$\pi_{3}^{(k)}(M)\leq(C(1+\log M))^{k}\pi_{3}(M)$ (4.33)

for all $k\geq 0$ and some constant $C>0$. For benefit of the reader, we note
that [21, Prop. 18] is stated for a fixed $k$ so that $\pi_{3}^{(k)}(M)\leq
C_{k}(\log M)^{k}\pi_{3}(M)$, but (4.33) follows from the proof, as there is
is shown that $C_{k}=C_{k-1}C^{\prime}$ for some constant $C^{\prime}$.

To show (4.32), we will split it into several cases. Let $A_{1}$ denote the
event that $e$ is on a segment of $\gamma_{n}$ that lies between two
successive open circuits, between $0$ and the first $\mathcal{I}_{1}$ or
between $\mathcal{I}_{Y_{2^{n+1}}}$ and $\partial B_{2^{n+1}}$, and there are
no closed edges along this segment. Let $A_{2}$ denote the event that $e$ lies
on one of the open circuits
$\mathcal{I}_{1},\ldots,\mathcal{I}_{Y_{2^{n+1}}}$. Let us also define
$A=A_{1}\cup A_{2}$.

First, on the event $A^{c}$, we show there must be a closed path from $e$ to
distance $M$. To simplify the notation, we will refer to $(0,0)$ and $\partial
B_{2^{n+1}}$ as open circuits. By Proposition 2.5(x), the dual edge
$e^{\star}$ lies on a closed circuit that encloses the origin, and this closed
circuit does not intersect the geodesic $\gamma_{n}$ at any other edge. Since
the dual circuit encloses the origin, it reaches to distance $M$. There are
two additional disjoint arms obtained by following $\gamma_{n}$ from each
endpoint of $e$ toward the two open circuits, then circling around the origin.
These paths contain at most $\mathfrak{X}_{n}$ many closed edges. Hence,

$\mathbb{P}(\\{e\in\gamma_{n}\\}\cap\\{\mathfrak{X}_{n}\leq
x_{n}^{\lambda}\\}\cap A^{c})\leq C\pi^{(x_{n}^{\lambda})}_{3}(M).$

Next, for the event $A_{1}$, the estimate below follows exactly from the same
proof for (3.6):

$\mathbb{P}(\\{e\in\gamma_{n}\\}\cap\\{\mathfrak{X}_{n}\leq
x_{n}^{\lambda}\\}\cap A_{1})\leq C\pi_{3}(M).$

And on the event $A_{2}$, we get

$\mathbb{P}(\\{e\in\gamma_{n}\\}\cap\\{\mathfrak{X}_{n}\leq
x_{n}^{\lambda}\\}\cap A_{2})\leq C\pi_{3}(M),$

which essentially follows from the proof of (3.7). Specifically, Proposition
2.5(v) ensures the existence of a dual path $\zeta_{e}$ from $e$ to $0$ which
consists of closed edges, except for the open edges where $\zeta_{e}$ crosses
each of the circuits $\mathcal{I}_{j}$. Here we choose the subsequence of
annuli $\\{A_{i_{j}}\\}_{j=0}^{J}$ to cover the open edges along $\zeta_{e}$
and the first defect encountered by $\gamma$ going from $e$ to $\mathcal{B}$.
Note that by Proposition 2.5(x), this first defect must be on a closed
circuit, thus the argument is the same as case (2) from the proof of (3.7).
Letting $E(B_{2^{n+1}})$ be the set of edges with at least one endpoint in the
interior of $B_{2^{n+1}}$, we have that

$\displaystyle\mathbb{P}\Big{(}|\gamma_{n}|\geq\lambda(2^{n+1})^{2+\varepsilon}\pi_{3}(2^{n+1}),\mathfrak{X}_{n}\leq
x_{n}^{\lambda}\Big{)}$ $\displaystyle\leq\mathbb{P}\Big{(}\sum_{e\in
B_{2^{n+1}}}\mathds{1}(e\in\gamma_{n},\mathfrak{X}_{n}\leq
x_{n}^{\lambda})\geq\lambda(2^{n+1})^{2+\varepsilon}\pi_{3}(2^{n+1})\Big{)}$
$\displaystyle\leq(\lambda(2^{n+1})^{2+\varepsilon}\pi_{3}(2^{n+1}))^{-1}\sum_{e\in
B_{2^{n+1}}}\mathbb{P}(e\in\gamma_{n},\mathfrak{X}_{n}\leq x_{n}^{\lambda})$
$\displaystyle=(\lambda(2^{n+1})^{2+\varepsilon}\pi_{3}(2^{n+1}))^{-1}\sum_{k=0}^{2^{n+1}}\sum_{\begin{subarray}{c}e\in
E(B_{2^{n+1}})\\\
\operatorname{dist}(e,\\{0\\})\vee\operatorname{dist}(e,\partial
B_{2^{n+1}})=k\end{subarray}}\mathbb{P}(e\in\gamma_{n},\mathfrak{X}_{n}\leq
x_{n}^{\lambda})$
$\displaystyle\leq(\lambda(2^{n+1})^{2+\varepsilon}\pi_{3}(2^{n+1}))^{-1}C\sum_{k=0}^{2^{n+1}}2^{n+1}\pi_{3}^{(x_{n}^{\lambda})}(k)$
$\displaystyle\leq(\lambda(2^{n+1})^{2+\varepsilon}\pi_{3}(2^{n+1}))^{-1}(Cn)^{x_{n}^{\lambda}}\sum_{k=1}^{2^{n+1}}2^{n+1}\pi_{3}(k)\qquad\text{(by
\eqref{defect_bd})}$
$\displaystyle\leq\dfrac{(Cn)^{x_{n}^{\lambda}}(2^{n+1})^{2}\pi_{3}(2^{n+1})}{\lambda(2^{n+1})^{2+\varepsilon}\pi_{3}(2^{n+1})}\qquad\text{(by
Lemma \ref{lem:sum_arm})}$
$\displaystyle\leq\dfrac{(Cn)^{x_{n}^{\lambda}}}{\lambda
2^{(n+1)\varepsilon}}=\frac{(e^{\log(Cn)})^{x_{n}^{\lambda}}}{\lambda
2^{(n+1)\varepsilon}}\leq\dfrac{e^{C(\log(\lambda)+n)^{2/3}}}{\lambda
2^{(n+1)\varepsilon}}\leq
C^{\prime}\exp\Bigl{(}-c^{\prime}(\log(\lambda)-n)\Bigr{)}.$

In the sixth line, we also absorbed the $k=0$ term into the $k=1$ term,
bounding $1$ by a constant times $\pi_{3}(1)$. On the other hand, we have

$\mathbb{P}(\mathfrak{X}_{n}\geq
x_{n}^{\lambda})\leq\mathbb{P}[\mathfrak{X}_{n}\geq
x_{n}^{\lambda},Y_{2^{n+1}}\leq(\log(\lambda)+n)^{2}]+\mathbb{P}[Y_{2^{n+1}}\geq(\log(\lambda)+n)^{2}]$
(4.34)

We bound each of the terms on the right-hand side of (4.34), starting with the
latter. As in the proof of Lemma 4.16, decompose $Y_{2^{n+1}}$ as
$\sum_{i=0}^{n}V_{i}$, where $V_{i}$ is the number of circuits
$\mathcal{I}_{j}$ in the sequence
$\mathcal{I}_{1},\ldots,\mathcal{I}_{Y_{2^{n+1}}}$ with
$\mathfrak{o}(\mathcal{I}_{j})=i$. By a union bound and (4.25),

$\displaystyle\quad\;\mathbb{P}(Y_{2^{n+1}}\geq(\log(\lambda)+n)^{2})\leq\sum_{i=0}^{n}\mathbb{P}\Bigl{(}V_{i}\geq\frac{(\log(\lambda)+n)^{2}}{n}\Bigr{)}$
$\displaystyle\leq n\exp\Bigl{(}-c\frac{(\log(\lambda)+n)^{2}}{n}\Bigr{)}\leq
Ce^{-c(\log(\lambda)+n)},$

where $C,c>0$ are constants changing from line to line. We decompose the other
term in (4.34) as

$\mathbb{P}(\mathfrak{X}_{n}\geq
x_{n}^{\lambda},Y_{2^{n+1}}\leq(\log(\lambda)+n)^{2})\leq\mathbb{P}(\max_{1\leq
i\leq Y_{2^{n+1}}}X_{i}\geq
x_{n}^{\lambda},Y_{2^{n+1}}\leq(\log(\lambda)+n)^{2})+\mathbb{P}(X^{n}\geq
x_{n}^{\lambda}).$

First, Proposition 4.18 implies that

$\mathbb{P}(X^{n}\geq x_{n}^{\lambda})\leq
C_{1}n^{5/4}e^{-c_{1}(\log(\lambda)+n)^{s}}\leq
C_{1}^{\prime}e^{-c_{1}^{\prime}(\log(\lambda)+n)^{s^{\prime}}}$

for constants $C_{1}^{\prime},c_{1}^{\prime},s^{\prime}>0$. On the other hand,
we use Proposition 4.17 to get, for constants $C,c,s$ that may change from
line to line,

$\displaystyle\quad\;\mathbb{P}(\max_{1\leq i\leq
Y_{2^{n+1}}}X_{i}\geq(\log(\lambda)+n)^{1/3},Y_{2^{n+1}}\leq(\log(\lambda)+n)^{2})$
$\displaystyle\leq\mathbb{P}(\max_{1\leq
i\leq(\log(\lambda)+n)^{2}}X_{i}\geq(\log(\lambda)+n)^{1/3})$
$\displaystyle\leq\sum_{1\leq
i\leq(\log(\lambda)+n)^{2}}\mathbb{P}(X_{i}\geq(\log(\lambda)+n)^{1/3})\leq
Ce^{-c(\log(\lambda)+n)^{s}}\sum_{1\leq i\leq(\log(\lambda)+n)^{2}}i^{5/3}$
$\displaystyle\leq C(\log(\lambda)+n)^{11/3}e^{-c(\log(\lambda)+n)^{s}}\leq
Ce^{-c(\log(\lambda)+n)^{s}}.\qed$

##  5\. Kesten’s separation results and consequences

In the remaining three sections of the paper, we develop the topological
theory needed to rigorously prove Proposition 2.5.

Kesten [13] developed a general theory of percolation on lattices embedded in
$\mathbb{R}^{d}$ and satisfying appropriate conditions. Here we review how
general separation theorems proved in [13] apply to the square lattice.

A mosaic (Definition 2 in Section 2.2 of [13]) is a graph $\mathcal{M}$
embedded in $\mathbb{R}^{2}$ satisfying the following three conditions. Here,
we consider each edge as a set of points in $\mathbb{R}^{2}$.

  1. (i)

$\mathcal{M}$ has no loops.

  2. (ii)

All edges of $\mathcal{M}$ are bounded, and every compact set of
$\mathbb{R}^{2}$ intersects only finitely many edges of $\mathcal{M}$.

  3. (iii)

Any two distinct edges of $\mathcal{M}$ are either disjoint or their
intersection consists of a single vertex of the graph.

An important fact, as noted by Kesten, is that any mosaic is a planar graph. A
mosaic $\mathcal{M}$ splits $\mathbb{R}^{2}$ into connected components called
faces. For a face $F$ of $\mathcal{M}$, to close-pack $F$ means to add an edge
between any pair of vertices on the boundary of $F$ that are not yet connected
([13], Definition 3 in Section 2.2). Given a subset $\mathcal{F}$ of faces,
let $\mathcal{G}$ and $\mathcal{G}^{\star}$ be the graphs obtained from
$\mathcal{M}$ by close-packing all faces in $\mathcal{F}$ and not in
$\mathcal{F}$, respectively. Then $(\mathcal{G},\mathcal{G}^{\star})$ is
called a matching pair; we consider a specific example in Lemma 5.3 which is
illustrated in Figures 6 and 7.

######  Remark 5.1.

As an example, let $\mathcal{M}$ be the triangular lattice. Every face in
$\mathcal{M}$ is a triangle, so no matter the choice of $\mathcal{F}$, the
induced matching pair is $(\mathcal{M},\mathcal{M})$. As we will see, this
self-similarity does not hold for the primal lattice. This is one reason why
working with the primal lattice is technically more challenging than the
triangular lattice.

The following theorem relates site percolation on the graph $\mathcal{G}$ to
site percolation on $\mathcal{G}^{\star}$. In site percolation, the open
cluster of an open vertex $v$ is the set of all vertices connected to $v$ by a
path whose every vertex is open.

######  Theorem 5.2.

[13, Cor. 2.2] Let $(\mathcal{G},\mathcal{G}^{\star})$ be a matching pair
constructed from a mosaic $\mathcal{M}$ and a set of faces $\mathcal{F}$.
Consider site percolation on the graph $\mathcal{G}$. Also, assume the
following.

  1. (i)

There exists $z<\infty$ such that there are at most $z$ edges of $\mathcal{G}$
incident to any vertex of $\mathcal{G}$.

  2. (ii)

All edges of $\mathcal{G}$ have finite diameter, and every compact set of
$\mathbb{R}^{d}$ intersects only finitely many edges of $\mathcal{G}$.

  3. (iii)

$\mathcal{G}$ is connected.

Then, if $W(v)$, the open cluster of the vertex $v$, is nonempty and bounded,
then there exists a closed circuit $J$ on $\mathcal{G}^{\star}$ surrounding
$v$.

In the setting of the present paper, instead of site percolation, we study
edge percolation on $\mathbb{Z}^{2}$. However, we may still apply Theorem 5.2
in the appropriate setting by noting that edge percolation on a graph
$\mathcal{G}$ is equivalent to site percolation on its covering graph
$\widetilde{\mathcal{G}}$ (The vertex set of $\widetilde{\mathcal{G}}$ is the
set of midpoints of the edges of $\mathcal{G}$, and two vertices in
$\widetilde{\mathcal{G}}$ are connected by an edge if the corresponding edges
in $\mathcal{G}$ share an endpoint). This observation was first made by Fisher
and Essam [8, 9], and is detailed in [13, Sec. 2.5].

