Proof. Assume $\mathcal {I}$ is not tall. Consequently, there exists an infinite set $A \subseteq \omega $ such that every infinite subset $B \subseteq A$ is in $\mathcal {I}^+$ . During the game $G_1(\mathcal {I})$ , player II can adopt a strategy of selecting, at each move, a part of the partition that intersects A infinitely and chooses a natural number $n_i \in A$ , ensuring it differs from all previously selected numbers. In the end $\{ n_i : i\in \omega \} \in \mathcal {I}^+$ .

Proof. Consider the coloring $c: [\omega ]^2 \to 2$ defined by

$$\begin{align*}c(\{n,m\}) = \begin{cases} 0 & \text{if } \{n,m\} \in E, \\ 1 & \text{otherwise}, \end{cases} \end{align*}$$

where E denotes the edge set of the random graph. In the first move of the game, player I partitions $\omega $ into two sets: $A_0^0 = \emptyset $ and $A_1^0 = \omega $ . Since $A_0^0$ is empty, player $II$ is compelled to choose $A_1^0$ and selects an element from it, denoted by $n_0$ .

On the kth move ( $k \geq 1$ ), let $n_{k-1}$ be the element chosen by player $II$ in the $(k-1)$ th move. Player I partitions the remaining set, $A_{i_{k-1}}^{k-1}$ , based on the coloring c with respect to $n_{k-1}$ . Specifically, player I partitions $A_{i_{k-1}}^{k-1}$ into two pieces:

$$\begin{align*}A_0^k = \{ m \in A_{i_{k-1}}^{k-1} \mid c(\{n_{k-1}, m\}) = 0 \} \end{align*}$$

and

$$\begin{align*}A_1^k = \{ m \in A_{i_{k-1}}^{k-1} \mid c(\{n_{k-1}, m\}) = 1 \}. \end{align*}$$

Player $II$ then chooses one of these pieces, either $A_0^k$ or $A_1^k$ , denoted as $A_{i_k}^k$ , and selects an element from it, denoted $n_k$ .

This process continues, ensuring that in each step, player I partitions the chosen piece in a way that enforces c-homogeneity relative to the previously chosen element. At the end of the game, the set of selected elements, $\{ n_k : k \in \omega \}$ , will be the union of two c-homogeneous sets (one consisting of cliques and the other of anti-cliques).

Since the union of a clique and an anti-clique from the random graph is an element of the ideal $\mathcal {R}$ , player I wins the game.

Proof. For a detailed proof, see [Reference Meza-Alcántara15].

Proof. This proof draws parallels with the proof of Proposition 2.2. Let $c: [\omega ]^2 \to 2$ be any given coloring. Consider the gameplay in $G_1(\mathcal {I})$ where player I and player II employ their strategies, with player II following their winning strategy.

In the initial move, player I partitions $\omega $ into two subsets: $A_0^0 = \emptyset $ and $A_1^0 = \omega $ . Because $A_0^0$ is empty, player II is compelled to choose $A_1^0$ , from which they select an element $n_0$ .

For each subsequent move $k \geq 1$ , assume $n_{k}$ is the element previously chosen by player II. Player I then partitions the remaining set $A_{i_{k}}^{k}$ , based on the coloring c with respect to $n_{k}$ , into:

$$\begin{align*}A_0^{k+1} = \{ m \in A_{i_{k}}^{k} : c(\{n_{k}, m\}) = 0 \} \end{align*}$$

and

$$\begin{align*}A_1^{k+1} = \{ m \in A_{i_{k}}^{k} : c(\{n_{k}, m\}) = 1 \}. \end{align*}$$

Player II, adhering to their winning strategy, chooses one of these sets, either $A_0^{k+1}$ or $A_1^{k+1}$ , designated as $A_{i_{k+1}}^{k+1}$ , and selects an element from it, denoted $n_{k+1}$ .

As player II is operating under a winning strategy, the set $\{n_k : k \in \omega \}$ forms an $\mathcal {I}$ -positive set. This set, comprising elements selected by player II during the game, satisfies that, for each $i\in \omega $ and for all $k, k'> i$ , $c(\{ n_i, n_k \}) = c(\{ n_i, n_{k'} \})$ , so $\{n_k : k \in \omega \}$ can be expressed as the union of two c-homogeneous subsets, $H_0\cup H_1 = \{n_k : k \in \omega \}$ . Since their union forms an $\mathcal {I}$ -positive set, at least one of these subsets, either $H_0$ or $H_1$ , must not belong to $\mathcal {I}$ . Therefore, this verifies that $\omega \to (\mathcal {I}^+)_2^2$ indeed holds.

Proof. We will prove that if a dominant set B is partitioned into $C_0 \cup C_1$ , at least one of the sets $C_0$ or $C_1$ is a dominant set. Without loss of generality, assume $B = \omega $ . For each ordinal $\alpha \in \omega _1 + 1$ , we will recursively define a function $\phi _\alpha : \omega ^{<\omega } \to 3$ as follows:

Base case: For $i \in 2$ , we define $\phi _0(s) = 1-i$ if $|A_s \cap C_i| < \omega $ and $\phi _0(s) = 2$ otherwise.

Recursive case: For $\alpha = \beta + 1$ and $i \in \{0, 1\}$ , we set $\phi _\alpha (s) = i$ if either $\phi _\beta (s) = i$ or ( $\phi _\beta (s) = 2$ and there are infinitely many n such that $\phi _\beta (s^\frown n) = i$ ). If there are infinitely many n such that $\phi _\beta (s^\frown n) = 0$ and infinitely many n such that $\phi _\beta (s^\frown n) = 1$ , we set $\phi _\alpha (s) = 0$ (this choice does not affect the proof). If $\phi _\beta (s) = 2$ and the set $\{ n \in \omega : \phi _\beta (s^\frown n) \in \{0, 1\}\}$ is finite, then we set $\phi _\alpha (s) = 2$ .

