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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.

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  • $\begingroup$ Try to prove that $F \subseteq \bigcup_{i \in I} U_i$, maybe by looking at some arbitrary member of $F$ and showing it is also a member of some $U_i$ $\endgroup$
    – Dan Simon
    Dec 18 '15 at 21:59
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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, $$ E=(E\setminus F)\cup\bigg(\bigcup_{i=1}^nU_i\bigg).$$

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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 $(2)$ $$ E=(E-F)\cup \bigcup_{i\in I} U_i $$

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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 $E=(E\backslash F)\cup W=(E\backslash F)\cup (\cup_iU_i).$

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