let S be set with subsets $ E_1,E_2,E_3,....$ Show that:
a) $F=\bigcup \limits_{n=1}^\infty \left(\bigcap \limits_{k=n}^\infty E_k\right)$ consists of all elements of S each of which belongs to all but a finite number of $E_j$.
b) $G=\bigcap \limits_{n=1}^\infty \left(\bigcup \limits_{j=n}^\infty E_j\right)$ consists of all elements of S each of which belongs to infinitely many $E_k$.
On
As to a): $x \in F$ iff there exists some $n(x)$ such that for all $k \ge n(x)$, $x \in E_k$. So $x$ is in eventually all (except maybe the first $n(x)$ many) $E_k$...
As to b): $x \in G$ iff for every $n$ there exists some $j \ge n$ such that $x \in E_j$, so we can always find "new" indices $j$ such that $E_j$ contains $x$.
On
If $x\in F$, then $x$ is element of one of the sets that are forming the union $F$, therefore $x\in \cap_{k = n}^\infty E_k$ for at least one $n\in\mathbb{N}$. Because $x$ is the element of the intersecion $\cap_{k = n}^\infty E_k$, $x$ is element of every $E_k$ where $k\geq n$. One can conclude that $x$ may be "missing" only in the first $n-1$ sets, which is a finite number.
Analogically for b).
The most straight-forward way to approach this is to show that $x \in F$ or $G$ if and only if it satisfies the conditions described, treating both directions of implication separately.
For example, for $F$: suppose that $x$ appears in all but finitely many $E_j$. Then there is an $n$ such that $x \in E_j$ whenever $j\geq n$. It follows that $ x \in \bigcap_{k=n}^\infty E_k$. It follows that $x \in F$.
Now, suppose that $x$ does not satisfy this condition. Then, there is an infinite sequence of sets $\{E_{n_k}\}$ such that $x \notin E_{n_k}$ (or, if you prefer: for every $n$, there is a $k\geq n$ such that $x \notin E_k$). It follows that for every $n$, $ x \notin \bigcap_{k=n}^\infty E_k$. We may therefore state $x \notin F$.
The description of $G$ can be proven in a similar fashion.