It is given as an exercise in two books of Nadler (Continuum Theory and Hyperspaces of Sets) that if $X$ is a continuum (compact, connected metric space), then $2^X$ (the set of non-empty closed subsets of $X$) and $C(X)$ (the connected subsets of $2^X$) in the Hausdorff metric have no cut points. It is known that these spaces are path-connected. I have had a hard time finding a reference for this. Does anyone know of one, or how to prove it? I have been wracking my brain on it for quite some time.
Thanks a lot!
Let $A \in C(X)$ be the point we remove.
First consider the set of all subcontinua not contained in $A$:
$\mathcal B = \{B \in C(X): B \not \subset A\}$
The order-arc theorem says each subcontinuum is at the bottom of an order-arc with $X$ as the top element. Since the arc is ordered by inclusion no element is contained in $A$. That shows $\mathcal B$ is not only connected but arc-connected.
Now let $a \in A$ be a boundary element. By definition we can find a sequence $x_n \to a$ with all $x_n \notin A$. It follows the sequence $\{x_n\} \to \{a\}$ in the hyperspace. But the sequence is contained in the component of $\mathcal B$ which is closed (in $C(X)-A$). Therefore $a$ shares a component with $\mathcal B$.
Nearly there. . . .
The subspace $\{\{b\}: b \in A\} \subset C(A)$ is a copy of $A$ and so we can throw all those $\{b\}$ into the component. Finally we can join an arbitrary subcontinuum $C \subset A$ to some $\{b\}$ by an order-arc by applying the order-arc-theoren to $C(A)$.
That proves $ \{B \in C(X): B \not \subset A\}$ and $\{C \in C(X): C \subset A\}$ share a component in $C(X)-A$.
But the two sets constitute all of $C(X)-A$. So that set is connected.
$\blacksquare$