144 (1994)

**Undetermined sets of point-open games**

### by

**Janusz P a w l i k o w s k i (Wrocław)**

**Abstract. We show that a set of reals is undetermined in Galvin’s point-open game** *iff it is uncountable and has property C*

^{00}### , which answers a question of Gruenhage.

*Let X be a topological space. The point-open game G(X) of Galvin [G]*

**is played as follows. Black chooses a point x**

0**is played as follows. Black chooses a point x**

**∈ X, then White chooses an** *open set U*

**∈ X, then White chooses an**

_{0}

*3 x*

_{0}

**, then B chooses a point x**

**, then B chooses a point x**

_{1}

**∈ X, then W chooses an open** *set U*

**∈ X, then W chooses an open**

_{1}

*3 x*

_{1}

**, etc. B wins the play (x**

**, etc. B wins the play (x**

_{0}

*, U*

_{0}

*, x*

_{1}

*, U*

_{1}

*, . . .) iff X =* S

*n*

*U*

_{n}## .

*Galvin [G] showed that the Continuum Hypothesis yields a Lusin set X* *which is undetermined (i.e. for which the game G(X) is undetermined). (A* *Lusin set is an uncountable set of reals which has countable intersection* with every meager set.)

## Recently Rec law [R] showed that every Lusin set is undetermined. Mo- tivated by Rec law’s result we prove the following.

**Theorem. Let X be a topological space in which every point is G**

**Theorem. Let X be a topological space in which every point is G**

*δ*

*. Then* *G(X) is undetermined iff X is uncountable and has property C*

^{00}*.*

*Property C*

^{00}## was introduced by Rothberger (see [M]). A topological space *X has property C*

^{00}*if for every sequence U*

*n*

*(n ∈ ω) of open covers of X* *there exist U*

_{n}*∈ U*

_{n}*such that X =* S

*n*

*U*

_{n}## . It is known (see [M] or [FM]) *that every Lusin set has property C*

^{00}## .

*Clearly, a space with property C*

^{00}## must be Lindel¨of. Martin’s Axiom im- plies that every Lindel¨of space of size less than 2

^{ℵ}^{0}

*has property C*

^{00}## and that there are sets of reals of size 2

^{ℵ}^{0}

*with property C*

^{00}## (see [M]). Thus, Martin’s Axiom yields undetermined sets of reals of size 2

^{ℵ}^{0}

## (Theorem 4 of [G]).

## On the other hand, in Laver’s [L] model for Borel’s conjecture all metric *spaces with property C*

^{00}## are countable (see Note 1). Thus, consistently, all metric spaces are determined.

*1991 Mathematics Subject Classification: 03E15, 54G15.*

### Supported by KBN grant PB 2 1017 91 01.

*The connection between property C*

^{00}## and point-open games is made *transparent by the following dual game G*

^{∗}*(X), due also to Galvin [G]. Now* **W chooses an open cover U**

**W chooses an open cover U**

_{0}

**of X, then B chooses a set U**

**of X, then B chooses a set U**

_{0}

*∈ U*

_{0}

**, then W** *chooses an open cover U*

1**of X, then B chooses a set U**

1**of X, then B chooses a set U**

*∈ U*

1## , etc. As before, **B wins if X =** S

**B wins if X =**

*n*

*U*

_{n}## .

*Galvin [G] showed that the games G(X) and G*

^{∗}*(X) are equivalent in* **the sense that W ↑ G(X) (has a winning strategy) iff W ↑ G**

