TITLE: Rudin Exercise 2.7: union of subsets of a metric space, and closure thereof
QUESTION [9 upvotes]: I am trying to solve exercise 7 in Chapter 2 of Rudin and was hoping someone could look over my proof.

Let $A_1, A_2, A_3, \ldots$ be subsets of a metric space. 
a) Let $B_n = \bigcup\limits_{i=1}^n A_i$, prove that $\overline{B}_n = \bigcup\limits_{i=1}^n \overline{A}_i$, for $n = 1, 2, 3, \ldots$
b) If $B = \bigcup\limits_{i=1}^{\infty} A_i$, prove that $\overline{B} \supset \bigcup\limits_{i=1}^{\infty} \overline{A}_i$. 

Show, by an example, that this inclusion can be proper. 
Here is my attempt. 

a) The closure of a set is the smallest closed set containing it. Thus, for all $i$, $\overline{A}_i$ is closed. Further, the finite union of closed sets is closed. Thus, $\bigcup\limits_{i=1}^n \overline{A}_i$ is closed. Furthermore, $\overline{A}_i = A_i \cup (A_i)'$, where $(A_i)'$ is the set of limit points of $A$. Thus, $A_i \subset \overline{A}_i$ for all $i$, which implies that $\bigcup\limits_{i=1}^n A_i \subset \bigcup\limits_{i=1}^n \overline{A}_i$, i.e., $B_n \subset \bigcup\limits_{i=1}^n \overline{A}_i$. By Theorem $2.27$, for any metric space $X$ where $E, F \subset X$, if $E \subset F$ where $F$ is closed, then $\overline{E} \subset F$. Therefore, we deduce that $\overline{B}_n \subset \bigcup\limits_{i=1}^n \overline{A}_i$. Furthermore, 
  $A_i \subset \bigcup\limits_{i=1}^n A_i$ for any $i$, meaning that $A_i \subset B_n$ for any $i$. But $B_n \subset \overline{B}_n$, meaning that
  \begin{align*}
A_i \subset B_n \subset \overline{B}_n,
\end{align*}
  i.e., $A_i \subset \overline{B}_n$, where $\overline{B}_n$ is closed. Thus, Theorem 2.27 gives that $\overline{A}_i \subset \overline{B}_n$ for any $i$, and hence that $\bigcup\limits_{i=1}^n \overline{A}_i \subset \overline{B}_n$. 
Thus, $\overline{B}_n \subset \bigcup\limits_{i=1}^n \overline{A}_i$ and $\bigcup\limits_{i=1}^n \overline{A}_i \subset \overline{B}_n$, so $\overline{B}_n = \bigcup\limits_{i=1}^n \overline{A}_i$. 
b) Let $B = \bigcup\limits_{i=1}^{\infty} A_i$. Since $A_i \subset \bigcup\limits_{i=1}^{\infty} A_i$, $A_i \subset B$. But $B \subset \overline{B}$, so 
  \begin{align*}
A_i \subset B \subset \overline{B},
\end{align*}
  hence, 
  \begin{align*}
A_i \subset \overline{B}.
\end{align*}
  But $\overline{B}$ is closed, so by Theorem $2.27$, we have
  \begin{align*}
\overline{A}_i \subset \overline{B},
\end{align*}
  for any $i$, which implies that
  \begin{align*}
\bigcup\limits_{i=1}^{\infty} \overline{A}_i \subset \overline{B},
\end{align*}
  which can be written as 
  \begin{align*}
\overline{B} \supset \bigcup\limits_{i=1}^{\infty} \overline{A}_i.
\end{align*}
As for an example to show that this inclusion can be proper, let us consider: 
  \begin{align*}
A_i = \left[\frac{1}{i}, 1\right],
\end{align*}
  in which case each $A_i$ is closed, so $\overline{A}_i = A_i$, meaning that their infinite unions are the same. That is, 
  \begin{align*}
\bigcup\limits_{i=1}^{\infty} A_i = \bigcup\limits_{i=1}^{\infty} \overline{A}_i. 
\end{align*}
  However, we have: 
  \begin{align*}
\bigcup\limits_{i=1}^{\infty} A_i = \bigcup\limits_{i=1}^{\infty} \left[\frac{1}{i}, 1\right] = (0,1]. 
\end{align*}
  However, for the left-hand side, we get: 
  \begin{align*}
B = \bigcup\limits_{i=1}^{\infty} A_i \implies \overline{B} = \overline{\bigcup\limits_{i=1}^{\infty} A_i} = \overline{(0,1]} = [0,1].
\end{align*}
  Since $(0,1] \subset [0,1]$, $\bigcup\limits_{i=1}^{\infty} \overline{A_i}$ is properly contained in $B$. 

Any feedback would be greatly appreciated.

REPLY [2 votes]: Your proof of $\text{a)}$ is acceptable, but you can also prove it without (directly) using the definition of limit points or Theorem 2.27. You only need two propositions that you mentioned and accept as true:
Proposition 1: The finite union of closed sets is closed.
Proposition 2: The closure $\overline{A}$ of a set $A$ satisfies
$\tag 1  \overline{A} \text{ is a closed set}$
$\tag 2  A \subset \overline{A}$
$\tag 3  \text{IF } A \subset L \text{ and } L \text{ is closed THEN } \overline{A} \subset L$
Using just these propositions and simple concepts from elementary set theory, you can demonstrate that 
$\tag 4 \overline{B}_n = \bigcup\limits_{i=1}^n \overline{A}_i$