# Prove that a closed contour in $\mathbb{C}$ is contained in the closure of the set of points with nonzero index

Let $A$ be a simply connected domain, $\gamma : I \to A$ a closed contour. Let $P$ be the set of all points $z \in \overline{A} - \gamma(I)$ such that $\text{Ind}_\gamma(z) \ne 0$. I want to show that that $\overline{P} = P \cup \gamma(I)$.

By continuity of $\text{Ind}_\gamma(z)$, we have $P \cup \gamma(I) = \overline{A} \cap \text{Ind}_\gamma^{-1}(\mathbb{Z} - \{0\}) = \overline{P \cup \gamma(I)} = \overline{P}\cup \gamma(I)$, so it is sufficient to show that $\gamma(I) \subset \overline{P}$.

Information that may be useful: $\text{Ind}_\gamma(z)$ is constant in each connected component of $\mathbb{C} - \gamma(I)$; $\text{Ind}_\gamma(z) = 0$ if $z \in \mathbb{C} - A$; $\overline{P}$ is compact.

• Maybe Jordan curve theorem is needed here? – Crostul Nov 10 '15 at 12:54
• @Crostul $\gamma(I)$ is not necessarily a simple closed curve. – Randy Randerson Nov 10 '15 at 17:56

This isn't true in general. For instance, suppose $\gamma$ goes around a circle once counterclockwise, then goes back around the same circle clockwise. Then $P$ is empty.
If you assume $\gamma$ only intersects itself finitely many times, then it is true. For in that case the non-self-intersection points of $\gamma$ are dense in $\gamma(I)$, so it suffices to show such points are all in $\bar{P}$. But at any such point $p$, $\gamma$ is locally injective; it then follows from the Jordan–Schoenflies theorem that there is a homeomorphism of a small ball around $p$ with a disk which turns $\gamma(I)$ into a diameter of the disk. The index of $\gamma$ on the two half-disks separated by the diameter will then differ by $1$, so at least one of them must have nonzero index.