In the setting of the primal lattice $\mathbb{Z}^{2}$, let $\mathcal{M}$ be
the mosaic defined as follows and depicted in Figure 6. We place a vertex of
$\mathcal{M}$ at the midpoint of each edge of the primal lattice; that is,
points of the form $(m,n+\frac{1}{2})$ and $(m+\frac{1}{2},n)$ for integers
$m,n$. Each vertex $v$ of $\mathcal{M}$ has edges connecting to four other
vertices, namely $v\pm(\frac{1}{2},\frac{1}{2})$ and
$v\pm(\frac{1}{2},-\frac{1}{2})$.

Figure 6. The vertices of the mosaic $\mathcal{M}$ are denoted with black
dots, and the edges are denoted with red/thick edges. The mosaic is overlaid
onto the primal lattice (black/thin edges).

Let $\mathcal{F}$ be the set of faces of $\mathcal{M}$ that contain a point of
$\mathbb{Z}^{2}$, and let
$(\widetilde{\mathcal{G}},\widetilde{\mathcal{G}}^{\star})$ be the associated
matching pair, shown in Figure 7.

Figure 7. On the left is the graph $\widetilde{\mathcal{G}}$, created by
adding an edge between any two vertices of $\mathcal{M}$ in faces that contain
a point of $\mathbb{Z}^{2}$. No additional vertices are added. For faces that
contain points of $(\mathbb{Z}^{2})^{\star}$, no edges are added. On the right
is the graph $\widetilde{\mathcal{G}}^{\star}$, which is created by adding
edges between vertices in $\mathcal{M}$ in faces that contain a point of
$(\mathbb{Z}^{2})^{\star}$. Again, no additional vertices are added. In the
right figure, we also see the edges from the primal lattice in black/thin for
reference, but these edges are not part of $\mathcal{G}^{\star}$.

Then, we have the following lemma.

######  Lemma 5.3.

$\widetilde{\mathcal{G}}$ is the covering graph for the primal lattice, and
$\widetilde{\mathcal{G}}^{\star}$ is the covering graph for the dual lattice.

###### Proof.

We prove that $\widetilde{\mathcal{G}}$ is the covering graph for the primal
lattice, and the other statement follows analogously. It is helpful to refer
to Figure 7. Both graphs have the same vertex set by definition. In
$\mathcal{M}$, each vertex $v$ is connected by an edge to four other vertices.
The corresponding edge $e$ in the primal lattice shares an endpoint with these
edges; they are the edges connected to $e$ that are perpendicular to $e$. The
two additional edges that share an endpoint with $e$ in the primal lattice
correspond to the vertices in $\mathcal{M}$ whose connections to $v$ are added
when constructing $\widetilde{\mathcal{G}}$. ∎

######  Definition 5.4.

For a set $W$ of primal edges, we define the dual boundary of $W$, denoted
$\operatorname{Bd}^{\star}(W)$, as the collection of dual edges $\hat{e}$ such
that the associated primal edge $e$ is not in $W$, but shares at least one
endpoint with an edge of $W$. Similarly, for a set $W^{\star}$ of dual edges,
the primal boundary of the open cluster is defined as the set of primal edges
whose dual edge shares at least one endpoint with an edge of
$\operatorname{Bd}(\hat{e})$.

We say that a connected set of open edges $W$ is a complete open cluster if no
other open edges in $\mathbb{Z}^{2}$ are connected to an endpoint of an edge
in $W$ by an open path.

######  Lemma 5.5.

If $W$ is a complete open (closed) cluster of primal edges, then
$\operatorname{Bd}^{\star}(W)$ consists entirely of dual closed (open) edges.
Similarly, if $W^{\star}$ is a closed (open) cluster of dual edges, then
$\operatorname{Bd}(W^{\star})$ consists entirely of primal open (closed)
edges.

###### Proof.

Let $W$ be an open cluster of primal edges, and assume to the contrary that
$\operatorname{Bd}^{\star}(W)$ has a dual open edge $\hat{e}$. Then, the
associate primal edge $e$ is open, and by definition $e$ is connected to an
edge of $W$. Hence, $e\in W$, a contradiction to the definition of
$\operatorname{Bd}^{\star}(W)$. ∎

Let $\mathcal{G}$ be a graph and $\widetilde{\mathcal{G}}$ its associated
covering graph. Let
$\widetilde{r}=(\widetilde{v}_{1},\widetilde{e}_{2},\ldots,\widetilde{e}_{k},\widetilde{v}_{k})$
be a path on $\widetilde{\mathcal{G}}$. For each $i$, let $e_{i}$ be the
associated edge on $\mathcal{G}$ corresponding to the vertex
$\widetilde{v}_{i}$ on $\widetilde{\mathcal{G}}$. As noted in Comment (iii) in
Section 2.5 of [13], for suitable choices $v_{0}$ and $v_{k}$ of endpoints of
$e_{1}$ and $e_{k}$, $r=(v_{0},e_{1},\ldots,e_{k},v_{k})$ is a path on
$\mathcal{G}$. However, $r$ is not necessarily a self-avoiding path, even if
$\widetilde{r}$ is. Nevertheless, if $\widetilde{r}$ is a circuit, then the
associated path $r$ on $\mathcal{G}$ contains a circuit. Finally, with
$\phi:\mathcal{G}\to\widetilde{\mathcal{G}}$ denoting the map sending edges of
$\mathcal{G}$ to vertices of $\widetilde{\mathcal{G}}$, and $W(e)$ the open
cluster of the edge $e$ in $\mathcal{G}$, [13, Prop. 3.1] gives
$\phi(W(e))=W(\phi(e))$. We therefore obtain the following lemma as a
corollary of Theorem 5.2 and Lemma 5.3. See Figure 8 for clarity.

Figure 8. On the left is a picture of an open cluster of open edges in
$\mathcal{G}$ (blue/dark). On the right is the associated cluster of open
vertices in $\widetilde{\mathcal{G}}$, connected by a blue/dark path. By
Theorem 5.2, there exists a closed circuit in
$\widetilde{\mathcal{G}}^{\star}$ surrounding this open cluster. This circuit
of closed vertices is shown on the right (red/light). The set of edges on the
left associated to the closed vertices on the right is not a circuit, but
contains a circuit surrounding the original open cluster of edges.

######  Lemma 5.6.

Let $W\subseteq\mathbb{Z}^{2}$ be a collection of vertices that are each
connected to each other by an open path and such that for $v\notin W$, $v$ is
not connected to any vertex in $W$ by an open path. Then, if the set $W$ is
bounded, there exists a closed circuit $\mathcal{D}$ on the dual lattice that
encloses $W$. If $W$ consists of more than a single point, then we can
consider $W$ as a complete open cluster of edges, and the circuit can be
chosen to consist only of edges of $\operatorname{Bd}^{\star}(W)$. By duality
and symmetry, the same holds with the roles of the primal and dual lattices
reversed and/or with the roles of “open” and “closed” reversed.

###### Proof.

If $W$ consists of a single vertex $v$, then there are no open edges incident
to $v$. Then, the dual edges to each of the four closed edges incident to $v$
forms a dual closed circuit surrounding $v$.

Now, assume that $W$ has more than one point so that we can consider it as a
complete open cluster of edges. We perform a modification of the environment
as follows. Set all edges whose dual does not lie on
$\operatorname{Bd}^{\star}(W)$ to open. We claim that in the modified
environment, $W$ is still a complete open cluster of edges. By definition, all
primal edges $e$ sharing an endpoint with an edge in $W$ are either already in
$W$, or their dual lies in $\operatorname{Bd}^{\star}(W)$. If the dual edge
$\widehat{e}$ lies in $\operatorname{Bd}^{\star}(W)$, then by Lemma 5.5, $e$
is closed. Thus, a vertex outside $W$ cannot connect to a vertex in $W$ by an
open path without taking a closed edge in $\operatorname{Bd}^{\star}(W)$.

Since $W$ is still a complete open cluster, the remarks above the theorem
imply that, in the modified environment, there exist a closed dual circuit
surrounding $W$. But these edges are also closed in the original environment.
∎

######  Lemma 5.7.

Let $\mathcal{D}$ be a closed circuit. Assume that there exists two disjoint
connected sets $\mathcal{A}$ and $\mathcal{B}$ of dual vertices lying in
$\mathsf{int}(\mathcal{D})$ such that all edges connecting vertices of
$\mathcal{A}$ are closed and the same with edges connecting vertices in
$\mathcal{B}$. Let $W_{\mathcal{A}}$ be the cluster of dual vertices connected
by a closed dual path to $\mathcal{A}$, and let $W_{\mathcal{B}}$ be the
cluster of closed dual vertices connected by a closed dual path to
$\mathcal{B}$. Then, if $W_{\mathcal{A}}\neq W_{\mathcal{B}}$, there exists an
open circuit $\mathcal{C}$ contained in $\mathsf{int}(\mathcal{D})$ that
contains $W_{\mathcal{A}}$ in its interior and $W_{\mathcal{B}}$ in its
exterior, or vise versa.

By duality, this lemma also holds by interchanging the roles of open/closed
and simultaneously interchanging the roles of primal/dual.

Figure 9 gives an illustration of Lemma 5.7.

Figure 9. Two disjoint clusters of closed edges, $W_{\mathcal{A}}$ and
$W_{\mathcal{B}}$, which lie inside a circuit $\mathcal{D}$ and are separated
by an open circuit $\mathcal{C}$ (blue/thin)

###### Proof.

We modify the environment so that all dual edges with at least one endpoint in
$\mathsf{ext}(\mathcal{D})$ (along with their primal counterparts) are set to
closed. Let $W_{\mathcal{D}}$ be the infinite cluster of points connected to
$\mathcal{D}$ by a closed dual path in this modified environment, and adjust
$W_{\mathcal{A}}$ and $W_{\mathcal{B}}$ to include these dual edges in the
case that one of the clusters connects to $\mathcal{D}$. We note that both
clusters cannot connect to $\mathcal{D}$ because then, there is a closed dual
path from $\mathcal{A}$ to $\mathcal{D}$, then from $\mathcal{D}$ to
$\mathcal{B}$, a contradiction. We consider two cases:

Case 1: $W_{\mathcal{A}}=W_{\mathcal{D}}$ or
$W_{\mathcal{B}}=W_{\mathcal{D}}$. Without loss of generality, say
$W_{\mathcal{B}}=W_{\mathcal{D}}$. Then, $W_{\mathcal{A}}$ must be bounded
because it cannot cross $\mathcal{D}$. Lemma 5.6 implies that there exists an
open circuit $\mathcal{C}$ containing $W_{\mathcal{A}}$ in its interior.
Vertices in $W_{\mathcal{B}}$ cannot lie on the circuit $\mathcal{C}$ because
$\mathcal{C}$ is open (and therefore primal). $W_{\mathcal{B}}$ contains
points in the exterior of $\mathcal{C}$ because
$W_{\mathcal{B}}=W_{\mathcal{D}}$ is unbounded by assumption. Furthermore,
$W_{\mathcal{B}}$ cannot contain points in the interior of $\mathcal{C}$;
otherwise, the closed dual path from the points of $W_{\mathcal{B}}$ in
$\mathsf{ext}(\mathcal{C})$ to the points of $W_{\mathcal{B}}$ in
$\mathsf{int}(\mathcal{C})$ crosses $\mathcal{C}$, a contradiction because
$\mathcal{C}$ is open. Hence, $\mathcal{C}$ contains $W_{\mathcal{A}}$ in its
interior and $W_{\mathcal{B}}$ in its exterior.

Case 2: $W_{\mathcal{B}}\neq W_{\mathcal{D}}$ and $W_{\mathcal{A}}\neq
W_{\mathcal{D}}$. We modify the environment further to obtain the previous
case. We first argue that there exists an infinite closed dual path from
either $W_{\mathcal{A}}$ or $W_{\mathcal{B}}$ that avoids the other. To see
this, take any dual path from $W_{\mathcal{A}}$. There is a last vertex in
$W_{\mathcal{A}}\cup W_{\mathcal{B}}$: the portion of the path starting from
that vertex is the desired path. Set all the edges on this path to closed.
Thus, we have reduced to the previous case. Since edges are only being changed
to closed in this modified environment, the open circuit we obtain is still
open in the original environment. ∎

For the benefit of Section 6, we note that Theorem 5.2 and Lemmas 5.6 and 5.7
are deterministic statements that hold for any given configuration on the
graphs.

##  6\. Topological details concerning circuits and paths

Here we provide rigorous justification of certain topological constructions
used throughout the paper. The first four lemmas will be applications of the
Jordan–Schönflies theorem, which says that for every Jordan curve
$\mathcal{J}\subset\mathbb{R}^{2}$, there is a homeomorphism
$f\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ such that $f(\mathcal{J})$ is the
unit circle. In particular, because homeomorphisms preserve path-
connectedness, the sets $\mathsf{int}(\mathcal{J})$ and
$\mathsf{ext}(\mathcal{J})$ are separately path-connected, and for every
$z\in\mathcal{J}$, the set
$\\{z\\}\cup\mathsf{int}(\mathcal{J})\cup\mathsf{ext}(\mathcal{J})$ is path-
connected.

Recall our terminology that $\mathcal{J}_{1}$ is enclosed by $\mathcal{J}_{2}$
if $\mathsf{int}(\mathcal{J}_{1})\subseteq\mathsf{int}(\mathcal{J}_{2})$.

######  Lemma 6.1.

Let $\mathcal{J}_{1},\mathcal{J}_{2}$ be two Jordan curves so that
$\mathcal{J}_{1}\subseteq\mathcal{J}_{2}\cup\mathsf{int}(\mathcal{J}_{2})$.
Then, $\mathcal{J}_{2}$ encloses $\mathcal{J}_{1}$.

