Admissible cycle around a compact plane set

Cell index inside. Let $Q$ be one of the closed grid rectangles, with $\mathrm{\partial }Q$ its positively oriented boundary, and let $p\in Q^{\circ }$. Along each of the four sides of $Q$ the point $\zeta -p$ has a continuous argument: by [L4] a continuous argument $\theta$ of $\mathrm{\partial }Q-p$ exists, and along a side the perpendicular foot from $p$ to the side's supporting line lies in the relative interior of the side, because both coordinates of $p$ lie strictly between the corresponding coordinates of $Q$; hence $\theta$ varies monotonically along each side by exactly the angle subtended at $p$ by that side, and the four such angles sum to $2\pi$ because the four triangles from $p$ to the sides tile $Q$ and their angles at $p$ cover one full turn. By [L4], $n(\mathrm{\partial }Q,p)=({\theta }_{\text{end}}-{\theta }_{\text{start}})\mathrm{/}2\pi =1$.

Cell index outside. Let $Q$ be one of the closed grid rectangles and let $p\mathrm{\notin }Q$. Then $p$ is strictly to the left of the left side, to the right of the right side, below the bottom side, or above the top side of $Q$; in each case the whole trace $\mathrm{\partial }{Q}^{\ast }$ lies in an open half-plane bounded by a line through $p$, so a continuous argument of $\mathrm{\partial }Q-p$ takes values in an interval of length $\pi$; its increment is a multiple of $2\pi$ by [L4], hence $0$, and $n(\mathrm{\partial }Q,p)=0$.

Cell index outside by Goursat. For $Q$ and $p\mathrm{\notin }Q$ the function $z\mapsto 1\mathrm{/}(z-p)$ is holomorphic on an open set containing $Q$, so [L6] gives ${\int }_{\mathrm{\partial }Q}dz\mathrm{/}(z-p)=0$ and hence $n(\mathrm{\partial }Q,p)=0$ again; the two computations agree and either may be used below.

Choose $J$ and its cells $\mathcal{Q}$ as in [L1]. For each cell $Q$, write its four positively oriented directed side contours separately. Form $\mathrm{\Gamma }$ directly as the finite list of those directed side occurrences that are not shared with another cell; a shared grid side has exactly two occurrences, with opposite directions, and neither is put in $\mathrm{\Gamma }$. This is a chain under [L2], without identifying it with or deleting terms from the list ${\sum }_Q\mathrm{\partial }Q$. Its trace is exactly the exposed grid sides, hence $\mathrm{\partial }J$. It is a cycle: the sum of the endpoint-boundary functions of all four sides of every cell is zero, while each omitted opposite pair also has zero endpoint-boundary function, so the remaining endpoint counts cancel at every grid vertex.

For $p$ lying on no grid line, $n(\mathrm{\Gamma },p)={\sum }_{Q\in \mathcal{Q}}n(\mathrm{\partial }Q,p)=1$ if $p\in J^{\circ }$ and $=0$ if $p\mathrm{\notin }J$: the first equality holds because the omitted opposite pairs contribute zero to the index by [L3] and the index is additive, and the second because exactly one cell contains $p$ in its interior when $p\in J^{\circ }$, while no cell contains $p$ when $p\mathrm{\notin }J$.

The index $n(\mathrm{\Gamma },\cdot )$ is continuous on $\mathbb{C}\setminus \mathrm{\partial }J$ by [L5]; since the points on no grid line are dense in $\mathbb{C}\setminus \mathrm{\partial }J$, [step 2.1] extends by continuity to $n(\mathrm{\Gamma },z)=1$ for every $z\in J^{\circ }$ and $n(\mathrm{\Gamma },z)=0$ for every $z\mathrm{\notin }J$.

Claim 1 follows with this $\mathrm{\Gamma }$: $K\subseteq \mathrm{int}J=J^{\circ }$ and $J\subseteq U$, so $n(\mathrm{\Gamma },z)=1$ for $z\in K$; and $z\mathrm{\notin }U$ implies $z\mathrm{\notin }J$, so $n(\mathrm{\Gamma },z)=0$; the trace of $\mathrm{\Gamma }$ is $\mathrm{\partial }J\subseteq J\setminus K\subseteq U\setminus K$.

For claim 2, choose a compact Jordan set $J_2$, again a finite union of grid rectangles, with $K\subseteq \mathrm{int}{J}_2\subseteq J_2\subseteq \mathrm{int}J$, and let $\beta$ be the cycle obtained from $J_2$ by the construction of [step 1.4]; then ${\beta }^{\ast }=\mathrm{\partial }{J}_2\subseteq \mathrm{int}J$ and ${\gamma }^{\ast }=\mathrm{\partial }J$ are disjoint, and $n(\gamma ,w)=1$ for every $w\in \mathrm{int}J$, in particular for every $w\in \mathrm{\partial }{J}_2={\beta }^{\ast }$.

With $\beta$ and $\gamma$ as in [step 4.2], also $n(\beta ,w)=0$ for every $w\in {\gamma }^{\ast }=\mathrm{\partial }J$: indeed $w\mathrm{\notin }{J}_2$ because $\mathrm{\partial }J\cap J_2=\mathrm{\varnothing }$, and [step 3.1] applied to the Jordan set $J_2$ gives $n(\beta ,w)=0$ for $w\mathrm{\notin }{J}_2$; moreover $n(\beta ,z)=1$ for $z\in K$ because $K\subseteq \mathrm{int}{J}_2$.

Claims 1 and 2 are established by [step 4.1] and [step 5.1] together with [step 4.2]: the cycle $\mathrm{\Gamma }$, and the nested pair $\beta ,\gamma$, have the stated index properties and traces.