definition of being closed under some operation how do we define a set to be closed under some operation? 
for example how is this theorem true?
Let G denote the family of all open subsets of the real numbers and F the family of all closed subsets of the real numbers.
1-G is closed under arbitrary unions and finite intersections.
 A: What the statement concerning $G$ means is:


*

*If $A$ is a set of open subsets of $\mathbb R$, then $\bigcup_{S\in A}S$ is an open subset of $\mathbb R$;

*If $A$ is a finite set of open subsets of $\mathbb R$, then $\bigcap_{S\in A}S$ is an open subset of $\mathbb R$.

A: Ooh... bad terminology.  
A set being $\color{blue}{open/closed}$ is an entirely different concept and has nothing whatsoever with an operation being $\color{red}{closed}$.
An operation, $\circ$ and a collection $C$,  is $\color{red}{closed}$ if for any $A, B\in C$ , then $A\circ B$ is defined and is in $C$.
Example:  the operation $+$ on $\mathbb N$ is $\color{red}{closed}$, because for and $a,b \in \mathbb N$ then $a+b \in \mathbb N$.  But the operation $-$ on $\mathbb N$ is not $\color{red}{closed}$, because for $5,7\in \mathbb N$ we have $5-7\not \in \mathbb N$.  But $-$ is $\color{red}{closed}$ on $\mathbb Z$ because for any $a,b \in \mathbb Z$ then $a-b \in \mathbb Z$.  But $\div$ is not closed on $\mathbb Z$ because for any integers $a,b$ it need not be that $a\div b \in \mathbb Z$.  But $\div$ is $\color{red}{closed}$ on $\mathbb Q\setminus\{0\}$ because if $a,b \in \mathbb Q$ and $a\ne 0, b\ne 0$ then $a\div b\in \mathbb Q$ and $a\div b \ne 0$.
Also an operation may be $\color{red}{closed}$ for a finite number of operations but not an infinite number.  For example $+$ on $\mathbb Z$ is closed under a finite number of opperations as for any $a_1, a_2, a_3,......., a_n\in \mathbb Z$ then $a_1 + a_2+a_3 + ...... + a_n \in \mathbb Z$.  But $+$ is not $\color{red}{closed}$ under an infinite number of operations as for $a_1, a_2, a_3,......, a_n$ is an infinite number of integers, it does not follow that $\sum_{k=1}^{\infty} a_k$ is an integer (or even defined).  Certainly $\sum_{k=1}^{\infty} 1$ is not an integer (it's not even a real number).
..... Any hoo....
So the question is $G = \{$ all $\color{blue}{open}$ sets of $\mathbb R\}$ and the operations are $\cup$ (finite or infinite) and $\cap$ (finitely many times.
The question is are those operations $\color{red}{closed}$?
Or in other words if $A_1, A_2, A_3, .....$ are $\color{blue}{open}$ sets then is $A_1\cup A_2\cup A_3 \cup.... $ always $\color{blue}{open}$?   And if $A_1, A_2, .... A_n$ are $\color{blue}{open}$ sets is $A_1\cap A_2 \cap ..... \cap A_n$ always $\color{blue}{open}$.
