A combinatorial property of Fréchet iterated filters

G. Debs, J. Saint Raymond / Topology and its Applications 159 (2012) 2205–2227

2217

(2) if n = 2p + 1 then

α∈A

⇔

  ∃ϕ ∈ Hom ω p ,

α ∈/ A

⇔

∃i ∈ ω ,

∀s ∈ ω p , ∀ j ∈ ω, α ∈ A (sϕ (s)) j ,  p  ∃ψ ∈ Hom ω / A i (t ψ(t )) . , ∀t ∈ ω p , α ∈

We recall that for any tree S, Hom( S ) denotes the set of all tree homomorphisms on S (see Section 2.3). For other undefined notation, in particular for the intertwining operation , we refer the reader to Section 2.1. Observe that in / A u ” which occur in the equivalences above the sequence u is indeed of the four formulas of the form “α ∈ A u ”, “α ∈ length n. n

n

Theorem 7.8. Consider the s.l. tree Σ = ω[ 2 ]  {0}n−[ 2 ]−1 and let ( T , ε ) be an ordered Ω -valued tree. For any Π n0 set A ⊂ 2Ω if Player II wins the game Γ A ( T (Σ) , ε ) for some s.l. enumeration of Σ then Player II wins the game Γ A ( T , ε ). Proof. Set p = [ n2 ] and q = n − 1 − p. So Σ = ω p  {0}q with p + q = n − 1 and p − 1  q  p; moreover since n  2 then p  1 and q  0. Let R be the partial ordering with domain E which generates the ordered Ω -valued tree T . Without loss of generality we may and shall assume that Ω = ω . Fix an arbitrary s.l. enumeration ν of the s.l. tree Σ , a Π n0 subset A of 2ω together with a family ( A u )u ∈ωn of 01 subsets of 2ω as in Lemma 7.7, and set: Γ = Γ A ( T , ε ) and Γ Σ = Γ ( T (Σ,ν ) , ε ). We now distinguish the two cases: Case 1. n = 2p so q = p − 1. Consider the finite game Γ ∗ of length n = 2p on ω × E in which Player I and Player II construct by alternative moves two sequences (ik , bk )1k p and ( jk , ck )1k p as follows

I: II:

(i 0 , b 0 )





(i 1 , b 1 )

( j0 , c0 )





( i p −1 , b p −1 )



...

( j p −1 , c p −1 )

with

(bk )0k p −1  (ck )0k p −1 ∈ T that is

b 0 R c 0 R b 1 R · · · R b p −1 R c p −1 . Then setting s = (ik )0k p −1 and t = ( jk )0k p −1 Player II wins the run in Γ ∗ iff

N ε(c p ) ⊂ A st . For all k  0 let 0¯ k denote the sequence of length k with constant value 0, then for any u ∈ Σ : – if |u | is even then u = s  0¯ |s| with s ∈ ω< p and then ν (u ) is even too, – if |u | is odd then u = s  0¯ |s|−1 with s ∈ ω p and then ν (u ) is odd too. In the rest of the proof we shall consider various strategies in the games Γ , Γ Σ and Γ ∗ . We shall view a strategy for a player as a mapping assigning to the sequence of the opponent player’s previous moves the player’s next move, so (a2 )0k → a2k+1 for Player I, and (a2−1 )1k → a2k for Player II. Also given any non-empty finite sequence s ∈ ω<ω we shall denote by s∗ its predecessor. Lemma 7.9. If Player II wins the game Γ ∗ then Player II wins the game Γ . Proof. Fix a bijection μ : ω p \ {0} → ω satisfying: if s ⊂ t then μ(s)  μ(t ). Fix also a winning strategy τ ∗ for Player II in Γ ∗ . We shall explicitly define from μ and τ ∗ a strategy τ for Player II in Γ . In fact during an infinite run (am )m∈ω compatible with τ Player II will construct simultaneously an additional infinite sequence (bm )m∈ω satisfying for all m ∈ ω and all s ∈ ω p : (i) dom(ε (bm )) ⊃ [0, m]; (ii) a2m R bm ; (iii) if μ(s) = m then τ ∗ (s, (bμ(s|k ) )k|s| ) = ( j , a2m+1 ).

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G. Debs, J. Saint Raymond / Topology and its Applications 159 (2012) 2205–2227

Here by condition (iii) we implicitly mean that (s, (bμ(s|k ) )k|s| ) is the sequence of Player I’s moves in some run in Γ ∗ which is compatible with τ ∗ , and a2m+1 ∈ E is the second coordinate of the response of τ ∗ in this run (we recall that Γ ∗ is a game on ω × E). We proceed by induction over m. Suppose that a0 , . . . , a2m , b0 , . . . , bm−1 are already defined. Let s ∈ ω p be such that μ(s) = m since μ is monotone then μ(s∗ ) < m and by condition (iii) τ ∗ (s∗ , (bμ(s|k ) )k|s∗ | ) = ( j ∗ , a2m−1 ) for some j ∗ ∈ ω. Now applying condition (c) of Definition 3.3 we can find some bm satisfying conditions (i) and (ii) above. But by the definition of the game Γ the sequence (a )2m is an R-chain. In particular we have a2μ(s∗ )+1 R a2m R bm , hence (s, (bμ(s|k ) )k|s| ) is indeed the sequence of Player I’s moves in a run compatible with τ ∗ . This achieves the inductive construction.  Consider now an infinite run (am )m∈ω in Γ compatible with τ . It follows from condition (i) that α = m∈ω ε (am ) is an element of 2ω and we shall now prove that α ∈ A by applying Lemma 7.7. For this let ϕ be the tree homomorphism on ω p defined inductively by ϕ (∅) = ∅ and ϕ (s) = ϕ (s∗ ) j if s = ∅ with μ(s) = m and τ ∗ (s, (bμ(s|k ) )k|s| ) = ( j , a2m+1 ). Since

τ ∗ is winning in Γ ∗ then for all s ∈ ω p we have α ∈ N ε(a2μ(s)+1 ) ⊂ A sϕ (s) . This proves by Lemma 7.7 that α ∈ A, hence the strategy τ is winning. 2 Lemma 7.10. If Player II wins the game Γ Σ then Player II wins the game Γ ∗ .

