Proof Let $\varphi $ be an lsc submeasure such that $\mathcal I=\mathrm {Fin}(\varphi )$ . A winning strategy for Player II can be described as follows. Start with $U_0=\mathcal I$ . Once $\mathcal {U}_{n-1}$ has been played, as $\mathcal {I}$ is not countably generated, choose $U_n\in \mathcal {U}_{n-1}$ which is not countably generated, and $I_n\in [U_n]^{<\omega }$ such that $\varphi (\bigcup I_n)>n$ .⊣

Proof Let $\delta $ be the Hausdorff metric on $\mathcal K(X)$ . Since $\mathcal J$ is $G_{\delta }$ , let $\{ F_n: n\in \omega \}$ be a sequences of closed subset of $\mathcal K(X)$ such that $\mathcal K(X)\setminus \mathcal J=\bigcup _{n\in \omega } F_n$ .

We can describe a winning strategy for Player I as follows. At Step $0$ , for each $A\in \mathcal J$ , choose $\epsilon _A>0$ such that $B_{ \delta }( A,\epsilon _A)\cap F_0=\emptyset $ . Since $(\mathcal K(X),\delta )$ is Lindelöf, there exists a countable subcover $\{ B_{\delta }(A^1_i, \epsilon _{A^1_i}): i\in \omega \}$ of $\mathcal J$ . Player I plays $\mathcal U_0=\{ B_{\delta }(A^1_i, \epsilon _{A^1_i}): i\in \omega \}$ . At Step $n+1$ , suppose Player II plays $B_{\delta }(A^n_{i_n},\epsilon _{A^n_{i_n}})$ and a finite subset $I_n\subseteq B_{\delta }(A^n_{i_n},\epsilon _{A^n_{i_n}})\cap \mathcal J $ . As in Step $1$ , for each $A\in B_{\delta }(A^n_{i_n},\epsilon _{A^n_{i_n}})\cap \mathcal J$ , there is an $\epsilon _A\in (0, \frac {1}{n+1})$ such that $B_{ \delta }( A,\epsilon _A)\cap F_n=\emptyset $ . Player I plays a countable subcover $\mathcal U_{n+1}=\{ B_{\delta }(A^{n+1}_i, \epsilon _{A^{n+1}_i}): i\in \omega \}$ of $B_{\delta }(A^n_{i_n},\epsilon _{A^n_{i_n}})\cap \mathcal J$ .

To see that the strategy is winning, note that the sequence $\{ A^n_{i_n}: n\in \omega \}$ is a Cauchy sequence in $\mathcal K(X)$ and hence converges to some $A\in \mathcal K(X)$ . By the construction, $A\not \in \bigcup _{n\in \omega } F_n$ , so $A\in \mathcal J$ . As each $I_n$ is a finite subset of $B_{\delta }(A^n_{i_n},\epsilon _{A^n_{i_n}})$ , the sequence $\bigcup _{n\in \omega } I_n$ also converges to A, and as, by [Reference Kechris8, Theorem 3], $\mathcal J$ is a $\sigma $ -ideal of compact sets, we get $A\cup \bigcup _{n\in \omega } \bigcup I_n \in \mathcal J$ . (Theorem 3 in [Reference Kechris8] is stated only for $G_{\delta }$ ideals of compact subsets of compact metric spaces, but its proof works for $G_{\delta }$ ideals of compact subsets of arbitrary Polish spaces.)⊣

Proof Let $\varphi $ be a flat submeasure with $\mathcal {I} = \mathrm {Fin}(\phi )$ . If

(1) $$ \begin{align} \forall x\in {\mathcal P}(\omega)\ \big(\sup_{n\in x} \phi(\{ n\}) <\infty \Rightarrow \phi(x)<\infty\big), \end{align} $$

then $\mathcal {I}$ is generated by the sets

$$ \begin{align*} \{ n\in \omega \colon \varphi(\{ n\}) <M\} \in \mathcal{I} \end{align*} $$

with $M\in \omega $ . Thus, $\mathcal {I}$ is countably generated.

