Proposition 26.2.3.label Let $\seq{X_j}$ be topological spaces, $X = \prod_{n \in \natp}X_{n}$, then:
- (1)
$\bigotimes_{n \in \natp}\cb_{X_n}\subset \cb_{X}$.
- (2)
If $\seq{X_n}$ are separable and metrisable, then $\bigotimes_{n \in \natp}\cb_{X_n}= \cb_{X}$.
Proof, [Proposition 1.5, Fol99] and [Proposition 8.1.7, Coh13]. (2): For each $J \subset \natp$ finite, $x \in X$, and $\eps > 0$, define
By Proposition 5.5.7, $X$ is also separable. Let $\seq{x_n}\subset X$ be a countable dense subset, and
For each $x \in X$ and $U \in \cn_{X}(x)$, there exists $J \subset X$ finite and $k \in \natp$ such that $x \in B_{J}(x, 1/k) \subset U$. By density of $\seq{x_n}$, there exists $n \in \natp$ such that $x_{n} \in B_{J}(x, 1/(2k))$. In which case, $x \in B_{J}(x_{n}, 1/(2k)) \subset B_{J}(x, 1/k) \subset U$. Thus $\fB$ is a countable base for the product topology, consisting of cylinder sets in $\bigotimes_{n \in \natp}\cb_{X_n}$.
As such, every open set in $X$ is a countable union of elements of $\fB$. Therefore $\bigotimes_{n \in \natp}\cb_{X_n}\supset \cb_{X}$.$\square$
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