Proof. Suppose by contradiction that Player II does not win the game Γ ∗ and fix a winning strategy σ ∗ for Player I in this Σ game (we recall that Γ ∗ is a finite game hence determined). Fix also a winning strategy τ Σ for Player  II in Γ . We shall construct an infinite run a¯ = (am )m∈ω in Γ Σ compatible with τ S and such that α = m∈ω ε (am ) ∈ / A which is a contradiction. Roughly speaking, and as far as this makes sense, a¯ will be the result of the “composition” of the two strategies σ ∗ and τ Σ . More precisely let (i 0 , a0 ) be the first move of Player I in the game Γ ∗ given by the strategy σ ∗ ; we define inductively (am )m∈ω satisfying: (i) a2m+1 = τ Σ ((a2 )m ), (ii) if 2m = ν (t  0|t | ) with t = ∅ then

σ ∗ (t , (aν (t|k 0k−1 ) )k|t | ) = (i , a2m ).

As above by condition (ii) we implicitly mean that (t , (aν (t|k 0k−1 ) )) is the sequence of Player II’s moves in some run in Γ ∗

which is compatible with σ ∗ . We leave  to the reader to check by induction that this definition is coherent. Since τ Σ is winning then α = m∈ω ε (am ) ∈ A, and again we shall prove that α ∈ / A by applying Lemma 7.7. For this we shall construct ψ ∈ Hom(ω< p ) such that for all t ∈ ω p −1 and all j ∈ ω , α ∈ / A i 0 (t ψ(t )) j , where i 0 is the integer already fixed by the first move given by σ ∗ . So let ψ be the homomorphism on ω< p defined inductively by ψ(∅) = ∅ and ψ(t ) = ψ(t ∗ )i if t = ∅ with ν (t  0|t | ) = 2m and σ ∗ (t , (aν (t|k 0k−1 ) )k|t | ) = (i , a2m ). Now fix t = ( jk )0k p−2 ∈ ω p−1 , set ψ(t ) = (ik )0k p−2 and let for all 1  k  p − 1, ik = ik −1 (we recall that i 0 is already defined). It follows then from condition (ii) that the sequence (ik , aν (t|k 0k ) )0k p −1  ( jk , aν (t|k 0k−1 ) )0k p −1 constitutes a run ρ in Γ ∗ of length 2p − 1, which is compatible with σ ∗ :

(i 0 , a 0 )





(i 1 , aν ( j 0 ,0 ) )

( j 0 , aν ( j 0 ) )



(i p −2 , aν (t | p−2 0 p−2 ) )

......

...... 

 ( j 1 , aν ( j 0 ,0, j 1 ) )





(i p −1 , aν (t 0 p−1 ) )

( j p −2 , aν (t 0 p−2 ) )

On the other hand since for all k ∈ ω , t k  0 p −1 ∈ S ∩ ω2p −1 it follows from condition (i) and the definition of the game Γ S that

a0 R aν ( j 0 ) R aν ( j 0 ,0 ) R . . . R aν (t 0 p−2 ) R aν (t 0 p−1 ) R aν (t k 0 p−1 ) . Hence for all j ∈ ω , ( j , aν (t k 0 p−1 ) ) is a legal move for Player II beyond the run ρ , and since the strategy σ ∗ is winning for Player I then by definition of the game Γ ∗ we have N ε(a ∩ A ci (t ψ(t )) j = ∅ and since k is arbitrary and p −1 ) ν (t k 0

)

0

A i 0 (t ψ(t )) j is open then α ∈ / A i 0 (t ψ(t )) j . This proves by Lemma 7.7 that α ∈ / A and finishes the proof of Lemma 7.10 and of Case 1 of Theorem 7.8.

2

Case 2. n = 2p + 1 so q = p. The proof is very similar to Case 1. One introduces a game Γ∗ of length n on goes as follows:

ω × E with the same rules. So a run in Γ∗

G. Debs, J. Saint Raymond / Topology and its Applications 159 (2012) 2205–2227

I: II:

(i 0 , b 0 )



······

( i p −1 , b p −1 )

( j0 , c0 )





2219

(i p , b p )

( j p −1 , c p −1 )

with

b 0 R c 0 R b 1 R · · · R b p −1 R c p −1 R b p s = (ik )0k p −1 ∈ ω p and t = ( jk )0k< p ∈ ω< p satisfying now |t | = |s| + 1. Notice however that in the present case the game Γ∗ being of odd length the last move in a maximal run is now made by Player I. This small advantage of Player I is then balanced by weakening slightly the win condition of Player II who wins a run in Γ∗ iff

N ε(b p ) ∩ A st = ∅. Then one proves in the same lines as Case 1: (1) If Player II wins the game Γ∗ then Player II wins the game Γ . (2) If Player II wins the game Γ Σ then Player II wins the game Γ∗ . Observing that the tree Σ in Theorem 7.8 is of rank n − 1 and gathering this result with Proposition 7.4 one then gets: Corollary 7.11. Let ( T , ε ) be an ordered Ω -valued tree. For any Π n0 subset A of 2Ω Player II wins the game Γ A ( T , ε ) if and only if Player II wins the game Γ A ( T (Σ) , ε ) for any e.s.l. tree Σ of rank < n. Notice that since the Borel games Γ A ( T , ε ) and Γ A ( T (Σ) , ε ) are determined (cf. [4]) one can replace everywhere in the statement of Corollary 7.11 “Player II” by “Player I”. It is however worth noting that one does not need to invoke Borel Determinacy for this. In fact by essentially the same arguments as in Lemma 7.9 one can show that if Player I wins the game Γ ∗ then Player I wins the game Γ . This shows in particular that the determinacy of the Borel games of the form Γ A ( T , ε ) follows from the determinacy of finite games which, by well-known results of H.M. Friedman [2], is far from being the case for arbitrary Borel games, even of finite rank. 8. Proof of condition (B) Fix m  1 and q  0 such that m = 1 + q and set ( T , ε ) = ( T q , εq ). We recall that ε is then an ωm -valuation hence the m game Γ A ( T , ε ) is defined for any set A ⊂ 2ω in particular for A = Nm . Our goal is to prove condition (B) of Theorem 4.1, that is: Fact 8.1. For any Π 02m set A ⊃ Nm Player II wins the game Γ A ( T , ε ). Let Σn denote the s.l. tree Fact 8.1 is equivalent to:

n

n

ω[ 2 ]  {0}n−[ 2 ]−1 endowed with an arbitrary s.l. tree enumeration; then by Theorem 7.8,