Assume, therefore, that (1) fails, which allows us to fix $x_0\in {\mathcal P}(\omega )$ such that

(2) $$ \begin{align} \sup_{n\in x_0}\varphi(\{ n\}) <\infty \;\text{ and }\; \varphi(x_0)=\infty. \end{align} $$

Set

(3) $$ \begin{align} M_0=\sup_{n\in x_0}\varphi(\{ n\}). \end{align} $$

Using (2), the definition (3), and the semicontinuity of $\varphi $ , we can find, for each $M>2M_0$ , a sequence $(a^M_i)_{i\in \omega }$ of finite subsets of $x_0$ such that, for all i,

(4) $$ \begin{align} \max a^M_i < \min a^M_{i+1}\;\text{ and }\; M-M_0\leq \varphi(a^M_i)<M. \end{align} $$

By flatness of $\varphi $ , we can find $N_M>0$ and a subsequence $(a^M_{i_j})$ of $(a^M_i)$ such that, for each $t\in \omega $ , $\varphi \big (\bigcup _{j\leq t} a^M_{i_j}\big ) < N_M$ , from which, by semicontinuity of $\varphi $ , we get

(5) $$ \begin{align} \varphi\big(\bigcup_j a^M_{i_j}\big) \leq N_M. \end{align} $$

For ease of notation, we assume that the whole sequence $(a^M_i)$ has property (5).

Towards a contradiction, suppose that there is a topology $\tau $ on $\mathcal {I}$ that makes $\mathcal {I}$ into a basic partial order. For $k\in \omega $ , let

$$ \begin{align*} x^M_k= \bigcup_{i\geq k} a^M_{i}. \end{align*} $$

Note that, by (5), we have that $x^M_k\in \mathcal {I}$ ; moreover, the sequence $(x^M_k)$ is bounded in $\mathcal {I}$ by $x^M_0$ . Thus, it has a subsequence $(x^M_{k_n})_n$ that is convergent with respect to $\tau $ . By the fact that $\tau $ is basic, for each k, the set ${\mathcal P}(x_k^M)$ is a $\tau $ -compact subset of $\mathcal {I}$ . It follows that the limit of $(x^M_{k_n})_n$ is an element of ${\mathcal P}(x_k^M)$ for each k; thus,

$$ \begin{align*} x_{k_n}^M \to \emptyset\text{ with respect to }\tau, \text{ as }n\to\infty, \end{align*} $$

since $\bigcap _k {\mathcal P}(x_k^M) =\{\emptyset \}$ . Since $\tau $ is a metric topology, we can find a diagonal sequence $(y_M)_{M\in \omega }$ convergent to $\emptyset $ , that is, for each M, there exists n with $y_M= x_{k_n}^M$ and $y_M\to \emptyset $ with respect to $\tau $ as $M\to \infty $ . Again, by using the fact that $\mathcal {I}$ is basic with $\tau $ , the convergent sequence $(y_M)_M$ has a bounded subsequence. But this is impossible, since by the second part of (4) and the definitions of $x^M_k$ and $y_M$ , we have $\phi (y_M)\geq M-M_0$ for each M.⊣

Proof First, we prove that, for each $K_1>0$ , there exists $K_2>0$ such that, for each $b\in {\mathcal P}(\omega )$ ,

(6) $$ \begin{align} \varphi(b)\leq K_1\Rightarrow \psi(b)\leq K_2\;\text{ and }\;\psi(b)\leq K_1\Rightarrow \varphi(b)\leq K_2. \end{align} $$

Otherwise, there exists $K>0$ such that, for each $n\in \omega $ , there exists a set $b_n\in {\mathcal P}(\omega )$ with either $\varphi (b_n)\leq K $ and $\psi (b_n)> n$ , or $\psi (b_n)\leq K $ and $\varphi (b_n)> n$ . By pigeonhole principle and compactness of ${\mathcal P}(\omega )$ , we can assume that

(7) $$ \begin{align} \psi(b_n)\leq K,\; \varphi(b_n)> n,\text{ for all }n, \text{ and }(b_n)_{n} \text{ converges to }b\in {\mathcal P}(\omega). \end{align} $$

Since $\psi (b_n)\leq K$ , for all n, by lower semicontinuous of $\psi $ , we have $\psi (b)\leq K$ , from which, by ${\mathcal I} =\mathrm {Fin}(\varphi )=\mathrm {Fin}(\psi )$ , we get

(8) $$ \begin{align} \varphi(b)<\infty. \end{align} $$

Let N witness flatness of $\psi $ for $K+1$ . Using (7), (8), and lower semicontinuity of $\varphi $ , one recursively constructs finite sets $c_n\in \mathcal P(b_{j_n}\setminus b)$ , $n\in \omega $ , for some increasing sequence $(j_n)_n$ , so that

$$ \begin{align*} \psi(\bigcup_{i\leq n}c_i)< N\text{ and }\varphi(c_n)> n,\text{ for all }n\in \omega. \end{align*} $$

The above inequalities give $\psi (\bigcup _{n\in \omega } c_n )\leq N$ and $\varphi (\bigcup _{n\in \omega }c_n)=\infty $ , contradicting $\mathrm {Fin}(\varphi )=\mathrm {Fin}(\psi )$ , and therefore proving (6).

