0
$\begingroup$

Here, we identify subsets of $\mathbb{N}$ with sequences obtained by listing the members of the set in strictly increasing order. Suppose that we have some set $\mathcal{F}$ of sets (sequences) of the form $(m_n, l_n)_{n=1}^t$. We let $\mathcal{U}$ denote the set of all $(m_n, l_n)_{n=1}^\infty$ such that $(m_n, l_n)_{n=1}^t\in \mathcal{F}$ for all $t\in\mathbb{N}$. We let $\mathcal{V}$ denote the set of all $(m_n)_{n=1}^\infty$ such that there exists $(l_n)_{n=1}^\infty$ such that $(m_n, l_n)_{n=1}^\infty \in \mathcal{U}$.

It is easy to see that $\mathcal{U}$ is closed and $\mathcal{V}$ is analytic (in either the Cantor or Ellentuck topology). Suppose also that for every infinite subset $K$ of $\mathbb{N}$, there exists a further, infinite subset $M$ of $L$ such that every infinite subset of $M$ lies in $\mathcal{V}$.

Is it known that the latter property is equivalent to a winning strategy for a certain player in some two player game?

$\endgroup$
3
  • 1
    $\begingroup$ I don't get it. In the first sentence you talk about subsets of $\mathbb N$. So how is $(m_n,l_n))_{n=1}^t$ a subset of $\mathbb N$? Isn't it a subset of $\mathbb N\times\mathbb N$? $\endgroup$
    – bof
    Commented Jun 24, 2019 at 4:24
  • $\begingroup$ The object of interest is $\mathcal{V}$, which consists of subsets of $\mathbb{N}$. $\endgroup$
    – user78375
    Commented Jun 24, 2019 at 11:08
  • $\begingroup$ Voted to close as unclear what you're asking because I had the same question as @bof and the OP refused to answer it. $\endgroup$ Commented Jun 24, 2019 at 17:21

1 Answer 1

2
$\begingroup$

If you mean completely Ramsey sets there is the following characterization using the Kastanas game. Let's start with some definitions :

  1. Here the letters $a, b, c, ...$ vary over finite subsets of $\mathbb{N}$ and $A, B, C, ...$ over infinite subsets of $\mathbb{N}$.

    • Let $a$ and $A$, we write $a<A$ if $a=\emptyset$ or $max(a)<min(A)$.

    • For $a<A$, let $$[a, A]=\{S\in [\mathbb{N}]^{\aleph_{0}} : a\subseteq S\subseteq a\cup S \}= \{a\cup C\in [\mathbb{N}]^{\aleph_{0}} : a<C\subseteq A \}$$ Note that $[\emptyset, A]=[A]^{\aleph_{0}}$

    • A set $X\subseteq [\mathbb{N}]^{\aleph_{0}}$ is called Ramsey if there is $A$ with $[\emptyset, A]\subseteq X$ or $[\emptyset, A]\subseteq\hspace{0.1cm}\sim X$, here $\sim X$ denotes the complement of $X$.

    • A set $X\subseteq [\mathbb{N}]^{\aleph_{0}}$ is called completely Ramsey if for every $a<A$ there is $B\subseteq A$ with $[a, B]\subseteq X$ or $[a, B]\subseteq\hspace{0.1cm} \sim X$.

  2. Let $A \subseteq [\omega]^{\omega}$ = the set of infinite sets of integers. Then $A$ has a homogeneous set $H$ if, by definition, $H \in [\omega]^{\omega}$ and either every infinite subset of $H$ belongs to $A$ or every infinite subset of H belongs to the complement of $A$. $A$ has the Ramsey property iff it has a homogeneous set.

  3. Kastanas's game. For $\varphi\subseteq [\omega]^{\omega}$ we define the game $G_{\varphi}$ as follows :enter image description here

Player I wins if $\{n_{0}, n_{1}, \cdots \}\in\varphi$

Theorem (Kastanas)

a.) Player I has a winning strategy in $G_{\varphi}$, iff there is a homogeneous set in $\varphi$ (i.e. an infinite $H$ such that every infinite subset of it belongs to $\varphi$)

b.) Player II has a winning strategy in $G_{\varphi}$ iff for every $A$ there is a subset of it homogeneous in $[\omega]^{\omega}\setminus \varphi$.

For the information of the bibliography we have the following article "On the Ramsey Property for Sets of Reals" Author: Ilias G. Kastanas.

$\endgroup$
2
  • $\begingroup$ This is about as far as I had gotten. The hypothesis that for every infinite subset $K$ of $\mathbb{N}$, there exists a further subset $M$ which is homogeneous in $\mathcal{V}$. If we write $M=(m_n)_{n=1}^\infty$, this means there exists $(l_n)_{n=1}^\infty$ such that $(m_n, l_n)_{n=1}^t\in \mathcal{F}$ for all $t$. However, this is only good enough to let me choose $(l_n)_{n=1}^\infty$ after all $(m_n)_{n=1}^\infty$ have been chosen. $\endgroup$
    – user78375
    Commented Jun 24, 2019 at 0:47
  • $\begingroup$ Since whether or not a sequence $(m_n, l_n)_{n=1}^\infty$ lies in $\mathcal{V}$ depends only on its initial segments, I would like to know if it is possible to choose the $l_n$ sequence in a game without future information. That is, if every $A$ has a subset homogeneous in $\mathcal{V}$, is it possible to write down game where Players take turns choosing infinite subsets and $m_n, l_n$ values, and one of the players wins if $(m_n, l_n)_{n=1}^\infty\in \mathcal{U}$ (equivalently, $(m_n, l_n)_{n=1}^t\in \mathcal{F}$ for all $t$? Basically, can we choose the $l_n$ concurrently with the $m_n$? $\endgroup$
    – user78375
    Commented Jun 24, 2019 at 0:51

You must log in to answer this question.