Limit case: For $i \in \{0, 1\}$ , we set $\phi _\alpha (s) = i$ if there exists $\beta < \alpha $ such that $\phi _\beta (s) = i$ , and $\phi _\alpha (s) = 2$ whenever for every $\beta < \alpha $ , $\phi _\beta (s) = 2$ .

Proof. This follows the structure of the proof for Proposition 2.5.

Proof. Assume $A \subseteq \omega $ has the property that for every natural number n, there is a string $s \in \omega ^n \cap T_A$ such that the set $\{ k \in \omega : s^\frown k \in \omega ^{n+1} \cap T_A \}$ has cardinality at least $(n+1)^2$ . Then A is $\mathcal {PC}$ -positive. To prove this, consider natural numbers k and l along with sets $ X_0, X_1, \dots X_k \in \mathcal {S}_0$ and $Y_0, Y_1, \dots Y_l \in \mathcal {S}_1$ . Define $X = \bigcup _{i\le k} X_i$ and $Y = \bigcup _{j \le l}Y_j$ . According to the definition of $\mathcal {S}_0$ , for all $s \in T_X$ , the inequality $|\{n\in \omega : s^\frown n \in T_X \}| \le k(|s| + 1)$ holds. From $\mathcal {S}_1$ , there exists some $N \in \omega $ such that for all $s \in \omega ^{<\omega }$ with $|s|> N$ , $|\{ n\in \omega : s^\frown n \in T_Y \}| \le l$ . Given that the function $(n+1)^2$ grows faster than $k(n + 1) + l$ ; thus, there is $M \in \omega $ where $(M+1)^2> k(M + 1) + l$ and for that reason A can not be contained in $X \cup Y$ , because there is a node in $T_A \cap \omega ^M$ possessing more descendants than any node in $T_{X \cup Y} \cap \omega ^M$ .

Player I starts the game by presenting a partition of $\omega $ into $B_0 \cup C_0$ . If $B_0$ is a dominant set, Player II chooses $D_0 = B_0$ ; otherwise, Player II chooses $D_0 = C_0$ . There exists a natural number n such that $D_0 \cap A_{(n)}$ is non-empty, and Player II selects $n_0 \in D_0 \cap A_{(n)}$ .

Player I then splits $D_0$ into two pieces, $B_1$ and $C_1$ . Player II chooses the dominant set among them and calls it $D_1$ . Player II then selects $n_1 \in D_1 \cap A_{(m)}$ where $n \neq m$ .

Player I presents the partition $B_2 \cup C_2$ , and Player II chooses the dominant set among them. Player II then looks for an $i_0 \in \omega $ such that $D_2$ is a dominant set below $(i_0)$ . Player II selects $n \in \omega $ such that $D_2 \cap A_{(i_0, n)}$ is non-empty, and takes $n_2 \in D_2 \cap A_{(i_0, n)}$ .

For the next four rounds, Player II chooses a part that is major below $(i_0)$ and selects $n_3, n_4, n_5$ and $n_6$ such that all of them are in different $A_{(i_0, n)}$ .

Player I continues giving partitions, and Player II stays in level k for $(k+1)^2$ rounds. After this, Player II chooses $i_k$ such that $D_m$ is major below $(i_0, i_1, \dots , i_k)$ . By construction and the initial observation, the final set $\{ n_j : j \in \omega \}$ is $\mathcal {PC}$ -positive.

Proof. To demonstrate that $\mathcal {PC}$ qualifies as a tall ideal, consider an arbitrary set $X \in [\omega ]^\omega $ . Let $\mathrm {suc}_{T_X}(s)$ denote the set of successors of s in $T_X$ . We proceed by considering two cases:

Case 1: Suppose there exists $s \in T_X$ such that $\mathrm {suc}_{T_X}(s)$ is infinite. In this scenario, let $S = \{ n \in \omega : s^\frown n \in T_X \}$ . For each $k \in S$ , select $n_k \in X \cap A_{s^\frown k}$ . Observe that the collection $\{ n_k : k \in S \}$ belongs to $\mathcal {S}_0$ , and consequently, it is also contained in $\mathcal {PC}$ .

Case 2: Assume for every $s \in T_X$ , the set $\mathrm {suc}_{T_X}(s)$ is finite. We construct recursively sequences $\{ n_k \in X : k \in \omega \}$ and $\langle s_k \in T_X : k \in \omega \rangle $ such that $n_i \neq n_j$ for $i \neq j$ , $s_m$ extends $s_n$ if $n < m$ , each $|A_{s_k} \cap X| = \omega $ , and each node t in the tree $T_{\{ n_k \in X : k \in \omega \}}$ has at most $|t| + 1$ immediate successors. Begin with $s_0 = (n) \in \omega ^1$ where $|A_{(n)} \cap X| = \omega $ and choose $n_0 \in A_{(n)} \cap X$ . Assuming $n_i$ and $s_i$ have been defined, and given $A_{s_i} \cap X$ is infinite, there exists an extension $s_{i+1} \in T_X$ of $s_i$ such that $|A_{s_{i+1}} \cap X| = \omega $ . Select $n_{i+1} \in A_{s_{i+1}} \cap X \setminus \{ n_j : j \leq i \}$ . The sequence $s_{i+1}$ has length at least $i+1$ , and the number of successors of each node in $T_{\{ n_j : j \leq i+1 \}}$ is limited to $i + 1$ , since the set size is $i+1$ . It is notable that any choice of $n_{i+1}$ does not introduce a new branch into the tree $T_{\{ n_k : k \leq i+1 \}} \cap \omega ^{\le |s_i|}$ . By construction, $\{ n_k : k \in \omega \} \in \mathcal {S}_1 \cap [X]^\omega $ .