Assume $\psi $ is not gradual. This assumption allows us to fix $M>0$ , for which the following condition holds. For each $N>0$ , we can find $l_N$ and a sequence $(B^N_n)_n$ of finite families of subsets of $\omega $ so that: $|B^N_n|\leq l_N$ , $\min \{ \min b\colon b\in B^N_n\} \to \infty $ as $n\to \infty $ , $\max _{b\in B^N_n}\psi (b)< M$ , and $\psi (\bigcup B^N_n)\geq N$ . It follows now from the second part of (6) that there exists $M'>0$ such that $\max _{b\in B^N_n}\varphi (b)< M'$ for all $n, N$ . From the first part of (6), it follows that there exists a function g with $g(N)\to \infty $ as $N\to \infty $ such that $\varphi (\bigcup B^N_n)> g(N)$ , for all $n, N$ . Thus, $\varphi $ is not gradual, and the lemma is proved.⊣

Proof Let $\mathcal {I} = \mathrm {Fin}(\varphi )$ for a flat submeasure $\varphi $ , and assume that $\varphi $ is not gradual. This assumption allows us to fix $M>0$ such that for each N there exists $\ell _N$ and a sequence $(B^N_n)$ of families of subsets of $\omega $ such that, for all $n, N\in \omega $ ,

(9) $$ \begin{align} |B^N_n| \leq \ell_N,\; n\leq \min_{b\in B^N_n} b,\; \max_{b\in B^N_n}\varphi(b)<M,\text{ and } \varphi(\bigcup B^N_n)\geq N. \end{align} $$

By semicontinuity of $\varphi $ , we can assume that the sets in $B^N_n$ are finite for each $n, N$ . By going to subsequences of $(B^N_n)$ , we can assume that, for each n,

$$ \begin{align*} |B^N_n| = \ell_N \end{align*} $$

and

$$ \begin{align*} (n,N)\not= (n', N')\Rightarrow \left(\max \bigcup B^N_n<\min \bigcup B^{N'}_{n'}\,\text{ or }\,\max \bigcup B^{N'}_{n'}<\min \bigcup B^N_n\right). \end{align*} $$

Let $K_M$ witness flatness of $\varphi $ for M. By the choice of $K_M$ and by (9), we see that, for each $p\in \omega $ , the following statement holds: for each N, for large enough n, if $a\subseteq p$ is such that $\varphi (a)<K_M$ and $b\in B^N_n$ , then $ \varphi (a\cup b) <K_M$ . Using this observation and the inequality $\varphi (\emptyset )=0<K_M$ , a recursive construction allows us to pick, for each N, an infinite subset $D^N$ of $\omega $ with the following property: for each finite sequence $(b_i)_{i<t}$ , with $t\in \omega $ , of sets picked from distinct $B^N_n$ with $N\in \omega $ and $n\in D^N$ , we have

$$ \begin{align*} \varphi( \bigcup_{i<t} b_i)<K_M. \end{align*} $$

Thus, by going to subsequences, we can assume that the above inequality holds for all finite sequences $(b_i)_{i<t}$ of sets picked from distinct $B^N_n$ with $n, N\in \omega $ . Now, lower semicontinuity of $\varphi $ gives that, for each infinite sequence $(b_i)_i$ of sets picked from distinct $B^N_n$ with $n, N\in \omega $ , we have

(10) $$ \begin{align} \varphi( \bigcup_i b_i)\leq K_M. \end{align} $$

From this point on, we follow the lines of the proof of [Reference Hrušák, Rojas-Rebolledo and Zapletal6, Theorem 2.4] with appropriate modifications. Let $C^k$ , $k\in \omega $ , be a partition of $\omega $ into infinite sets. For each N, let $(f_p^N)_p$ be a sequence of functions

$$ \begin{align*} f_p^{k,N}\colon C^k\to \bigcup_{n\in C^k} B^N_n \end{align*} $$

such that $f_p^{k,N}(n)\in B^N_n$ for all $n\in C^k$ and, for all k and $a\in [\omega ]^{l_N}$ ,

