The question is stated as: Consider a nonempty set $W$. Let $F$ be the set of all nonempty finite subsets of W. Find $\bigcup_{A \in F}A$ and $\bigcap_{A \in F}A$.
So what i have done: knowing that $(\forall x)(x \in W \Leftrightarrow \{x\} \subseteq W)$, we have that for any $x$ in $W$ the singleton set $\{x\}$ is also in $F$, thus the union of all these singleton sets is all the elements of $W$, then $\bigcup_{A \in F}A = W$.
For the intersection part i think its possible to have two different results, if we consider $W$ as a singleton like $W=\{a\}$, then $F=\{\{a\}\}$ and in this case $\bigcap_{A \in F}A = \{a\} = W$. But if we consider $W$ having two or more elements we will have an obvious intersection with disjoint sets and therefore $\bigcap_{A \in F}A = \emptyset$.
My results are correct? and should i did this with less words?