Next, we aim to show that $\mathcal {PC}$ is an $F_\sigma $ set. To do so, we will prove that $\mathcal {PC}$ is generated by compact sets. We define the following subset of $\mathcal {PC}$ :

$$\begin{align*}\mathcal{S}_2 = \{ A \subseteq \omega : \exists s \in \omega^{<\omega} \ ( A \subseteq A_s \text{ and } \forall n \in \omega \ (|A \cap A_{s^{\frown} n}| \le 1 ) ) \}. \end{align*}$$

It is clear that $\mathcal {S}_2 \supseteq \mathcal {S}_0$ .

By Claims 2.3 and 2.4, $\mathcal {PC}$ is generated by compact sets, so it is an $F_\sigma $ set.

Proof. The proof follows a similar argument as the proof of Proposition 2.7.

Proof. It is sufficient to show that for any $D \in \mathcal {H}^+$ , Player II has a winning strategy in $G_1(\mathcal {H} \upharpoonright D)$ . This assertion is corroborated by Proposition 2.4, which ensures that any such D satisfies $D \to (\mathcal {TC}(f)^+)_2^2$ .

Suppose $B \subseteq D$ satisfies the condition that for each $n \in \omega $ , there exists a sequence $s \in \omega ^n \cap T_B$ such that $| \{ k \in \omega : s^\frown k \in \omega ^{n+1} \cap T_B \} | \ge 2n \cdot f(n) + 1$ . Then $B\in \mathcal {TC}(f)^+$ . This line of reasoning aligns with the argumentation outlined in the proof of Theorem 2.1.

For the sake of contradiction, suppose $B \subseteq X \cup Y$ , where $X = \bigcup _{i<j} X_i$ with each $X_i \in \mathcal {S}_0$ and $Y = \bigcup _{i<k} Y_i$ with each $Y_i \in \mathcal {S}_3(f)$ . There exists $N \in \omega $ such that $|suc_{T_X}(s)| \le j$ for each $s \in T_X$ if $|s| \ge N$ , and $|suc_{T_Y}(s)| \le k \cdot f(|s|)$ for each $s \in T_Y$ . This is because the total number of nodes in $T_Y$ at level $|s|$ is at most $k \cdot f(|s|)$ . Let $M = \max \{N, j, k\}$ . Since f is increasing, we have $f(M) \ge M$ . For each $s \in \omega ^M \cap T_B$ , it holds that $|suc_{T_B}(s)| \le |suc_{T_X}(s)| + |suc_{T_Y}(s)| \le j \cdot M + k \cdot f(M) \le M \cdot f(M) + M \cdot f(M) = 2M \cdot f(M)$ . Hence, B cannot satisfy the property described above.

The winning strategy for Player II is analogous to that in the proof of Theorem 2.1; the only difference is the number of nodes Player II needs to select at each level. This variation poses no issue, as whenever we partition a dominating set, there are infinitely many nodes from which to choose.

Proof. The construction of $R_k^n$ will be done by recursion. Let $\{X_l : l \in \omega \}$ be an increasing sequence of finite subsets of natural numbers, with $X_0 = \{0, 1, \dots , n-~2\}$ .

• • Base case: The only possibility for k subsets $(A_1, A_2, \dots , A_k)$ of $[X_0]^{n-1}$ , such that at least one of the $A_i$ is non-empty and $A_1, A_2, \dots , A_k$ have empty pairwise intersections, is: $A_1 = X_0$ and $A_i = \emptyset $ (for $i \neq 1$ ), or $A_2 = X_0$ and $A_i = \emptyset $ (for $i \neq 2$ ), etc. In this case, define $R_k^n(0, 1, 2, \dots , n-1) = 0$ , $R_k^n(0, 1, 2, \dots , n) = 1$ , $\dots $ , $R_k^n(0, 1, \dots , n+k-2) = k-1$ and let $X_1 = \{0, 1, \dots , n+k-2\}$ .

• • Successor case: Suppose that we have defined a finite subset $X_l$ of natural numbers (which by construction is also a natural number, i.e., $|X_l| = X_l$ ). For every family of subsets $A_1, A_2, \dots , A_k \subseteq [X_l]^{n-1}$ such that at least one of the $A_i$ is non-empty and the sets $A_i$ are disjoint, we take the first natural number j not used in the construction and set $R_k^n(j, j_2, j_3, \dots , j_n) = i$ for any $\{j_2, j_3, \dots , j_n\} \in A_i$ . Since there are only a finite number of such families due to $X_l$ being finite, we let $X_{l+1}$ be such that $X_l \subseteq X_{l+1}$ and $j \in X_{l+1}$ (for each family of subsets).

Note that $[\omega ]^n = \bigcup _{l \in \omega } [X_l]^n$ , so at the end, we have defined $R_k^n$ on the entire domain $[\omega ]^n$ , and by construction, $R_k^n$ has the $(n,k)$ -star property.

Proof. The proof of this proposition is done by recursion. We start by defining the function $f(i) = i$ for $i \in \{0,1,\dots ,n-2\}$ . Then, we proceed with the following steps:

• • Base case: Assume that $c(\{0, 1, \dots , n-1\}) = i$ , where $\{0, 1, \dots , n-1\}$ are the first n natural numbers. Using the $(n,k)$ -star property of $R^n_k$ , there exists an $x \in \omega $ such that $R^n_k(\{0, 1, \dots , n-2, x\}) = i$ . Thus, we set $f(n-1) = x$ .

• • Successor case: Assume that we have already defined f up to the natural number l. We define $X_i$ as the set of all $x \in [l+1]^n \setminus [l]^n$ such that $c(x) = i$ , and $X_i'$ as the set of all $x \setminus \{l+1\}$ where $x \in X_i$ . We also define $A_i$ as the set of all $f(j)$ such that $j \in X_i'$ , for every $i \in k$ . Since $R^n_k$ satisfies the $(n,k)$ -star property with the sets $A_i$ , there exists an $x \in \omega $ such that $R^n_k(\{x, j_2, \dots , j_n\}) = i$ whenever $j_2, \dots , j_n \in A_i$ are distinct natural numbers. In this case, we assign $f(l+1) = x$ .