(11) $$ \begin{align} \{ f^{k,N}_p(n)\colon p\in a\} = B^N_n,\text{ for some }n\in C^k. \end{align} $$

Such sequences can be found by [Reference Hrušák, Rojas-Rebolledo and Zapletal6, Claim 2.5, p. 33]. Now, for each p, we let

$$ \begin{align*} J^k_p= \bigcup_N \bigcup_{n\in C^k} f^{k,N}_p(n). \end{align*} $$

From the definition of $J^k_p$ , by (11) and (9), we get that, for each N,

(12) $$ \begin{align} \varphi(\bigcup_{p\in a} J^k_p)\geq N, \text{ for all }k\text{ and }a\in [\omega]^{l_N}. \end{align} $$

On the other hand, observe that, for each $g\in \omega ^{\omega }$ , the set $\bigcup _k J^k_{g(k)}$ is the union of a sequence $(b_i)$ , where sets $b_i$ are chosen from distinct $B^N_n$ with $n, N\in \omega $ . Thus, (10) implies that

(13) $$ \begin{align} \varphi(\bigcup_k J^k_{g(k)} ) \leq N_M, \text{ for each }g\in \omega^{\omega}. \end{align} $$

Let ${\mathcal A}\subseteq \omega ^{\omega }$ be a perfect family of eventually different elements of $\omega ^{\omega }$ . For $g\in {\mathcal A}$ , define

$$ \begin{align*} G(g)= \bigcup_k J^k_{g(k)}. \end{align*} $$

It is clear that $G\colon {\mathcal A}\to 2^{\omega }$ is a continuous injection and (13) implies that $G(g)\in \mathcal {I}$ for each $g\in {\mathcal A}$ .

We check that the family $\{ G(g)\colon g\in {\mathcal A}\}$ is strongly unbounded in $\mathcal {I}$ . Let $g_i$ , $i<l_N$ , be distinct elements of $\mathcal A$ . Then, there exists $k\in \omega $ such that $g_i(k)$ are all distinct for $i<l_N$ . Therefore, by (12), we get

$$ \begin{align*} \varphi\left(\bigcup_{i<l_N} G(g_i)\right)\geq \varphi(\bigcup_{i<l_N} J^k_{g_i(k)})\geq N. \end{align*} $$

Thus, if $g_i\in {\mathcal A}$ , $i\in \omega $ , are all distinct, then

$$ \begin{align*} \bigcup_i G(g_i)\not\in \mathcal{I}, \end{align*} $$

and the conclusion follows.⊣

Question 5.2. Let $\mathcal I$ be an $F_{\sigma }$ ideal of subsets of $\omega $ .

1. (1) Is $\mathcal {I}\equiv _T [\mathfrak c]^{<\omega }$ or $\mathcal {I}\leq _T l_1$ ?

2. (2) Assume $\omega ^{\omega }\leq _T \mathcal {I}$ . Is $l_1\leq _T\mathcal {I}$ ?

3. (3) [Reference Louveau and Veličković10] Assume $\omega ^{\omega }\not \leq _T \mathcal {I}$ . Is $\mathcal {I}\ \sigma $ -weakly bounded?

Proof Let $\varphi $ be a lower semicontinuous submeasure such that $\mathcal {I}=\text {Fin}(\varphi )$ , and let $\tau $ be a winning strategy for Player I in $\Gamma _1(\mathcal {I})$ . Assume, in order to reach a contradiction, that $\mathcal {I}$ is not $\sigma $ -weakly bounded. We will play against $\tau $ and produce a play in which Player II wins: $\mathcal {I}=U_0$ is by assumption not $\sigma $ -weakly bounded. After Move $n-1$ has been made, assume that $U_{n-1}$ is chosen so that the restriction of I to $U_{n-1}$ is not $\sigma $ -weakly bounded. Now, in the nth move, if $\mathcal {U}_n=\tau (U_{n-1})$ is a covering for $U_{n-1}$ , there must be $U_n\in \mathcal {U}_n$ that is not $\sigma $ -weakly bounded, we play such $U_n$ together with a strongly unbounded sequence $(x^n_i)_i$ in $U_n$ . Once $\tau $ has chosen $A_n\in [\omega ]^{\omega }$ , as any subsequence of $(x^n_i)_i$ is strongly unbounded, we can find $m_n$ large enough so that $\varphi (b_n)>n$ , where $b_n$ is as in (14). At the end, this play against $\tau $ would have produced the sequence $(b_n)_n$ which is unbounded in $\mathcal {I}$ .⊣

