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This very usual statement needs to be proven:

$$\overline{\bigcup_{i} A_i} = \bigcup_i\overline{A_i}$$

Does this make sense:

$\overline{A_i}$ is closed, so $\bigcup_i\overline{A_i}$ is closed and $\overline{A_i} = A_i$ so:

$$\bigcup_i\overline{A_i} = \overline{\bigcup_i\overline{A_i}} = \overline{\bigcup_i A_i}$$

EDIT:

Sorry, my question was laid carelessly. It's a finite union of subsets.

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closed as unclear what you're asking by Rob Arthan, Silvia Ghinassi, Harish Chandra Rajpoot, T. Bongers, hardmath Feb 23 at 15:57

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

6  
Arbitrary union of closed sets is not usually closed. – Joanpemo Feb 22 at 23:55
5  
Even beyond all the infinitary issues mentioned, why is $\bar{A_i}=A_i$? – Steven Stadnicki Feb 23 at 0:30
2  
Please fix the question. Don't just add a comment saying you got it wrong. – Rob Arthan Feb 23 at 0:36

This is true only for finite case. For infinite case it is true for only $$ \bigcup_i\overline{A_i}\subset \overline{\bigcup_{i} A_i} $$ Edit: The proof is like, since $$ A\subset B\implies \overline{A} \subset \overline{B} \quad \text{and}\quad A_i\subset \bigcup_{i=1}^{\infty}A_i $$ for any $i$. So we have $$ \overline{A}_i\subset \overline{\bigcup_{i=1}^{\infty}A_i}\quad \text{and}\quad \bigcup_{i=1}^{\infty}\overline{A}_i\subset \overline{\bigcup_{i=1}^{\infty}A_i}\tag{1} $$ The converse is not true as a counter example is given in other post.

Now we prove that finite union of closure of $A_i$ always equals the closure of finite union of $A_i$. By definition, closure is the smallest close set that contains the set, i.e. $\bigcup_{i=1}^{n}A_i\subset \overline{\bigcup_{i=1}^{n}A_i}$. Since each $\overline{A}_i$ is closed, $\bigcup_{i=1}^{n}\overline{A}_i$ is also closed. Since $\bigcup_{i=1}^{n}A_i\subset \bigcup_{i=1}^{n}\overline{A}_i$, we have $\overline{\bigcup_{i=1}^{n}A_i}\subset \bigcup_{i=1}^{n}\overline{A}_i$. Together with $(1)$, we have $$ \overline{\bigcup_{i=1}^{n}A_i}=\bigcup_{i=1}^{n}\overline{A}_i $$

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It is in general not true that $\overline{\bigcup_{i \in I} A_i} = \bigcup_{i \in I} \overline{A_i}$.

For example. Consider $\mathbb R$ with its usual topology and let, for $n \in \mathbb N$, $A_n = (-1 + \frac{1}{n}, 1 - \frac{1}{n})$. Then $\overline{A_n} = [-1 + \frac{1}{n}, 1 - \frac{1}{n}]$. Now $\overline{ \bigcup_{n \in \mathbb N} A_n} = \overline{(-1,1)} = [-1,1]$, but $\bigcup_{n \in \mathbb N} \overline{A_n} = \bigcup_{n \in \mathbb N} [-1 + \frac{1}{n}, 1 - \frac{1}{n}] = (-1,1)$

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On one hand, for each $k$, we have $A_k \subseteq{\bigcup_{i} A_i}$.
Hence $\overline{A_k} \subseteq\overline{\bigcup_{i} A_i}$.
Hence $\bigcup_{k}\overline{A_k} \subseteq\overline{\bigcup_{i} A_i}$.
But $\bigcup_{k}\overline{A_k}=\bigcup_{i}\overline{A_i}$, hence we proved $\bigcup_{i}\overline{A_i} \subseteq\overline{\bigcup_{i} A_i}$.
This direction works for any index set $I$, finite or infinite.

For the other direction assume that $I$ is finite, we need to show that $ \overline{\bigcup_{i} A_i}\subseteq \bigcup_{i}\overline{A_i}$.
We need to show that if $x\in\overline{\bigcup_{i} A_i}$, then $x\in\bigcup_{i}\overline{A_i}$.
Equivalently, using the contrapositive, it is enough to show that, for any $x$,
if $x\notin\bigcup_{i}\overline{A_i}$ then $x\notin\overline{\bigcup_{i} A_i}$.
So, fix any $x\notin\bigcup_{i}\overline{A_i}$, and say $I=\{i_1,i_2,...,i_n\}$. For each $k=1,2,..,n$, since $x\notin\overline{A_{i_k}}$, we conclude that there is a neighborhood $U_k$ of $x$ missing $A_{i_k}$. Let $U=\bigcap_{k=1}^n U_k$. Note that $U$ is open, as the intersection of finitely many open sets. Hence $U$ is a neighborhood of $x$ that misses $\bigcup_{k=1}^n A_{i_k}$, that is $U$ misses (in your notation) $\bigcup_i A_i$. Hence $x\notin\overline{\bigcup_{i} A_i}$, which completes the proof.

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This is a proof for the finite index set case, wlog it is equivalent to show that for each $n\in\Bbb N$, then $$\cup_{i=1}^n \overline {A_i}=\overline{\cup_{i=1}^n A_i}.$$ Recalling by definition that $\bar A= A\cup A'$ where $A'$ denotes the set of limit points of $A$, we write $$\cup_{i=1}^n \overline{A_i}=A_1\cup\cdots\cup A_n\cup{A'_1}\cup\cdots\cup{A'_n},$$ whereas $$\overline{\cup_{i=1}^n A_i}=A_1\cup\cdots\cup A_n\cup(A_1\cup\cdots\cup A_n)'.$$ In fact, we can show $$(A_1\cup\cdots\cup A_n)'={A'_1}\cup\cdots\cup{A'_n}.$$ First, suppose $x\in A'_k$ for some $k\in\{1,2,\cdots,n\}$, then there exists a sequence $\{x_m\}\subset A_k-\{x\}$ such that $x_m\to x$ as $m\to\infty$. Obviously $\{x_m\}\subset A_1\cup\cdots\cup A_n-\{x\}$, therefore it follows that $x\in (A_1\cup\cdots\cup A_n)'$, and hence ${A'_1}\cup\cdots\cup{A'_n}\subset(A_1\cup\cdots\cup A_n)'$. (In fact this direction holds for arbitrary index sets.)

For the other direction of inclusion, finiteness is crucial: Pick any $x\in(A_1\cup\cdots\cup A_n)'$, then there exists a sequence $\{x_m\}\subset A_1\cup\cdots\cup A_n-\{x\}$ such that $x_m\to x$ as $m\to\infty$. Then by pigeonhole principle we can pick at least one subsequence $\{x_{m,l}\}\subset A_k$ for some $k\in\{1,2,\cdots,n\}$, which shows that $x\in A'_k$ and hence $(A_1\cup\cdots\cup A_n)'\subset{A'_1}\cup\cdots\cup{A'_n}$.

This completes the proof.

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