By construction, the function f satisfies the required property.

Proof. To prove that c satisfies the $(n,k)$ -star property in W, we must establish that for any finite disjoint sets $A_0', A_1', \dots , A_{k-1}' \subseteq [W]^{n-1}$ , there exists $x \in W$ such that $c(x, j_2, \dots , j_n) = i$ for all $\{j_2, \dots , j_n\} \in A_i'$ . Let $B_i = A_i \cup A_i'$ for each i. Since c fulfills the $(n,k)$ -star property in A, there must exist a witness $x \in A$ effective for the sets $B_i$ . Given $x \in A \setminus \bigcup _{i < k} \bigcup A_i$ , by definition, x also resides in W, thus serving as the required witness for the sets $A_i'$ .

Proof. Suppose it is false for the partition $A = B \cup C$ . Then in $B^{n-1}$ , there exist finite disjoint sets $B_0, \dots , B_{k-1}$ such that no $x \in B$ satisfies $f(x, j_2, \dots , j_n) = i$ for any $\{j_2, \dots , j_n\} \in B_i$ and some $i \in k$ . Similarly, there exist $C_0, \dots , C_{k-1}$ subsets of $C^{n-1}$ that satisfy an equivalent property. Therefore, in A, there are no witnesses for the sets $A_i = B_i \cup C_i$ , which contradicts the fact that A satisfies the $(n,k)$ -star property.

Proof. First, we show that $\mathcal {R}_{k,l}^n$ is generated by a closed set. The ideal $\mathcal {R}_{k,l}^n$ is generated by the set $\{ A \subseteq \omega : |colors(R^n_k,A)| \le l \}$ , which can be represented as the intersection of closed sets of the form $\{ A \subseteq \omega : |colors(R^n_k,A)| \neq i \}$ , where $i \in \{0, 1, \dots , l-1\}$ . Therefore, $\mathcal {R}_{k,l}^n$ is generated by a closed set.

Next, we show that $\mathcal {R}_{k,l}^n$ is tall. This follows from the classical Ramsey theorem, which states that for any positive integers n and k, there exists a positive integer $R(n,k)$ such that for any n-coloring of the edges of a complete graph on $R(n,k)$ vertices, there exists a monochromatic complete subgraph on k vertices. In this case, the set of vertices in the monochromatic subgraph corresponds to a set in $\mathcal {R}_{k,l}^n$ , showing that it is tall.

Proof. Suppose $\mathcal {R}^n_{k,l} \le _K \mathcal {I}$ . Let $f: \omega \to \omega $ be a Katětov function, that is, $A \in \mathcal {R}^n_{k,l}$ implies $f^{-1}[A] \in \mathcal {I}$ . Define a coloring $c: [\omega ]^n \to k$ by

$$\begin{align*}c(\{a_0, \dots, a_{n-1}\}) = R^n_k(\{f(a_0), \dots, f(a_{n-1})\}) \end{align*}$$

when $|\{f(a_0), \dots , f(a_{n-1})\}| = n$ . If $|\{f(a_0), \dots , f(a_{n-1})\}| \le n-1$ , assign $c(\{a_0, \dots , a_{n-1}\}) = 0$ . We aim to show that if $A \subseteq \omega $ and $|colors(c, A)| \le l$ , then $A \in \mathcal {I}$ , thus $\omega \to (\mathcal {I}^+)^n_{k,l}$ is false for the coloring c. Assume $A \subseteq \omega $ such that $|colors(c, A)| \le l$ .

Since $|colors(R^n_k, f[A])| \le l$ , $f[A] \in \mathcal {R}^n_{k,l}$ . As f is a Katětov function, $f^{-1}[f[A]] \in \mathcal {I}$ , and $A \subseteq f^{-1}[f[A]]$ , therefore $A \in \mathcal {I}$ .

For the converse implication, suppose $\omega \not \to (\mathcal {I}^+)^n_{k,l}$ . There exists a coloring $c: [\omega ]^n \to k$ such that if $|c[[A]^n]| \le l$ , then $A \in \mathcal {I}$ . By proposition 3.2, there is an injective function $f: \omega \to \omega $ such that for every $a_1, a_2, \dots , a_n$ , distinct natural numbers, $c(\{a_1, a_2, \dots , a_n\}) = R^n_k(\{f(a_1), f(a_2), \dots , f(a_n)\})$ . This function f serves as a Katětov function proving that $\mathcal {R}^n_{k,l} \le _K \mathcal {I}$ .

Proof. In the game $G_1(\mathcal {ED})$ , Player I can effectively utilize a simple strategy. Let $n_{-1} = 0$ . Player I initiates the game by choosing $A_0 = \{n_{-1}\} \times \omega $ and $B_0 = \{n \in \omega : n> 0\} \times \omega $ . If Player II selects $A_0$ , Player II immediately loses. Thus, Player II must choose $B_0$ and pick a pair $(n_0, m_0) \in B_0$ .

Subsequently, at each round i, Player I presents $A_i = \{n \in \omega : n_{i-2} < n \leq n_{i-1}\} \times \omega $ and $B_i = \{n \in \omega : n> n_{i-1}\} \times \omega $ . Player II is forced to consistently choose $B_i$ and select a pair $(n_i, m_i)$ . Ultimately, the set $\{(n_i, m_i) : i \in \omega \}$ forms a subset with at most one element in each column of $\omega ^2$ , which leads to Player II’s defeat.

