product of the bonding maps af(a)g(a):Xg(a)→Xf(a). We get the inverse system{Yf,pfg,Γ} which is κ-complete, bonding maps are skeletal and w(Yf)≤κ. So, we can take Y=lim←{Yf,pfg,Γ}.
Now, define a maph:X→Y by the following formula:
(4.5)h({xs}s∈S)={xf}f∈Γ,
where xf={xf(a)}a∈dom(f)∈Yf and f∈∏‍{Σa:a∈dom(f)} and dom(f)∈[S]κ. By the property
(4.6){xs}s∈S={ts}s∈S⟺∀s∈S∀σ∈Σs,xσ=tσ⟺∀f∈Γ,xf=tf,
the map h is well defined and it is injection.
The maph is surjection. Indeed, let {bf}f∈Γ∈Y. For each s∈S and each σ∈Σs we fix fσs∈Γ such that s∈dom(fσs) and fσs(s)=σ. Let πf(s):Yf→Xf(s) be a projection for each f∈Γ.
For eacht∈S let define bt={bσ}σ∈Σt, where bσ=πfσt(t)(bfσt). We will prove that an element bt is a thread of the space Xt. Indeed, if σ≥ρ and σ,ρ∈Σt, then take functions fσt and gρt. For abbreviation, denote f=fσt and g=gρt. Define a function h:dom(f)∪dom(g)→⋃‍{Σt:t∈dom(f)∪dom(g)} in the following way:
(4.7)h(s)={g(s),ifs∈dom⁡(g)∖dom⁡(f),f(s),ifs∈dom⁡(f).
The function h is element of Γ and f,g⪯h. Note that h∣dom(f)=f and h∣dom(g)∖{t}=g∣dom(g)∖{t}. Since
(4.8){bg(s)}s∈dom⁡(g)=bg=pgh(bh)=(∏s∈dom⁡(g)sg(s)h(s))(πdom⁡(g)dom⁡(h)(bh))=(∏s∈dom⁡(g)sg(s)h(s))({bh(s)}s∈dom⁡(g))={sg(s)h(s)(bh(s))}s∈dom⁡(g),
we get
(4.9)bρ=bg(t)=sg(t)h(t)(bh(t))=sg(t)f(t)(bf(t))=sρσ(bσ).
It is clear that h({at}t∈S)={bf}f∈Γ.
We shall prove that the maph is continuous. Take an open subset U=∏s∈dom(f)‍Af(s)⊆Yf such that
(4.10)Af(s)={V,ifs=s0,Xf(s),otherwise,
where V⊆Xf(s0) is open subset. A map pf is projection from the inverse limit Y to Yf. It is sufficient to show that
(4.11)h-1((pf)-1(U))=∏s∈SBs,
where
(4.12)Bs={W,ifs=s0,Xs,otherwise,
and W=πf(s0)-1(V) and πf(s0):Yf→Xσ0 is the projection and f(s0)=σ0. We have
(4.13){xs}s∈S∈h-1((pf)-1(U))⟺pf(h({xs}s∈S))∈U⟺pf({xf}f∈Γ)=xf∈U⟺xf(s0)∈V⟺xs0∈W⟺x∈∏s∈SBs⊆∏s∈SXs=X.
Since the maph is bijection and
(4.14)(pf)-1(U)=h(h-1((pf)-1(U)))=h(∏s∈SBs)
for any subbase subset ∏s∈S‍Bs⊆X, the map h is open.In the caseκ=ω 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.IfD is a set and κ is cardinal number then we denote ⋃α<κ‍Dα by D<κ.The following result probably is known but we give a proof for the sake of completeness.Theorem 4.3.
Letκ be an infinite cardinal and let T be a set such that |T|≥κκ. If A∈[T]κ and fδ:T<κ→T for all δ<κ<κ then there exists a set B⊆T such that |B|≤τ and A⊆B and fδ(C)∈B for every C∈B<κ and every δ<κ<k, where
(4.15)τ={κ<κ,forregularκ,κκ,otherwise.Proof.
Assume thatκ is regular cardinal. Let A∈[T]κ and let fδ:⋃α<κ‍Tα→T for δ<κ<κ. Let A0=A. Assume that we have defined Aα for α<β such that |Aα|≤κ|α|. Put
(4.16)Aβ=(⋃α<β‍Aα)∪{fδ(C):C∈(⋃α<β‍Aα)<β,δ<κ|β|}.
Calculate the size of the set Aβ:
(4.17)|Aβ|≤|(⋃α<β‍Aα)||κ|β|||(⋃α<β‍Aα)<β|≤κ|β||(κ|β|)|β||≤κ|β|.
LetB=⋃β<κ‍Aβ, so we get |B|≤κ<κ. Fix a sequence 〈bα:α<β〉⊆B and fγ. Since cf(κ)=κ there exists δ<κ such that C={bα:α<β}⊆Aδ and fγ(C)∈Aσ+1 for some σ<κ.
In the second casecf(κ)<κ, we proceed the above induction up to β=κ. Let B=Aκ, so we get |B|≤κκ and B=⋃β<κ+‍Aβ. Similarly to the first case we get that B is closed under all function fδ, δ<κ<κ.Theorem 4.4.