Fact 8.2. For any Π 02m set A ⊃ Nm Player II wins the game Γ A ( T (Σ2m ) , ε ). In fact we shall prove the following stronger result: Fact 8.3. Player II wins the game ΓNm ( T (Σ2m ) , ε ). Let us mention here for completeness the following result that we shall not need and which can be proved by a straightforward induction: Fact 8.4. Player I wins the game ΓNm ( T , ε ). Of course there is no contradiction between the two latter results since Nm is precisely a Σ 02m set but not a Π 02m set. Unfortunately we do not have a direct proof of Fact 8.3 which will actually be obtained as a consequence of a more technical result. Before we state this result let us recall that if ( T , ε ) is an ordered Ω -valued tree say T = T R then for any tree Σ and any set A ⊂ 2Ω , Γ A ( T (Σ) , ε ) is a game on the tree Tε ∩ T (Σ) , hence a sequence (ak )k∈ω is an infinite run in this game iff:

/ ν (Σ), ak−1 = ak . (0) For all k ∈

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G. Debs, J. Saint Raymond / Topology and its Applications 159 (2012) 2205–2227

(1) For all k > 0, ε (ak−1 ) ⊂ ε (ak ). (2) For all t ∈ Σ \ {∅}, aν (t ∗ ) R aν (t ) . Notice that since Σ is s.l., if ν (t ) = 2 then ν (t ∗ ) = 2 − 1 so by condition (2) a2−1 R a2 and since ε is R-monotone then ε(a2−1 ) ⊂ ε(a2 ). Hence the conjunction of (1) and (2) above is equivalent to the conjunction of the following conditions (I) and (II): (I) For all t ∈ Σ \ {∅}, if



(II)

ν (t ) = 2 then a2−1 R a2 .

For all k, ε (a2k ) ⊂ ε (a2k+1 ); For all t ∈ Σ \ {∅}, if

ν (t ) = 2 + 1 and ν (t ∗ ) = 2k then a2k R a2+1 ,

which express the restrictions in the game on the moves of Player I and Player II respectively (in addition to the trivial condition (0)). In particular when describing a strategy for Player II in such a game we shall always implicitly assume that condition (I) is already fulfilled. We can now state the result from which Fact 8.3 will be derived: (Σ,ν )

Theorem 8.5. For all q  0, and any s.l.e. tree (Σ, ν ) of rank  1 + 2q, Player II has a strategy τ in the game tree Tεq ∩ T q that any maximal run a¯ compatible with τ is infinite, so a¯ = (an )n∈ω , and satisfies for all s, t ∈ Σ . If |s| = |t | = 1 and ν (s) < ν (t ) are both odd then aν (s) R q1 aν (t ) .

such

8.6. Proof of Fact 8.3 from Theorem 8.5. Again set m = 1 + q, n = 2m and let Σn be as in 8.2 above. Since the rank (Σ ) of Σn is n − 1 = 2q + 1 one can apply Theorem 8.5 to Σ = Σn which provides a strategy τ for Player II in Tεq ∩ T q n . Since T q is εq -free then without loss of generality we can assume that for any maximal run (ak )k∈ω compatible with τ ,   m ωm . k∈ω dom(ε (ak )) = ω that is α = k∈ω ε (ak ) is an element of 2 (Σn )

We shall check that such a strategy is winning in the game ΓNm ( T q of the set



ν (s): s ∈ Σn and |s| = 1

, εq ). Let (nk )k∈ω be the increasing enumeration



which is infinite by the definition of Σn . Since the strategy τ is given by Theorem 8.5 then for any infinite run (ak )k∈ω (Σ ) in Tεq ∩ T q n compatible with τ , the sequence (ank )k∈ω forms an R q1 -chain. It follows then from Proposition 4.9 that

α=



k ∈ω

ε(ak ) is an element of Nm , which shows that τ is winning in ΓNm ( T p(Σ,qn ) , ε p,q ).

In fact the inductive proof of Theorem 8.5 will force us to construct strategies with even stronger properties which we will explicit later on. We introduce now a notion which will play a central role in the proof of Theorem 8.5. 8.7. A derivation on e.s.l. tree s. If Σ is a tree on some set E then given any element s ∈ Σ one can consider

  Σs = u: su ∈ Σ

which is also a tree on E of smaller rank. This derivation on trees is a classical tool for defining inductive procedures on trees. We introduce now, in the frame of enumerated trees, a variation of the previous derivation. Let (Σ, ν ) be an enumerated tree. For any s ∈ Σ of even length we define

    Σ s = Σs ∪ t : t ∈ Σ, ν (t ) is odd and > ν (s) ∪ t ∗ , t : t ∈ Σ, ν (t ) is even and > ν (s) which is a tree on E ∪ Σ . So any element of Σ s \ Σs is a sequence on Σ of length at most 2. Notice that, unless t  = t ∗ , two elements t  and t ∗ , t of length 1 and 2 in Σ s \ Σs are incomparable, though t  and t might be comparable as elements of Σ . In particular if Σ is s.l. then so is Σ s . Moreover Σ s is endowed with a natural tree enumeration ν s defined by



ν s (u ) =

ν (su ) − ν (s) if u ∈ Σs , ν (t ) − ν (s) if u = t ∈ Σ s \ Σs

and if the enumeration ν is s.l. on Σ then so is ν s on Σ s . As usual when there is no ambiguity on ν we shall denote (Σ s , ν s ) more simply by Σ s . Finally let us recall that for all s, rk(Σs ) + |s|  rk(Σ). In particular if s = ∅ then rk(Σs ) < rk(Σ); but this is no more the case with the new derivation, though we always have rk(Σ s )  rk(Σ). For example as one can easily check if Σ is an infinite tree of rank  2 then rk(Σ s ) = rk(Σ) for all s. But fortunately these pathologies disappear whenever rk(Σ)  3. More precisely: Lemma 8.8. For any s = ∅ either rk(Σ s )  2 or rk(Σ s ) + 2  rk(Σ).

