Showing a characterization of Ramsey ideals

Question

I'm studying the Filipow and Szuca's article "Ideal version of Ramsey's theorem", and I'm having some problems showing Theorem 3.11. Given a sequence $(x_n)_{n \in \mathbb{N}}$ of on a topological space $X$ and $\mathcal{I}$ an ideal on $\mathbb{N}$ containing the ideal $\mathsf{Fin}$ of the finite subsets of $\mathbb{N}$, we say that the sequence $(x_n)$ is $\textit{$\mathcal{I}$-convergent}$ to some $x \in X$ if $\{ n \in \mathbb{N}\,:\,x_n \notin U \} \in \mathcal{I}$, for every open set $U$ with $x \in U$. When $\mathcal{I} = \mathsf{Fin}$, we have the usual notion of convergence.

Proposition: Let $\mathcal{I}$ be a $q^+$-ideal, which means that for any $A \in \mathcal{I}^+$ and every partition $(F_n)_{n \in \mathbb{N}}$ of $A$ into finite sets, there is $S \in \mathcal{I}^+$ such that $S \subseteq A$ and $|S \cap F_n| \leq 1$, for each $n$. Then the following are equivalent:

1. $\mathcal{I}$ is Ramsey at $\mathbb{N}$, i.e., for any coloring $c: [\mathbb{N}]^2 \to \{ 0,1 \}$ there is a $c$-homogeneous set $A \in \mathcal{I}^+$, where $\mathcal{I}^+$ denotes the coideal of $\mathcal{I}$.
2. $\mathcal{I}$ is monotone (an ideal $\mathcal{I}$ is $\textit{monotone}$ if for any sequence $(x_n)_{n \in \mathbb{N}}$ of real numbers, there is some $A \in \mathcal{I}^+$ such that the subsequence $(x_n)_{n \in A}$ is monotone).
3. $\mathcal{I}$ has the finite Bolzano-Weierstrass property (an ideal $\mathcal{I}$ has the $\textit{finite Bolzano-Weierstrass property}$ if for any bounded sequence $(x_n)_{n \in \mathbb{N}}$ of real numbers, there is some $A \in \mathcal{I}^+$ such that the subsequence $(x_n)_{n \in A}$ is convergent).

Proof: (1) $\Rightarrow$ (2) Suppose $\mathcal{I}$ is Ramsey at $\mathbb{N}$ and let $(x_n)$ be a sequence of real numbers. Since $\mathcal{I}$ is Ramsey at $\mathbb{N}$, given a coloring $c: [\mathbb{N}]^2 \to \{ 0,1 \}$ there is a $c$-homogeneous $A \in \mathcal{I}^+$, and since $\mathcal{I}$ is $q^+$, for every partition $(F_n)_{n \in \mathbb{N}}$ of $A$ into finite sets, there is $S \in \mathcal{I}^+$ such that $S \subseteq A$ and $|S \cap F_n| \leq 1$, for each $n$. So, for each $n$, let $k_n \in S \cap F_n$ (I guess I can choose $\mathcal{I}$-convergent of $(x_n)$ using these $k_n$, but I don't know how to do it).

(2) $\Rightarrow$ (3) Not difficult.

(3) $\Rightarrow$ (1) Suppose $\mathcal{I}$ has the finite Bolzano-Weierstrass property and let $c: [\mathbb{N}]^2 \to \{ 0,1 \}$ be a coloring. For any bounded sequence $(x_n)$ of real numbers, let $A \in \mathcal{I}^+$ such that the subsequence $(x_n)_{n \in A}$ is convergent. (Here, I don't know how to find a $c$-homogeneous set using the subsequence )

Answer 1

The implication ``$(1)\to (2)$'' does not need the assumption that $\mathcal{I}$ is a $q^+$-ideal. It follows from the well-known argument: let's color all pairs of natural numbers according to the scheme: $$c(\{a,b\})=\left\{\begin{array}{lllll} 0 & \textrm{iff} & a<b & \textrm{and} & x_a\leq x_b,\\ 1 & \textrm{iff} & a<b & \textrm{and} & x_a> x_b. \end{array} \right.$$ Then for any $c$-homogeneous set $A\in\mathcal{I}^+$, $(x_n)_{n\in A}$ is monotone.