**the sense that W ↑ G(X) (has a winning strategy) iff W ↑ G**

^{∗}*(X); similarly* **for B. In particular, G(X) is determined iff G**

**for B. In particular, G(X) is determined iff G**

^{∗}*(X) is.*

*Let G*

^{σ}*(X) and G*

^{∗σ}*(X) be games that are played as G(X) and G*

^{∗}*(X)* **are, but in which B wins if X =** T

**are, but in which B wins if X =**

*n*

## S

*m>n*

*U*

*m*

## . These games are again *equivalent (see [G], Theorem 1). Clearly, |X| ≤ ℵ*

_{0}

**⇒ B ↑ G**

**⇒ B ↑ G**

^{∗σ}**(X) ⇒ B ↑** *G*

**(X) ⇒ B ↑**

^{∗}**(X), and it is not hard to see that if each point of X is G**

**(X), and it is not hard to see that if each point of X is G**

_{δ}**, then B ↑** *G*

**, then B ↑**

^{∗}*(X) ⇒ |X| ≤ ℵ*

0 *(see [G], Theorem 2). Also, X 6∈ C*

^{00}**⇒ W ↑ G**

**⇒ W ↑ G**

^{∗}*(X) ⇒* **W ↑ G**

**W ↑ G**

^{∗σ}**(X) (for the first implication W plays covers that witness X 6∈ C**

**(X) (for the first implication W plays covers that witness X 6∈ C**

^{00}## ).

**We shall prove that W ↑ G**

**We shall prove that W ↑ G**

^{∗σ}*(X) ⇒ X 6∈ C*

^{00}## .

*First let us play one more game. The game M*

^{∗}*(X) is defined as G*

^{∗}*(X)* **is but B chooses finite subsets V**

**is but B chooses finite subsets V**

*n*

*⊆ U*

*n*

## . He wins if S

*n*

## S *V*

*n*

*= X. The σ is* introduced as before.

*The game is motivated by property M of Menger (see [FM]). A topolog-* *ical space has property M if for every sequence U*

_{n}*(n ∈ ω) of open covers* *of X there exist finite V*

*n*

*⊆ U*

*n*

## such that S

*n*

## S *V*

*n*

*= X. Clearly, property* *C*

^{00}*implies property M .*

**Lemma 1. Suppose that X has property M. Then W has no winning** *strategy in M*

**Lemma 1. Suppose that X has property M. Then W has no winning**

^{∗σ}*(X).*

*P r o o f. X is clearly Lindel¨of. Without loss of generality, we can assume* **that W plays increasing sequences from which B chooses single sets. Then** **a strategy for W can be identified with a family {U**

**a strategy for W can be identified with a family {U**

_{σ}*: σ ∈*

^{<ω}*ω} such that* *for every σ, {U*

_{σ}

^{_}

_{i}*: i < ω} is an increasing open cover of X. We seek s ∈*

^{ω}*ω* *such that ∀x ∈ X ∃*

^{∞}*n x ∈ U*

_{s|n}## .

*For integers j > 0 and k, m ≥ 0 let* *V*

*k*

*(m, j) =* \

*τ ∈*^{m}*j*

*U*

*τ*

^{_}*k*

*.*

*Note that {V*

_{k}*(m, j) : k ∈ ω} is an increasing open cover of X: given x ∈ X,* *find k*

_{τ}*(τ ∈*

^{m}*j) with x ∈ U*

_{τ}

^{_}

_{k}

_{τ}*and let k = max*

_{τ}*k*

_{τ}*; then x ∈ V*

_{k}*(m, j).*

*For integers j > 0, k ≥ j and m, n ≥ 0 let* *W*

_{k}^{n}*(m, j) =* [

*j=k*0*≤k*1*≤...≤k**n+1**=k*

## \

*n*

*i=0*

*V*

*k*

_{i+1}*(m + i, k*

*i*

*) .*

*Again {W*

_{k}^{n}*(m, j) : k ≥ j} is an increasing open cover of X: given x ∈ X,*

*find k*

_{1}

*≥ k*

_{0}

*with x ∈ V*

_{k}_{1}

*(m, k*

_{0}

*), next k*

_{2}

*≥ k*

_{1}

*with x ∈ V*

_{k}_{2}

*(m + 1, k*

_{1}

## ),

*etc. So, {W*

_{k}^{n}*(m, j) : k ≥ j} (n ∈ ω) is a sequence of open covers, and, since* *X ∈ M , there is t*

_{m,j}*∈*

^{ω}*ω such that*

*∀x ∈ X ∃*

^{∞}*n x ∈ W*

_{t}^{n}

_{m,j}

_{(n)}*(m, j) .*

*Let s ∈*

^{ω}*ω be strictly increasing such that ∀m, j ∀*

^{∞}*n s(m + n) ≥ t*

*m,j*

*(n).*

## Then

*∀x ∈ X ∀m , j ∃n x ∈ W*

_{s(m+n)}^{n}*(m, j) .* *Claim. ∀x ∈ X ∃*

^{∞}*n x ∈ U*

_{s|n}## .