###### Proof.

Suppose not. Then, there exists
$x\in\mathsf{int}(\mathcal{J}_{1})\cap\mathsf{int}(\mathcal{J}_{2})^{c}$, and
there exists an infinite path from $x$ that stays entirely in
$\mathsf{ext}(\mathcal{J}_{2})$ (except for possibly the initial point). Since
$x\in\mathsf{int}(\mathcal{J}_{1})$, $x$ must cross
$\mathcal{J}_{1}\subseteq\mathcal{J}_{2}\cup\mathsf{int}(\mathcal{J}_{2})$, a
contradiction. ∎

######  Lemma 6.2.

Let $\mathcal{J}_{1}$ and $\mathcal{J}_{2}$ be Jordan curves, neither of which
encloses the other. If
$\mathsf{int}(\mathcal{J}_{1})\cap\mathsf{int}(\mathcal{J}_{2})$ is nonempty,
then the following statements hold:

  1. (a)

Both $\mathsf{int}(\mathcal{J}_{1})\cap\mathcal{J}_{2}$ and
$\mathsf{ext}(\mathcal{J}_{1})\cap\mathcal{J}_{2}$ are nonempty.

  2. (b)

$\mathcal{J}_{1}\cap\mathcal{J}_{2}$ contains at least two distinct points.

###### Proof.

By assumption, we can find
$x_{1}\in\mathsf{int}(\mathcal{J}_{1})\setminus\mathsf{int}(\mathcal{J}_{2})$,
$x_{2}\in\mathsf{int}(\mathcal{J}_{2})\setminus\mathsf{int}(\mathcal{J}_{1})$,
and $y\in\mathsf{int}(\mathcal{J}_{1})\cap\mathsf{int}(\mathcal{J}_{2})$.
Since $x_{1}$ and $y$ both belong to $\mathsf{int}(\mathcal{J}_{1})$, there is
a continuous curve in the plane starting at $y$, ending at $x_{1}$, and
remaining entirely in $\mathsf{int}(\mathcal{J}_{1})$. But since $y$ belongs
to $\mathsf{int}(\mathcal{J}_{2})$ while $x$ does not, this curve must contain
a point lying on $\mathcal{J}_{2}$. On the other hand, because
$x_{2}\notin\mathsf{int}(\mathcal{J}_{1})$, there is a continuous curve in the
plane starting at $x_{2}$, extending to infinity, and remaining entirely in
$\mathsf{ext}(\mathcal{J}_{1})\cup\\{x_{2}\\}$. Given that
$x_{2}\in\mathsf{int}(\mathcal{J}_{2})$, this curve must intersect
$\mathcal{J}_{2}$, hence $\mathcal{J}_{2}\cap\mathsf{ext}(\mathcal{J}_{1})$ is
nonempty. We have now seen that $\mathcal{J}_{2}$ intersects both
$\mathsf{int}(\mathcal{J}_{1})$ and $\mathsf{ext}(\mathcal{J}_{1})$, thereby
proving part (a). Now take some
$z_{\mathrm{int}}\in\mathcal{J}_{2}\cap\mathsf{int}(\mathcal{J}_{1})$ and
$z_{\mathrm{ext}}\in\mathcal{J}_{2}\cap\mathsf{ext}(\mathcal{J}_{1})$. Any
continuous curve between $z_{\mathrm{int}}$ and $z_{\mathrm{ext}}$ must
intersect $\mathcal{J}_{1}$, and by definition $\mathcal{J}_{2}$ offers two
such curves which are disjoint (except at $z_{\mathrm{int}}$ and
$z_{\mathrm{ext}}$, of course). This observation implies part (b). ∎

In the following results, $\mathsf{Ball}_{\varepsilon}(z)$ denotes the open
ball centered at $z\in\mathbb{R}^{2}$ with radius $\varepsilon>0$.

######  Lemma 6.3.

Let $\mathcal{J}$ be a Jordan curve. Consider two adjacent boxes on the square
lattice, and let $e$ be the edge connecting them. Assume that all of $e$ lies
on $\mathcal{J}$, and further assume that $\mathsf{int}(\mathcal{J})$ has
empty intersection with the interiors of each the two boxes. Then, exactly one
of the dual neighbors of $e$ lies in $\mathsf{int}(\mathcal{J})$, and the
other dual neighbor lies in $\mathsf{ext}(\mathcal{J})$.

###### Proof.

For such an edge $e$, let $z$ be its center point. It is immediate that the
set $\mathsf{Ball}_{1/4}(z)\setminus\mathcal{J}$ has two connected components.
$\mathcal{J}$ is the boundary of its interior and exterior, so any open set
intersecting $\mathcal{J}$ has nonempty intersection with both
$\mathsf{int}(\mathcal{J})$ and $\mathsf{ext}(\mathcal{J})$. Hence, one of the
components of $\mathsf{Ball}_{1/4}(z)\setminus\mathcal{J}$ must be contained
in $\mathsf{int}(\mathcal{J})$ and the other is contained in
$\mathsf{ext}(\mathcal{J})$. Let $z_{1}^{\star}$ and $z_{2}^{\star}$ be the
dual neighbors of $e$. The dual edge $e^{\star}$ connects $z_{1}^{\star}$ to
one of these connected components without crossing $\mathcal{J}$, and also
connects $z_{2}^{\star}$ to the other connected component without crossing
$\mathcal{J}$. ∎

The following is similar to Lemma 6.3 but applies in a more general setting.

######  Lemma 6.4.

For any Jordan curve $\mathcal{J}$, any $z\in\mathcal{J}$, and any
$\varepsilon>0$, there exists an open set
$U\subseteq\mathsf{Ball}_{\varepsilon}(z)$ containing $z$ such that
$U\setminus\mathcal{J}$ has exactly two connected components.

###### Proof.

Let $f\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ be a homeomorphism such that
$f(\mathcal{J})$ is the unit circle. Then $f(\mathsf{Ball}_{\varepsilon}(z))$
is an open set containing $f(z)$. Therefore, we can find $\delta\in(0,2)$
small enough that $\mathsf{Ball}_{\delta}(f(z))\subseteq
f(\mathsf{Ball}_{\varepsilon}(z))$. Clearly
$\mathsf{Ball}_{\delta}(f(z))\setminus f(\mathcal{J})$ has two connected
components, and so the claim is satisfied by taking
$U=f^{-1}\big{(}\mathsf{Ball}_{\delta}(f(z))\big{)}$. ∎

######  Lemma 6.5.

Let $\mathcal{J}$ be a Jordan curve, and let
$\varphi\colon[0,1]\to\mathbb{R}^{2}$ be a continuous curve such that
$\varphi(0)$ and $\varphi(1)$ are distinct points on $\mathcal{J}$, and
$\varphi(t)\in\mathsf{int}(\mathcal{J})$ for all $t\in(0,1)$. Denote by
$\mathcal{J}_{1}$ the Jordan curve starting at $\varphi(0)$, proceeding in
some direction along $\mathcal{J}$ until reaching $\varphi(1)$, and then
following $\varphi$ back to $\varphi(0)$. Let $\mathcal{J}_{2}$ be the same
but proceeding along $\mathcal{J}$ in the other direction. We can then
decompose the interior of $\mathcal{J}$ as the disjoint union

$\displaystyle\begin{split}\mathsf{int}(\mathcal{J})=\mathsf{int}(\mathcal{J}_{1})\uplus\mathsf{int}(\mathcal{J}_{2})\uplus\\{\varphi(t):\,t\in(0,1)\\}.\end{split}$
(6.1)

Figure 10 gives a pictoral representation of Lemma 6.5.

Figure 10. Lemma 6.5: The curve $\varphi$ spits $\mathsf{int}(\mathcal{J})$
into three disjoint sets

###### Proof.

First we argue that the right-hand side of (6.1) is contained in the left-hand
side. As a trivial first step, we assumed that
$\varphi(t)\in\mathsf{int}(\mathcal{J})$ for all $t\in(0,1)$. Next, consider
any $x\in\mathsf{int}(\mathcal{J}_{1})$ and any continuous curve $\phi$
starting at $x$ and extending to infinity. We wish to show that $\phi$ must
intersect $\mathcal{J}$. Now, $\phi$ must intersect $\mathcal{J}_{1}$; if this
intersection is a point on $\mathcal{J}$, then we are done. Otherwise the
point of intersection is $\phi(t)$ for some $t\in(0,1)$; since this point
belongs to $\mathsf{int}(\mathcal{J})$, we again conclude that $\phi$ must
eventually intersect $\mathcal{J}$. We have thus argued that
$\mathsf{int}(\mathcal{J}_{1})\subseteq\mathsf{int}(\mathcal{J})$, and of
course $\mathsf{int}(\mathcal{J}_{2})\subseteq\mathsf{int}(\mathcal{J})$ also
holds by symmetric reasoning. Furthermore, $\mathsf{int}(\mathcal{J}_{1})$ and
$\mathsf{int}(\mathcal{J}_{2})$ must be disjoint by the following logic.

Since $\mathcal{J}_{1}$ and $\mathsf{int}(\mathcal{J}_{1})$ are disjoint, we
have

$\displaystyle(\mathcal{J}_{1}\cup\mathcal{J})\cap\mathsf{int}(\mathcal{J}_{1})=\mathcal{J}\cap\mathsf{int}(\mathcal{J}_{1})\subseteq\mathcal{J}\cap\mathsf{int}(\mathcal{J})=\varnothing.$

Meanwhile, because $\mathcal{J}$ and $\mathsf{ext}(\mathcal{J})$ are disjoint,
we have

$\displaystyle(\mathcal{J}_{1}\cup\mathcal{J})\cap\mathsf{ext}(\mathcal{J})$
$\displaystyle=(\mathcal{J}_{1}\setminus\mathcal{J})\cap\mathsf{ext}(\mathcal{J})$
$\displaystyle=\\{\gamma(t):\,t\in(0,1)\\}\cap\mathsf{ext}(\mathcal{J})\subseteq\mathsf{int}(\mathcal{J})\cap\mathsf{ext}(\mathcal{J})=\varnothing.$

Considering that $\mathcal{J}_{2}\subseteq\mathcal{J}_{1}\cup\mathcal{J}$, it
follows from these two observations that

$\displaystyle\begin{split}\mathcal{J}_{2}\cap\mathsf{int}(\mathcal{J}_{1})=\varnothing=\mathcal{J}_{2}\cap\mathsf{ext}(\mathcal{J}).\end{split}$
(6.2)

Now take any $x\in\mathsf{int}(\mathcal{J}_{1})$ and
$z\in\mathcal{J}\setminus\mathcal{J}_{2}$. Since $z$ necessarily belongs to
$\mathcal{J}_{1}$, there exists a continuous path $\phi$ from $x$ to $z$
remaining entirely in $\mathsf{int}(\mathcal{J}_{1})\cap\\{z\\}$. By the first
equality in (6.2), this path never intersects $\mathcal{J}_{2}$. Next we
extend $\phi$ from $z$ to infinity using a path remaining entirely in
$\mathsf{ext}(\mathcal{J})\cap\\{z\\}$; by the second equality in (6.2), this
extension also avoids $\mathcal{J}_{2}$. We have thus constructed a path
disjoint from $\mathcal{J}_{2}$ that starts at $x$ and extends to infinity,
implying that $x\in\mathsf{ext}(\mathcal{J}_{2})$. In particular, $x$ does not
belong to $\mathsf{int}(\mathcal{J}_{2})$.

All that remains to show is that the left-hand side of (6.1) is contained in
the right-hand side. So suppose $x\in\mathsf{int}(\mathcal{J})$ but belongs to
neither $\mathsf{int}(\mathcal{J}_{2})$ nor $\\{\varphi(t):\,t\in(0,1)\\}$. It
follows that $x\notin\mathcal{J}_{2}$ since
$\mathcal{J}_{2}\cap\mathsf{int}(\mathcal{J})$ is equal to
$\\{\varphi(t):\,t\in(0,1)\\}$. This leaves only the possibility that
$x\in\mathsf{ext}(\mathcal{J}_{2})$, meaning there is a continuous curve
$\phi$ starting at $x$ and extending to infinity which never intersects
$\mathcal{J}_{2}$. But because $x\in\mathsf{int}(\mathcal{J})$, this curve
must intersect $\mathcal{J}$, necessarily at some point belonging to
$\mathcal{J}\setminus\mathcal{J}_{2}\subseteq\mathcal{J}_{1}$. Let $z$ be the
first intersection of $\phi$ with $\mathcal{J}$ upon leaving $x$, and let
$\phi_{x\to z}$ be portion of $\phi$ from $x$ to $z$. In particular, $z$ is
the unique intersection point of $\phi_{x\to z}$ with $\mathcal{J}$.
Furthermore, $\phi_{x\to z}$ can only intersect $\mathcal{J}_{1}$ at
$\mathcal{J}_{1}\setminus\mathcal{J}_{2}\subseteq\mathcal{J}$, meaning $z$ is
also the unique intersection point of $\phi_{x\to z}$ with $\mathcal{J}_{1}$.

