The aim of this paper is to investigate the open-open
game of uncountable length. We introduce a cardinal number μ(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.
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. If κ 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 length κ. 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
Let X 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 family ℬ 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 space X 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.
Suppose c(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 is skeletal 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 of κ-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.
For ω-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 Y⊆X is dense, then X is κ-favorable if and only if Y is κ-favorable.
Proof.
Let a function σX 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 that σY 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.
Let X 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 Lemma 3.2, 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
if f(α)=(β,ζ), then β,ζ≤α;
f-1(β,γ)<f-1(β,ζ) if and only if γ<ζ;
f-1(γ,β)<f-1(ζ,β) if and only if γ<ζ.
One can take as f 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 function s: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({Intclπσ1(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.
If α<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.
If f(γ)=(θ,λ) and β<λ, then take
(3.12)Dβθ=Intclπσθ+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 have W∩Intclπσξ+1(Bδ+1)=∅, a contradiction.
Corollary 3.4.
If X 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.
Let X=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 that sup{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. Classes 𝒞κ
Let κ 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 from 𝒞κ belongs to 𝒞κ.
Proof.
Let X=∏{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.
If f∈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 map h: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 map h 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 each t∈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 map h 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 map h 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 case κ=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.
If D 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.
Let κ 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 that κ 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|.
Let B=⋃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 case cf(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.
If X belongs to the class 𝒞κ then c(X)≤κ.
Proof.
If X∈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 Lemma 1 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 if 𝒫 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.
If 𝒫 is a set of open subset of topological space X such that
is closed under κ-winning strategy, finite union and intersection,
has property (seq),
then X/𝒫 with topology 𝒯𝒫 is completely regular space and q:X→X/𝒫 is skeletal.
If a topological space X 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 space X 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.
Let ℬ 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
|𝒫|≤τ, where
(4.19)τ={κ<κ,forregularκ,κκ,otherwise,
ℛ⊆𝒫,
𝒫 is closed under κ-winning strategy σ, function of finite intersection property and finite union property,
𝒫 is closed under σn, n<ω, hence 𝒫 holds property (seq).
Therefore by Theorem 4.5 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].
Theorem 4.6 suggests question.
Does each space X 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.
If X is topological space with μ(X)=κ then c(XI)≤τ and
(4.20)τ={κ<κ,forregularκ,κκ,otherwise.
Proof.
By Theorem 4.3 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.
Acknowledgment
The author thanks the referee for careful reading and valuable suggestions.
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