Corollary of the De Morgan's Laws People, I am trying to do my exercise list of the course of Measure Theory and I had a doubt about one "corollary" from the De Morgan's Laws. The De Morgan's law says that: 

For a family $\{E_\alpha; \alpha \in \Gamma \}$, where $\Gamma$ is some indexing set, of subsets of a universal set $\mathfrak{U}$, the following properties hold: 
  
  
*
  
*$\displaystyle\left(\bigcup\limits_{\alpha \in \Gamma} E_\alpha\right)^c=\bigcap\limits_{\alpha \in \Gamma} E_\alpha^c$ 
  
*$\displaystyle\left(\bigcap\limits_{\alpha \in \Gamma} E_\alpha\right)^c = \bigcup\limits_{\alpha \in \Gamma} E_\alpha^c$
  

I would like to know it is true that if $\Gamma$ a indexing set and $\{E_\alpha; \alpha \in \Gamma \}$ and $\{F_\alpha; \alpha \in \Gamma \}$ are families of subsets of a set $X$, then 


*

*$\displaystyle\bigcup\limits_{\alpha \in \Gamma} (E_\alpha - F_\alpha) = \bigcup\limits_{\alpha \in \Gamma} E_\alpha - \bigcap\limits_{\alpha \in \Gamma} F_\alpha$ and 

*$\displaystyle\bigcap\limits_{\alpha \in \Gamma} (E_\alpha - F_\alpha) = \bigcap\limits_{\alpha \in \Gamma} E_\alpha - \bigcup\limits_{\alpha \in \Gamma} F_\alpha.$


I have already done the first side of que item 1. Have someone some idea if the other side holds?
Thank you very much, everybody!
 A: If $\Gamma$ is a indexing set and $\{E_\alpha; \alpha \in \Gamma \}$ and $\{F_\alpha; \alpha \in \Gamma \}$ are families of subsets of a set $X$, then 

1. $\phantom{mm}$ $\displaystyle\bigcup\limits_{\alpha \in \Gamma} (E_\alpha - F_\alpha) \subseteq \bigcup\limits_{\alpha \in \Gamma} E_\alpha - \bigcap\limits_{\alpha \in \Gamma} F_\alpha$ 

Proof: If $x\in \displaystyle\bigcup\limits_{\alpha \in \Gamma} (E_\alpha - F_\alpha)$ then, there is an $\alpha_0 \in \Gamma$ such that  $x\in E_{\alpha_0}$ and  $x\notin F_{\alpha_0}$. So $x\in \bigcup\limits_{\alpha \in \Gamma} E_\alpha $ and $x\notin  \bigcap\limits_{\alpha \in \Gamma} F_\alpha$. So we have $x\in \bigcup\limits_{\alpha \in \Gamma} E_\alpha - \bigcap\limits_{\alpha \in \Gamma} F_\alpha$. 
The reverse inclusion does NOT hold. 
Counterexemple: Let $A$ be a non empty set. Let $\Gamma=\{a,b\}$ and $E_a=F_a=A$ and $E_b=F_b=\emptyset$. Then
$\displaystyle\bigcup\limits_{\alpha \in \Gamma} (E_\alpha - F_\alpha)=\emptyset \neq A = \bigcup\limits_{\alpha \in \Gamma} E_\alpha - \bigcap\limits_{\alpha \in \Gamma} F_\alpha$ 
On the other hand, we have 

2. $\phantom{mm}$ $\displaystyle\bigcap\limits_{\alpha \in \Gamma} (E_\alpha - F_\alpha) = \bigcap\limits_{\alpha \in \Gamma} E_\alpha - \bigcup\limits_{\alpha \in \Gamma} F_\alpha$

Proof: ($\subseteq$)If $x \in \displaystyle\bigcap\limits_{\alpha \in \Gamma} (E_\alpha - F_\alpha)$, then, for all $\alpha \in \Gamma$, $x \in E_\alpha$ and $x\notin F_\alpha$. So $x \in \bigcap\limits_{\alpha \in \Gamma} E_\alpha$ and $x \notin \bigcup\limits_{\alpha \in \Gamma} F_\alpha$. So we have $x\in \bigcap\limits_{\alpha \in \Gamma} E_\alpha - \bigcup\limits_{\alpha \in \Gamma} F_\alpha$. 
($\supseteq$)If $x\in \bigcap\limits_{\alpha \in \Gamma} E_\alpha - \bigcup\limits_{\alpha \in \Gamma} F_\alpha$ then, $x \in \bigcap\limits_{\alpha \in \Gamma} E_\alpha$ and $x \notin \bigcup\limits_{\alpha \in \Gamma} F_\alpha$. So, 
for all $\alpha \in \Gamma$, $x \in E_\alpha$ and $x\notin F_\alpha$. So, for all $\alpha \in \Gamma$, $x \in E_\alpha - F_\alpha$. So we have $x \in \displaystyle\bigcap\limits_{\alpha \in \Gamma} (E_\alpha - F_\alpha)$.
Remark: In the proof of item 2, note that the proof of the ($\supseteq$) part is essentially the reverse implications used in the proof of the ($\subseteq$) part. 
The proof of item 1 does not allow a similar "reversal", because, from $x\in \bigcup\limits_{\alpha \in \Gamma} E_\alpha $ and $x\notin  \bigcap\limits_{\alpha \in \Gamma} F_\alpha$, we can get only that there are $\alpha_0, \alpha_1 \in \Gamma$ such that  $x\in E_{\alpha_0}$ and  $x\notin F_{\alpha_1}$, but we have no way to conclude that $\alpha_1=\alpha_0$.
