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.

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    $\begingroup$ 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). $\endgroup$ Commented Mar 5 at 16:33
  • $\begingroup$ Never heard about Zoretti theorem... $\endgroup$
    – Jean Marie
    Commented Mar 5 at 18:43

1 Answer 1


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.


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