Is there an uncountable extension of the Ramsey set $[\omega]^2$?

Question

We say that a family ${\cal A}\subseteq {\cal P}(\omega)$ is Ramsey if for every map $c:{\cal A}\to\{0,1\}$ there is an infinite set $X\subseteq \omega$ with the following properties:

1. ${\cal A}\cap {\cal P}(X)$ is infinite, and
2. the restriction $c|_{{\cal A}\cap ({\cal P}(X))}:{{\cal A}\cap ({\cal P}(X))}\to \{0,1\}$ is constant.

Let $[\omega]^2= \big\{\{n, m\}: n\neq m \in \omega\big\}$. Ramsey's Theorem states that $[\omega]^2$ is a Ramsey family.

Is there an uncountable Ramsey family ${\cal A}\subseteq {\cal P}(\omega)$ such that $[\omega]^2\subseteq {\cal A}$?

Answer 1

Yes. Just take $\mathcal{A}$ to be $[\omega]^2$ together with the powerset of the even integers.

Answer 2

If $N$ is an infinite proper subset of $\omega$ then the family $\mathcal A=[\omega]^2\cup\{A\subseteq\omega:A\not\subseteq N\}$ is a Ramsey family and $|\mathcal A|=2^{\aleph_0}$.

On the other hand, if $[\omega]^2\subseteq\mathcal A\subseteq\mathcal P(\omega)$ and $\mathcal A$ is a Ramsey family, then there is an infinite set $N\subseteq\omega$ such that $\mathcal A\cap\mathcal P(N)=[N]^2$; just consider the map $c:\mathcal A\to\{0,1\}$ such that $c(A)=0\iff|A|=2$.