Proposition 15.2.3. Let $\seqf{X_j}$ be topological spaces, $X = \prod_{j = 1}^{n} X_{j}$, then:
$\bigotimes_{j = 1}^{n} \cb_{X_j}\subset \cb_{X}$.
If $\seq{X_j}$ are separable, then $\bigotimes_{j = 1}^{n} \cb_{X_j}= \cb_{X}$
Proof. (2): By Proposition 4.5.7, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j}\subset X$ such that $\overline{\bracs{x_j|j \in \natp}}\supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k}\in \cn_{X_k}(\pi_{k}(x_{j}))$ for each $1 \le k \le n$ and $x_{j} \in \bigcap_{k = 1}^{n} \pi_{k}^{-1}(U_{j, k}) \subset U$.
Since $\bigcap_{k = 1}^{n} \pi_{k}^{-1}(U_{j, k}) \in \bigotimes_{j = 1}^{n} \cb_{X_j}$ and
\[U = \bigcup_{j \in \natp}\bigcap_{k = 1}^{n} \pi_{k}^{-1}(U_{j, k})\]
The open set $U$ is in $\bigotimes_{j = 1}^{n} \cb_{X_j}$.$\square$