######  Claim 6.6.

For all $\varepsilon>0$ small enough (depending on $z$), we have

$\displaystyle\begin{split}\mathcal{J}\cap\mathsf{Ball}_{\varepsilon}(z)=\mathcal{J}_{1}\cap\mathsf{Ball}_{\varepsilon}(z).\end{split}$
(6.3)

###### Proof.

Being a Jordan curve, $\mathcal{J}_{2}$ is closed, and so
$\mathbb{R}^{2}\setminus\mathcal{J}_{2}$ is open. Considering that
$z\in\mathbb{R}^{2}\setminus\mathcal{J}_{2}$, we can thus choose
$\varepsilon_{0}>0$ such that $B_{\varepsilon}(z)\cap\mathcal{J}_{2}$ is empty
whenever $\varepsilon\leq\varepsilon_{0}$. For such $\varepsilon$, using the
fact that
$\mathcal{J}\setminus\mathcal{J}_{2}=\mathcal{J}_{1}\setminus\mathcal{J}_{2}$,
we now have

$\displaystyle\mathcal{J}\cap\mathsf{Ball}_{\varepsilon}(z)$
$\displaystyle=(\mathcal{J}\setminus\mathcal{J}_{2})\cap\mathsf{Ball}_{\varepsilon}(z)=(\mathcal{J}_{1}\setminus\mathcal{J}_{2})\cap\mathsf{Ball}_{\varepsilon}(z)=\mathcal{J}_{1}\cap\mathsf{Ball}_{\varepsilon}(z),$

where the first and last equalities use
$\mathsf{Ball}_{\varepsilon}(z)\cap\mathcal{J}_{2}=\varnothing$. ∎

Now let $\varepsilon>0$ be small enough that (6.3) holds, and then take $U$ as
in Lemma 6.4. In particular, we have

$\displaystyle\begin{split}\mathcal{J}\cap
U=\mathcal{J}\cap\mathsf{Ball}_{\varepsilon}(z)\cap
U=\mathcal{J}_{1}\cap\mathsf{Ball}_{\varepsilon}(z)\cap U=\mathcal{J}_{1}\cap
U.\end{split}$ (6.4)

Because $U$ has nonempty intersection with both $\mathsf{int}(\mathcal{J})$
and $\mathsf{ext}(\mathcal{J})$, we conclude that the two connected components
of $U\setminus\mathcal{J}$ are precisely $U\cap\mathsf{int}(\mathcal{J})$ and
$U\cap\mathsf{ext}(\mathcal{J})$. But
$U\setminus\mathcal{J}_{1}=U\setminus\mathcal{J}$ by (6.4), and so by the
exact same logic, these two connected components are also expressible as
$U\cap\mathsf{int}(\mathcal{J}_{1})$ and $U\cap\mathsf{ext}(\mathcal{J}_{1})$.
In particular, $U\cap\mathsf{int}(\mathcal{J}_{1})$ is equal to either
$U\cap\mathsf{int}(\mathcal{J})$ or $U\cap\mathsf{ext}(\mathcal{J})$.
Considering that
$\mathsf{int}(\mathcal{J}_{1})\subset\mathsf{int}(\mathcal{J})$, it must be
the former:

$\displaystyle\begin{split}U\cap\mathsf{int}(\mathcal{J}_{1})=U\cap\mathsf{int}(\mathcal{J}).\end{split}$
(6.5)

To complete the proof, we return to our consideration of $\phi_{x\to z}$. This
curve starts in $\mathsf{int}(\mathcal{J})$ and only encounters $\mathcal{J}$
at the terminal point $z$, and so it must otherwise remain in
$\mathsf{int}(\mathcal{J})$. Since $\phi_{x\to z}$ intersects
$U\cap\mathsf{int}(\mathcal{J})$, it follows from (6.5) that $\phi_{x\to z}$
intersects $\mathsf{int}(\mathcal{J}_{1})$, meaning that $x$ is connected to
some element of $\mathsf{int}(\mathcal{J}_{1})$ by a continuous curve avoiding
$\mathcal{J}_{1}$. Hence $x\in\mathsf{int}(\mathcal{J}_{1})$, as needed. ∎

######  Lemma 6.7.

Let $\mathcal{J}_{1}$ be a Jordan curve, and let
$\Phi\colon[0,1]\to\mathbb{R}^{2}$ be a continuous curve such that $\Phi(0)$
and $\Phi(1)$ are distinct points on $\mathcal{J}_{1}$, and
$\Phi(t)\in\mathsf{ext}(\mathcal{J}_{1})$ for all $t\in(0,1)$. Let
$\varphi_{a}$ denote the portion of $\mathcal{J}_{1}$ proceeding clockwise
from $\Phi(0)$ to $\Phi(1)$, and let $\varphi_{b}$ be the same but proceeding
counterclockwise. We parameterize these functions
$\varphi_{a},\varphi_{b}\colon[0,1]\to\mathbb{R}^{2}$ such that
$\varphi_{a}(0)=\varphi_{b}(0)=\Phi(0)$ and
$\varphi_{a}(1)=\varphi_{b}(1)=\Phi(1)$. Denote by $\mathcal{J}_{a}$ the
Jordan curve starting at $\Phi(0)$, following $\varphi_{a}$ until reaching
$\Phi(1)$, and then following $\Phi$ back to $\Phi(0)$. Let $\mathcal{J}_{b}$
be the same but using $\varphi_{b}$. Then either

$\displaystyle\begin{split}\mathsf{int}(\mathcal{J}_{a})=\mathsf{int}(\mathcal{J}_{1})\uplus\mathsf{int}(\mathcal{J}_{b})\uplus\\{\varphi_{b}(t):\,t\in(0,1)\\}\end{split}$
(6.6a) or
$\displaystyle\begin{split}\mathsf{int}(\mathcal{J}_{b})=\mathsf{int}(\mathcal{J}_{1})\uplus\mathsf{int}(\mathcal{J}_{a})\uplus\\{\varphi_{a}(t):\,t\in(0,1)\\}.\end{split}$
(6.6b)

Figure 11 gives a graphical depiction of Lemma 6.7.

Figure 11. A depiction of Lemma 6.7: The curve $\mathcal{J}_{1}$ is depicted
in black/thick and the curve $\Phi$ is depicted in gray/thin. In this case,
(6.6a) holds.

###### Proof.

We divide our argument into four steps.

Step 1: Show that
$\mathsf{int}(\mathcal{J}_{a})\cap\mathsf{ext}(\mathcal{J}_{1})$ is nonempty.

For any $y\in\mathcal{J}_{a}\setminus\mathcal{J}_{1}$, we have
$y\in\mathsf{ext}(\mathcal{J}_{1})$ by assumption. Since
$\mathsf{ext}(\mathcal{J}_{1})$ is open, there is $\varepsilon>0$ small enough
that $\mathsf{Ball}_{\varepsilon}(y)\subseteq\mathsf{ext}(\mathcal{J}_{1})$.
Because $y\in\mathcal{J}_{a}$, $\mathsf{Ball}_{\varepsilon}(y)$ has nonempty
intersection with $\mathsf{int}(\mathcal{J}_{a})$. This completes the first
step.

Step 2: Show that
$\mathsf{int}(\mathcal{J}_{a})\cap\mathsf{int}(\mathcal{J}_{b})$ is nonempty.

By the first step, there exists
$x\in\mathsf{int}(\mathcal{J}_{a})\cap\mathsf{ext}(\mathcal{J}_{1})$. In
particular, there exists a path $\phi$ starting at $x$, extending to infinity,
and remaining entirely in $\mathsf{ext}(\mathcal{J}_{1})$. But because
$x\in\mathsf{int}(\mathcal{J}_{a})$, this path must intersect
$\mathcal{J}_{a}$. Denote by $z\in\mathcal{J}_{a}$ the last intersection
point, and let $\phi_{z\to\infty}$ be the portion of $\phi$ starting at $z$
and extending to infinity; with this choice, $\phi_{z\to\infty}$ never again
intersects $\mathcal{J}_{a}$. Now note that

$\displaystyle\begin{split}\mathcal{J}_{1}\cap\mathsf{ext}(\mathcal{J}_{1})=\mathcal{J}_{2}\cap\mathsf{ext}(\mathcal{J}_{1})=\\{\Phi(t):\,t\in(0,1)\\}.\end{split}$
(6.7)

By construction we have $z\in\mathsf{ext}(\mathcal{J}_{1})$, and so we can
choose $\varepsilon>0$ sufficiently small that
$\mathsf{Ball}_{\varepsilon}(z)\subseteq\mathsf{ext}(\mathcal{J}_{1})$, so as
to guarantee

$\displaystyle\begin{split}\mathcal{J}_{1}\cap\mathsf{Ball}_{\varepsilon}(z)=\mathcal{J}_{1}\cap\mathsf{ext}(\mathcal{J}_{1})\cap\mathsf{Ball}_{\varepsilon}(z)=\mathcal{J}_{2}\cap\mathsf{ext}(\mathcal{J}_{1})\cap\mathsf{Ball}_{\varepsilon}(z)=\mathcal{J}_{2}\cap\mathsf{Ball}_{\varepsilon}(z).\end{split}$
(6.8)

By Lemma 6.4, there exists a open set
$U\subseteq\mathsf{Ball}_{\varepsilon}(z)$ containing $z$ such that
$U\setminus\mathcal{J}_{1}$ has exactly two components. Moreover, (6.8) can be
specialized this open set:

$\displaystyle\begin{split}\mathcal{J}_{a}\cap
U=\mathcal{J}_{a}\cap\mathsf{Ball}_{\varepsilon}(z)\cap
U=\mathcal{J}_{b}\cap\mathsf{Ball}_{\varepsilon}(z)\cap U=\mathcal{J}_{b}\cap
U.\end{split}$ (6.9)

Because $U$ has nonempty intersection with both
$\mathsf{int}(\mathcal{J}_{a})$ and $\mathsf{ext}(\mathcal{J}_{a})$, we
conclude that the two connected components of $U\setminus\mathcal{J}_{a}$ are
precisely $U\cap\mathsf{int}(\mathcal{J}_{a})$ and
$U\cap\mathsf{ext}(\mathcal{J}_{a})$. But
$U\setminus\mathcal{J}_{a}=U\setminus\mathcal{J}_{b}$ by (6.9), and so by the
exact same logic, these two connected components are also expressible as
$U\cap\mathsf{int}(\mathcal{J}_{b})$ and $U\cap\mathsf{ext}(\mathcal{J}_{b})$.
In particular, $U\cap\mathsf{ext}(\mathcal{J}_{b})$ is equal to either
$U\cap\mathsf{int}(\mathcal{J}_{a})$ or $U\cap\mathsf{ext}(\mathcal{J}_{a})$.
We claim it must be the latter. Indeed, since $\phi_{z\to\infty}$ extends to
infinity yet never intersects
$\mathcal{J}_{a}\cup\mathcal{J}_{1}\supseteq\mathcal{J}_{b}$ except at its
starting point $z$, it must otherwise remain in
$\mathsf{ext}(\mathcal{J}_{a})\cap\mathsf{ext}(\mathcal{J}_{b})$. Hence
$U\cap\mathsf{ext}(\mathcal{J}_{a})$ and $U\cap\mathsf{ext}(\mathcal{J}_{b})$
have nonempty intersection, forcing the two sets to be equal. This leaves us
to conclude that
$U\cap\mathsf{int}(\mathcal{J}_{a})=U\cap\mathsf{int}(\mathcal{J}_{b})$; in
particular, $\mathsf{int}(\mathcal{J}_{a})\cap\mathsf{int}(\mathcal{J}_{b})$
is nonempty.

Step 3: Argue that one of $\mathcal{J}_{a}$ and $\mathcal{J}_{b}$ must enclose
the other.

Consider the two portions of $\mathcal{J}_{b}$ between $\Phi(0)$ and
$\Phi(1)$. The portion formed by the curve $\Phi$ is shared entirely with
$\mathcal{J}_{a}$. The other portion of $\mathcal{J}_{b}$ intersects
$\mathcal{J}_{a}$ only at the endpoints $\Phi(0)$ and $\Phi(1)$, and so this
portion otherwise remains either entirely in $\mathsf{int}(\mathcal{J}_{a})$
or entirely in $\mathsf{ext}(\mathcal{J}_{a})$. Putting these two observations
together, we determine that either
$\mathcal{J}_{b}\subseteq\mathcal{J}_{a}\cup\mathsf{int}(\mathcal{J}_{a})$ or
$\mathcal{J}_{b}\subseteq\mathcal{J}_{a}\cup\mathsf{ext}(\mathcal{J}_{a})$. In
either case, by (the contrapositive of) Lemma 6.2(a), one of $\mathcal{J}_{a}$
and $\mathcal{J}_{b}$ encloses the other. For concreteness, let us assume
$\mathcal{J}_{a}$ encloses $\mathcal{J}_{b}$, since the reverse scenario would
use exactly the same argument.

Step 4: Appeal to Lemma 6.5.

Recall that $\varphi_{b}$ is the portion of $\mathcal{J}_{b}$ which is not
shared with $\mathcal{J}_{a}$ except at the endpoints $\varphi_{b}(0)=\Phi(0)$
and $\varphi_{b}(1)=\Phi(1)$. Consider any $y=\varphi_{b}(t)$ for $t\in(0,1)$.
Since $y\notin\mathcal{J}_{a}$, we must have either
$y\in\mathsf{int}(\mathcal{J}_{a})$ or $y\in\mathsf{ext}(\mathcal{J}_{a})$. We
claim the former is true. Indeed, because $y\in\mathcal{J}_{b}$, every open
neighborhood of $y$ must intersect both $\mathsf{int}(\mathcal{J}_{b})$ and
$\mathsf{ext}(\mathcal{J}_{b})$. But we have assumed
$\mathsf{int}(\mathcal{J}_{b})\subseteq\mathsf{int}(\mathcal{J}_{a})$, and so
every open neighborhood of $y$ must intersect $\mathsf{int}(\mathcal{J}_{a})$,
forcing $y\in\mathsf{int}(\mathcal{J}_{a})$. We have thus shown that

$\displaystyle\\{\varphi_{b}(t):\,t\in(0,1)\\}\subseteq\mathsf{int}(\mathcal{J}_{a}).$

Therefore, we are in the setting of Lemma 6.5 with
$\mathcal{J}=\mathcal{J}_{a}$, $\mathcal{J}_{2}=\mathcal{J}_{b}$, and
$\varphi=\varphi_{b}$. Indeed, in this notation, $\mathcal{J}_{1}$ is the
union of $\varphi_{b}$ with some portion of $\mathcal{J}_{a}$ (namely the
clockwise arc of $\mathcal{J}_{1}$ between $\Phi(0)$ and $\Phi(1)$), while
$\mathcal{J}_{2}=\mathcal{J}_{b}$ is the union of $\varphi_{b}$ with the
complementary portion of $\mathcal{J}_{a}$ (namely $\Phi$ itself). See Figure
11 for reference. The desired conclusion now follows from (6.1). ∎

Recall Definition 2.1 for circuits.