G. Debs, J. Saint Raymond / Topology and its Applications 159 (2012) 2205–2227

2221

Proof. Observe that Σs∗ = {∅} ∪ (Σ s \ Σs ) is a tree of rank  2. Since Σ s = Σs ∪ Σs∗ then rk(Σ s ) = max(rk(Σs ), rk(Σs∗ )). Hence either rk(Σ s ) = rk(Σs∗ )  2 or rk(Σ s ) = rk(Σs ), and in the latter case rk(Σ s ) + 2  rk(Σs ) + |s|  rk(Σ) since s is non-empty and of even length. 2 Definition 8.9. Let (Σ, ν ) be an e.s.l. tree and R 1 be a partial ordering on a set E. We shall say that a finite sequence a¯ = (a0 , . . . , an−1 ) ∈ E <ω is Σ, R 1 -coherent if for all t  , t ∈ Σ with 2k − 1 = ν (t  ) < ν (t ) = 2 + 1 < n: (a) If ν (t ∗ ) = 2m then a2m R 1 a2+1 . (b) If |t  | = |t | = 1 then a2k−1 R 1 a2+1 . (c) If |t  | = |t | = 1 and a2k−1 R 1 a2k then a2k R 1 a2+1 . Notice that we did not impose any restriction on n = |¯a|; however being Σ, R 1 -coherent depends only on the largest beginning of a¯ of even length that is (a0 , . . . , a2m−1 ) where m = [ n2 ]. Definition 8.10. Let (T R , ε ) be an ordered Ω -valued tree and (Σ, ν ) an e.s.l. tree. Given any partial ordering R 1 ⊂ R, we (Σ,ν ) in an R 1 -strategy if any run compatible with τ is Σ, R 1 -coherent. shall say that a strategy τ for Player II in Tε ∩ T R These definitions are of course motivated by Theorem 8.5. Indeed we shall prove this theorem by proving the existence (Σ,ν ) of an R q1 -strategy for Player II in Tεq ∩ T q . The proof of this latter result will actually be by induction and is behind the precise definition of Σ, R 1 -coherent sequences, namely the technical condition (c) of Definition 8.9 whose unique role is to ensure inside the inductive procedure the more crucial condition (b). Finally notice that, since R 1 ⊂ R, condition (a) is (Σ,ν ) . In particular we have the following a strengthening of one of the rules imposed on Player II in the game tree Tε ∩ T R obvious fact. Lemma 8.11. Let ( R 1 , R , ε , Σ) be as in Definition 8.10 and a¯ = (a0 , . . . , a2+1 ) be a Σ, R 1 -coherent sequence of even length in the (Σ,ν ) (Σ,ν ) then a¯ ∈ Tε ∩ T R . common domain of R 1 , R , ε . If ε (a2+1 ) ⊂ ε (a2 ) and a¯ ∗ = (a0 , . . . , a2 ) ∈ Tε ∩ T R In all the sequel we shall view a strategy for Player II as a mapping of the form τ : (a0 , . . . , a2 ) → a2+1 where as usual the argument (a0 , . . . , a2 ) is implicitly assumed to be a run in the game and compatible with τ . We start by a simple result. (Σ)

Lemma 8.12. For all q ∈ ω and for any e.s.l. tree (Σ, ν ) of rank  2, Player II has an R q1 -strategy in Tεq ∩ T R q for which any maximal compatible run is infinite. Proof. Set ( R 1 , R , ε ) = ( R q1 , R q , εq ) and observe that if (Σ, ν ) be an e.s.l. tree of rank  2 then for any t ∈ Σ if

ν (t ) is odd

then necessarily t ∗ = ∅, hence condition (a) of Definition 8.9 reduces to a0 R 1 a2+1 . We then define inductively a2+1 = τ (a0 , . . . , a2 ) as follows:

– If either  = 0 or a2−1 R 1 a2 , let a2+1 = a2 . – If not, observe first that by condition (I) a2−1 R a2 ; then applying Proposition 4.8 take for a2+1 any element such that ε(a2 ) = ε(a2+1 ) and a2−1 R 1 a2+1 . By an easy induction one checks that a0 R 1 a2k+1 R 1 a2+1 for all k < , from which it follows clearly that (Σ) R -strategy for Player II in Tε ∩ T R . Finally since a2+1 is always defined any maximal run compatible with τ is clearly infinite. 2

τ is an

1

We can now state the stronger version of Theorem 8.5 that we will in fact prove. (Σ,ν )

Theorem 8.13. For any e.s.l. tree (Σ, ν ) of rank  1 + 2q, Player II has an R q1 -strategy in Tεq ∩ T R q

for which any maximal compatible

run is infinite. Proof. The proof is by induction over q. For q = 0 the result is a particular instance of Lemma 8.12. So assume that the result holds for q. To prove that it holds for q + 1 we will have to start with a fixed e.s.l. tree Σ of rank  1 + 2(q + 1) and try to apply the induction hypothesis to some (actually infinitely many) e.s.l. trees of the form Σ s with s = ∅. But observe that by Lemma 8.8 either rk(Σ s )  2 in which case one can apply Lemma 8.12, or else rk(Σ s )  rk(Σ) − 2  1 + 2(q + 1) − 2 = 2q + 1 and in this case one can apply the induction hypothesis.

