Incremental update.

src/topology/main/interiorclosureboundary.tex

@@ -51,6 +51,15 @@
     Since $f$ is continuous, $f^{-1}(\ol{f(A)})$ is closed and contains $A$.
 \end{proof}
 
+\begin{proposition}
+\label{proposition:closure-finite-union}
+    Let $X$ be a topological space and $\seqf{E_j} \subset 2^X$, then $\bigcup_{j = 1}^n \ol{E_j} = \ol{\bigcup_{j = 1}^n E_j}$.
+\end{proposition}
+\begin{proof}
+    Since $\bigcup_{j = 1}^n \ol{E_j}$ is closed, $\bigcup_{j = 1}^n \ol{E_j} \supset \ol{\bigcup_{j = 1}^n E_j}$. On the other hand, $\ol{\bigcup_{j = 1}^n E_j} \supset E_j$ for each $1 \le j \le n$. Thus $\ol{\bigcup_{j = 1}^n E_j} \supset \ol{E_j}$ for each $1 \le j \le n$, and $\ol{\bigcup_{j = 1}^n E_j} \supset \bigcup_{j = 1}^n\ol{E_j}$.
+\end{proof}
+
+
 
 \begin{definition}[Dense]
 \label{definition:dense}