Proof. To demonstrate that $\mathcal {ED}_m$ is an $F_\sigma $ ideal, it suffices to construct a lower semicontinuous submeasure $\varphi _m$ such that $\text {Fin}(\varphi _m) = \mathcal {ED}_m$ . Define $\varphi _m: \mathcal {P}(\omega ) \to \omega + 1$ as follows:

$$\begin{align*}\varphi_m(E) = \min \{ \alpha \in \omega + 1 : (\exists \mathcal{X} \in [\mathcal{A}_m \cup \mathcal{S}_0]^\alpha)(E \subseteq \bigcup \mathcal{X}) \}. \end{align*}$$

It is easily verified that $\varphi _m$ is a lower semicontinuous submeasure such that $\text {Fin}(\varphi _m) = \mathcal {ED}_m$ . The ideal $\mathcal {PC}$ introduced in the second section satisfies $\mathcal {PC} \subset \mathcal {ED}_m$ and is a tall ideal. Consequently, $\mathcal {ED}_m$ is also a tall ideal.

Proof. Let $ f_n : \omega \to \omega $ be a bijective function defined for every $ n \in \omega $ , mapping $ A_{(n)} $ to $ \omega $ . Assign to each $ k \in \omega $ an index $ i_k \in \omega $ such that $ k \in A_j $ if and only if $ i_k = j $ . We then define $ f: \omega \to \omega \times \omega $ by $ f(k) = (i_k, f_{i_k}(k)) $ . It’s not hard to see that $ f $ meets the desired condition; this function is simply a translation between the two definitions.

Similarly, assign a bijective function $ g_s : \omega \to \omega $ to each $ s \in \omega ^2 $ , mapping $ A_s $ to $ \omega $ . For every $ k \in \omega $ , define $ i_k \in \omega ^2 $ such that $ k \in A_j $ if and only if $ i_k = j $ . The function $ g: \omega \to \omega \times \omega \times \omega $ is defined as $ g(k) = (i_k(0), i_k(1), g_{i_k}(k)) $ . Once again, it is straightforward to verify that the function $ g $ demonstrates the equivalence of the two ideals.

Proof. For each $n \in \omega $ , let $x_n \in \omega ^\omega $ be a sequence such that $n \in A_{x_n \upharpoonright i}$ for every $i \in \omega $ . We define the function $split: [\omega ]^2 \to \omega $ by setting $split(\{n,k\}) = \min \{ i \in \omega : x_n(i) \neq x_k(i) \}$ , which is well-defined due to the third property in Definition 2.3.

Define a coloring $c: [\omega ]^2 \to m+2$ as follows:

$$\begin{align*}c(\{n, k\}) = \begin{cases} split(\{n,k\}) & \text{if } split(\{n,k\}) \le m, \\ m + 1 & \text{if } split(\{n,k\}) \ge m+1. \end{cases} \end{align*}$$

Note that if $A \subseteq \omega $ forms a monochromatic set for some color $i \le m$ , then there exists a set $B \supseteq A$ such that $B \in \mathcal {S}_0$ . Conversely, if $Y \subseteq \omega $ forms a monochromatic set for color $m+1$ , then there exists $s \in \omega ^{m+1}$ such that $Y \subseteq A_s \in \mathcal {A}_m$ . This coloring c demonstrates that $\mathcal {ED}_m$ does not satisfy the Ramsey property with $m+2$ colors. Furthermore, the set $\{ A \subseteq \omega : |colors(c, A)| = 1 \}$ forms a base for the ideal $\mathcal {ED}_m$ .

Proof. Firstly, observe that for any $n, m \in \omega $ , the set $A_{(n)}$ is a positive set within the ideal $\mathcal {ED}_{m+1}$ , confirming that $\mathcal {ED}_{m+1} \upharpoonright A_{(n)}$ forms an ideal.

Consider any sequence s in $\omega ^{m+1}$ . Let $(n)^\frown s$ denote the extended sequence in $\omega ^{m+2}$ where $(n)^\frown s(0) = n$ and the subsequent entries are given by $(n)^\frown s(i) = s(i-1)$ for $0 < i \le m+1$ . For any sequence $s \in \omega ^{m+1}$ , assign to s an arbitrary bijective function $f_s: A_{(n)^\frown s} \to A_s$ , mapping elements between these two infinite sets.

For any $k \in A_{(n)}$ , identify the unique sequence $s_k$ such that k belongs to $A_{(n) ^\frown s_k}$ . Construct the function $f: A_{(n)} \to \omega $ by assigning $f(k) = f_{s_k}(k)$ . Note that the function f is bijective and certifies that those two ideals are equivalent.

Proof. Let be an order isomorphism with respect to the tree ordering (the tree $T_M$ from Definition 2.5). For each $s \in T_M$ , let $f_s$ be a bijective function between the infinite set $A_s \cap M$ and $A_{g(s)}$ . For each $n \in M$ , there exists a unique $s_n \in T_M \cap \omega ^{m+1}$ such that $n \in A_s \cap M$ . We define the function $f: M \to \omega $ given by $f(n) = f_{s_n}(n)$ . It is not difficult to see that f, as defined, demonstrates the equivalence between the ideals $\mathcal {ED}_m \upharpoonright M$ and $\mathcal {ED}_m$ .

Proof. Let $\mathcal {I}$ be an ideal on $\omega $ , and suppose $X \subseteq \omega $ is a positive set for $\mathcal {I}$ . If $Y \subseteq X$ and $Y \notin \mathcal {I} \upharpoonright X$ , then $Y \notin \mathcal {I}$ . This argument substantiates the first two claims of the lemma.