*P r o o f. Suppose not. Fix x and m with ∀n x 6∈ U*

_{s|(m+n)}## . By the choice *of s there is n with x ∈ W*

_{s(m+n)}^{n}*(m, s(m)). So, there are integers*

*s(m) = k*

0*≤ k*

1*≤ . . . ≤ k*

*n+1*

*= s(m + n)* such that

*x ∈*

## \

*n*

*i=0*

*V*

_{k}

_{i+1}*(m + i, k*

_{i}*).*

*Now, s|m ∈*

^{m}*k*

0*, x ∈ V*

*k*

_{1}

*(m, k*

0*) and x 6∈ U*

_{s|m}

^{_}

_{s(m)}*yield k*

1*> s(m). Next,* *s|(m + 1) ∈*

^{(m+1)}*k*

1*, x ∈ V*

*k*

_{2}

*(m + 1, k*

1*) and x 6∈ U*

_{s|(m+1)}

^{_}

_{s(m+1)}## yield *k*

_{2}

*> s(m + 1). Proceeding in this way we get k*

_{n+1}*> s(m + n), which is a* contradiction.

**It follows that if W plays according to {U**

**It follows that if W plays according to {U**

_{σ}*: σ ∈*

^{<ω}**ω} and B according** **to s, then B wins.**

**ω} and B according**

**to s, then B wins.**

*Now we prove that if X ∈ C*

^{00}**then B can spoil each strategy of W in** *G*

^{∗σ}*(X). The idea of diagonalization used in the proof is taken from [FM],* Lemma 5.1.

*Lemma 2. Suppose that X has property C*

^{00}**. Then W has no winning** *strategy in G*

**. Then W has no winning**

^{∗σ}*(X).*

**P r o o f. Again X is Lindel¨of and we can identify a strategy for W with** *a family {U*

**P r o o f. Again X is Lindel¨of and we can identify a strategy for W with**

_{σ}*: σ ∈*

^{<ω}*ω} of open sets such that ∀σ X =* S

*i*

*U*

_{σ}*_*

*i*

## . We seek *s ∈*

^{ω}*ω such that ∀x ∈ X ∃*

^{∞}*n x ∈ U*

_{s|n}## .

*For integers j > 0, m ≥ 0 and for σ : j*

^{m}*7→ ω let* *U*

_{σ}*(m, j) =* \

*τ ∈*^{m}*j*

## [ *{U*

_{τ}*_*

*σ|i*

*: 0 < i ≤ j*

^{m}*}.*

**B is sure to cover this set if from round m on he plays according to σ,** *provided so far he has played numbers < j.*

**B is sure to cover this set if from round m on he plays according to σ,**

*Claim 1. ∀m, j U*

_{σ}*(m, j)’s form an open cover of X.*

*P r o o f. Fix m and j. Let x ∈ X be given. Let hτ*

_{k}*: k < j*

^{m}*i be an*

## enumeration of

^{m}*j. Define σ by induction: choose σ(0) so that x ∈ U*

_{τ}_{0}

^{_}

_{σ(0)}## ,

*next choose σ(1) so that x ∈ U*

_{τ}_{1}

*_*

*σ(0)*

^{_}*σ(1)*

## , etc.

*Claim 2. There are increasing sequences hj*

*n*

*: n < ωi, hm*

*n*

*: n < ωi of* *integers such that*

*∀x ∈ X ∃*

^{∞}*n ∃ σ : (m*

_{n+1}*− m*

_{n}*) 7→ j*

_{n+1}*x ∈ U*

_{σ}*(m*

_{n}*, j*

_{n}*) .*

*P r o o f. Let j*

_{0}

*= 1, m*

_{0}

*= 0. We start a game. At the nth round, j*

_{n}## and *m*

_{n}**are given and W plays an open cover**

*U*

*σ*

*(m*

*n*

*, j*

*n*

## ) *(σ : j*

_{n}^{m}

^{n}*7→ ω) .*

**B responds with an integer j**

**B responds with an integer j**

_{n+1}*≥ j*

_{n}## , but really thinks about S

*{U*

_{σ}*(m*

_{n}*, j*

_{n}## ) : max

_{i}*σ(i) < j*

_{n+1}*}. Then he declares m*

_{n+1}*= m*

_{n}*+ j*

_{n}^{m}

^{n}## .

