Me when I forget to commit.

src/topology/main/interiorclosureboundary.tex

@@ -78,6 +78,15 @@
     Let $x \in \prod_{i \in I}X_i$ and $U \in \cn(x)$, then there exists $J \subset I$ finite and $\seqf{U_j}$ such that $U_j \in \cn(\pi_j(x))$ for each $j \in J$, and $x \in \bigcap_{j \in J}\pi_j^{-1}(U_j) \subset U$. By density of each $A_j$, there exists $y \in X$ such that $y_j \in U_j$ for each $j \in J$. Thus $y \in U$, and $A$ is dense.
 \end{proof}
 
+\begin{proposition}
+\label{proposition:separable-product}
+    Let $\seqf{X_j}$ be separable topological spaces, then $\prod_{j = 1}^n X_j$ is also separable.
+\end{proposition}
+\begin{proof}
+    Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \ref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
+\end{proof}
+
+
 \begin{lemma}
 \label{lemma:closurecomplement}
     Let $X$ be a topological space and $A \subset X$, then $(A^o)^c = \overline{A^c}$.