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If $A,B \in \mathbb{R}$, then $\overline{A \cup B} = \overline {A} \cup \overline {B}$?

$\overline{A \cup B} = (A \cup B) \cup (A \cup B)' = \text{ }...$

I can't prove it. because i am not sure if $(A \cup B)' = A' \cup B'$.

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  • $\begingroup$ Look at deMorgan's laws which state $(A\cup B)^\prime = A^\prime \cap B^\prime$. $\endgroup$ – Alain Remillard Oct 20 '15 at 0:59
  • $\begingroup$ @AlainRemillard With $A'$ the OP denotes the set of all limit points of $A,$ not its complement $\endgroup$ – CIJ Oct 20 '15 at 1:00
  • $\begingroup$ Oups my bad, i'll try to ne more carefull next time $\endgroup$ – Alain Remillard Oct 20 '15 at 1:03
  • $\begingroup$ I feel like this exact question was asked a couple of days ago. $\endgroup$ – Cameron Williams Oct 20 '15 at 1:35
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Another way that avoids worrying about limit points:

$\overline A \cup \overline B$ is a closed set which contains $A \cup B$, and $\overline{A \cup B}$ is the smallest closed set which contains $A \cup B$, so $\overline{A \cup B} \subseteq \overline A \cup \overline B$.

On the other hand, $\overline{A \cup B}$ is a closed set containing $A$, and $\overline A$ is the smallest closed set containing $A$, so $\overline A \subseteq \overline{A \cup B}$. Similarly, $\overline B \subseteq \overline{A \cup B}$. Therefore, $\overline A \cup \overline B \subseteq \overline{A \cup B}$.

Since we have proved containment in both directions, we conclude that $\overline{A \cup B} = \overline A \cup \overline B$.

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Can show easily enough that $A \subset B \Rightarrow A' \subset B'$ which means $A' \cup B' \subset (A \cup B)'$.

If $e \in (A \cup B)'$ then $\exists c\neq e \in N_\epsilon(e) \cap (A \cup B)$ $\forall \epsilon > 0$. If $e \not\in A'$ i.e. $\not\exists$ any such $a \in A$ for some $\epsilon>0$ then must exist some $b \in B$ so $e \in B'$, so $e \in A' \cup B'$. Therefore $(A \cup B)' \subset A' \cup B'$ and so $A' \cup B' = (A \cup B)'$.

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(1).$x\subset y\implies \bar x \subset \bar y$ . Therefore $$\bar A\subset \overline {A\cup B} \text { and } \bar B\subset \overline {A\cup B}$$ $$ \text { so } \bar A \cup \bar B \subset \overline {A \cup B}.$$ (2) $\bar x \cup \bar y=\overline {\bar x \cup \bar y}. $ (The union of 2 closed sets is closed.) Since, by (1) we have $$A\cup B\subset \bar A \cup B\subset \bar A \cup \bar B,$$ $$\text { therefore }\overline {A\cup B}\subset \overline {\bar A \cup \bar B} = \bar A\cup \bar B.$$

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