Proof Fix a gradually flat submeasure $\varphi $ such that $\mathcal {I}=\mathrm {Fin}(\varphi )$ . Given $k\in \omega $ , there exists $M_k>0$ as in the definition of gradual submeasure (with $k+1$ playing the role of M and $M_k$ playing the role of N), for which there is a function $h_k\in \omega ^{\omega }$ such that

$$ \begin{align*} \forall l\ \forall B\in [{\mathcal P}(\omega\setminus h_k(l))]^{\leq l} \big( \max_{b\in B}\varphi(b)< k+1\Rightarrow \varphi(\bigcup B)<M_k\big). \end{align*} $$

Given this $M_k$ , we find $N_k>0$ as in the definition of a flat submeasure, for which there is a function $h_k'\in \omega ^{\omega }$ such that for each n and each $a\in {\mathcal P}(n)$ with $\varphi (a)<N_k$ , we have

$$ \begin{align*} \forall b\in {\mathcal P}(\omega\setminus h_k'(n)) \big(\varphi(b)<M_k\Rightarrow \varphi(a\cup b)<N_k\big). \end{align*} $$

For $k\in \omega $ , set

$$ \begin{align*} {\mathcal X}_k = \{ x\in {\mathcal P}(\omega) \colon \varphi(x)<k+1\}\;\text{ and }\; {\mathcal Y}_k = \{ x\in {\mathcal P}(\omega) \colon \varphi(x)<N_k\}. \end{align*} $$

Point (i) is obvious from the definition of ${\mathcal X}_k$ . Point (ii) follows from the set $\{ x\in {\mathcal P}(\omega )\colon \varphi (x)\leq N_k\}$ being a closed subset of ${\mathcal P}(\omega )$ , by semicontinuity of $\varphi $ , and from

$$ \begin{align*} {\mathcal Y}_k \subseteq \{ x\in {\mathcal P}(\omega)\colon \varphi(x)\leq N_k\}\subseteq \mathcal{I}. \end{align*} $$

Let

$$ \begin{align*} g_{k,l}(n) =\max( h_k(l), h_k'(n)). \end{align*} $$

Observe that $g_{k,l}$ has the following property:

(15) $$ \begin{align} \begin{array}{ll} \forall n &\!\!\!\!\!\! \forall a\in {\mathcal P}(n) \Big( \phi(a)<N_k \Rightarrow\\ &\forall B\in [{\mathcal P}(\omega\setminus g_{k,l}(n))]^{\leq l} \big( \max_{b\in B}\phi(b)< k+1\Rightarrow \varphi(a\cup \bigcup B)<N_k\big)\Big). \end{array} \end{align} $$

Property (15) implies (iii). The lemma follows.⊣

Proof of Theorem 5.1 $\textrm{(i)}\Rightarrow \textrm{(ii)}\;$ We will use the following ideal introduced in [Reference Louveau and Veličković10]. Let $\alpha \in \omega ^{\omega }$ , and define

$$ \begin{align*} \mathcal{J}^{\alpha}=\{A\subseteq\omega\times\omega :\forall n\ A(n)\subseteq\alpha(n) \text{ and }|A(n)|/2^n\rightarrow 0\}, \end{align*} $$

where $A(n) = \{ m\in \omega \colon (n,m)\in A\}$ . (Such an ideal is called ${\mathcal J}^{\alpha , (2^n)}( c_0)$ in [Reference Louveau and Veličković10]. For convenience, we shortened this piece of notation.) It was proved in [Reference Louveau and Veličković10, Proposition 4(b) and the Remark that follows pp. 186–187] that

