How to show that $g:2^M\to 2^\mathbb{N}$ defined by $g(A) = X\cup A$ is continuous?

Question

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

Definition $5$: A set $S\subseteq 2^\mathbb{N}$ 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 $2^\mathbb{N}?$

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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 $g:2^M\to 2^\mathbb{N}$ by $$g(A) = X\cup A.$$ 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.

Answer 1

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.

Answer 2

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