Determine if $\mathbb{R}^{\omega}$ with the product topology is second countable

Question

Determine if $\mathbb{R}^{\omega}$ with the product topology is second countable.

I do not know if this space is second countable or not, I have noticed this post Product of infinite discrete space is second countable , and I realize that if $\mathbb{R}$ is discrete then the result is obtained but I am not sure of the discrete definition and if this post is useful for me What I want, could someone help me please? Thank you.

Answer 1

Let $(X_n, \mathcal{T}_n)$ be a second countable space with countable base $\mathcal{B}_n$. Let $X = \prod_n X_n$ in the product topology. Then $X$ is second countable.

Define $$\mathcal{B} = \{\prod_n O_n: \exists F \subseteq \mathbb{N} \text { finite }: (\forall n \in F: O_n \in \mathcal{B}_n) \land (\forall n \in \mathbb{N}\setminus F: O_n = X_n)\}$$

and note that all members of $\mathcal{B}$ are basic open sets for the standard base of the product (products of open sets, all but finitely many of which are the whole space). It's easy to show that because all $\mathcal{B}_n$ are themselves bases for their respective factor space, $\mathcal{B}$ is a base for $X$, and it's countable, as for each fixed $F$, we have $\aleph_0^{|F|} = \aleph_0$ many sets of the required form, and we have $\aleph_0$ many finite subsets of $\mathbb{N}$. So $|\mathcal{B}| = \aleph_0 \times \aleph_0= \aleph_0$, as required.

As $\mathbb{R}$ is second countable (rational-ended open intervals), so is $\mathbb{R}^\omega$ in the product topology.

For your linked post (countable discrete spaces) we use the base of singletons in each factor to see the space is second countable.

Answer 2

An open base set of that space is of the form (call it an n form)
U = U1×U2×..×U_n×R×R×... for some open U1,.. U_n.
Each Uj is a union of open base sets U_jk, k = 1,2,...

Thus U is a union of open sets of the form
U_1,k1 x U_2,k2 ×..× U_n.kn × R × R × ...

Consequently, as there are only countably many of those forms,
every open set of the n form is a countable union of open sets of the form
U_1,k1 x U_2,k2 ×..× U_n.kn × R × R × ...

Each of the open base sets for R^omega is in the n form for some n in N, thus a countable union of the above described product of open sets. As omega × omega = omega, I have described a countable collection of open sets for which every open set of R^omega is a union of some of them.