# Removing curve from simply connected domain's boundary : does it stay connected?

Let $$U \subset \mathbb R^2$$ be an open, regular (meaning $$U$$ is the interior of its closure), bounded and simply connected set. Let $$\gamma : [0,1] \rightarrow \partial U$$ be a simple curve such that $$\gamma(0) \neq \gamma(1)$$.

Is $$\partial U \setminus \gamma((0,1))$$ still connected ?

I think this should be true. Assume for a contradiction that this is not the case. Let $$K$$ be the connected component $$\{\gamma(0)\} \subset K \subsetneq \partial U \setminus \gamma((0,1))$$, where we remark that $$K$$ and $$\partial U \setminus \gamma((0,1))$$ are compact. By Zoretti's Theorem, for any $$\varepsilon > 0$$, we can find a simple closed curve $$J : [0,1] \rightarrow \mathbb R^2$$ for which $$K \subset \mathrm{int}(J), \quad J \cap \left( \partial U \setminus \gamma((0,1)) \right) = \emptyset, \quad \mathrm{dist}(K, j) \leq \varepsilon \quad \forall j\in J$$

Note that $$J$$ cannot be contained in $$U$$ (otherwise, $$K \subset \mathrm{int}(J) \subset U$$ which contradicts the regularity of $$U$$) and $$J$$ can only cross $$\partial U$$ through $$\gamma((0,1))$$.

I think we should be able to deduce from this that $$\gamma(0)$$ would actually lie on the exterior of $$J$$ but I am not able to complete my proof.

Any help is welcomed.

EDIT : This is related to the notion of cross-cut. A cross-cut $$\phi$$ leads to a decomposition of $$U$$ into two simply connected domains, whose boundaries are then connected and contain $$\phi$$. What would happen to these boundaries if one removed $$\phi$$ ?

EDIT 2 : My above argument is incorrect because $$\gamma((0,1))$$ is not necessarily open in $$\partial U$$ topology and this is what leads to the counter-example below.

• Curious question. To those glancing at this for the first time, under the given assumptions the boundary will be connected, it will not be a Jordan curve generically. If there is an example it must be a boundary that's not a Jordan curve (else we get a homeomorphic image of $S^1$ and the question is immediate). Commented Mar 5 at 16:33
• Never heard about Zoretti theorem... Commented Mar 5 at 18:43

What you are trying to prove is false. Let $$C_1, C_2$$ be two nested round circles in the plane $$E^2$$ which are tangent to each other at a point $$p$$. Take $$C=C_1\cup C_2$$. The complement, $$E^2 \setminus C$$, consists of three components, one of them is the "exterior" of the outer circle $$C_1$$, the second is the "interior" of the inner circle $$C_2$$. Let $$U$$ be the third component. It is bounded, regular and simply connected, $$\partial U=C$$. Take any open arc $$a\subset C$$ containing $$p$$. Then $$C\setminus a$$ is disconnected.