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G. Debs, J. Saint Raymond / Topology and its Applications 159 (2012) 2205–2227

For the rest of the proof we set

R 1 = R q1 ,

R = Rq, R˜ = R q+1 ,

ε = εq ,

R˜ 1 = R q1+1 ,

Ω = ω1+q ,

E = dom(ε ),

ε˜ = εq+1 ,

E˜ = dom(˜ε ),

Ω˜ = ω × Ω.

˜ -valuation on E˜ and E˜ = Φ( E ) = Fin(ω, E × 2) is viewed as a subset We recall that ε is an Ω -valuation on E and ε˜ is an Ω of Fin(ω, E ) × Fin(ω, 2). Hence an element a of E˜ is represented under the form a = ((a(i ) )i ∈Ja , (Ja0 , Ja1 )) where Ja = dom(a) is a finite subset of ω and (Ja0 , Ja1 ) is a partition of Ja ; in particular any pair of coordinates of the triple (Ja , Ja0 , Ja1 ) determines the third one. We also recall that the relations R˜ and R˜ 1 are defined by

 a R˜ b

⇔

Ja0 ⊂ Jb0 Ja1

 a R˜ 1 b

⇔

⊂

Jb1

and

∀i ∈ Ja0 ,

a (i ) R b (i ) ,

and

∀i ∈

a (i ) R 1 b (i ) ,

Ja1 ,

Ja0 = Jb0

and

∀i ∈ Ja0 ,

a (i ) R b (i ) ,

Ja1 ⊂ Jb1

and

∀i ∈ Ja1 ,

a (i ) R 1 b (i )

(see Section 4 for more details). (Σ) So let (Σ, ν ) be an e.s.l. tree of a rank  1 + 2(q + 1). We have to construct an R˜ 1 -strategy τ in the game tree Tε˜ ∩ T ˜ . R We shall not define this strategy immediately, but exhibit first partially some of its properties. In all the sequel we consider a sequence a¯ = (a0 , a1 , . . . , a2 , a2+1 ) ∈ E˜ <ω and set for all n  2 + 1:

J n0 = Ja0n ;

J n1 = Ja1n ;

J n = Jan

and for all m  :



K 00 = J 00 ;

K 01 = J 01 ,

0 0 Km = J 2m \ J 2m−1 ;

1 1 Km = J 2m \ J 2m−1

if 0 < m  .

We shall consider then the following condition:

⎧ (Σ) ⎪ a¯ ∗ = (a0 , a1 , . . . , a2 ) ∈ Tε˜ ∩ T ˜ , ⎪ R ⎨ ():

⎪ ⎪ ⎩ ∀m  ,

J 2m+1 = J 2m

and



0 J 2m +1

=

0 J 2k

if 2m + 1 = ν (t ) with ν (t ∗ ) = 2k,

0 J 2m

if not. (Σ)

0 1 We emphasize that we do not suppose a priori that a¯ ∈ Tε˜ ∩ T ˜ . We recall the sets J 2m +1 = J 2m+1 \ J 2m+1 are also R determined by ().

Lemma 8.14. If a¯ satisfies () then: (a) If 2 + 1 ∈ / ν (Σ) then J 20+1 = J 20 ; J 21+1 = J 21 and J 2+1 = J 2 .  0  0 1 1 (b) J 2+1 = J 2 = { K m ∪ Km ; m  , 2m ∈ ν (Σ)} = { K m ∪ Km ; m  }.

/ ν (Σ) then J 20+1 = J 20 , hence J 21+1 = J 21 . Proof. (a) By () we always have J 2+1 = J 2 and if moreover 2 + 1 ∈ (Σ) we have a2m−1 R˜ a2m hence (b) Notice that for all m > 0 if m   then 2m  2 and by the definition of Tε˜ ∩ T ˜ R

0 0 1 1 J 2m −1 ⊂ J 2m , J 2m−1 ⊂ J 2m and a fortiori J 2m−1 ⊂ J 2m .



0 1 0 1 Then by () we have J 2m−1 ⊂ J 2m = J 2m+1 , hence J 2m+1 \ J 2m−1 = K {Km ∪ Km ; m ∪ K m and by induction J 2+1 = 0 1 m  }. Moreover by (a) if 2m ∈ / Σ then J 2m+1 \ J 2m−1 = ∅ hence J 2+1 = { K m ∪ K m ; m  , 2m ∈ ν (Σ)}. 2

Lemma 8.15. If a¯ satisfies () then: (a) If 2 + 1 = ν (t ) ∈ ν (Σ) then:

J 20+1 = J 21+1 =

 



0 Km ; m  , 2m = ν (s), s ⊂ t ,



0 Km ; m  , 2m = ν (s), s ⊂ t ∪

m

1 Km .

G. Debs, J. Saint Raymond / Topology and its Applications 159 (2012) 2205–2227

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(b) If 2 = ν (s) ∈ ν (Σ) then:



J 20 =



J 21 =

   0 Km ; m  , 2m = ν s , s ⊂ s ,

   0 Km ; m  , 2m = ν s , s ⊂  s ∪



1 Km .

m 0 0 0 0 0 0 0 1 1 Proof. (a) If 2 + 1 = ν (t ) with ν (t ∗ ) = 2k then by () J 2k −1 ⊂ J 2k = J 2+1 and J 2k −1 ⊂ J 2k . Hence J 2+1 = J 2k = K k ∪ J 2k−1 0 ∗∗ 0 with 2k − 1 = ν (t ); and repeating the argument inductively we get J 2+1 = { K m ; m  , 2m = ν (s), s ⊂ t } hence  0  1 J 21+1 = J 2+1 \ J 20+1 = { K m ; m  , 2m = ν (s), s ⊂ t } ∪ m K m .

