Proposition 4.5.4. Let $X$ be a topological space and $\seqf{E_j}\subset 2^{X}$, then $\bigcup_{j = 1}^{n} \ol{E_j}= \ol{\bigcup_{j = 1}^n E_j}$.
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 $\ol{\bigcup_{j = 1}^n E_j}\supset \bigcup_{j = 1}^{n}\ol{E_j}$.$\square$