Let $\gamma: \mathbf{S}^1 \to \mathbf{R}^2$ be a simple, closed curve. Assume additionally that $\gamma$ is regular, of class $C^2$ at least. The complement of $\gamma$ in $\mathbf{R}^2$ has two connected components, of which we denote $\Omega \subset \mathbf{R}^2$ that which is bounded. (This has boundary $\partial \Omega = \gamma$.)

We define the *cut locus* of $\Omega$ to be the closure of the set of points $x \in \Omega$ for which there exist two or more points $z_1,z_2 \in \partial \Omega$ with $\lvert x-z_i \rvert = \mathrm{dist}(x,\Omega)$. We denote this set $C \subset \Omega$. (This coincides with the definition of the cut locus of $\partial \Omega$ in terms of the exponential map, when endowing $\Omega$ with the Euclidean metric.)

It is claimed in a paper of [Panov and Petrunin](https://arxiv.org/abs/1312.6856) that an arbitrarily small perturbation of $\gamma$ guarantees that the cut locus $C$ is a finite graph, embedded inside $\Omega$. In fact, once the graph structure of $C$ is established one can show that $C$ is a tree. The authors seem to attribute these observations to [Ionin and Pestov](https://mathscinet-ams-org.libproxy.ucl.ac.uk/mathscinet-getitem?mr=107214), but unfortunately this latter paper looks to be available only in Russian.

**Question**: What is the formal justification of this perturbation argument? It looks like [Sard's theorem](https://en.wikipedia.org/wiki/Sard%27s_theorem) might apply, but then that is mainly concerned with regular sets, which $C$ is not. What `pathological' shapes of $\gamma$ lead to a cut locus $C$ which is *not* a finite graph?