Let $G$ be the set of non-empty closed bounded sets of $\mathbb{R}^{p}$. Suppose that $(F_{n})_{n}$ is a nested sequence in $G$ and let $F= \cap_{n \in \mathbb{N}}F_{n}$. Show that $(a)$ $F \in G$ and $(b)$ $(F_{n})_{n}$ converges to $F$ for the Hausdorff metric $h$ on $G$.
I'm not sure how to start on this. I also haven't learned about compact sets yet. Can someone help?
(a) Each $F_n$ is a compact non-empty set. A property of such sets is that every decreasing (by inclusion) sequence of compact sets has non-empty intersection. So $\bigcap_{n\in\mathbb N}F_n$ is compact (since it is closed and bounded) and non-empty. Therefore, $\bigcap_{n\in\mathbb N}F_n\in G$.
(b) Suppose otherwise. Then there is a $\varepsilon>0$ such that$$(\forall n\in\mathbb{N}):h\left(F_n,\bigcap_{n\in\mathbb N}F_n\right)\geqslant\varepsilon.$$For each natural $n$, let $x_n\in F_n$ be such that $d\bigl(x_n,\bigcap_{n\in\mathbb N}F_n\bigr)\geqslant\varepsilon$. Each $x_n$ is in $F_1$, wich is compact. Therefore $(x_n)_{n\in\mathbb{N}}$ has a convergente subsequence. We can, without loss of generality, assume that the sequence $(x_n)_{n\in\mathbb{N}}$ itself converges to some $x$. Each element of the sequence belongs to $F_1$, which is closed; therefore $x\in F_1$. If $n\geqslant2$, then $x_n\in F_2$ and so $x\in F_2$, and so on. So, $x\in\bigcap_{n\in\mathbb N}F_n$. Bt this contradicts the assumption that the distance from every $x_n$ to $\bigcap_{n\in\mathbb N}F_n$ is greater than or equal to $\varepsilon$.