(16) $$ \begin{align} \mathcal{J}^{\alpha}\leq_T {\mathcal Z}_0. \end{align} $$

Now, given a sequence $\alpha _k\in \omega ^{\omega }$ , $k\in \omega $ , we consider the ideal

$$ \begin{align*} \bigoplus_k {\mathcal J}^{\alpha_k} = \{ A\subseteq \omega\times\omega\times\omega \colon \big( \forall k \ A(k) \in {\mathcal J}^{\alpha_k}\big) \text{ and } \big(\exists k_0\forall k\geq k_0 \ A(k)=\emptyset\big) \}, \end{align*} $$

where $A(k) = \{ (n,m)\in \omega \colon (k, n,m)\in A\}$ . We argue that

(17) $$ \begin{align} \bigoplus_k {\mathcal J}^{\alpha_k}\leq_T{\mathcal Z}_0. \end{align} $$

Indeed, for each k, fix a Tukey map $f_k\colon {\mathcal J}^{\alpha _k} \to {\mathcal Z}_0$ that exists by (16). For $A\in \bigoplus _k {\mathcal J}^{\alpha _k}$ , let $k_A$ be the smallest element of $\omega $ such that $A(k)=\emptyset $ for all $k\geq k_A$ . Define

$$ \begin{align*} F(A)= k_A\cup \bigcup_{k< k_A} f_k\big(A(k)\big). \end{align*} $$

Obviously, $F\colon \bigoplus _k {\mathcal J}^{\alpha _k}\to {\mathcal Z}_0$ , and it is easy to check that it is a Tukey map.

By (17), it suffices to show that, for appropriately chosen $\alpha _k$ , $k\in \omega $ , we have

(18) $$ \begin{align} \mathcal{I}\leq_T \bigoplus_k {\mathcal J}^{\alpha_k}. \end{align} $$

We define $\alpha _k\in \omega ^{\omega }$ , for $k\in \omega $ . Let $g_{k,l}$ , ${\mathcal X}_k$ , and ${\mathcal Y}_k$ , $k,l\in \omega $ , be as in the conclusion of Lemma 5.4. Set

$$ \begin{align*} m^k_0=0 \;\text{ and }\; m^k_{l} = g_{k, 2^{l+1}}(m^k_{l-1}), \text{ for }l>0. \end{align*} $$

Then, by Lemma 5.4(iii), the sequence $(m^k_l)$ has the following property for each $l>0$ :

(19) $$ \begin{align} \begin{array}{ll} \forall a\in {\mathcal P}(m^k_{l-1}) \ \forall B\in [{\mathcal P}(\omega\setminus m^k_{l})]^{\leq 2^{l+1}} \big( ( a\in {\mathcal Y}_k\; &\!\!\!\!\!\! \text{and }B\subseteq {\mathcal X}_k)\\ &\Rightarrow a\cup \bigcup B \in {\mathcal Y}_k\big). \end{array} \end{align} $$

Let $B_l$ , for $l\in \omega $ , and $l_0$ be such that

(20) $$ \begin{align} B_l\subseteq {\mathcal X}_k\cap {\mathcal P}\big([m^k_l, m^k_{l+1})\big)\text{ and } |B_l|\leq 2^{l+1}, \text{ for all }l\geq l_0. \end{align} $$

By using (19) recursively and then taking the union, we get that

(21) $$ \begin{align} \bigcup_{l>l_0/2} \bigcup B_{2l} \in \overline{{\mathcal Y}_k}\;\text{ and }\; \bigcup_{l\geq l_0/2} \bigcup B_{2l+1} \in \overline{{\mathcal Y}_k}. \end{align} $$

Since the closure of $\mathcal Y_k$ is contained in $\mathcal I$ , we have $\bigcup _{l} \bigcup B_{l} \in \mathcal {I}$ , from which we draw the following immediate conclusion:

(22) $$ \begin{align} \{ x\in {\mathcal P}(\omega)\colon \forall l \ \big( x\cap [m^k_l, m^k_{l+1}) \in B_l\big)\} \text{ is bounded in }\mathcal{I}. \end{align} $$

For $l\in \omega $ , let

$$ \begin{align*} I^k_l= [m^k_l, m^k_{l+1}). \end{align*} $$

Fix a one-to-one function $\pi _k:\bigcup _l\mathcal {P}(I^k_l)\to \omega $ , and define $\alpha _k\in \omega ^{\omega }$ by letting

$$ \begin{align*} \alpha_k(l)=1+\max \pi_k\big(\mathcal{P}(I^k_l)\big). \end{align*} $$