######  Lemma 6.8.

If $\mathcal{J}_{1}$ and $\mathcal{J}_{2}$ are Jordan curves such that
$\mathsf{int}(\mathcal{J}_{1})\subsetneq\mathsf{int}(\mathcal{J}_{2})$, then
the area of $\mathsf{int}(\mathcal{J}_{1})$ is strictly less than that of
$\mathsf{int}(\mathcal{J}_{2})$. Furthermore, if $\mathcal{C}$ is a circuit,
then the area of $\mathsf{int}(\mathcal{C})$ is a positive integer.

###### Proof.

For the first claim of the lemma, it suffices to find some open set
$U\subseteq\mathsf{ext}(\mathcal{J}_{1})\cap\mathsf{int}(\mathcal{J}_{2})$. By
assumption there exists
$x\in\mathsf{int}(\mathcal{J}_{2})\setminus\mathsf{int}(\mathcal{J}_{1})$. If
$x\in\mathsf{ext}(\mathcal{J}_{1})$, then take $U$ to be any open set which
contains $x$ and remains in
$\mathsf{ext}(\mathcal{J}_{1})\cap\mathsf{int}(\mathcal{J}_{2})$. Otherwise we
must have $x\in\mathcal{J}_{1}$, and then we take some open set
$V\subseteq\mathsf{int}(\mathcal{J}_{2})$ containing $x$. This set $V$ must
contain some $y\in\mathsf{ext}(\mathcal{J}_{1})$, and so we take $U$ to be any
open set containing $y$ which remains in
$V\cap\mathsf{ext}(\mathcal{J}_{1})\subseteq\mathsf{ext}(\mathcal{J}_{1})\cap\mathsf{int}(\mathcal{J}_{2})$.

For the second claim, let us consider the case when $\mathcal{C}$ is a primal
circuit; the case of a dual circuit is entirely analogous. For every
$x\in\mathbb{Z}^{2}$, the set $U_{x}=x+(0,1)^{2}$ can have no intersection
with $\mathcal{C}$. (This is because $\mathcal{C}$ consists entirely of edges
between nearest-neighbor vertices.) And clearly $U_{x}$ is connected, and so
it must lie either entirely in $\mathsf{int}(\mathcal{C})$ or entirely in
$\mathsf{ext}(\mathcal{C})$. Since each $U_{x}$ has area $1$ and the area of
$\mathbb{R}^{2}\setminus\biguplus_{x\in\mathbb{Z}^{2}}U_{x}$ is equal to $0$,
we conclude that the area of $\mathsf{int}(\mathcal{C})$ is precisely the
number of $x$ such that $U_{x}\subseteq\mathsf{int}(\mathcal{C})$. This number
is positive because the interior of any Jordan curve is open. ∎

######  Lemma 6.9.

Let $\mathcal{P}$ be an open primal circuit and $\mathcal{D}$ a closed dual
circuit. If $\mathsf{int}(\mathcal{P})\cap\mathsf{int}(\mathcal{D})$ is
nonempty, then one must enclose the other.

###### Proof.

Suppose the conclusion were false, in which case Lemma 6.2(b) would guarantee
that the two circuits intersect. But since $\mathcal{P}$ is primal and
$\mathcal{D}$ is dual, this intersection must be the midpoint of some edge
$e$. Hence $e$ is both open and closed, a contradiction. ∎

######  Lemma 6.10.

Consider any circuits $\mathcal{C}_{1},\dots,\mathcal{C}_{k}$ such that
$\bigcap_{i=1}^{k}\mathsf{int}(\mathcal{C}_{k})$ is nonempty. There exists a
unique circuit $\mathcal{C}$ such that

$\displaystyle\begin{split}\bigcup_{i=1}^{k}\mathsf{int}(\mathcal{C}_{i})\subseteq\mathsf{int}(\mathcal{C})\quad\text{and}\quad\mathcal{C}\subseteq\bigcup_{i=1}^{k}\mathcal{C}_{i}.\end{split}$
(6.10)

###### Proof.

First we prove uniqueness. Consider two circuits
$\mathcal{C},\widetilde{\mathcal{C}}$ that both satisfy (6.10) with the given
collection $(\mathcal{C}_{i})_{i=1}^{k}$. Suppose towards a contradiction that
$\mathcal{C}\neq\widetilde{\mathcal{C}}$. Then we claim there is some edge $e$
belonging to one of the circuits whose midpoint is in the exterior of the
other circuit. To see this claim, first consider the case that one circuit
encloses the other, say
$\mathsf{int}(\widetilde{\mathcal{C}})\subseteq\mathsf{int}(\mathcal{C})$.
Then every edge of $\mathcal{C}$ either belongs to $\widetilde{\mathcal{C}}$
and has its midpoint in $\mathsf{ext}(\widetilde{\mathcal{C}})$. Since we
assumed $\mathcal{C}\neq\widetilde{\mathcal{C}}$, there must be at least one
edge with the latter property. On the other hand, if neither circuit encloses
the other, then we can simply appeal to Lemma 6.2(a).

From the claim, we find some $z\in\mathsf{ext}(\widetilde{\mathcal{C}})$ which
is the midpoint of an edge $e$ belonging to $\mathcal{C}$. The edge $e$ must
belong to $\mathcal{C}_{i}$ for some $i$, and so every open neighborhood of
$e$ has nonempty intersection with both $\mathsf{int}(\mathcal{C}_{i})$ and
$\mathsf{ext}(\mathcal{C}_{i})$. In particular,

$\displaystyle\mathsf{Ball}_{1/2}(z)\cap\mathsf{int}(\mathcal{C}_{i})\neq\varnothing.$

Next notice that because the midpoint $z$ is in the exterior of
$\widetilde{\mathcal{C}}$, the open set $\mathsf{Ball}_{1/2}(z)$ has no
intersection with $\widetilde{\mathcal{C}}$; it follows that

$\displaystyle\mathsf{Ball}_{1/2}(z)\subseteq\mathsf{ext}(\widetilde{\mathcal{C}}).$

But when read together, the two previous displays imply that
$\mathsf{int}(\mathcal{C}_{i})$ has nonempty intersection with
$\mathsf{ext}(\widetilde{\mathcal{C}})$, which contradicts the hypothesis that
$\mathsf{int}(\mathcal{C}_{i})\subseteq\mathsf{int}(\widetilde{\mathcal{C}})$.
We are left to conclude that $\widetilde{\mathcal{C}}$ must be equal to
$\mathcal{C}$.

Having established uniqueness, we look to prove existence. We begin with just
two circuits. That is, we wish to find some circuit $\mathcal{C}$ such that

$\displaystyle\begin{split}\mathsf{int}(\mathcal{C}_{1})\cup\mathsf{int}(\mathcal{C}_{2})\subseteq\mathsf{int}(\mathcal{C})\quad\text{and}\quad\mathcal{C}\subseteq\mathcal{C}_{1}\cup\mathcal{C}_{2}.\end{split}$
(6.11)

If $\mathsf{int}(\mathcal{C}_{1})\subseteq\mathsf{int}(\mathcal{C}_{2})$ or
$\mathsf{int}(\mathcal{C}_{2})\subseteq\mathsf{int}(\mathcal{C}_{1})$, then we
simply take $\mathcal{C}$ to be $\mathcal{C}_{2}$ or $\mathcal{C}_{1}$,
respectively. So let us assume neither containment is true, and then Lemma
6.2(a) implies the existence of some
$z_{\mathrm{int}}\in\mathsf{ext}(\mathcal{C}_{1})\cap\mathcal{C}_{2}$. From
$z_{\mathrm{ext}}$ we follow the circuit $\mathcal{C}_{2}$ in both directions;
by Lemma 6.2(b), in each direction we will encounter a distinct first
intersection with $\mathcal{C}_{1}$. Let $\Phi$ denote the subpath of
$\mathcal{C}_{2}$ between these intersection points and containing
$z_{\mathrm{ext}}$, so that $\Phi$ is entirely in
$\mathsf{ext}(\mathcal{C}_{1})$ except at its endpoints. We are thus in the
setting of Lemma 6.7, with the possibility to complete $\Phi$ to a full Jordan
curve by following $\mathcal{C}_{1}$ either clockwise or counterclockwise. The
circuit resulting from one of these directions will enclose the circuit
resulting from the other direction; let $\mathcal{C}$ be the enclosing circuit
and $\mathcal{C}_{b}$ the enclosed circuit, so that (6.6) gives

$\displaystyle\begin{split}\mathsf{int}(\mathcal{C})=\mathsf{int}(\mathcal{C}_{1})\uplus\mathsf{int}(\mathcal{C}_{b})\uplus\Phi.\end{split}$
(6.12)

Notice that the edges in $\mathcal{C}$ belong to either $\mathcal{C}_{1}$ or
to $\Phi$, which was a subpath of $\mathcal{C}_{2}$. So if $\mathcal{C}$
encloses $\mathcal{C}_{2}$, then $\mathcal{C}$ satisfies (6.11). Otherwise, we
proceed inductively with $\mathcal{C}$ replacing $\mathcal{C}_{1}$. To
complete the proof, we just need to argue that the procedure just performed
can only be repeated finitely many times. Indeed, in light of (6.12), Lemma
6.8 tells us that the area of $\mathsf{int}(\mathcal{C})$ is at least 1 unit
greater than that of $\mathsf{int}(\mathcal{C}_{1})$. At the same time, the
area of a circuit using only edges in $\mathcal{C}_{1}\cup\mathcal{C}_{2}$ has
a finite upper bound, and so the argument can only be repeated finitely many
times. Once it terminates, the resulting circuit $\mathcal{C}$ must satisfy
(6.11).

Our final step is to prove existence for general $n$ from the $n=2$ case just
handled. Given any two circuits $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ such
that $\mathsf{int}(\mathcal{C}_{1})\cap\mathsf{int}(\mathcal{C}_{2})$ is
nonempty, let $\mathcal{C}_{1}\vee\mathcal{C}_{2}$ denote the unique circuit
$\mathcal{C}$ satisfying (6.11). Then given a sequence
$(\mathcal{C}_{i})_{i=1}^{k}$ such that
$\bigcap_{i=1}^{k}\mathsf{int}(\mathcal{C}_{i})\neq\varnothing$, we
inductively define

$\displaystyle\mathcal{C}_{1}^{\prime}$ $\displaystyle=\hbox
to0.0pt{$\mathcal{C}_{1}$,\hss}\phantom{\mathcal{C}_{i-1}^{\prime}\vee\mathcal{C}_{i}}\quad\text{and}$
$\displaystyle\mathcal{C}_{i}^{\prime}$
$\displaystyle=\mathcal{C}_{i-1}^{\prime}\vee\mathcal{C}_{i}\quad\text{for
$i\in\\{2,\dots,n\\}$}.$

By simple induction, the final circuit $\mathcal{C}_{n}^{\prime}$ satisfies
(6.10). ∎

######  Lemma 6.11.

Let $S\subset\mathbb{R}^{2}$ be a nonempty connected subset of
$\mathbb{R}^{2}$, and consider any collection of circuits
$(\mathcal{C}_{i})_{i\in I}$ such that
$S\subseteq\mathsf{int}(\mathcal{C}_{i})$ for every $i$. There exists a unique
circuit $\mathcal{C}$ such that

$\displaystyle\begin{split}S\subseteq\mathsf{int}(\mathcal{C})\subseteq\bigcap_{i\in
I}\mathsf{int}(\mathcal{C}_{i})\quad\text{and}\quad\mathcal{C}\subseteq\bigcup_{i\in
I}\mathcal{C}_{i}.\end{split}$ (6.13)

###### Proof.

First we prove uniqueness. Consider two circuits
$\mathcal{C},\widetilde{\mathcal{C}}$ that both satisfy (6.13) with the given
collection $(\mathcal{C}_{i})_{i\in I}$. Suppose towards a contradiction that
$\mathcal{C}\neq\widetilde{\mathcal{C}}$. Then we claim there is some edge $e$
belonging to one of the circuits whose midpoint is in the interior of the
other circuit. To see this claim, first consider the case that one circuit
encloses the other, say
$\mathsf{int}(\mathcal{C})\subseteq\mathsf{int}(\widetilde{\mathcal{C}})$.
Then every edge of $\mathcal{C}$ either belongs to $\widetilde{\mathcal{C}}$
and has its midpoint in $\mathsf{int}(\widetilde{\mathcal{C}})$. Since we
assumed $\mathcal{C}\neq\widetilde{\mathcal{C}}$, there must be at least one
edge with the latter property. On the other hand, if neither circuit encloses
the other, then we can simply appeal to Lemma 6.2(a).