*We view this as the M*

^{∗σ}**(X) game played by W according to a fixed** *strategy. Since C*

**(X) game played by W according to a fixed**

^{00}**⇒ M , by Lemma 1, B can spoil this strategy.**

**⇒ M , by Lemma 1, B can spoil this strategy.**

*For k*

1*< . . . < k*

*n*

*< ω and σ*

*i*

*: (m*

*k*

*+1*

_{i}*− m*

*k*

_{i}*) 7→ j*

*k*

*+1*

_{i}## define *W (k*

1*, . . . , k*

*n*

*; σ*

1*, . . . , σ*

*n*

## ) =

## \

*n*

*i=1*

*U*

*σ*

_{i}*(m*

*k*

_{i}*, j*

*k*

_{i}*) .*

*By Claim 2 we see that for every n, W (k*

_{1}

*, . . . , k*

_{n}*; σ*

_{1}

*, . . . , σ*

_{n}## )’s form an *open cover of X. Since X ∈ C*

^{00}*, there are σ*

^{n}_{i}*, k*

^{n}_{i}*(n = 1, 2, . . . ; i = 1, . . . , n)* such that

*∀x ∈ X ∃*

^{∞}*n x ∈ W (k*

^{n}_{1}

*, . . . , k*

_{n}^{n}*; σ*

_{1}

^{n}*, . . . , σ*

_{n}^{n}*) .*

*Let l*

*n*

*∈ {k*

_{1}

^{n}*, . . . , k*

_{n}^{n}*} \ {k*

^{n−1}_{1}

*, . . . , k*

^{n−1}_{n−1}*} and let τ*

*n*

*be the σ*

_{i}^{n}## corre- *sponding to l*

_{n}*. Then l*

_{n}*’s are distinct and, by the definition of W ’s, we* get

*∀x ∈ X ∃*

^{∞}*n x ∈ U*

_{τ}

_{n}*(m*

_{l}

_{n}*, j*

_{l}

_{n}*) .* *Now define s ∈*

^{ω}*ω by*

*s(m*

_{l}

_{n}*+ i) = τ*

_{n}*(i) ,* *for n ∈ ω and i ∈ dom(τ*

*n*

## ), and put 0 elsewhere.

*Claim 3. ∀x ∈ X ∃*

^{∞}*n x ∈ U*

_{s|n}## .

*P r o o f. If x ∈ U*

_{τ}

_{n}*(m*

_{l}

_{n}*, j*

_{l}

_{n}*) then, since s|m*

_{l}

_{n}*: m*

_{l}

_{n}*7→ j*

_{l}

_{n}## , we get *x ∈* [

*{U*

_{s|m}

_{ln}*_*

*τ*

_{n}*|i*

*: 0 < i ≤ m*

_{l}

_{n}_{+1}

*− m*

_{l}

_{n}*} .* *But s|m*

_{l}

_{n}

^{_}*τ*

_{n}*|i = s|(m*

_{l}

_{n}*+ i).*

**It follows that if W plays according to {U**

**It follows that if W plays according to {U**

_{σ}*: σ ∈*

^{<ω}**ω} and B plays** **according to s, then B wins.**

**ω} and B plays**

**according to s, then B wins.**

*Call an open cover U of X strong if for each U ∈ U, the family {V ∈ U :* **U ⊆ V } covers X. Galvin showed that for a regular space X, W ↑ G(X)** *iff no strong open cover contains an increasing subcover {U*