Now we produce a function $\Psi \colon \mathcal {I} \to \bigoplus _k {\mathcal J}^{\alpha _k}$ . For $x\in \mathcal {I}$ , we have $\varphi (x)<\infty $ , which allows us to define $k_x\in \omega $ to be the smallest k with $x\in {\mathcal X}_k$ . Let $A_x\subseteq \omega \times \omega $ be defined by

$$ \begin{align*} (l,m)\in A_x\Leftrightarrow x\cap I^{k_x}_l\not=\emptyset\text{ and }m=\pi(x\cap I^{k_x}_l). \end{align*} $$

Observe that for every x, $A_x\in \mathcal {J}^{\alpha _{k_x}}$ . Define $\Psi $ by $\Psi (x)= \{ k_x\}\times A_x$ . It is clear that $\Psi \colon \mathcal {I}\to \bigoplus _k {\mathcal J}^{\alpha _k}$ .

We claim that $\Psi $ is Tukey. Let $A\in \bigoplus _k\mathcal {J}^{\alpha _k}$ and

$$ \begin{align*} \mathcal{A}=\{x\in\mathcal{I}:\Psi(x)\subseteq A\}. \end{align*} $$

We need to see that $\bigcup {\mathcal A}\in \mathcal {I}$ . As $A\in \bigoplus _k\mathcal {J}^{\alpha _k}$ , we have that $A(k)=\emptyset $ for all but finitely many k. Thus, it suffices to see that for a fixed k

$$ \begin{align*} \bigcup \{ x\in {\mathcal X}_k \colon A_x\subseteq A(k)\}\in \mathcal{I}. \end{align*} $$

Fix k, and set $B=A(k)$ , $I_l = I^k_l$ , and ${\mathcal B}= \{ x\in {\mathcal X}_k\colon A_x\subseteq B\}$ . We need to see that

(23) $$ \begin{align} {\mathcal B}\text{ is bounded in }\mathcal{I}. \end{align} $$

Since $B\in {\mathcal J}^{\alpha _k}$ , there is $l_0$ such that for all $l\geq l_0$ , $|B(l)|<2^l$ . So, for each $l\geq l_0$ ,

$$ \begin{align*} |\{x\cap I_l: x\in {\mathcal B}\}|=|\{m: \exists x\in {\mathcal B}\ \pi(x\cap I_l)=m \}|\leq|B(l)|<2^l. \end{align*} $$

Hence, for each $l\geq l_0$ , if $|\{x\cap I_l: x\in {\mathcal B}\}|<2^l$ , and, therefore, by (22), we get (23), as required.

$\textrm{(ii)}\Rightarrow \textrm{(iii)}\;$ follows from $\mathcal Z_0\leq _T l_1$ , see [Reference Louveau and Veličković10].

$\textrm{(iii)}\Rightarrow \textrm{(iv)}\;$ follows from $l_1<_T [\mathfrak c]^{<\omega }$ , see [Reference Louveau and Veličković10].

$\textrm{(iv)}\Rightarrow \textrm{(i)}\; $ Since $\mathcal {I}$ is flat, by Theorem 4.2, $\mathcal {I}$ is gradually flat.

$\textrm{(i)}\Rightarrow \textrm{(v)}\;$ We describe a winning strategy for Player I in $\Gamma _1(\mathcal I)$ . Let the sets $\mathcal {X}_k$ and ${\mathcal Y}_k$ , $k\in \omega $ , be as in the conclusion of Lemma 5.4. By Lemma 5.4(i), the family $\{ {\mathcal X}_k\colon k\in \omega \}$ forms a covering of $\mathcal {I}$ . In Move $0$ , Player I plays this covering. Player II responds by picking a set ${\mathcal X}_k$ and a sequence $(x_i)$ of elements of ${\mathcal X}_k$ . So k is fixed by Player II. For the fixed k, we perform the following analysis. Let $g_{k,l}$ , $k,l\in \omega $ , be as in the conclusion of Lemma 5.4. Since k is fixed, we set $g_l=g_{k,l}$ . Define

$$ \begin{align*} m_0=0 \;\text{ and }\; m_{l} = g_{(l+1)^2+1}(m_{l-1}), \text{ for }l>0. \end{align*} $$

Then, by Lemma 5.4(iii), the sequence $(m_l)$ has the following property for each $l>0$ :