From the claim, we find some $z\in\mathsf{int}(\widetilde{\mathcal{C}})$ which
is the midpoint of an edge $e$ belonging to $\mathcal{C}$. The edge $e$ must
belong to $\mathcal{C}_{i}$ for some $i$, and so every open neighborhood of
$e$ has nonempty intersection with both $\mathsf{int}(\mathcal{C}_{i})$ and
$\mathsf{ext}(\mathcal{C}_{i})$. In particular,

$\displaystyle\mathsf{Ball}_{1/2}(z)\cap\mathsf{ext}(\mathcal{C}_{i})\neq\varnothing.$

Next notice that because the midpoint $z$ is in the interior of
$\widetilde{\mathcal{C}}$, the open set $\mathsf{Ball}_{1/2}(z)$ has no
intersection with $\widetilde{\mathcal{C}}$; it follows that

$\displaystyle\mathsf{Ball}_{1/2}(z)\subseteq\mathsf{int}(\widetilde{\mathcal{C}}).$

But when read together, the two previous displays imply that
$\mathsf{int}(\widetilde{\mathcal{C}})$ has nonempty intersection with
$\mathsf{ext}(\mathcal{C}_{i})$, which contradicts the hypothesis that
$\mathsf{int}(\widetilde{\mathcal{C}})\subseteq\mathsf{int}(\mathcal{C}_{i})$.
We are left to conclude that $\widetilde{\mathcal{C}}$ must be equal to
$\mathcal{C}$.

Having established uniqueness, we look to prove existence. We begin with just
two circuits. That is, we wish to find some circuit $\mathcal{C}$ such that

$\displaystyle\begin{split}S\subseteq\mathsf{int}(\mathcal{C})\subseteq\mathsf{int}(\mathcal{C}_{1})\cap\mathsf{int}(\mathcal{C}_{2})\quad\text{and}\quad\mathcal{C}\subseteq\mathcal{C}_{1}\cup\mathcal{C}_{2}.\end{split}$
(6.14)

If $\mathsf{int}(\mathcal{C}_{1})\subseteq\mathsf{int}(\mathcal{C}_{2})$ or
$\mathsf{int}(\mathcal{C}_{2})\subseteq\mathsf{int}(\mathcal{C}_{1})$, then we
simply take $\mathcal{C}$ to be $\mathcal{C}_{1}$ or $\mathcal{C}_{2}$,
respectively. let us assume neither containment is true, and then Lemma 6.2(a)
implies the existence of some
$z_{\mathrm{int}}\in\mathsf{int}(\mathcal{C}_{1})\cap\mathcal{C}_{2}$. From
$z_{\mathrm{int}}$ we follow the circuit $\mathcal{C}_{2}$ in both directions;
by Lemma 6.2(b), in each direction we will encounter a distinct first
intersection with $\mathcal{C}_{1}$. Let $\varphi$ denote the subpath of
$\mathcal{C}_{2}$ between these intersection points and containing
$z_{\mathrm{int}}$, so that $\varphi$ is entirely in
$\mathsf{int}(\mathcal{C}_{1})$ except at its endpoints. We are thus in the
setting of Lemma 6.5, with the possibility to complete $\varphi$ to a full
Jordan curve by following $\mathcal{C}_{1}$ either clockwise or
counterclockwise. We choose the direction that contains $S$ in its interior
and call the resulting circuit $\mathcal{C}$. Because $S$ is connected, the
decomposition (6.1) shows that such a choice exists and is unique; we thus
have
$S\subseteq\mathsf{int}(\mathcal{C})\subsetneq\mathsf{int}(\mathcal{C}_{1})$.
Notice that the edges in $\mathcal{C}$ belong to either $\mathcal{C}_{1}$ or
to $\varphi$, which was a subpath of $\mathcal{C}_{2}$. So if $\mathcal{C}$ is
enclosed by $\mathcal{C}_{2}$, then we are done. Otherwise, we proceed
inductively with $\mathcal{C}$ replacing $\mathcal{C}_{1}$. To complete the
proof, we just need to argue that the procedure just performed can only be
repeated finitely many times. Indeed, Lemma 6.8 tells us that the area of
$\mathsf{int}(\mathcal{C})$ is at least 1 unit less than that of
$\mathsf{int}(\mathcal{C}_{1})$. Since $\mathsf{int}(\mathcal{C})$ has finite
area, it is now evident that the argument can only be repeated finitely many
times. Once it terminates, we are left with a circuit $\mathcal{C}$ satisfying
(6.14).

Our final step is to prove existence for an arbitrary index set $I$. There are
only countably many distinct circuits in the lattice, and so it suffices to
assume the index set $I$ is the set of positive integers. Given any two
circuits $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ such that
$S\subseteq\mathsf{int}(\mathcal{C}_{1})\cap\mathsf{int}(\mathcal{C}_{2})$,
let $\mathcal{C}_{1}\wedge\mathcal{C}_{2}$ denote the unique circuit
$\mathcal{C}$ satisfying (6.14). Then given a sequence
$(\mathcal{C}_{i})_{i=1}^{\infty}$ such that
$S\subseteq\mathsf{int}(\mathcal{C}_{i})$ for every $i$, we inductively define

$\displaystyle\mathcal{C}_{1}^{\prime}$ $\displaystyle=\hbox
to0.0pt{$\mathcal{C}_{1}$,\hss}\phantom{\mathcal{C}_{i-1}^{\prime}\wedge\mathcal{C}_{i}}\quad\text{and}$
$\displaystyle\mathcal{C}_{i}^{\prime}$
$\displaystyle=\mathcal{C}_{i-1}^{\prime}\wedge\mathcal{C}_{i}\quad\text{for
$i\geq 2$}.$

By definition, we have

$\displaystyle\begin{split}S\subseteq\mathsf{int}(\mathcal{C}_{i}^{\prime})&\subseteq\mathsf{int}(\mathcal{C}_{i-1}^{\prime})\cap\mathsf{int}(\mathcal{C}_{i})\\\
&\subseteq\mathsf{int}(\mathcal{C}_{i-2}^{\prime})\cap\mathsf{int}(\mathcal{C}_{i-1})\cap\mathsf{int}(\mathcal{C}_{i})\\\
&\hskip 5.59721pt\vdots\\\
&\subseteq\mathsf{int}(\mathcal{C}_{1})\cap\cdots\cap\mathsf{int}(\mathcal{C}_{i}).\end{split}$
(6.15)

Since there are only finitely many circuits enclosed by $\mathcal{C}_{1}$, we
must eventually have $\mathcal{C}_{i}^{\prime}$ equal to some $\mathcal{C}$
for all large $i$. For this circuit $\mathcal{C}$, (6.13) follows from the
fact that (6.15) holds for every $i$. ∎

##  7\. Proof of Proposition 2.5 (construction of geodesic)

For the benefit of the reader, we recall the conventions of Definitions 2.2
and 2.4 and the statement of Proposition 2.5. We say a circuit $\mathcal{C}$
encloses a set of vertices $\mathcal{A}$ (i.e. either
$\mathcal{A}\subseteq\mathbb{Z}^{2}$ or $A\subset\widehat{\mathbb{Z}}^{2}$) if
$\mathcal{A}\subseteq\mathsf{int}(\mathcal{C})$. We say a circuit
$\mathcal{C}$ encloses another circuit $\mathcal{C}^{\prime}$ if
$\mathsf{int}(\mathcal{C}^{\prime})\subseteq\mathsf{int}(\mathcal{C})$.
Definition 2.4 states that primal path $\gamma$ is open if all its edges are
open. We also recall the conventions for dual neighbors: every edge $e$ has
two dual neighbors–the endpoints of the dual edge $e^{\star}$. On the other
hand, every vertex $v$ has four dual neighbors. In both cases, dual neighbors
are vertices on the dual lattice.

  * •

If $A\subset\mathbb{Z}^{2}$, then we say $\gamma$ starts (ends) at
$\mathcal{A}$ if its first (last) vertex is an element of $\mathcal{A}$.

  * •

If $\mathcal{E}$ is a collection of primal edges, then we say $\gamma$ starts
(ends) at $\mathcal{E}$ if its first (last) vertex is an endpoint of some
element of $\mathcal{E}$.

  * •

If $\widehat{\mathcal{E}}$ is a collection of dual edges, then we say $\gamma$
starts (ends) at $\widehat{\mathcal{E}}$ if its first (last) vertex is a dual
neighbor of some element of $\widehat{\mathcal{E}}$.

A dual path $\zeta$ is closed if all its edges are closed.

  * •

If $\mathcal{A}\subset\mathbb{Z}^{2}$, then we say $\zeta$ starts (ends) at
$\mathcal{A}$ if its first (last) vertex is equal to
$x\pm(\frac{1}{2},\frac{1}{2})$ or $x\pm(\frac{1}{2},-\frac{1}{2})$ for some
$x\in\mathcal{A}$.

  * •

If $\mathcal{E}$ is a collection of primal edges, then we say $\gamma$ starts
(ends) at $\mathcal{E}$ if its first (last) vertex is a dual neighbor of some
element of $\mathcal{E}$.

  * •

If $\widehat{\mathcal{E}}$ is a collection of dual edges, then we say $\gamma$
starts (ends) at $\widehat{\mathcal{E}}$ if its first (last) vertex is an
endpoint of some element of $\widehat{\mathcal{E}}$.

Proposition 2.5 states the following: let $\mathcal{A}$ and $\mathcal{B}$ be
finite disjoint connected subsets of $\mathbb{Z}^{2}$. On the event
$\Omega_{\infty}$, there exists a (possibly empty) sequence of edge-disjoint
open circuits
$\mathcal{I}_{1},\ldots,\mathcal{I}_{L},\mathcal{I}_{L+1},\ldots,\mathcal{I}_{P}$
satisfying the following:

  1. (i)

$\mathcal{A}\subseteq\mathsf{int}(\mathcal{I}_{1})\subseteq\mathsf{int}(\mathcal{I}_{2})\subseteq\cdots\subseteq\mathsf{int}(\mathcal{I}_{L})\subseteq\mathsf{int}(\mathcal{I}_{L})\cup\mathcal{I}_{L}\subseteq\mathcal{B}^{C}$.

  2. (ii)

$\mathsf{int}(\mathcal{I}_{L})\cap\mathsf{int}(\mathcal{I}_{L+1})=\varnothing$.

  3. (iii)

$A^{C}\supseteq\mathsf{int}(\mathcal{I}_{L+1})\cup\mathcal{I}_{L+1}\supseteq\mathsf{int}(\mathcal{I}_{L+1})\supseteq\mathsf{int}(\mathcal{I}_{L+2})\supseteq\cdots\supseteq\mathsf{int}(\mathcal{I}_{P})\supseteq
B$.

  4. (iv)

For $j\in\\{1,\ldots,P-2\\}$, the circuits $\mathcal{I}_{j}$ and
$\mathcal{I}_{j+2}$ are vertex-disjoint.

  5. (v)

For $j\in\\{1,\ldots,P\\}$ and every $e\in\mathcal{I}_{j}$, there exists a
dual path $\zeta_{e}$ to $\mathcal{A}$ that has exactly $j-1$ open edges–one
crossing each of the circuits $\mathcal{I}_{1},\ldots,\mathcal{I}_{j-1}$.

Furthermore, there exists a geodesic $\gamma$ from $\mathcal{A}$ to
$\mathcal{B}$ and a disjoint dual path $\zeta$ from $\mathcal{A}$ to
$\mathcal{B}$ satisfying the following properties (here we note that the paths
$\zeta_{e}$ in Item (v) are not necessarily disjoint from $\gamma$):

  1. (vi)

$\zeta$ has exactly $P$ open edges–one crossing each of the circuits
$\mathcal{I}_{1},\ldots,\mathcal{I}_{L},\mathcal{I}_{L+1},\ldots,\mathcal{I}_{P}$.

  2. (vii)

For each circuit $\mathcal{I}_{j}$, let $x$ and $y$ be the first and last
vertices from $\gamma$ on that circuit. Then, the portion of $\gamma$ between
$x$ and $y$ lies entirely on $\mathcal{I}_{j}$. If $\mathcal{I}_{j}$ and
$\mathcal{I}_{j+1}$ are not vertex-disjoint, then the path $\gamma$ crosses
from $\mathcal{I}_{j}$ to $\mathcal{I}_{j+1}$ via a common vertex, and
$\gamma$ contains no edges lying strictly between the circuits.

  3. (viii)

If $\gamma_{R}$ is the geodesic corresponding to $\mathcal{A}=\\{0\\}$ and
$\mathcal{B}_{R}=\partial B_{R}$ for $R\geq 0$, then the sequence $\gamma_{R}$
can be chosen so that the portion of $\\{\gamma_{R}\\}_{R\in\mathbb{Z}}$
between successive open circuits does not depend on $R$ (while the portion
between the last circuit in $\partial B_{R}$ and the boundary does depend on
$R$).

  4. (ix)

From every open edge $e\in\gamma$ with
$e\notin\mathcal{I}_{1}\cup\cdots\cup\mathcal{I}_{L}\cup\mathcal{I}_{L+1}\cup\cdots\cup\mathcal{I}_{P}$,
there exists a closed path from $e$ to $\zeta$.

  5. (x)

The dual of each closed edge along $\gamma$ belongs to a closed circuit
$\mathcal{U}$ that either contains $\mathcal{A}$ in its interior and
$\mathcal{B}$ in its exterior, or vice versa. The circuit $\zeta$ does not
contain the dual of any other edges along $\gamma$.

  6. (xi)

With $\\{0,1\\}$-valued edge-weights, the closed circuits $\mathcal{U}$ from
Item (x) can be chosen to form a edge-disjoint collection
$\mathcal{U}_{1},\ldots,\mathcal{U}_{V}$. The union of the circuits
$\mathcal{I}_{1},\ldots,\mathcal{I}_{P}$ and the circuits
$\mathcal{U}_{1},\ldots,\mathcal{U}_{V}$ forms a sequence
$\mathcal{C}_{1},\ldots,\mathcal{C}_{K}$, which is ordered so that, for some
index $W\in\\{0,\ldots,K\\}$,
$\mathsf{int}(\mathcal{C}_{W})\cap\mathsf{int}(\mathcal{C}_{W+1})=\varnothing$
(with the convention
$\mathsf{int}(\mathcal{C}_{0})=\mathsf{int}(\mathcal{C}_{K+1})=\varnothing)$,
and we have the following inclusions

$\mathcal{A}\subseteq\mathsf{int}(\mathcal{C}_{1})\subseteq\cdots\subseteq\mathsf{int}(\mathcal{C}_{W}),\quad\text{and}\quad\mathsf{int}(\mathcal{C}_{W+1})\supseteq\cdots\mathsf{int}(\mathcal{C}_{K})\supseteq
B.$

We recall that, on the event $\Omega_{\infty}$, there exists a closed circuit
$\mathcal{D}$ enclosing both $\mathcal{A}$ and $\mathcal{B}$. Choose such a
circuit $\widehat{\mathcal{D}}$ in a deterministic fashion. One way to do this
is to let $\widehat{\mathcal{D}}$ be the unique innermost such circuit from
Lemma 6.11, but it is not necessary to make this particular choice.

