A Ramsey not Completely Ramsey Set of $[\omega]^\omega$. Let $a \in [\omega]^{<\omega}$ (a finite subset), $A \in [\omega]^{\omega}$ (an infinite subset). Let us define $$[a, A] = \\{a \cup B: B \in [A]^{\omega} \wedge max(a) < min(B) \\}.$$ These sets forms a basis for the Ellentuck topology on $[\omega]^{\omega}$. A set $X \subset [\omega]^{\omega}$ is said to be **Ramsey** if there is $B \in [\omega]^{\omega}$ such that either $[B]^\omega \subset X$ or $X \cap [B]^\omega = \emptyset$. A set $X \subset [\omega]^{\omega}$ is said to be **completely Ramsey** if for all $a \in [\omega]^{<\omega}$ and $A \in [\omega]^{\omega}$, there is a $B \in [A]^{\omega}$ such that either $[a, B] \subset X$ or $[a, B] \cap X = \emptyset$. Now, it is obvious that completely Ramsey implies Ramsey. I tried to find the counterexample to the converse but I did not succeed. PS.: I know there are alternative definitions. Thanks.

Define an equivalence relation $\sim$ on $[\omega]^\omega$ by $A\sim B$ iff $|A\mathop{\triangle}B|<\omega$, where the operation is symmetric difference. In each $\sim$-equivalence class $\mathscr{E}$ fix a representative $A_{\mathscr{E}}$. For each $B\in[\omega]^\omega$ let $\mathscr{E}(B)$ be the $\sim$-class of $B$, and let

$$c(B)=\begin{cases} 0,&\text{if }|B\mathop{\triangle}A_{\mathscr{E}(B)}|\text{ is even}\\\ 1,&\text{if }|B\mathop{\triangle}A_{\mathscr{E}(B)}|\text{ is odd}\;. \end{cases}$$

It’s not hard to check that $c$ is a $2$-coloring of $[\omega]^\omega$ with no infinite monochromatic set.

Let $Y=\left\\{B\in[\omega]^\omega:c(B)=0\right\\}$, and let $X=Y\cup\left\\{B\in[\omega]^\omega:1\
otin B\right\\}$. Then $[\omega\setminus 2]^\omega\subseteq X$, so $X$ is Ramsey. However, there is no $B\in[\omega]^\omega$ such that $[\\{1\\},B]\subseteq X$ or $[\\{1\\},B]\cap X=\varnothing$, so $X$ is not completely Ramsey.