**U ⊆ V } covers X. Galvin showed that for a regular space X, W ↑ G(X)**

_{n}*: n ∈ ω}*

## ([GT], Theorem 4). Combining Galvin’s theorem with ours we can give a

*characterization of regular C*

^{00}*spaces. By a covering tree we mean a family*

*T of finite sequences σ = hU*

0*, . . . , U*

*n−1*

*i (n ∈ ω) of open subsets of X such* *that ∀σ ∈ T ∀k < |σ| σ|k ∈ T and ∀σ ∈ T {U : σ*

^{_}*U ∈ T } covers X.*

*Proposition 1. Let X be a regular topological space. Then the following* *are equivalent.*

*(a) X has property C*

^{00}*.*

*(b) In every covering tree there exists a branch hU*

_{n}*: n ∈ ωi with* S

*n*

*U*

*n*

*= X (equivalently, with* T

*m*

## S

*n>m*

*U*

*n*

*= X).*

*(c) In every strong open cover there exists an increasing subcover {U*

_{n}## : *n ∈ ω}.*

*P r o o f (cf. [GT], Theorem 4). (a)⇒(b) follows by Lemma 2. (b)⇒(c)* is easy: given a strong open cover, use increasing finite sequences of its members as a covering tree.

*We shall show (c)⇒(a). Assume (c). First, X is Lindel¨of. Otherwise any* finitely additive open cover with no countable subcover violates (c). Also, *X is zerodimensional (has a clopen base). Indeed, there is no continuous* *function f from X onto [0, 1] (otherwise {f*

^{−1}*[V ] : V ⊆ [0, 1] open with* *[0, 1]\V uncountable} violates (c)). For completely regular spaces this means* *having a clopen base, and X, being Lindel¨of regular, is completely regular.*

*Now, let U*

*n*

*(n ∈ ω) be a sequence of open covers of X. Since X is* *Lindel¨of and zerodimensional, we can assume that each U*

_{n}## is countable and *consists of pairwise disjoint clopens (see [K], §26).*

*Let U*

_{n}*= {U*

_{i}^{n}*: i < ω} (some U*

_{i}^{n}*’s may be empty). Define f : X 7→*

^{ω}*ω* by

*f (x)(n) = i iff x ∈ U*

_{i}^{n}*.* *Let V =* S

*{U*

_{i}^{n}*: |f [U*

_{i}^{n}*]| ≤ ℵ*

_{0}

*}. Then |f [V ]| ≤ ℵ*

_{0}

*and since ∀x ∈ X x ∈* T

*n*

*U*

_{f (x)(n)}^{n}*, it is not hard to see that there exist t*

*2n+1*

*∈ ω (n ∈ ω) with* *V ⊆* T

*n*

*U*

_{t}^{2n+1}

_{2n+1}*. Suppose that X \ V 6= ∅ (otherwise we are done). We can* *easily refine the covers U*

*2n*

*(with the help of U*

*2m*

*’s, m ≥ n) to covers W*

*n*

*so that W*

_{n+1}*is a refinement of W*

_{n}*and each W ∈ W*

_{n}*which meets X \ V* *contains at least two sets from W*

*n+1*

*which meet X \ V .*

*Let V*

_{σ}*= V ∪* S

*n<|σ|*

*W*

_{σ(n)}^{n}*for σ ∈*

^{<ω}*ω such that each W*

_{σ(n)}^{n}## meets *X \ V . Then V*

_{σ}*’s constitute a strong open cover of X. Also, if V*

_{σ}*⊆ V*

_{τ}## then *σ ⊆ τ (because no W*

_{k}^{n}*which meets X \ V can be covered by finitely many* *sets taken from different W*

*m*

*(m > n)).*

*It follows that from an increasing sequence of V*

_{σ}*’s covering X, which* *exists by (c), we get s ∈*

^{ω}*ω with X \ V ⊆* S

*n*

*W*

_{s(n)}^{n}*. Since there are t*

*2n*

*(n ∈ ω) with W*

_{s(n)}^{n}*⊆ U*

_{t}^{2n}

_{2n}*, we get X ⊆* S

*n*

*U*

_{t}^{n}

_{n}## .