For the third claim, note that the ideal $\mathcal {ED}_m$ is generated by $\mathcal {A}_m \cup \mathcal {S}_0$ . Proceeding by contraposition, assume $X \subseteq X_0 \cup \dots \cup X_k \cup Y_0 \cup \dots \cup Y_l$ , with $X_i \in \mathcal {A}_m$ and $Y_i \in \mathcal {S}_0$ . For each $0 \leq i \leq k$ , there exists $n_i$ such that $X_i \subseteq A_{(n_i)}$ . For each $0 \leq j \leq l$ and each $Y_j$ , consider two scenarios:

• • There exists $n_j$ such that $Y_j \subseteq A_{(n_j)}$ ,

• • For every $n \in \omega $ , $|A_n \cap Y_j| = 1$ .

Let N be a natural number greater than all $n_j$ and $n_i$ . Then, for each $k> N$ , we have $|A_{(k)} \cap X| \leq l$ , thus X does not satisfy the property posited by the proposition.

Proof. In both directions, consider a coloring $c: [\omega ]^2 \to 2$ . By applying the hypothesis to c and its complement $1 - c$ , one ensures the existence of either a positive set for one of the colors or arbitrarily large monochromatic subsets for both colors.

Proof. We aim to find a sequence $ s \in \omega ^{\leq m} $ such that $ A_s \cap X $ belongs to $ \mathcal {ED}_m^+ $ and for each $ n \in \omega $ , $ A_{s^\frown n} \cap X $ belongs to $ \mathcal {ED}_m $ . This is achieved recursively starting with the empty sequence $ \emptyset $ . If for every $ n \in \omega $ , $ A_{(n)} \cap X $ belongs to $ \mathcal {ED}_m $ , then we set $ s = \emptyset $ . Otherwise, if there exists an $ n_0 $ such that $ A_{(n_0)} \cap X $ is in $ \mathcal {ED}_m^+ $ , we initiate $ s_0 = (n_0) $ .

Continuing this process at step $ k $ , assuming $ n_0, n_1, \ldots , n_k $ are defined such that the finite sequence $ s_k = (n_0, n_1, \ldots , n_k) $ ensures $ A_{s_k} \cap X $ belongs to $ \mathcal {ED}_m^+ $ . If there exists $ n_{k+1} $ such that $ A_{s_k ^\frown n_{k+1}} \cap X $ remains in $ \mathcal {ED}_m^+ $ , the recursion continues; otherwise, we set $ s = s_k $ . The recursion cannot proceed indefinitely since for each $ t \in \omega ^{m+1} $ , $ A_t \cap X $ must belong to $ \mathcal {ED}_m $ , indicating a finite conclusion within at most $ m+1 $ steps.

Define $ X' = X \cap A_s $ . We shall prove that for each $ k \in \omega $ , the set $ I_k = \{ n \in \omega : |X' \cap A_{s^\frown n}| \geq k \} $ is infinite. Assuming the contrary for some $ k $ , that $ I_k $ is finite, leads to $ X' $ being contained in a finite union of $ X' \cap A_{s^\frown n} \in \mathcal {ED}_m $ and sets $ S_i \in \mathcal {S}_0 \subset \mathcal {ED}_m $ , placing $ X' $ in $ \mathcal {ED}_m $ .

Let $ (k_n)_{n \in \omega } $ be a sequence of distinct natural numbers, each $ k_n $ belonging to $ I_k $ , and select $ y_n \in [X' \cap A_{s^\frown k_n}]^n $ . Define $ Y $ as $ \bigcup _{n \in \omega } y_n $ . According to Lemma 3.5, $ Y $ constitutes a positive set for the ideal $ \mathcal {ED}_m $ . Furthermore, the function $ f: Y \to \Delta $ , which bijectively maps each $ y_n $ to $ \{(i, n) : i \in n\} \subset \Delta $ , verifies the equivalence of $ \mathcal {ED}_m \upharpoonright Y $ with $ \mathcal {ED}_{fin} $ .

Proof. The proof proceeds by induction on $n \in \omega $ to demonstrate the simultaneous validity of the following two Ramsey properties:

1. 1. $\omega \to (\mathcal {ED}_n^+)^2_{n+1}$ ,

2. 2. $\omega \to (<\omega , \dots , <\omega \mid \mathcal {ED}_n^+)^2_{n+2}$ .

The base case $n=0$ is straightforward, where the first property simply involves a one-color partitioning. The second property for $n=0$ has been previously established by D. Meza for the ideal $\mathcal {ED}$ and it is known that $\mathcal {ED}_0 \approx \mathcal {ED}$ . It can also be demonstrated using [Reference Hrušák, Meza-Alcántara, Thümmel and Uzcátegui9, Theorem 4.14], which states that $\omega \to (<\omega , \mathcal {ED}_{fin}^+)^2_2$ , hence particularly $\omega \to (<\omega , \mathcal {ED}^+)^2_2$ . By Proposition 3.9 and Remark 3.2 we have that $\omega \to (<\omega , <\omega \mid \mathcal {ED}_0^+)^2_2$ .

Assuming the properties hold for some $n \in \omega $ as our induction hypothesis, consider a coloring $c: [\omega ]^2 \to n+2$ . The objective is to identify a subset $X \subseteq \omega $ such that $X \in \mathcal {ED}_{n+1}^+$ and X is monochromatic under c.

For each $k \in \omega $ , applying Lemma 3.3 and Remark 3.2, we utilize the induction hypothesis for the ideal $\mathcal {ED}_{n+1} \upharpoonright A_{(k)} \approx \mathcal {ED}_n$ , ensuring that $A_{(k)} \to (<\omega , \dots , <\omega \mid (\mathcal {ED}_{n+1} \upharpoonright A_{(k)})^+)^2_{n+2}$ holds. This property suggests two scenarios:

• • There exists some $k \in \omega $ and $X \subseteq A_{(k)}$ such that X is monochromatic and not in $\mathcal {ED}_{n+1} \upharpoonright A_{(k)}$ . This case is straightforward, as Lemma 3.5 confirms $X \notin \mathcal {ED}_{n+1}$ .

