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:

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


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


closed as unclear what you're asking by Rob Arthan, Silvia Ghinassi, Harish Chandra Rajpoot, user296602, hardmath Feb 23 '16 at 15:57

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    $\begingroup$ Arbitrary union of closed sets is not usually closed. $\endgroup$ – DonAntonio Feb 22 '16 at 23:55
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    $\begingroup$ Even beyond all the infinitary issues mentioned, why is $\bar{A_i}=A_i$? $\endgroup$ – Steven Stadnicki Feb 23 '16 at 0:30
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    $\begingroup$ Please fix the question. Don't just add a comment saying you got it wrong. $\endgroup$ – Rob Arthan Feb 23 '16 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 $$ \overline{\bigcup_{i=1}^{n}A_i}=\bigcup_{i=1}^{n}\overline{A}_i $$


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 $\bigcup_{n \in \mathbb N} \overline{A_n} = \bigcup_{n \in \mathbb N} [-1 + \frac{1}{n}, 1 - \frac{1}{n}] = (-1,1)$


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.


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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