Proof Check: Closure of Union Contains Union of Closures

[Mandelbroth]

I intend to show, for a set $X$ containing $A_i$ for all $i$, $$\overline{\bigcup A_i}\supseteq \bigcup \overline{A_i}.$$
^//Proof: We proceed to prove that $\forall x\in X,~x\in\bigcup\overline{A_i}\implies x\in\overline{\bigcup A_i}$. Equivalently, $\forall x\in X,~x\not\in\overline{\bigcup A_i}\implies x\not\in\bigcup\overline{A_i}$. Thus, let $x\not\in\overline{\bigcup A_i}$.

From this, we conclude $\exists N\ni x: N\cap(\bigcup A_i)=\emptyset\implies\bigcup (N\cap A_i)=\emptyset\implies\forall i,~N\cap A_i=\emptyset$.

$\therefore \forall i,~x\not\in\overline{A_i}$. Thus, $x\not\in\bigcup\overline{A_i}$, and we conclude $\overline{\bigcup A_i}\supseteq \bigcup \overline{A_i}$. Halmos.

This proof is correct, right?

 

[mmwave]

I would like to commend you on your proof and confirm that it is indeed correct. Your use of logical equivalences and set notation is clear and concise, making it easy to follow your reasoning. Keep up the good work!