(b) Observe that since a2−1 R˜ a2 then by definition J 20 = K 0 ∪ J 20−1 and J 21 = K 1 ∪ J 21−1 and conclude by applying (a). 2 Definition of τ . For all s ∈ Σ \ {∅} such that ν (s) is even consider the e.s.l. tree (Σ s , ν s ) (see Section 8.7). We recall that by Lemma 8.8 either rk(Σ s )  2 or rk(Σ s )  rk(Σ) − 2  1 + 2q. Hence by applying Lemma 8.12 in the first case or the induction hypothesis in the second case, we can fix an R 1 -strategy a2+1 = τ (a0 , a1 , . . . , a2 ) inductively by

a2+1 =





(i )



a2+1 i ∈ J , J 20+1 , J 20+1 2+1

s

) τ s in the game tree Tε ∩ T(Σ . Then we define R



where J 2+1 , J 20+1 , J 21+1 are defined by () and for all i ∈ J 2+1 :

 (i ) a2+1

=

i) (i ) (i ) 0 τ s (a(2m , a2m+1 , . . . , a2 ) if i ∈ K m with 2m = ν (s) > 0,

(i )

a 2

if not. (i )

0 We have to check that in the first alternative a2+1 is well defined. So fix i ∈ K m such that 2m = ν (s) > 0 for some s (i ) (i ) (i ) (Σ ) s ∈ Σ ; we have to prove that (a2m , a2m+1 , . . . , a2 ) ∈ Tε ∩ T R , which we do by induction on . For  = 0 this is obvious.

(i )

(i )

(Σ s )

(i )

So suppose that (a2m , a2m+1 , . . . , a2−2 ) ∈ Tε ∩ T R ( j)

( j)

( j)

and is compatible with

τ s . Since a2−1 R˜ a2 then for all j ∈ J 2−1 we

( j)

have either a2−1 R a2 or a2−1 R 1 a2 depending whether j is in J 20−1 or in J 21−1 ; but since R 1 ⊂ R then in both cases

( j) ( j) (i ) (i ) (i ) 0 we have a2−1 R a2 . In particular since i ∈ K m ⊂ J 2−1 then a2−1 R a2 which shows that a2 is a legal move for Player I s s (Σ ) (i ) (i ) (i ) (Σ ) (i ) in Tε ∩ T R hence (a2m , a2m+1 , . . . , a2 ) is indeed a run in Tε ∩ T R and a2+1 is well defined.

Lemma 8.16. If a¯ is compatible with τ then: (i )

(i )

(i )

(a) For all i ∈ K 00 , (a0 , a1 , . . . , a2+1 ) is an R-chain. (i )

(i )

(i )

1 , (a2m , a2m+1 , . . . , a2+1 ) is an R 1 -chain. (b) For all i ∈ K m

(i ) (i ) 0 ˜ Proof. (a) If i ∈ K 00 then by Lemma 8.15, since 0 = ν (∅), i ∈ J 2n −1 , and since a2n−1 R a2n then a2n−1 R a2n ; and by the (i ) (i ) definition of τ we have a2n = a2n+1 . (i )

(i )

1 1 1 (b) Similarly if i ∈ K m then by Lemma 8.15 for all n > m, i ∈ J 2n −1 hence a2n−1 R a2n and again by the definition of

(i )

(i ) we have a2n = a2n+1 .

τ

2

0 If a¯ = (a0 , a1 , . . . , a2+1 ) and i ∈ K m (notice that m is entirely determined by i) we define



(i )

(i )

(i )



a¯ (i ) = a2m , a2m+1 , . . . , a2+1 = (b0 , b1 , . . . , b2+1−2m ). We now come to the main lemma. Lemma 8.17. If a¯ is compatible with τ , a¯ ∗ is Σ, R˜ 1 -coherent, and a¯ (i ) is Σ s , R 1 -coherent for all possible i and s, then a¯ is Σ, R˜ 1 coherent. Proof. The proof is rather long and we will split it into three parts corresponding to the three conditions (a), (b), (c) of Definition 8.9 that we will prove in the following order: (b), (c), (a). Proof of condition (b): Let k   be such that ν (t  ) = 2k − 1 < 2 + 1 = ν (t ), with |s| = |t | = 1, we have to prove that a2k−1 R˜ 1 a2+1 .

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0 0 0 1 1 Recall that by Lemma 8.15, J 2k−1 ⊂ J 2+1 and by () J 2k −1 = J 2+1 = J 0 , hence J 2k−1 ⊂ J 2+1 ; so what we only need to prove is that a2k−1 R˜ a2+1 , that is for i ∈ J 2k−1 :



(i ) (i ) 0 (1) If i ∈ J 2k −1 then a2k−1 R a2+1 ; (i ) 1 1 (i ) (2) If i ∈ J 2k −1 then a2k−1 R a2+1 .

So let i ∈ J 2k−1 : (i )

(i )

0 0 0 (1) If i ∈ J 2k −1 = J 0 = K 0 then by Lemma 8.16, a2k−1 R a2+1 . 1 (2) If i ∈ J 2k −1 then by Lemma 8.15:

(i )

(i )

1 for some m  k and then by Lemma 8.16, a2k−1 R 1 a2+1 ; – either i ∈ K m 0 – or else i ∈ K m for some 0 < m  k such that 2m = ν (s) < ν (t  ) < ν (t ) for some s ∈ Σ , and a¯ (i ) is compatible with τ s hence is (Σ s , R 1 )-coherent. Since ν (s) = 2m > 0 and |t  | = |t | = 1 then s is necessarily incomparable with t  and t, hence t  , t ∈ Σ s ; and since t  = t then t  , t are incomparable elements of Σ s of length 1. Moreover 2k − 1 − 2m = ν s (t  ) < ν s (t ) = 2 + 1 − 2m hence by condition (b), applied to the (Σ s , R 1 )-coherent sequence a¯ (i ) , (i ) (i ) we have a2k−1 = bν s (t  ) R 1 bν s (t ) = a2+1 .

Proof of condition (c): Let k   be such that ν (t  ) = 2k − 1 < 2 + 1 = ν (t ), with |t  | = |t | = 1, and suppose in addition 0 = J 00 = J 20+1 , so as above all we need to that a2k−1 R˜ 1 a2k . We have to prove that a2k R˜ 1 a2+1 . Since a0 R˜ 1 a2k then J 2k

prove is that a2k R˜ a2+1 .  0 0 0 0 1 0 = J 00 = J 2k Notice also that since J 2k 0
(i )

0 (1) If i ∈ J 2k = K 00 then by Lemma 8.16, a2k R a2k+1 .