• • For each $i \in n+2$ , and for all $k, N \in \omega $ , there exists $X_k \subseteq A_{(k)}$ of size N that is monochromatic in color i. We will construct a positive set analogous to the one defined in the third part of Lemma 3.5. This necessitates the introduction of the following lemma.

Using Lemma 3.7, we construct an increasing sequence $\langle k_j : j \in \omega \rangle $ and sets $a_j$ , $j \in \omega $ such that $k_j < k_{j+1}$ for all $j \in \omega $ . For each $j \in \omega $ , we choose a subset $a_j \subseteq A_{(k_j)}$ such that $|a_j| = j$ , $c(x, y) = i_j$ for all $x, y \in a_j$ with $x \neq y$ , and $c(x, y) = i_j$ for all $x in a_j$ and $y \in a_N$ for all $N> j$ . The set $\bigcup _{i_j = i} a_j$ is an i-monochromatic positive set for the ideal $\mathcal {ED}_{n+1}$ because there are only a finite number of colors and there exists $i \in n+2$ such that the set of indices $\{ j \in \omega : i_j = i \}$ is infinite. This completes the first part of the proof: $\omega \to (\mathcal {ED}_{n+1}^+)^2_{n+2}$ .

The proof that $\omega \to (<\omega , \dots , <\omega \mid \mathcal {ED}_{n+2}^+)^2_{n+3}$ holds is as follows. Consider a coloring $c: [\omega ]^2 \to n+3$ . Define $c': [\omega ]^2 \to n+2$ as:

$$\begin{align*}c'(\{ a, b \}) = \begin{cases} 0 & \text{if } c(\{a,b\}) = 0 \\ c(\{a,b\}) - 1 & \text{if } c(\{a,b\})> 0. \end{cases} \end{align*}$$

By the previous proof, there exists $B \subseteq \omega $ which is a monochromatic positive set for the coloring $c'$ . If B is monochromatic for some color $i> 0$ , then we are done because that i-monochromatic set for $c'$ is also an $i+1$ -monochromatic set for c.

Suppose $B \in \mathcal {ED}_{n+1}^+$ is monochromatic positive for color $0$ under the coloring $c'$ . For the coloring $c'$ restricted to $[A_{(0)}]^2$ , by the induction hypothesis applied to the ideal $\mathcal {ED}_{n+1} \upharpoonright A_{(0)} \approx \mathcal {ED}_n$ , there are two possibilities: there exists a color $i \in n+2$ and $C \subseteq A_{(0)}$ a monochromatic positive set for the ideal $\mathcal {ED}_{n+1} \upharpoonright A_{(0)}$ , by Lemma 3.5, C is also positive for the ideal $\mathcal {ED}_{n+1}$ . If the color of the monochromatic is different from $0$ , we conclude again. Finally, suppose that is not the case, then for each $N \in \omega $ and each $i> 0$ , there exists a $I_N$ monochromatic positive of size N and color i, for the coloring $c'$ ; that is, for the coloring c there are monochromatic sets of arbitrarily large sizes of all colors from $2$ onwards; we just need to find a positive monochromatic set of color $0$ , of color $1$ or well arbitrarily large finite monochromatic sets for both colors.

By Lemma 3.6, there exists a subset $Y \subseteq B$ such that $\mathcal {ED}_{n+1} \upharpoonright Y \approx \mathcal {ED}_{fin}$ , by [Reference Hrušák, Meza-Alcántara, Thümmel and Uzcátegui9, Theorem 4.14], it is established that $\omega \to (<\omega , \mathcal {ED}_{fin}^+)^2_2$ . By Proposition 3.9 and Remark 3.2 it is shown that $Y \to (<\omega , <\omega \mid (\mathcal {ED}_{n+1} \upharpoonright Y)^+)^2_2$ , that is, for the coloring c restricted to $[Y]^2$ , there exists a positive monochromatic of one of the two colors $0$ or $1$ , in which case we finish because a positive for the ideal $\mathcal {ED}_{n+1} \upharpoonright Y$ , is also a positive for the ideal $\mathcal {ED}_{n+1}$ , or there are arbitrarily large finite homogeneous sets, which also concludes the proof because we already knew that there were arbitrarily large finite monochromatic sets for all other colors (in this case).

Proof. For any $ n \in \omega $ , it holds that $ \mathcal {ED}_\omega \subset \mathcal {ED}_n $ , $ \mathcal {PC}(f) \subset \mathcal {ED}_n $ , and $ \mathcal {TC}(f) \subset \mathcal {ED}_n $ . This implies that any positive set for $ \mathcal {ED}_n $ is also a positive set for the other three ideals.

Question 3.1. Does the property $\omega \to (\mathcal {PC}^+)^n_k$ hold for $n> 2$ ?

Question 4.1. Does player II have a winning strategy in $ G_1(\mathcal {S}) $ ?

Proof. Suppose for the sake of contradiction that $B \notin \mathcal {S}^+_{n+1}$ and $C \notin \mathcal {S}^+_{n+1}$ . Then, there exist clopen sets $V_1$ and $V_2$ of measure $\frac {1}{2^{n+1}}$ each, such that for every $U_1 \in B$ and $U_2 \in C$ , we have $V_1 \cap U_1 \neq \emptyset $ and $V_2 \cap U_2 \neq \emptyset $ . Therefore, $V = V_1 \cup V_2$ is a clopen set of measure at most $\frac {1}{2^n}$ that intersects every $U \in A$ , which contradicts the assumption that $A \in \mathcal {S}^+_n$ . This completes the proof.

Proof. By the previous lemma, if $A \in \mathcal {S}^+_\omega $ and A can be written as the union of B and C, then either $B \in \mathcal {S}^+_\omega $ or $C \in \mathcal {S}^+_\omega $ . Additionally, it can be observed that $\Omega \in \mathcal {S}^+_\omega $ .

