DeMorgan's Law regarding unindexed families of sets I am not sure if I understand proving results regarding unindexed families of sets and would appreciate some help.
(i) Suppose that $A$ is a set and $F$ is a family of sets.Prove that $A$\ $\bigcup$$F=\bigcap${$A$\ $B:B \in F $}.
My attempt:
$x \in A$\ $\bigcup F$

*

*iff $x \in A$ and $x \notin \bigcup F$

*iff $x \in A$ and $x \notin B $ for every $B \in F$.

*iff $x \in A$\ $B$ for every $B \in F$

*iff $x \in \bigcap A$\ $B$

*iff $x \in$ $\bigcap${$A$\ $B:B \in F $}

Therefore : $A$\ $\bigcup$$F=\bigcap${$A$\ $B:B \in F $}.
(ii) Let $F$ and $G$ be two families of sets.Prove that $\bigcup(F\cup G)=(\bigcup F)\cup(\bigcup G)$
My attempt:
$x \in(F\cup G)$

*

*iff $x\in (F \cup G)$ for some $F \in A$ and $F \in B$

*iff $x \in F$ or $x \in G$ for some $F \in A$ and $F \in B$

*iff $x \in \bigcup F$ or $x \in \bigcup G$

*iff $x \in (\bigcup F) \cup (\bigcup G)$
Therefore: $\bigcup(F\cup G)=(\bigcup F)\cup(\bigcup G)$ with $A$ and $B $ being some families?? Does seperating $F \in A$ and $F \in B$ make sense?..originally i thought i should have done :$(F \cup G) \in A$..would that have been wrong?
Thank you for your time.
 A: In your answer on (i) the fourth bullet is wrong and should be left out or  interchanged with:


*

*iff $x\in c$ for every $c\in\{A\setminus B\mid B\in F\}$
By answering (ii) two families/sets $A$ and $B$ "fall from the sky". 
They are not mentioned in what you are asked to prove.
Equivalent are:


*

*$x\in\bigcup(F\cup G)$

*$x\in a$ for some $a\in F\cup G$

*$x\in a$ for some $a$ that satisfies $a\in F$ or satisfies $a\in G$

*$x\in a$ for some $a\in F$ or $x\in a$ for some $a\in G$

*$x\in\bigcup F$ or $x\in\bigcup G$

*$x\in\left(\bigcup F\right)\cup\left(\bigcup G\right)$

In this answer I stay in line with the notation that you practicize, but that is not how I would do it.
For me personally if $A$ is a set then $\cup A$ is again a set and this with: $$x\in\cup A\iff x\in a\text{ for some }a\in A$$ using the small cup. In that sense $\cup$ is an operator on sets.
Further $F\cup G$ is then an abbreviation of $\cup\{F,G\}$ and $\bigcup_{\lambda\in\Lambda}A_{\lambda}$ (using bigcup) is an abbreviation of $\cup\{A_{\lambda}\mid\lambda\in\Lambda\}$.
Similar story for cap and bigcap
