I believe the space must be compact and connected in order to guarantee its hyperspace with the Vietoris topology is path connected, but I cannot come up with or find any counterexamples.
2026-02-22 19:10:12.1771787412
Example of path connected metric space whose hyperspace with Vietoris topology is not path connected?
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This paper starts with some useful references. Borsuk and Mazurkiewicz already showed in 1931 that if $X$ is a compact and connected metric space, $H(X)$, the hyperspace of all non-empty closed (or equivalently, compact) in the Vietoris (or Hausdorff-metric, if you prefer) topology is path-connected (and still compact connected and metrisable of course). The same holds for the hyperspace of subcontinua $C(X)$, as well. Of course if $H(X)$ is path-connected, it is connected and then $X$ is connected as well, so connectedness of $X$ is necessary. If $X$ is not compact, $H(X)$ would not be metrisable (not even normal).
So to find a path-connected metric $X$ with non-path-connected $H(X)$, we need a non-compact $X$. Then the question becomes: do we look at $H(X)$, the closed sets, or $K(X)$, the subspace of compact non-empty sets of $X$. The latter space is metrisable (using the Hausdorff metric which induces the same topology on this set as its subspace topology from the Vietoris topology), the former is not.
The paper I linked to, answers precisely these questions in theorems 3.1 and 3.3:
For metrisable $X$, the hyperspace $K(X)$ of compacta in the Vietoris topology is path-connected iff each compact subset of $X$ is contained in a continuum (compact connected subset). Example 3.2 provides an example of a path-connected subset of the plane that fails this property.
Also, $H(X)$ (the closed subsets) is path-connected iff $X$ is continuum-wise connected (any two points $x \neq y$ lie in a subcontinuum of $X$), so there a path-connected $X$ will have a path-connected $H(X)$.