Krdytkyu

In Galvin and Prikry's paper, they inroduce completely Ramsey sets.

Definition $5$: A set $Ssubseteq 2^mathbb{N}$ is completely Ramsey if $f^{-1}(S)$ is Ramsey for every continuous mapping $f:2^mathbb{N}to 2^mathbb{N}.$

Question: What is a topology in $2^mathbb{N}?$

As mentioned by @Patrick Stevens below, the authors consider product topology on $2^mathbb{N}.$

I have another question:

Question: Suppose that $X$ is a finite subset of $mathbb{N}$ and $M$ is a countable subset of $mathbb{N}.$
Define $g:2^Mto 2^mathbb{N}$ by
$$g(A) = Xcup A.$$
How to show that $g$ is continuous?

The function $g$ is defined in the proof of Lemma $7$ of the paper above.
The authors do not prove it.

A map into a product is continuous iff all the compositions with the projections from that product are continuous.

Now $pi_n circ g$ is the constant $1$ map for $n in X$, the identity for all other $n$, so always continuous, so $g$ is too:

Note that $A$ is just identified with $chi_A: mathbb{N} to 2$, so $n in A$ is just the same as $pi_n(A)=1$. So $pi_n(A cup X)$ is $1$ for all $n in X$ and all $n in A$ and $0$ for all $ n notin A cup X$. So when $n notin X$, $pi_n(g(A))= pi_n(A cup X) = pi_n(A)$ and for all $n in X$, $pi_n(X cup A) = 1$.

Finiteness of $X$ nor infiniteness of $M$ plays any part.

As the paper says in the first paragraph, it's the product topology.