(24) $$ \begin{align} \begin{array}{ll} \forall a\in {\mathcal P}(m_{l-1}) \ \forall B\in [{\mathcal P}(\omega\setminus m_{l})]^{\leq (l+1)^2+1} \big( ( a\in {\mathcal Y}_k\; &\!\!\!\!\!\! \text{and }B\subseteq {\mathcal X}_k)\\ &\Rightarrow a\cup \bigcup B \in {\mathcal Y}_k\big). \end{array} \end{align} $$

Let $B_l$ , for $l\in \omega $ , be such that

(25) $$ \begin{align} B_l\subseteq {\mathcal X}_k\cap {\mathcal P}\left([m_l, m_{l+1})\right)\text{ and } |B_l|\leq (l+1)^2+1. \end{align} $$

By an argument similar to the one justifying (21), it follows from (24) that

$$ \begin{align*} \bigcup_{l>0} \bigcup B_{2l} \in \overline{{\mathcal Y}_k}\;\text{ and }\; \bigcup_{l} \bigcup B_{2l+1} \in \overline{{\mathcal Y}_k}. \end{align*} $$

Thus, by Lemma 5.4(ii), we have $\bigcup _{l} \bigcup B_{l} \in \mathcal {I}$ , from which we get

(26) $$ \begin{align} \{ x\in {\mathcal P}(\omega)\colon \forall l \ \big( x\cap [m_l, m_{l+1}) \in B_l\big)\} \text{ is bounded in }\mathcal{I}. \end{align} $$

Now Player I plays a convergent subsequence $(y^0_i)_{i<\omega }$ of $(x_i)_{i<\omega }$ so that

(27) $$ \begin{align} \forall l\ |\{ y^0_i\colon i\in \omega\} \cap [m_l, m_{l+1})| \leq l+1. \end{align} $$

Assume Players I and II are about to make Move $n+1$ . Assume we have a sequence of sets $a_l\subseteq [m_l, m_{l+1})$ , $l<n$ , such that the set played by Player II in Move p with $1\leq p\leq n$ is

$$ \begin{align*} U_p= \{ x\in {\mathcal X}_k\colon \forall l<p\ x\cap [m_l, m_{l+1}) = a_l\}. \end{align*} $$

Now, in Move $n+1$ , Player I plays the family of sets

$$ \begin{align*} V_a = \{ x\in U_n\colon x\cap [m_n, m_{n+1}) = a\}. \end{align*} $$

Then, still in Move $n+1$ , Player II picks one of these sets, which amounts to picking $a_{n+1} \subseteq [m_n, m_{n+1})$ . We let $U_{n+1}= V_{a_{n+1}}$ . Further, Player II plays an arbitrary sequence $(x_i)$ in $U_{n+1}$ . Then Player I picks a convergent subsequence $(y^{n+1}_i)_{i<\omega }$ of $(x_i)_{i<\omega }$ so that

(28) $$ \begin{align} \forall l\geq n+1 \ |\{ y^{n+1}_i\colon i\in \omega\} \cap [m_l, m_{l+1})| \leq l+1. \end{align} $$

This concludes our description of a strategy for Player I. We claim this is a winning strategy.

Indeed, after a run of the game is finished, we set

$$ \begin{align*} B_l = \{ y^{n}_i\colon n, i\in \omega\} \cap [m_l, m_{l+1}). \end{align*} $$

Clearly, $B_l\subseteq {\mathcal X}_k$ . It follows from (27) and (28) that $|B_l|\leq l^2+1$ . Thus, condition (25) is fulfilled. Therefore, the set

$$ \begin{align*} \{ y^n_i\colon n,i\in\omega\} \end{align*} $$

is bounded in $\mathcal {I}$ as it is included in

$$ \begin{align*} \{ x\in {\mathcal P}(\omega)\colon \forall l \ \big( x\cap [m_l, m_{l+1}) \in B_l\big)\}, \end{align*} $$

which is bounded in $\mathcal {I}$ by (26). It follows that Player I wins.

$\textrm{(v)}\Rightarrow \textrm{(vi)}$ follows from Lemma 5.3.

$\textrm{(vi)}\Rightarrow \textrm{(vii)}$ follows directly from [Reference Louveau and Veličković10, Theorem 1].

$\textrm{(vii)}\Rightarrow \textrm{(i)}$ follows from Theorem 4.2.⊣