Proof. To demonstrate that $\mathcal {S}_\omega $ is an $F_{\sigma \delta }$ ideal, it suffices to show that $\mathcal {S}_n^+$ is a $G_\delta $ set for every $n \in \omega $ . Given that $\Omega _n$ is countable, for each $V \in \Omega _n$ , the set

$$\begin{align*}X_V = \{ A \subseteq \Omega : (\exists U \in A) (U \cap V = \emptyset) \} \end{align*}$$

is open in $2^\Omega $ . Consequently, $\mathcal {S}_n^+ = \bigcap _{V \in \Omega _n} X_V$ is a $G_\delta $ set.

Moreover, for every $n>0$ and for each $A \in \mathcal {S}_n^+$ , we have that $A \in \mathcal {S}^+$ .

Proof. This proof leverages the following claim, which will be applied repeatedly:

Claim: If $A \in \mathcal {S}^+_n$ and $B \in \mathcal {S}_\omega $ , then $A \setminus B \in \mathcal {S}^+_{n+1}$ . This follows directly from Lemma 4.1.

Consider a positive set $A \in \mathcal {S}_\omega ^+$ and a coloring $c: [A]^2 \to 2$ . We aim to recursively define a sequence of sets $\{A_n \in \mathcal {S}_\omega ^+ : n \in \omega \}$ and elements $x_n \in A_n$ such that:

• • $A_0 = A$ ,

• • $x_n \in A_n$ ,

• • $A_{n+1} = \{ y \in A_n : c(\{x_n, y\}) = 0 \}$ .

If the recursive construction continues indefinitely, then $\{x_n : n \in \omega \}$ forms an infinite $0$ -homogeneous set, fulfilling our requirement. If the process halts, then there exists an $n \in \omega $ such that for every $x \in A_n$ , the set $\{y : c(\{x, y\}) = 0\} \subseteq \mathcal {S}_\omega $ . Given $A_n \in \mathcal {S}_\omega ^+$ , there is some $m \in \omega $ such that $A_n \in \mathcal {S}_m^+$ .

Let $\{V_i : i \in \omega \}$ enumerate $\Omega _{m+1}$ . Choose $x_0 \in A_n$ such that $x_0 \cap V_0 = \emptyset $ , possible because $A_n \in \mathcal {S}_m^+$ and thus in $\mathcal {S}_{m+1}^+$ . Define

$$\begin{align*}B_1 = \{y \in A_n : c(\{x_0, y\}) = 1\},\end{align*}$$

and note, by the claim, $B_1 \in \mathcal {S}_{m+1}^+$ .

Recursively, for each $ i \in \omega $ , define the following:

• • $ B_i \in \mathcal {S}_{m+1}^+ $ ,

• • an element $x_i \in B_i$ such that $x_i \cap V_i = \emptyset $ ,

• • $B_{i+1} = A_n \setminus \bigcup _{j < i} \{y \in B_j : c(\{x_i, y\}) = 0\}$ .

Here, $A_n \in \mathcal {S}_m^+$ and $\bigcup _{j < i} \{y \in B_j : c(\{x_i, y\}) = 0\} \in \mathcal {S}_\omega $ , hence by the claim, $B_{i+1} \in \mathcal {S}_{m+1}^+$ for each $i \in \omega $ ( $m+1$ is a fixed number). By construction, $\{x_i : i \in \omega \}$ is a $1$ -homogeneous $\mathcal {S}_\omega $ -positive set, concluding the proof.

Proof. Let $\mathcal {J}$ be an $F_\sigma $ ideal such that $\mathcal {I} \subseteq \mathcal {J}$ , and let $\phi $ be a lower semicontinuous submeasure such that $\text {Fin}(\phi ) = \mathcal {J}$ . A winning strategy for player II is as follows: in the nth move, player I partitions a positive set of the ideal $\mathcal {J}$ . Then, player II chooses one of the partitions, which remains a positive set of the ideal $\mathcal {J}$ , and selects $a_n$ to be a finite set such that $\phi (a_n) = n$ . At the end of the game, we have that $\phi \left (\bigcup _{n \in \omega } a_n\right ) = \infty $ , thus player II wins the game.

Proof. Let $ a_0 = 0 $ and $ b_0 = 1 $ . To begin, Player I plays the partition $ [0, \frac {1}{2}] \cup (\frac {1}{2}, 1] $ (note that these intervals are in the rational numbers, not the real numbers). Player II then selects one of the intervals, either $ [a_0, \frac {a_0 + b_0}{2}] $ or $ (\frac {a_0 + b_0}{2}, b_0] $ , and provides a finite set $ F_0 $ contained within the chosen interval.

Recursively, we construct two sequences of rational numbers, $ \{a_n : n \in \omega \} $ and $ \{b_n : n \in \omega \} $ , such that:

• • $ a_0 = 0 $ and $ b_0 = 1 $ ,

• • if, in the $ i $ th move, Player II chooses $ [a_i, \frac {a_i + b_i}{2}] $ , then $ a_{i+1} = a_i $ and $ b_{i+1} = \frac {a_i + b_i}{2} $ ,

• • if, in the $ i $ th move, Player II chooses $ (\frac {a_i + b_i}{2}, b_i] $ , then $ a_{i+1} = \frac {a_i + b_i}{2} $ and $ b_{i+1} = b_i $ .

Player I’s winning strategy is as follows: in the $ i $ th move, Player I partitions $ [a_i, b_i) $ as $ [a_i, \frac {a_i + b_i}{2}] \cup (\frac {a_i + b_i}{2}, b_i] $ . Player II must then choose one of these two intervals and provide a finite set $ F_i $ within the chosen interval.By the end of the game, the sequence of finite sets chosen by Player II will have a unique accumulation point.

Proof. The reason Player II has a winning strategy in $G_{fin}(\mathcal {S})$ is that $\mathcal {S}$ is an $F_\sigma $ ideal.