I'm looking for any theorems and proofs for connectedness for hyperspaces exp(X). I would like to take a look for especially this theorem:
$$ X \textit{ is connected } \leftrightarrow exp(X) \textit{ is connected } $$
The proof below by Henno Brandsma is nice, but how i can check if finite set $\bigcup F_{n}$ is dense in $exp(X)$ ?
The Vietoris topology $\exp(X)$ of a ($T_1$) space $X$ has as a subbase all sets of the form $[U]=\{F \in \exp(X): F \cap U \neq \emptyset\}$ and $\langle U \rangle=\{F \in \exp(X): F \subset U\}$, where $U$ ranges over the non- empty open subsets of $X$.
Suppose $\exp(X)$ is connected. To see that $X$ is connected, suppose $X$ can be written as a union of two non-empty open disjoint sets $U$ and $V$. Then $\exp(X)$ can be written as the disjoint union of the open non-empty sets $\langle U \rangle$, $\langle V \rangle$ and $[U] \cap [V]$ (a closed set is either a subset of $U$, a subset of $V$, or it intersects both of them, as the sets are disjoint and cover $X$). Non-emptyness uses finite subsets of $X$. This contradicts the connectedness of $X$.
The reverse is more involved. Assume $X$ is connected. Define $F_n \subset \exp(X)$ as the set of all subsets of $X$ that have at most $n$ elements. $F_1$ is homeomorphic to $X$ so connected as well. in fact, there is a continuous map from $X^n$ onto $F_n$ that sends a tuple to the set of its elements (check this is indeed continuous). This implies, as powers of a connected space are connected, that all $F_n$ are connected subsets, and so their union, the set of all finite subsets of $\exp(X)$, is also connected, as the subsets are increasing in $n$. And one checks easily that the finite sets are dense in $\exp(X)$, so the latter space is also connected, as the closure of a connected set.
Added: to see that $\cup_n F_n$, the set of all finite subsets of $X$, is dense in $\exp(X)$: let $O$ be a basic subset of $\exp(X)$. This is a finite intersection of subbasic open sets. All of these are of the form $\langle U_1,U_2,\ldots,U_n \rangle = \{F \in \exp(X): F \subseteq \cup_{i=1}^n U_i, \forall_i: U_i \cap F \neq \emptyset \}$, where all $U_i$ are non-empty and open in $X$. This is not too hard too check. For every such basic set, pick $x_1 \in U_1, \ldots, x_n \in U_n$. Then $F = \{x_1, \ldots, x_n\}$ is finite, is in $\exp(X)$, as $X$ is $T_1$, and is a member of $\langle U_1,U_2,\ldots,U_n \rangle$. So the set of finite sets intersects all basic open sets in $\exp(X)$, so is dense.