###  7.1. Construction of the circuits and Proof of Items (i)–(iv)

Assuming that the set of circuits

$S_{0}=\\{\text{Open circuits
}\mathcal{C}:\mathcal{A}\subseteq\mathsf{int}(\mathcal{C})\text{ and
}\mathcal{B}\subseteq\mathsf{ext}(\mathcal{C})\\}$

is nonempty, let $\mathcal{I}_{1}$ be the unique innermost circuit from Lemma
6.11 consisting of edges of circuits in $S_{0}$ and satisfying

$\mathcal{A}\subseteq\mathsf{int}(\mathcal{I}_{1})\subseteq\bigcap_{\mathcal{C}\in
S_{0}}\mathsf{int}(\mathcal{C}).$

Then, also,

$\mathcal{B}\subseteq\bigcup_{\mathcal{C}\in
S_{0}}(\mathsf{ext}(\mathcal{C})\cup\mathcal{C})\subseteq\mathsf{ext}(\mathcal{I}_{1})\cup\mathcal{I}_{1},$

but points of $\mathcal{B}$ cannot lie on $\mathcal{I}_{1}$ because they do
not lie on any of the $\mathcal{C}\in S_{0}$ by assumption. Hence,
$\mathcal{B}\subseteq\mathsf{ext}(\mathcal{I}_{1})$.

Inductively, for $j\geq 1$, assume that $\mathcal{I}_{j}$ has been
constructed, and assume that the set

$S_{j}=\\{\text{Open circuits }\mathcal{C}\text{ edge disjoint from and
enclosing }\mathcal{I}_{i}:\mathcal{B}\subseteq\mathsf{ext}(\mathcal{C})\\}$
(7.1)

is nonempty. Then, let $\mathcal{I}_{j+1}$ be the unique innermost circuit
chosen from the set $S_{j}$ from Lemma 6.11. Again,
$\mathcal{A}\subseteq\mathsf{int}(\mathcal{I}_{j+1})$ and
$\mathcal{B}\subseteq\mathsf{ext}(\mathcal{I}_{j+1})$.

Let $L$ be the smallest index $j$ such that $S_{j}=\varnothing$. We see from
induction that

$\mathcal{A}\subseteq\mathsf{int}(\mathcal{I}_{1})\subseteq\mathsf{int}(\mathcal{I}_{2})\subseteq\cdots\subseteq\mathsf{int}(\mathcal{I}_{L})\subseteq\mathsf{int}(\mathcal{I}_{L})\cup\mathcal{I}_{L}\subseteq\mathcal{B}^{C},$

so Item (i) holds. We now argue that $L$ must be finite. Take $\zeta$ to be
any finite dual path from $\mathcal{A}$ to $\mathcal{B}$. The path $\zeta$
must cross any circuit that encloses $\mathcal{A}$ and keeps $\mathcal{B}$ in
its exterior. The path $\zeta$ can only cross $\mathcal{I}_{j}$ at the
midpoint of an edge, and since the circuits $\mathcal{I}_{j}$ are pairwise
edge-disjoint, $L$ can be no greater than the finite number of edges in
$\zeta$.

Next, define $S_{L}^{\prime}$ to be

$S_{L}^{\prime}=\\{\text{Open circuits }\mathcal{C}\text{ edge disjoint from
}\mathcal{I}_{L}:\mathsf{int}(\mathcal{I}_{L})\subseteq\mathsf{ext}(\mathcal{C})\text{
and }\mathcal{B}\subseteq\mathsf{int}(\mathcal{C})\\}.$

In the case $L=0$, we replace $\mathsf{int}(\mathcal{I}_{L})$ with
$\mathcal{A}$ in the definition of $S_{L}^{\prime}$. In the proof below, we
often refer to $\mathsf{int}(\mathcal{I}_{L})$, but the reader should replace
this with $\mathcal{A}$ for the case $L=0$.

Assuming that $S_{L}^{\prime}$ is nonempty, we show that $S_{L}^{\prime}$ is
finite. Lemma 6.9 implies that any open circuit then either encloses or is
enclosed by $\widehat{\mathcal{D}}$. If an open circuit encloses
$\widehat{\mathcal{D}}$, it encloses both $\mathcal{A}$ and $\mathcal{B}$ and
is therefore not in $S_{L}^{\prime}$. Thus, $S_{L}^{\prime}$ is a subset of
open circuits enclosed by $\widehat{\mathcal{D}}$ and is therefore finite. Let
$\mathcal{I}_{L+1}$ be the unique circuit from Lemma 6.10 satisfying
$\bigcup_{\mathcal{C}\in
S_{L}^{\prime}}\mathsf{int}(\mathcal{C})\subseteq\mathsf{int}(\mathcal{I}_{L+1})$
and $\mathcal{I}_{L+1}\subseteq\bigcup_{\mathcal{C}\in
S_{L}^{\prime}}\mathcal{C}$ . It follows immediately that
$\mathcal{B}\subseteq\mathsf{int}(\mathcal{I}_{L+1})$ and that
$\mathcal{I}_{L+1}$ is edge disjoint from $\mathcal{I}_{L}$. We now prove the
nontrivial fact that
$\mathsf{int}(\mathcal{I}_{L})\subseteq\mathsf{ext}(\mathcal{I}_{L+1})$, and
in particular,
$\mathsf{int}(\mathcal{I}_{L})\cap\mathsf{int}(\mathcal{I}_{L+1})=\varnothing$,
proving Item (ii). Refer to Figure 12 for an explanation of why this is
nontrivial and a visual representation of the contradiction derived in what
follows.

Figure 12. In this figure, $\mathcal{I}_{L}$ is the circuit enclosing
$\mathcal{A}$ denoted in black/thin. Consider the red/medium thickness and
blue/thickest circuits. These both enclose $\mathcal{B}$ and both keep
$\mathcal{I}_{L}$ in their exteriors. However, the unique maximal circuit that
joins the two circuits as in Lemma 6.10 (the outermore dashed gray circuit)
does not keep $\mathcal{I}_{L}$ in the exterior. We show that these pairs of
circuits cannot exist by showing the existence of the innermore dashed gray
circuit built from the two circuits. This contradicts the maximality of
$\mathcal{I}_{L}$.

.

Pick arbitrary $\mathcal{C}_{1},\mathcal{C}_{2}\in S_{L}^{\prime}$, and let
$\mathcal{C}$ be the unique circuit in Lemma 6.10 satisfying
$\mathsf{int}(\mathcal{C}_{1})\cup\mathsf{int}(\mathcal{C}_{2})\subseteq\mathsf{int}(\mathcal{C})$
and $\mathcal{C}\subseteq\mathcal{C}_{1}\cup\mathcal{C}_{2}$. By induction, it
suffices to show that
$\mathsf{int}(\mathcal{I}_{L})\subseteq\mathsf{ext}(\mathcal{C})$. Assume, to
the contrary, that this is false. The set $\mathsf{int}(\mathcal{I}_{L})$
cannot contain points on $\mathcal{C}$ because $\mathcal{C}$ consists of edges
of $\mathcal{C}_{1},\mathcal{C}_{2}$, and $\mathsf{int}(\mathcal{I}_{L})$ is
contained in the exterior of both of these. Then, since
$\mathsf{int}(\mathcal{I}_{L})$ is path connected, it cannot contain points in
both $\mathsf{int}(\mathcal{C})$ and $\mathsf{ext}(\mathcal{C})$ (If $L=0$, we
replace $\mathsf{int}(\mathcal{I}_{L})$ with the path connected set consisting
of vertices of $\mathcal{A}$ and all edges connecting them). Hence, our
assumption implies that
$\mathsf{int}(\mathcal{I}_{L})\subseteq\mathsf{int}(\mathcal{C})$. Thus, the
set

$\widetilde{S}=\\{\text{Open circuits
}\widetilde{\mathcal{C}}\subseteq\mathcal{C}_{1}\cup\mathcal{C}_{2}\text{
enclosing }\mathsf{int}(\mathcal{I}_{L})\\}$

is nonempty. Let $\widetilde{\mathcal{C}}_{1}$ be the unique circuit from
Lemma 6.11 satisfying
$\mathsf{int}(\widetilde{\mathcal{C}}_{1})\subseteq\bigcup_{\widetilde{\mathcal{C}}\in\widetilde{S}}\mathsf{int}(\widetilde{\mathcal{C}})$
and
$\widetilde{\mathcal{C}}_{1}\subseteq\bigcup_{\widetilde{\mathcal{C}}\in\widetilde{S}}\widetilde{\mathcal{C}}$.
By similar reasoning as before, $\mathcal{B}$ must be contained in exactly one
of $\mathsf{int}(\widetilde{\mathcal{C}}_{1})$ or
$\mathsf{ext}(\widetilde{\mathcal{C}}_{1})$. If
$\mathcal{B}\subseteq\mathsf{ext}(\widetilde{\mathcal{C}}_{1})$, then
$\widetilde{\mathcal{C}}_{1}\in S_{L}$, a contradiction because $S_{L}$ is
empty. Therefore,
$\mathcal{B}\subseteq\mathsf{int}(\widetilde{\mathcal{C}}_{1})$. Consider two
cases:

Case 1: $\mathcal{C}_{1}$ and $\widetilde{\mathcal{C}}_{1}$ have one or fewer
vertices in common. Then, since $\widetilde{\mathcal{C}}_{1}$ consists only of
edges of $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$, we have
$\widetilde{\mathcal{C}}_{1}=\mathcal{C}_{2}$. This is a contradiction because
$\widetilde{\mathcal{C}}_{1}$ encloses $\mathsf{int}(\mathcal{I}_{L})$, but
$\mathsf{int}(\mathcal{I}_{L})\subseteq\mathsf{ext}(\mathcal{C}_{2})$.

Case 2: $\mathcal{C}_{1}$ and $\widetilde{\mathcal{C}}_{1}$ have at least two
vertices in common. We show here that there exists a point along
$\mathcal{C}_{1}$ in $\mathsf{int}(\widetilde{\mathcal{C}}_{1})$. Once this is
shown, we follow $\mathcal{C}_{1}$ in each direction along this point until
hitting the two vertices of $\widetilde{\mathcal{C}}_{1}$. Lemma 6.5 allows us
to construct a smaller circuit built from edges of
$\mathcal{C}_{1},\mathcal{C}_{2}$ that encloses $\mathcal{I}_{L}$, a
contradiction to the minimality of $\widetilde{\mathcal{C}}_{1}$.

We know $\mathcal{C}_{1}\neq\widetilde{\mathcal{C}}_{1}$ by the same reasoning
as in Case 1. Furthermore, the circuit $\mathcal{C}_{1}$ cannot enclose
$\widetilde{\mathcal{C}}_{1}$ because then, it would enclose
$\mathcal{I}_{L}$. We now have two subcases:

Subcase 2.1: $\widetilde{\mathcal{C}}_{1}$ encloses $\mathcal{C}_{1}$. Then,
all points of $\mathcal{C}_{1}$ are on or in the interior of
$\widetilde{\mathcal{C}}_{1}$. Not all points on $\mathcal{C}_{1}$ can be on
$\widetilde{\mathcal{C}}_{1}$ because no proper subset of a Jordan curve is a
Jordan curve and $\widetilde{\mathcal{C}}_{1}\neq\mathcal{C}_{1}$.

Subcase 2.2: Neither of the circuits
$\widetilde{\mathcal{C}}_{1},\mathcal{C}_{1}$ encloses the other. Both
circuits contain $\mathcal{B}$ in their interior, so Lemma 6.2 provides the
desired point along $\mathcal{C}_{1}$ in the interior of
$\widetilde{\mathcal{C}}_{1}$.

Lastly, we construct the circuits $\mathcal{I}_{L+2},\ldots,\mathcal{I}_{P}$.
Given $\mathcal{I}_{j}$ for $j\geq L+1$, consider the set

$S_{j}^{\prime}=\\{\text{Open circuits }\mathcal{C}\text{ enclosing
}\mathcal{B}\text{ that are also enclosed by and edge-disjoint from
}\mathcal{I}_{j}.\\}$

If $S_{j}^{\prime}$ is nonempty, let $\mathcal{I}_{j+1}$ be the unique
outermost circuit constructed in Lemma 6.10 from the circuits in
$S_{j}^{\prime}$. Then, $\mathcal{I}_{j+1}$ encloses $\mathcal{B}$ and is
edge-disjoint from $\mathcal{I}_{j}$. The points along the edges of
$\mathcal{I}_{j+1}$ all lie on or in the interior of $\mathcal{I}_{j}$, so
$\mathcal{I}_{j}$ must enclose $\mathcal{I}_{j+1}$ by Lemma 6.1.

The index $P$ is the smallest index $j$ so that $S_{j}^{\prime}=\varnothing$.
Then,

$\mathcal{A}^{C}\supseteq\mathcal{I}_{L+1}\cup\mathsf{int}(\mathcal{I}_{L+1})\supseteq\mathsf{int}(\mathcal{I}_{L+1})\supseteq\mathsf{int}(\mathcal{I}_{L+2})\supseteq\cdots\supseteq\mathsf{int}(\mathcal{I}_{P})\supseteq\mathcal{B},$

thus proving Item (iii).