IfX belongs to the class 𝒞κ then c(X)≤κ.Proof.
IfX∈𝒞κ then by Theorems 3.3 and Proposition 2.2 we get c(X)≤μ(X)≤κ.We apply some facts from the paper [3]. Let 𝒫 be a family of open subset of topological space X and x,y∈X. We say that x~𝒫y if and only if x∈V⇔y∈V for every V∈𝒫. The family of all sets [x]𝒫={y:y~𝒫x} we denote by X/𝒫. Define a map q:X→X/𝒫 as follows q[x]=[x]𝒫. The set X/𝒫 is equipped with topology 𝒯𝒫 generated by all images q[V] where V∈𝒫.Recall Lemma1 from paper [3]:  if 𝒫 is a family of open set of X and 𝒫 is closed under finite intersection then the mapping q:X→X/𝒫 is continuous. Moreover if X=⋃‍𝒫 then the family {q[V]:V∈𝒫} is a base for the topology 𝒯𝒫.Notice that if𝒫 has a property
(seq)∀(W∈P)∃({Vn:n<ω}⊆P)∃({Un:n<ω}⊆P),W=⋃n<ωUn,∀(n<ω)Un⊆X∖Vn⊆Un+1,
then ⋃‍𝒫=X and by [3, Lemma 3] the topology 𝒯𝒫 is Hausdorff. Moreover if 𝒫 is closed under finite intersection then by [3, Lemma 4] the topology 𝒯𝒫 is regular. Theorem 5 and Lemma 9 [3] yeild.Theorem 4.5.
If𝒫 is a set of open subset of topological space X such that  (1)
is closed underκ-winning strategy, finite union and intersection,(2)
has property(seq), then X/𝒫 with topology 𝒯𝒫 is completely regular space and q:X→X/𝒫 is skeletal.If a topological spaceX has the cardinal number μ(X)=ω then X∈𝒞ω, but for μ(X) equals for instance ω1 we get only X∈𝒞ω1ω.Theorem 4.6.
Each Tichonov spaceX with μ(X)=κ can be dense embedded into inverse limit of a system {Xσ,πϱσ,Σ}, where all bonding map are skeletal, indexing set Σ is τ-complete each Xσ is Tichonov space with  w(Xσ)≤τ and
(4.18)τ={κ<κ,forregularκ,κκ,otherwise.Proof.
Letℬ be a π-base for topological space X consisting of cozero sets and σ:⋃‍{ℬα:α<κ}→ℬ be a κ-winning strategy. We can define a function of finite intersection property and finite union property as follows: g({B0,B1,…,Bn})=B0∩B1∩⋯∩Bn and h({B0,B1,…,Bn})=B0∪B1∪⋯∪Bn. For each cozero set V∈ℬ fix a continuous function fV:X→[0,1] such that V=fV-1((0,1]). Put σ2n(V)=fV-1((1/n,1]) and σ2n+1(V)=fV-1([0,1/n)). By Theorem 4.3 for each ℛ∈[ℬ]κ and all functions h,g,σn,σ there is subset 𝒫⊆ℬ such that  (1)
|𝒫|≤τ, where
(4.19)τ={κ<κ,forregularκ,κκ,otherwise,(2)
ℛ⊆𝒫,(3)
𝒫 is closed under κ-winning strategy σ, function of finite intersection property and finite union property,(4)
𝒫 is closed under σn, n<ω, hence 𝒫 holds property (seq).
Therefore by Theorem4.5 we get skeletal mapping q𝒫:X→X/𝒫. Let Σ⊆[ℬ]≤τ be a set of families which satisfies above condition (1), (2), (3) and the (4). If Σ is directed by inclusion. It is easy to check that Σ is τ-complete. Similar to [3, Theorem 11] we define a function f:X→Y as follows f(x)={f𝒫(x)}, where f(x)𝒫=q𝒫(x) and Y=lim←{X/ℛ,q𝒫ℛ,𝒞}. If ℛ,𝒫∈𝒞 and 𝒫⊆ℛ, then q𝒫ℛ(f(x)ℛ)=f(x)𝒫. Thus f(x) is a thread, that is, f(x)∈Y. It easy to see that f is homeomorphism onto its image and f[X] is dense in Y, compare [3, proof of Theorem 11].Theorem4.6 suggests question.Does each spaceX belong to 𝒞μ(X)?Fleissner [13] proved that there exists a space Y such that c(Y)=ℵ0 and c(Y3)=ℵ2. Hence, we get μ(Y)=ℵ1, by Theorem 3.3 and Corollary 4.2. Suppose that Y∈𝒞μ(X) then c(Y3)≤ℵ1, by Theorem 4.4, a contradiction.Corollary 4.7.
IfX is topological space with μ(X)=κ then c(XI)≤τ and
(4.20)τ={κ<κ,forregularκ,κκ,otherwise.Proof.
By Theorem4.3 we get XI∈𝒞τ. Hence by Theorems 4.4 and 4.1 we have c(XI)≤τ.By above Corollary we get the following.Corollary 4.8 (see [14, Kurepa]).
If{Xs:s∈S} is a family of topological spaces and c(Xs)≤κ for each s∈S, then c(∏‍{Xs:s∈S})≤2κ.



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