# Union of closure of sets is the closure of the union

Let $A_i$ be a subset of a metric space for $i\in \mathbb{N}$. $$\text{Let }B_n := \bigcup_{i=1}^n A_i. \text{Prove (for any } n\in \mathbb{N} \text{) that } \bar B_n = \bigcup_{i=1}^n \bar A_i.$$$$\text{If } B = \bigcup_{i=1}^\infty A_i \text{, prove that } \bar B \supseteq \bigcup_{i=1}^\infty \bar A_i. \\\text{ Give an example to show that this containment might be proper.}$$ 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?

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A couple of hints:

1. $A \subseteq B$ implies $\overline{A} \subseteq \overline{B}$.
2. From the above it should be clear that $\overline{A_1} \cup \overline{A_2} \subseteq \overline{ A_1 \cup A_2 }$. Suppose now $x \notin \overline{A_1} \cup \overline{A_2}$: can you show it is not in $\overline{ A_1 \cup A_2 }$? (What does it mean that $x \notin \overline{A_1}$ and $x \notin \overline{A_2}$?)
3. The set of all rational numbers is countable, yet dense in the real line.
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 Thanks! That helped me get started. – Andrew C Sep 13 '12 at 17:49

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 $\overline{\bigcup\limits_{i=1}^nA_i}\subset\overline{\bigcup\limits_{i=1}^\infty A_i}$

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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 $0 \in \overline{B} \setminus \cup_{i=1}^\infty \overline{A_i}$.

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