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

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 $$2^\mathbb{N}?$$

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.

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.

• It seems that from your argument above, any function $g:2^M\to 2^\mathbb{N}$ is continuous. May I know where is my mistake? In particular, I do not see the definition of $g$ being used anywhere in your argument. Commented Nov 17, 2018 at 9:56
• @Idonknow No, I use the specific definition of $g$ to see the identities on the projections. There are many discontinuous maps from $2^M$ to $2^\mathbb{N}$. Commented Nov 17, 2018 at 9:58
• How do you obtain that $\pi_n\circ g$ is the constant function $1$? Commented Nov 17, 2018 at 10:00
• @Idonknow I expanded the argument. $X$ is always a subset of $g(A)$. Commented Nov 17, 2018 at 10:04

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

• I have modified my question. In particular, how do we show that a map $g:2^M\to 2^\mathbb{N}$ is continuous where $M$ is a countable subset of $\mathbb{N}?$ Commented Nov 17, 2018 at 9:47