Union of closure of sets is the closure of the union: true for finite, false for infinite unions 
Let $A_i$ be a subset of a metric space for each $i\in \mathbb{N}$.
  
  
*
  
*Let $B_n := \bigcup_{i=1}^n A_i$. Prove (for any) $n \in \mathbb{N}$ that $\overline{B_n} = \bigcup_{i=1}^n \overline{A_i}$.
  
*If $B = \bigcup_{i=1}^\infty A_i$, prove that $\overline{B} \supseteq \bigcup_{i=1}^\infty \overline{A_i}$. Give an example to show that this containment might be proper.
  

If $A_i$ is closed then $A_i = \bar A_i$, but I'm stuck as to how to prove $B=\bar B_n$. If I prove the first statement for when $A_i$ is closed does that mean it is also true for when $A_i$ is open because I can construct a closed set containing $A_i$?
For the example, would constructing a sequence of closed segments between $0$ and $1$ that gets arbitrarily close to $1$ and taking the union of the segments be considered a proper containment?
 A: To prove the true statements we'll make use of the following facts:


*

*If $A \subseteq B$, then $\overline{A} \subseteq \overline{B}$. 

*$x \in \overline{A}$ iff $U \cap A \neq \varnothing$ for every open neighbourhood $U$ of $x$.  (This is in the more general context of general topological spaces. For metric spaces it suffices to restrict the $U$s to the open $\varepsilon$-balls centred at $x$.)



To see that $\bigcup_{i=1}^n \overline{A_i} \subseteq \overline{B_n}$ observe that $A_i \subseteq B_n$ for each $i \leq n$, and apply the first fact above.
To see that $\overline{B_n} \subseteq \bigcup_{i=1}^n \overline{A}_i$ note that this is equivalent to showing $X \setminus \bigcup_{i=1}^n \overline{A}_i \subseteq X \setminus \overline{B_n}$. If $x \in X \setminus \bigcup_{i=1}^n \overline{A_i}$, then by the second fact above it follows that for each $i \leq n$ there is an open neighbourhood $U_i$ of $x$ which is disjoint from $A_i$. Now note that $U := U_1 \cap \cdots \cap U_n$ is an open neighbourhood of $x$ which is disjoint from $B_n$.

Note that the first fact above also shows that $\bigcup_{i=1}^\infty \overline{A_i} \subseteq \overline{\bigcup_{i=1}^\infty A_i} = \overline{B}$.
To give an example of where the equality does not hold, just enumerate the set $\mathbb{Q}$ of rational numbers as $\{ a_i : i \in \mathbb{N} \}$, and for each $i$ define $A_i = \{ a_i \}$. Working in $\mathbb{R}$ with the usual (metric/order) topology, we can show that
$$
{\textstyle \bigcup_{i=1}^\infty} \overline{A_i} = \mathbb{Q} \neq \mathbb{R} = 
\overline { {\textstyle \bigcup_{i=1}^\infty} A_i }.
$$
A: Hint 1: for $\overline{A_1}\cup\overline{A_2}\subset\overline{A_1\cup A_2}$
Suppose that $x\in\overline{A_1}$. This means that for any open $U$ containing $x$, $U\cap A_1\not=\varnothing$. Obviously, for any $U$ containing $x$, $U\cap(A_1\cup A_2)\supset U\cap A_1\not=\varnothing$. Thus, if $x\in\overline{A_1}$, then $x\in\overline{A_1\cup A_2}$; that is, $\overline{A_1}\subset\overline{A_1\cup A_2}$.
Hint 2: for $\overline{A_1\cup A_2}\subset\overline{A_1}\cup\overline{A_2}$
Suppose that $x\in\overline{A_1\cup A_2}$. This mean that for any open $U$ containing $x$, $U\cap(A_1\cup A_2)\not=\varnothing$. Suppose that $x\not\in\overline{A_1}$; that is, for some open $U_1$ containing $x$, we have that $U_1\cap A_1=\varnothing$, and that $x\not\in\overline{A_2}$; that is, for some open $U_2$ containing $x$, we have $U_2\cap A_2=\varnothing$. Consider $U=U_1\cap U_2$. Obviously, $U$ is open and $x\in U$. Find a contradiction.
Hint 3: for $\bigcup\limits_{i=1}^n\overline{A_i}=\overline{\bigcup\limits_{i=1}^nA_i}$
Use induction on $\overline{A_1}\cup\overline{A_2}=\overline{A_1\cup A_2}$.
Hint 4: for $\bigcup\limits_{i=1}^\infty\overline{A_i}\subset\overline{\bigcup\limits_{i=1}^\infty A_i}$
If $x\in\bigcup\limits_{i=1}^\infty\overline{A_i}$ then there is some $n$ so that $x\in\overline{A_n}$. Use Hint 3 to show $\overline{A_n}\subset\overline{\bigcup\limits_{i=1}^nA_i}$ and Hint 1 to show that $\overline{\bigcup\limits_{i=1}^nA_i}\subset\overline{\bigcup\limits_{i=1}^\infty A_i}$
A: Notice that $A_i \subset B_n \subset \overline{B_n}$. Since $\overline{B_n}$ is closed it follows that $\overline{A_i}\subset \overline{B_n}$, hence it follows that $\cup_{i=1}^n \overline{A_i}\subset \overline{B_n}$. This is true even with arbitrary unions.
Furthermore, the union of a finite number of closed sets is closed, and we have $B_n \subset \cup_{i=1}^n \overline{A_i}$, so it follows that $\overline{B_n} \subset \cup_{i=1}^n \overline{A_i}$. This is not necessarily true with infinite unions.
Another example illustrating the latter point is $A_i = \frac{1}{i} = \overline{A_i}$, then $0 \in \overline{B} \setminus \cup_{i=1}^\infty \overline{A_i}$.
A: $x \in \bar{A} \cup \bar{B}$


*

*iff $\forall \delta >0,B_\delta(x) \,\cap A \neq \varnothing$ or
$B_\delta(x)\,\cap B \neq \varnothing$

*iff $B_\delta(x) \,\cap A\cup B \neq \varnothing$

*iff $x \in \overline{A\cup B}$
$x\in \overline{A\cup B}$ 


*

*iff $\forall \delta >0, B_\delta(x) \cap A\cup B \neq \varnothing$

*iff $B_\delta(x)\cap A \neq \varnothing$ or $B_\delta(x) \cap B \neq \varnothing$

*iff $x\in \overline{A}$ or $x \in \overline{B}$

*iff $x\in \overline{A} \cup \overline{B}$
