# Proof Check: Closure of Union Contains Union of Closures

1. Oct 5, 2013

### 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?