How to check that finite sets are dense in exp(X)? How i can check if  finite set $\bigcup F_{n}$ is dense in $exp(X)$, where $exp(X)$ is
$$exp(X)=  \{ A\in X ; A\not= \emptyset ; A \textit{ compact in } X\} $$ 
($exp(X)$ is hyperspace, so it is set $\{ A\in X ; A\not= \emptyset ; A \textit{ compact in } X\}$ with Wietoris topology )

Fisrt I'll just include some notations.  Given a space $X$, let $\exp(X)$ denote the collection of all (nonempty) compact subse of $X$. The collection of all nonempty sets of the form $$\langle U ; V_1 , \ldots , V_n \rangle := \{ K \in \exp(X) : K \subseteq U, K \cap V_i \neq \varnothing \text{ for all }i \leq n \}$$ (where $U , V_1 , \ldots , V_n \subseteq X$ are open) can be shown to be a base for the Vietoris topology on $\exp(X)$.
So it suffices to show that whenever $\langle U ; V_1 , \ldots , V_n \rangle$ is nonempty, it contains a finite subset of $X$.
Well, suppose that $U , V_1 , \ldots , V_n \subseteq X$ are open such that $\langle U; V_1 , \ldots , V_n \rangle$ is nonempty.  Notice that this means that (actually, is equivalent to) $U \cap V_i \neq \varnothing$ for each $i \leq n$. For each $i \leq n$ we then simply pick some $x_i \in U \cap V_i$.  Then $\{ x_1 , \ldots , x_n \}$ is easily seen to be an element of $\langle U ; V_1 , \ldots , V_n \rangle$.