1 (2) If i ∈ J 2k then:

(i )

(i )

1 – either i ∈ K m for some m  k and then by Lemma 8.16, a2k R 1 a2k+1 ;

for some 0 < m < k such that 2m = ν (s) < ν (t  ) < ν (t ) and a¯ (i ) is compatible with τ s hence is (Σ s , R 1 )– or else i ∈ (i ) 0 0 1 0 1 (i ) ˜1 coherent. Since J 2k −1 = K 0 then K m ⊂ J 2k−1 and since a2k−1 R a2k then a2k−1 R a2k . But as we already showed (i ) (i ) t  , t are incomparable elements of length 1 in Σ s . Hence by condition (c) applied to a¯ (i ) we have a2k R 1 a2+1 . 0 Km

0 J 2k

Proof of condition (a): Suppose that ν (t ) = 2 + 1 and ν (t ∗ ) = 2k; we have to prove that a2k R˜ 1 a2+1 . Again by (), = J 20+1 , and we only need to prove that a2k R˜ a2+1 . So let i ∈ J 2k :

0 (1) If i ∈ J 2k then:

(i )

(i )

– either i ∈ K 00 and then by Lemma 8.16, a2k R a2+1 ;

for some 0 < m  k such that 2m = ν (s) < ν (t ) for s ∈ Σ with s ⊂ t, and a¯ (i ) is compatible with – or else i ∈ s s τ hence is (Σ , R 1 )-coherent. Moreover if we set t = su with u ∈ Σ s then t ∗ = su ∗ and ν s (u ) = 2 + 1 − 2m and ) (i ) (i ) ν s (u ∗ ) = 2k − 2m. Hence by condition (a) applied to a¯ (i) we have a(2ki) R 1 a(2i+ 1 and a fortiori a2k R a2+1 . 0 Km

1 (2) If i ∈ J 2k then:

(i )

(i )

1 – either i ∈ K m for some m  k and then by Lemma 8.16, a2k R 1 a2k+1 ;

– or else i ∈

for some 0 < m  k such that 2m = ν (s) < ν (t ) for s ∈ Σ with s ⊂ t, hence m < k so K m ⊂ J 2k−1 , and a¯ (i ) is compatible with τ s hence is (Σ s , R 1 )-coherent. Let t ∗∗ be the predecessor of t ∗ then ν (t ∗∗ ) = 2k − 1 and s ⊂ t ∗∗ . Then as above t , t ∗∗ , t ∗∗ , t ∗ ∈ Σ s with ν s (t ∗∗ ) = 2k − 1 − 2m, ν s (t ∗∗ , t ∗ ) = 2k − 2m and ν s (t ) = 2 + 1 − 2m. 0 Km

0

1

(i ) ) 1 0 1 (i ) ˜1 ¯ (i ) we have a(2ki ) R 1 a(2i+ Since K m ⊂ J 2k 1. −1 and a2k−1 R a2k then a2k−1 R a2k . Hence by condition (c) applied to a

This finishes the proof of Lemma 8.17.

2

To finish the proof of Theorem 8.13 observe that if a¯ is compatible with ) ε(a(2i) ) ⊂ ε(a(2i+ 1 ) and applying Definition 4.3 we get

ε˜ (a2 ) =

i ∈ J 2

   )  ˜ {i } × ε a(2i) ⊂ {i } × ε a(2i+ 1 = ε (a2+1 ). i ∈ J 2

τ then for all i ∈ J 2+1 , a¯ (i) ∈ Tε ∩ T(Σ) hence R

G. Debs, J. Saint Raymond / Topology and its Applications 159 (2012) 2205–2227

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(Σ)

It follows then from Lemma 8.11 that under the same hypothesis as in Lemma 8.17 one can conclude that a¯ ∈ Tε˜ ∩ T ˜ . R (Σ) Applying this observation inductively one gets that τ is a strategy for Player II in Tε˜ ∩ T ˜ , and Lemma 8.17 shows that τ R

is an R˜ 1 -strategy.

This ends up the proof of condition (B) of Theorem 4.1. 9. Optimality In this section we prove that the number m + 1 in Theorem 1.1 is optimal. Theorem 9.1. For all m  2 there exists a 02m set Hm containing Nm with the property that the intersection of any family of m elements in Hm is non-empty. Proof. Let H1 be the set of infinite subsets of

ω, so

H1 = { M ⊂ ω: ∀ p , ∃q  p , q ∈ M }. We then define inductively for m  1:

    Hm+1 = M ∈ P ωm+1 : ∀ p , ∃q  p , M (q) ∈ Nm and ∃ p , ∀q  p , M (q) ∈ Hm . We recall that for any M ⊂ ωm+1 = ω × ωm and any j ∈ ω we denote by M ( j ) the section M ( j ) = {k¯ ∈ ωm : ( j , k¯ ) ∈ M }. Lemma 9.2. For all m  2 the set Hm is 02m . Proof. The proof is by induction on m. H1 is clearly a Π 02 subset of P (ω). Assume that the set Hm is Π 02m . Since for all j ∈ ω the mapping M → M ( j ) is continuous from P (ωm+1 ) to P (ωm ), then the set



M ∈P





ωm+1 : ∃ p , ∀q  p , M (q) ∈ Hm



is Σ 02m+1 ; and since the set Nm is Σ 02m , then the set



M ∈P





ωm+1 : ∀ p , ∃q  p , M (q) ∈ Nm



is Π 02m+1 . It follows that Hm+1 is difference of two Σ 02m+1 sets, hence a 02m+2 (and in particular Π 02m+2 ) set.

