Why $E_1=E\cup\bigcup_{i=1}^\infty G_i$?

Question

Let $E_k\supset E_{k+1}$ and $E=\bigcap_{i=1}^\infty E_i$. We set $$G_k=E_k\setminus E_{k+1}.$$

I don't understand why $$E_1=E\cup\bigcup_{i=1}^\infty G_i.$$ To me we simply have $E_1=\bigcup_{i=1}^\infty G_i$.

Answer 1

Suppose that $x\in E$. Then for each $k\in\Bbb Z^+$ we have $x\in E_{k+1}$, and therefore $x\notin E_k\setminus E_{k+1}$, i.e., $x\notin G_k$. In other words, $x$ is not in any of the sets $G_k$, so $x\notin\bigcup_{k\ge 1}G_k$. On the other hand $x\in E_1$. Thus, $x\in E_1\setminus\bigcup_{k\ge 1}G_k$. The same is true of every member of $E$.