*Rothberger also considered property C*

^{0}*, which is defined as C*

^{00}## is but

*the covers U*

_{n}*are finite. We can define a game corresponding to C*

^{0}## by

*introducing to G*

^{∗}**the requirement that the covers played by W are finite.**

*Then an analogue of Lemma 2 is true (see the proof of (a)⇒(b) below). We* *also have the following. (A tree is finitely branching if the set of immediate* successors of any node is finite.)

*Proposition 2. Let X be a regular topological space. Then the following* *are equivalent.*

*(a) X has property C*

^{0}*.*

*(b) In every finitely branching covering tree there exists a branch hU*

_{n}## : *n ∈ ωi with* S

*n*

*U*

*n*

*= X (equivalently, with* T

*m*

## S

*n>m*

*U*

*n*

*= X).*

*(c) In every strong open cover U such that for each U ∈ U, a finite* *subfamily of {V ∈ U : U ⊆ V } covers X, there exists an increasing subcover* *{U*

_{n}*: n ∈ ω}.*

*P r o o f. We sketch (a)⇒(b); (b)⇒(c)⇒(a) are proved as in Proposition 1.*

*Suppose that X ∈ C*

^{0}*. Let T be a finitely branching covering tree. For each* *n ∈ ω, let V*

*n*

*be a common finite refinement of all covers U*

*σ*

## =

df*{U :* *σ*

^{_}*U ∈ T } (σ ∈ T and |σ| = n). Such a refinement exists because there* *are only finitely many covers to refine. Since X ∈ C*

^{0}## there is a sequence *V*

*n*

*∈ V*

*n*

*such that X =* T

*m*

## S

*n>m*

*V*

*n*

*. Define a branch hU*

*n*

*: n ∈ ωi of T by* *U*

_{n}*= any U ⊇ V*

_{n}*such that hU*

_{0}

*, . . . , U*

_{n−1}*, U i ∈ T .*

*N o t e. 1. It is folklore that every metric space X ∈ C*

^{00}## is homeomorphic *to a subspace of R (the reals). Such an X is zerodimensional (it cannot be* *continuously mapped onto [0, 1] as C*

^{00}## is preserved by continuous images and *[0, 1] 6∈ C*

^{00}*; [K], §40). Being Lindel¨of, X is separable, and so homeomorphic* to a subset of

^{ω}*ω. Since every C*

^{00}## set of reals has strong measure zero, Borel’s *conjecture implies that every C*

^{00}## metric space is countable.

*2. In the spirit of [R], X ⊆ R with property C*

^{00}## can be characterized by any of the following (see [P]):

**(a) for any F**

_{σ}*(equivalently, closed) A ⊆ R*

^{2}

## with all vertical sections *A*

*x*

*(x ∈ X) meager,* S

*x∈X*

*A*

*x*

*6= R;*

*(b) for any closed A ⊆ R*

^{2}

*with all vertical sections A*

_{x}*(x ∈ X) null,* S

*x∈X*

*A*

*x*

## is null.

*Also, X ⊆ R has strong measure zero iff any of the following holds:*

**(a) for any F**

_{σ}*(equivalently, closed) A ⊆ R*

^{2}

## with all vertical sections *A*

*x*

*(x ∈ R) meager,* S

*x∈X*

*A*

*x*

*6= R ([AR]);*

*(b) for any closed A ⊆ R*

^{2}

*with all vertical sections A*

_{x}*(x ∈ R) null,* S

*x∈X*

*A*

*x*

## is null ([P]);

*(c) for any closed null D ⊆ R, X + D is null ([P]).*

*3. There exists (in ZFC) an uncountable C*

^{00}## space in which every point *is G*

_{δ}## . Todorˇcević [T] has an example of a zerodimensional first countable *Hausdorff space of size ℵ*

_{1}

## whose every continuous image into any second *countable space (in particular, into ω*

^{ω}## ) is countable.

## Question. In Propositions 1 and 2, can one remove the assumption that *X is regular (it is used in (c)⇒(a))?*

**Acknowledgements. Thanks to Irek Rec law for inspiring correspon-** dence and for communicating Gruenhage’s question.

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