Union an intersection of sets I have that $F\subset E~$ where $F=\cup_{i\in I} V_i~$ and $~V_i=U_i\cap F$
and $U_i\subset E$
Why $$E= (E\setminus F)\cup (\cup_{i\in I} U_i)$$
Thank you.
 A: We can see that because 
$$ F=\bigcup_{i=1}^{n}V_i$$
and each $V_i$ satisfies 
$$ V_i\subset U_i$$
from $V_i=U_i\cap F,$ that 
$$ F=\bigcup_{i=1}^nV_i\subset\bigcup_{i=1}^nU_i\implies F\subset \bigcup_{i=1}^nU_i.$$
Now, we should consider the meaning of 
$$ (E\setminus F)\cup\bigg(\bigcup_{i=1}^nU_i\bigg).$$
$$ E\setminus F:=\{x\in E:x\notin F\}.$$
Because we know that 
$$ F\subset \bigcup_{i=1}^nU_i$$
when we take the union of these two previous sets, we find that every missing element removed from set subtraction is added back in, along with some possible duplicates. But, we don't care about duplicates- they merge together, in essence. And so,
$$ E=(E\setminus F)\cup\bigg(\bigcup_{i=1}^nU_i\bigg).$$
A: Since $F\subset E$, $F=E\cap F$. So 
$$
F=E\cap\bigcup_{i\in I} V_i=\bigcup_{i\in I} (E \cap V_i)=\bigcup_{i\in I} (E \cap (U_i\cap F))=\bigcup_{i\in I} (U_i\cap F)=\bigcup_{i\in I} U_i\cap F
$$
$E \cap (U_i\cap F)=U_i\cap F$ because $U_i\cap F\subset E$. 
So
$$
F\subset \bigcup_{i\in I} U_i
$$
Hence
$$
E=(E-F)\cup F= (E-F)\cup (\bigcup_{i\in I} U_i\cap F)\subset (E-F)\cup \bigcup_{i\in I} U_i\tag1
$$
On the other hand
$$
(E-F\subset E)\land (\bigcup_{i\in I} U_i\subset E)\implies ((E-F)\cup \bigcup_{i\in I} U_i)\subset E\tag2
$$
Thus by $(1)$ and $(2)$
$$
E=(E-F)\cup \bigcup_{i\in I} U_i
$$
A: Let $W=\cup_i U_i.$.... We have $E\supset W$, and $$W\supset (W\cap F)=\cup_i(U_i\cap F)=\cup_iV_i=F.$$ So $E\supset W\supset F$..... So $$E=(E\backslash F)\cup (E\cap F)=(E\backslash F)\cup  F \subset (E\backslash F)\cup W\subset (E\backslash F)\cup E=E.$$  Therefore $E=(E\backslash F)\cup W=(E\backslash F)\cup (\cup_iU_i).$
