How to prove the closure $\bigcup_{n\ge 1} F_{n}$ is totally bounded and closed Let $(X,p)$ be a metric space. Write $F$ for the set of subsets of $X$ which are closed, bounded, and non-empty. For each integer $n\geq1$, write $F_{n}$ for the set of subsets of X which are finite and non-empty, and have no more than n elements. Certainly $F_{n}\subset F$ for all $n\geq 1$.

Question:
Let $A$ and $B$ lie in $F$. Also suppose $$d(x,B)=\inf_{y\in B} (p(x,y))\;.$$
Also, define $$\alpha (A,B)=\sup_{x\in A}d(x,B)$$ and $$\beta(A,B)=\max(\alpha(A,B), \alpha(B,A))$$ for all $A,B\in F$. Thus, we need to prove $\beta$ is a metric on $F$ (I have proved it now), and we still need to prove $F_{n}$ is a closed set in metric space $(F,\beta$), and the closure of $\bigcup_{n\ge 1} F_{n}$ is the set of non-empty, totally bounded subsets of $X$.

I have proved  $\beta$ is a metric on $F$, but still have no idea how to prove the rest of the questions. This problem is given by my professor.
 A: Let $A$ be an element of $F\setminus F_n$ (i.e., $A$ has at least $n+1$ elements and may even be infinite). Pick $n+1$ distinct points $a_0,\ldots, a_{n}\in A$ and let $r=\min_{0\le i<j\le n}p(a_i,a_j)>0$. Now let $B$ be any element of $F_n$. Then for any $b\in B$ there exits at most one $i$ with $p(a_i,b)<\frac r2$, hence by pigeon-hole there exists at least one $i$ with $p(a_i,b)\ge\frac r2$ for all points $b\in B$. That is, for this $i$ we have $d(x_i,B)\ge \frac r2$. We conclude $\beta(A,B)\ge\alpha(A,B)\ge \frac r2$. Inother words, the open $\frac r2$-ball around $A$ is disjoint to $F_n$, thus showing that $F\setminus F_n$ is $\alpha$-open, equivalently, $F_n$ is closed.

Now let $A$ be totally bounded. First we have to show that $A$ is in the closure of $\bigcup F_n$. Given $\epsilon>0$, the total boundedness gives us a finte set of points $x_1,\ldots x_n$ such that the $\epsilon$-balls around these cover $A$. Hence if we let $B=\{x_1,\ldots, x_n\}$, we have $d(x,B)\le \epsilon$ for all $x\in A$. Hence $\alpha(A,B)\le\epsilon$. On the other hand, we may assume wlog that taking away any of the $x_i$ from $B$ destroys the $\epsilon$-cover property. This implies that $d(x_i,A)<\epsilon$ for all $x_i\in B$, hence $\alpha(B,A)\le \epsilon$ as well. So finally $\beta(A,B)\le\epsilon$. As $\epsilon$ was arbitrary, we see that $A\in\overline{\bigcup_n F_n}$.
On the other hand let $A$ be any element of $\overline{\bigcup_n F_n}$.
Then total boundedness of $A$ follows because for given $\epsilon>0$ we find a finite set $B$ that is less than $\epsilon$ apart from $A$ (with respect to $\alpha$), which just say that the union of the $\epsilon$-balls around the elements of $B$ cover $A$.
A: To prove that $F_n$ is closed. Take an element $K$ not in $F_n$, then we can take $n+1$ points $x_1,x_2,...,x_{n+1}$ in $K$. Let $2r$ be the minimum distance between any two of these $n+1$ points. Then $$B(K,r)$$ is disjoint from $F_n$. The reason is that any element $J$ of $B(K,r)$ ought to have a point in the $n+1$ balls (in $X$) with centers $x_1,...,x_{n+1}$ and radius $r$. By the choice of $r$ these balls are disjoint. Therefore $J$ has at least $n+1$ points, i.e. $J\notin F_n$.