2

Lemma 9.3. For all m  1 the set Hm contains Nm . Proof. Since every member of N1 is cofinite, hence infinite, we have N1 ⊂ H1 . Assume Nm ⊂ Hm , and let M ∈ Nm+1 . Then for all but finitely many j we have M ( j ) ∈ Nm ⊂ Hm , hence M ∈ Hm+1 .

2

Lemma 9.4. For all m  1 the intersection of m elements of Hm is compatible with Nm , hence non-empty. Proof. Again the proof is by induction on m. For m = 1 every element M of H1 is infinite: thus it meets every cofinite N ∈ N1 . Assume that the intersection of

any m elements of Hm is compatible with Nm . Let ( M 0 , M 1 , . . . , Mm ) ∈ Hm+1 and N ∈ Nm+1 : we have to prove that 0km M k ∩ N = ∅. By definition of Hm+1 , we know that there exists some p such that M k (q) ∈ Hm for all q  p and every k  m. Then by the induction hypothesis we have

∀q  p ,

M ∗ (q) =



 M k (q) ∈ N m

0k
(q, k¯ ) ∈ N ∩ 0km M k . 2

This finishes the proof of Theorem 9.1.

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10. Extensions We shall discuss in this section several possible extensions of Theorem 1.1. The first extension concerns the definability feature of this result. For this we need to deal with concepts from Effective Descriptive Set Theory and we shall assume the reader to be familiar with this theory, for example as exposed in [5]. We can then state the following: 0 subset of P (ωm ) containing Nm one can find m + 1 11 elements with empty intersection. Theorem 10.1. In any Π2m

For a formal proof of this result one has to repeat the whole arguments of the paper and check step by step that all constructions and definitions which occur in this proof are indeed 11 . This is obvious in almost all the proof since most of the definitions are given by explicit arithmetical formulas. There is only one exception to this, which is the proof of Lemma 7.9. Indeed this proof makes use of an arbitrary winning strategy τ ∗ for Player II in the game Γ ∗ . But one can observe that since the win condition for Player II in this game is Π11 and open (actually clopen) then by the effective version of Gale-Stewart’s theorem (see [5]) one can find a 11 such winning strategy. The second natural direction for extending Theorem 1.1 is to look for a similar conclusion starting with an arbitrary Π n0 set containing Nm . Notice that since Nm−1 embeds obviously in Nm then without loss of generality one may suppose that m < n. On the other hand since Nm is itself a Σ 02m set and is closed under finite intersections then one has to impose that n  2m. Conjecture A0 . If 1  m < n  2m then in any Π n0 set containing Nm one can find a family of n − m + 1 elements with empty intersection. So Theorem 1.1 treats the case n = 2m. But we are also able to prove that this conjecture holds for n = 2m − 1 and n = 2m − 2 (with n > m). The proofs, which combine the method of proof of Theorem 1.1 with some additional non-trivial work, are unfortunately quite tedious. Before we go further in our discussion let us recall some basic facts from [1]. If F is an analytic filter with countable domain – say ω for simplicity – then by the Separation Theorem, any pointwise limit f = limF f n along F of a sequence of continuous functions f n is automatically Borel. But it is a non-trivial fact, proved in [1], that the set of all such functions f on 2ω is exactly the set of all Borel functions of class ξ for some uniquely determined ordinal ξ = rk(F ), that we call the rank of the filter F . We also point out that the construction of the filters Nm can be extended to define inductively for any countable ordinal ξ  1, a Borel filter Nξ of rank ξ : for successor ordinals the construction is due to Katˇetov and is quite similar to the finite case, while for limit ordinals one takes a sort of inductive limit of previous filters (see [1] for more details). We conjecture the following generalization of A0 : Conjecture A. If λ is a limit ordinal and 0  m  n  2m are finite integers then in any Π 02+λ+n set containing N1+λ+m one can find a family of n − m + 2 elements with empty intersection. (Notice that for λ = 0 the condition 0  m  n  2m in A is equivalent to the A0 -condition 1  m + 1 < n + 2  2m + 2.) Now since all the filters Nξ are Borel the following question arises naturally: What is the minimal Baire class containing a filter of a given rank? By easy computations (see [1]) we have the following: – if λ is a limit ordinal then Nλ is a Σ 0λ set; – if ξ = λ + n is a successor ordinal (with λ limit and n  1 finite) then Nξ is in the Baire class Σ 01+λ+2n−1 which gives a lower bound for the possible complexity of a Borel filter of rank ξ , and we conjecture that it is the best bound (in a strong sense). Observe that if λ is a limit ordinal then by [1] Theorem 8.4 there is no Π 0λ filter of rank λ nor a Π 0λ+1 filter of rank λ + 1, hence in these cases the question above is already settled. Conjecture B. If ξ = λ + n is a successor ordinal (with λ limit and n  1 finite) then there is no Π 01+λ+2n−1 filter of rank  ξ . As we shall see now this latter conjecture is actually related to Conjecture A. Notice first that any analytic filter refining

Nξ is obviously of rank  ξ , and it was conjectured in [1] that the converse is also true: Conjecture C. Any analytic filter of rank  ξ contains a copy of Nξ . and obviously if Conjecture C were true then Conjecture B would follow from Conjecture A. To finish let us point out that Conjectures A, B, C are simple consequences of Baire theorem in the case ξ = 1, and in [1] it is proved that they all three hold for ξ = 2 too.

G. Debs, J. Saint Raymond / Topology and its Applications 159 (2012) 2205–2227

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G. Debs, J. Saint Raymond, Filter descriptive classes of Borel functions, Fund. Math. 204 (2009) 189–213. H.M. Friedman, Higher set theory and mathematical practice, Ann. Math. Logic 2 (3) (1970/1971) 325–357. M. Katˇetov, Product of filters, Comment. Math. Univ. Carolin. 9 (1968) 173–184. D.A. Martin, A purely inductive proof of Borel determinacy, in: Recursion Theory, Ithaca, NY, 1982, in: Proc. Sympos. Pure Math., vol. 42, 1985, pp. 303– 308. [5] Y.N. Moschovakis, Descriptive Set Theory, North-Holland, 1980.