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src/topology/main/interiorclosureboundary.tex

@@ -91,10 +91,16 @@
 
 \begin{proposition}
 \label{proposition:separable-product}
-    Let $\seqf{X_j}$ be separable topological spaces, then $\prod_{j = 1}^n X_j$ is also separable.
+    Let $\seq{X_n}$ be separable topological spaces, then $\prod_{n \in \natp}X_n$ is 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 \autoref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
+    For each $n \in \nat$, let $A_n \subset X_n$ be a countable dense subset. For each $N \in \nat$, let $B_N \subset X$ such that:
+    \begin{enumerate}
+        \item $B_N$ is countable.
+        \item For each $\bracsn{a_n}_1^N \subset \prod_{n = 1}^N A_n$, there exists $b \in B_N$ such that $a_n = b_n$ for all $1 \le j \le N$. 
+    \end{enumerate}
+
+    Then, $\bigcup_{N \in \natp}B_N$ is a countable dense subset of $\prod_{n \in \natp}X_n$.
 \end{proof}