To get Item (iv), note that successive circuits
$\mathcal{I}_{j},\mathcal{I}_{j+1}$ are edge disjoint and either (i) one
encloses the other or (ii) they have disjoint interiors. Hence, they can only
touch at a corner (see Figure 3), so the circuits $\mathcal{I}_{j}$ and
$\mathcal{I}_{j+2}$ must be vertex-disjoint.

###  7.2. Construction of the dual paths $\zeta_{e}$ and $\zeta$ (Proof of
Items (v)–(vi))

We argue here that there exists a dual path from $\mathcal{A}$ to
$\mathcal{B}$ that has exactly $P$ open edges—one crossing each of the
circuits
$\mathcal{I}_{1},\ldots,\mathcal{I}_{L},\mathcal{I}_{L+1},\ldots,\mathcal{I}_{P}$.
This is done by starting at $\mathcal{B}$ and moving backwards to
$\mathcal{A}$. There are five stages of the construction:

  1. (a)

Show the existence of a closed dual path from $\mathcal{B}$ to
$\mathcal{I}_{P}$.

  2. (b)

For $j\in\\{L+1,\ldots,P-1\\}$, given a fixed dual neighbor, $x^{\star}$, of
$\mathcal{I}_{j+1}$, lying in $\mathsf{ext}(\mathcal{I}_{j+1})$, show the
existence of a closed dual path from $x^{\star}$ to $\mathcal{I}_{j}$.

  3. (c)

Given a fixed dual neighbor $x^{\star}$ of $\mathcal{I}_{L+1}$ lying in
$\mathsf{ext}(\mathcal{I}_{L+1})$ show the existence of a closed dual path
from $x^{\star}$ to $\mathcal{I}_{L}$ (or to $\mathcal{A}$ if $K=0$). If $P=L$
so that there is no circuit $\mathcal{I}_{L+1}$, then the path is from
$\mathcal{B}$ (without specifying the vertex) to $\mathcal{I}_{L}$ (or to
$\mathcal{A}$ if $L=0$).

  4. (d)

For $j\in\\{1,\ldots,L-1\\}$, given a fixed dual neighbor $x^{\star}$ of
$\mathcal{I}_{j+1}$ lying in $\mathsf{int}(\mathcal{I}_{j+1})$, show the
existence of a closed dual path from $x^{\star}$ to $\mathcal{I}_{j}$.

  5. (e)

Given a fixed dual neighbor $x^{\star}$ of $\mathcal{I}_{1}$ lying in
$\mathsf{int}(\mathcal{I}_{1})$, show the existence of a closed dual path from
$x^{\star}$ to $\mathcal{A}$.

These steps give us a map for constructing the path $\zeta$ in Item (vi):
Starting from $\mathcal{B}$, we travel to a dual neighbor of
$\mathcal{I}_{P}$. Then, we can cross to the exterior of $\mathcal{I}_{P}$
with a single open edge. From the endpoint of that edge in
$\mathsf{ext}(\mathcal{I}_{P})$, we take a path to a dual neighbor of
$\mathcal{I}_{P-1}$, then cross $\mathcal{I}_{P-1}$ with a single edge.
Continue this procedure until we reach $\mathcal{A}$. The construction of the
paths $\zeta_{e}$ from Item (v) is the same, except that we now start at the
edge $e$ and go backwards to $\mathcal{A}$. The choice of the geodesic
$\gamma$ will come later. In Section 7.5, we show that $\gamma$ and $\zeta$
can be chosen to be disjoint.

Part (a): If $P=L$ (meaning the set
$\\{\mathcal{I}_{L+1},\ldots,\mathcal{I}_{P}\\}$ is empty), then this case is
deferred to Part (c). Otherwise, we start by showing there is a closed path
from $\mathcal{B}$ to $\mathcal{I}_{P}$. Here, we interpret $\mathcal{B}$ as a
collection of vertices, and $\mathcal{I}_{P}$ as a collection of edges, so
this means that there exists a closed dual path from some dual neighbor of a
vertex in $\mathcal{B}$ to some dual neighbor of an edge in $\mathcal{I}_{P}$.

Since $\mathcal{B}\subseteq\mathsf{int}(\mathcal{I}_{P})$, then every dual
neighbor of a point $x\in B$ lies in $\mathsf{int}(\mathcal{I}_{P})$. To see
this, note that dual vertices cannot lie on $\mathcal{I}_{P}$ so they must
either lie in $\mathsf{int}(\mathcal{I}_{P})$ or
$\mathsf{ext}(\mathcal{I}_{P})$. If $x\in\mathsf{int}(\mathcal{I}_{P})$ and a
dual neighbor $x^{\star}$ lies in $\mathsf{ext}(\mathcal{I}_{P})$, then the
diagonal path connecting the two vertices must cross $\mathcal{I}_{P}$. As
$\mathcal{I}_{P}$ lies on the primal lattice, this is not possible. Now, we
claim that the set of dual neighbors of vertices in $\mathcal{B}$ is a
connected set (connected by dual edges). The set of northwest dual neighbors
is connected, since this is just a shifted version of the set $\mathcal{B}$.
Likewise, the sets of southwest, northeast, and southeast dual neighbors are
all connected. But each of these sets is connected to each other, so the
entire set of dual neighbors is connected.

We perform a modification of the environment as follows. Figure 13 gives an
illustation. Set all primal edges (along with their dual counterparts) that
lie on or have at least one vertex in the exterior of $\mathcal{I}_{P}$ to
closed. Set all dual edges connecting vertices dual to $\mathcal{B}$ to closed
(along with their primal counterparts). This latter set of edges is connected,
and we let $W$ be the cluster of closed edges that contains these. Then, there
is no closed dual path from $\mathcal{B}$ to $\mathcal{I}_{P}$ in the original
environment if and only if $W$ is bounded in the modified environment. If $W$
is bounded in this environment, then Lemma 5.6 guarantees the existence of an
open circuit $\mathcal{C}$ that contains $W$ in its interior. The circuit
$\mathcal{C}$ is enclosed by and edge disjoint from $\mathcal{I}_{P}$ because
all edges on or in the exterior of $\mathcal{I}_{P}$ are closed. The open
circuit $\mathcal{C}$ must also be open in the original environment.

Figure 13. Closed edges are denoted by red/thick, and open edges are denoted
by blue/thin. The edges on $\mathcal{I}_{P}$ have been set to closed, as well
as all edges outside $\mathcal{I}_{P}$ and all edges connecting vertices of
$\mathcal{B}$. In this example, the closed component $W$ of $\mathcal{B}$ is
bounded, so $B$ is not connected to $\mathcal{I}_{P}$ by a closed path, and
there exists an open circuit enclosing $W$.

Since $\mathcal{I}_{P}$ is the last circuit in the sequence, we derive a
contradiction once we show that $\mathcal{C}$ also contains $\mathcal{B}$ in
its interior. Any vertex $x\in B$ cannot lie in $\mathsf{ext}(\mathcal{C})$
because otherwise, the diagonal path from $x$ to any of its dual neighbors
(which lie in $\mathsf{int}(\mathcal{C})$) must cross $\mathcal{C}$. But
$\mathcal{C}$ consists of primal edges and so this is not possible. The only
remaining possibility for $\mathcal{B}\not\subseteq\mathsf{int}(\mathcal{C})$
comes when $x$ lies on the circuit $\mathcal{C}$. Then, exactly two edges of
$\mathcal{C}$ share the vertex $x$. Take one of these edges $e$. By Lemma 6.3,
one of the two dual neighbors of $e$ lies in $\mathsf{ext}(\mathcal{C})$. But
this dual neighbor of the edge $e$ is a dual neighbor of the vertex $x$, which
is contained in $\mathsf{int}(\mathcal{C})$ by assumption.

Part (b): Similarly as in Part (a), modify the environment so that all edges
along $\mathcal{I}_{j}$ or with at least one endpoint in
$\mathsf{ext}(\mathcal{I}_{j})$ are set to closed (along with their dual
edges). Further, since $x^{\star}$ is a dual neighbor of $\mathcal{I}_{j+1}$,
there exists a dual edge $e^{\star}$ crossing $\mathcal{I}_{j+1}$. Set
$e^{\star}$ and its corresponding primal edge to closed. If there is no closed
path from $x^{\star}$ to $\mathcal{I}_{j}$ in the original environment, then
in the modified environment, the closed cluster of dual edges connected to
$x^{\star}$ is bounded. Lemma 5.6 impies the existence of an open circuit
$\mathcal{C}$ that contains $x^{\star}$ in its interior. $\mathcal{C}$ is
enclosed by and edge-disjoint from $\mathcal{I}_{j}$ because all edges on or
in the exterior of $\mathcal{I}_{j}$ were set to closed. The other endpoint of
$e^{\star}$ also lies in $\mathsf{int}(\mathcal{C})$ because $e^{\star}$ was
set to closed and therefore cannot cross $\mathcal{C}$. But since
$x^{\star}\in\mathsf{ext}(\mathcal{I}_{j+1})$, this other endpoint lies in
$\mathsf{int}(\mathcal{I}_{j+1})$ by Lemma 6.3. By Lemma 6.10, there exists a
circuit $\mathcal{C}^{\prime}$ which (i) consists entirely of edges of
$\mathcal{I}_{j+1}$ and $\mathcal{C}$ (and therefore is edge disjoint and
enclosed by $\mathcal{I}_{j}$) and (ii) encloses $\mathcal{I}_{j+1}$. This
contradicts the definition of $\mathcal{I}_{j+1}$ as the outermost next
circuit in the sequence after $\mathcal{I}_{j}$.

Part (c): Recall the definition of the circuit $\widehat{\mathcal{D}}$ at the
beginning of this section and the observation that all the circuits
$\mathcal{I}_{1},\ldots,\mathcal{I}_{L},\mathcal{I}_{L+1},\ldots,\mathcal{I}_{P}$
are contained in $\mathsf{int}(\widehat{\mathcal{D}})$.

We consider a modification of the environment as follows. If $L=P$ (meaning
there is no circuit $\mathcal{I}_{L+1}$), then consider the connected set of
dual vertices that are neighbors of some vertex in $\mathcal{B}$. Set all the
dual edges connecting this set to closed (along with their primal
counterparts). Call the cluster of closed dual edges connected to these edges
$W_{\mathcal{B}}$. If $P>L$, then we are given a fixed dual neighbor
$x^{\star}$ of $\mathcal{I}_{L+1}$ lying in $\mathsf{ext}(\mathcal{I}_{L+1})$.
The vertex $x^{\star}$ is the endpoint of an edge crossing
$\mathcal{I}_{L+1}$: set that edge to closed. Set all primal edges with at
least one endpoint in $\mathsf{int}(\mathcal{I}_{L+1})$ to closed. Then,
$W_{\mathcal{B}}$ is the cluster of dual closed edges that connects to these
edges inside $\mathcal{I}_{L+1}$.

We construct another closed cluster, denoted $W_{\mathcal{A}}$, as follows. If
$L=0$ so that there is no circuit $\mathcal{I}_{L}$, then we set all dual
edges connecting dual neighbors of $\mathcal{A}$ to closed and let
$W_{\mathcal{A}}$ be the closed dual cluster of these edges. If $L>0$, then we
set all edges that lie on the circuit $\mathcal{I}_{L}$ or have at least one
endpoint in $\mathsf{int}(\mathcal{I}_{L})$ to closed, and we let
$W_{\mathcal{A}}$ be the closed cluster of the associated dual edges in the
modified environment. We also set all edges with at least one endpoint in
$\mathsf{ext}(\widehat{\mathcal{D}})$ to closed, and let
$W_{\widehat{\mathcal{D}}}$ be the infinite cluster of closed edges connected
to $\widehat{\mathcal{D}}$ in the modified environment.

Now, assume, by way of contradiction, that in the orignial environment, there
is no closed dual path from $x^{\star}$ (or from $\mathcal{B}$ if $L=P$) to
$\mathcal{I}_{L}$ (or to $\mathcal{A}$ if $L=0$). Then, in the modified
environment, $W_{\mathcal{A}}\neq W_{\mathcal{B}}$. Lemma 5.7 implies the
existence of an open circuit $\mathcal{C}$, lying in the interior of
$\mathcal{D}$ that contains $W_{\mathcal{A}}$ in its interior and
$W_{\mathcal{B}}$ in its exterior, or vise versa (It may be helpful to refer
back to Figure 9). The circuit $\mathcal{C}$ must also be open in the original
environment.

In the first case, $\mathcal{C}$ is edge disjoint from $\mathcal{I}_{L}$
because all the edges on $\mathcal{I}_{L}$ were set to closed in the modified
environment. $\mathcal{C}$ contains the point $x^{\star}$, which lies in
$\mathsf{ext}(\mathcal{I}_{L+1})$ by assumption. $\mathcal{C}$ and
$\mathcal{I}_{L+1}$ also share points in their interior because they both
enclose $\mathcal{B}$. Lemma 6.10 implies the existence of another open
circuit $\mathcal{C}^{\prime}$ that is edge disjoint from $\mathcal{I}_{L}$
and so that
$\mathsf{int}(\mathcal{I}_{L+1})\cup\mathsf{int}(\mathcal{C})\subseteq\mathcal{I}(\mathcal{C}^{\prime})$.
This is a contradiction to our construction of the outermost circuit
$\mathcal{I}_{L+1}$ in Section 7.1. In the case that $\mathcal{C}$ contains
$W_{\mathcal{B}}$ in its interior and $W_{\mathcal{A}}$ in its exterior, since
all edges on or inside $\mathcal{I}_{L}$ were set to closed, $\mathcal{C}$ is
edge disjoint from and encloses $\mathcal{I}_{L}$. This contradicts the
definition of $\mathcal{I}_{L}$ as the last disjoint innermost circuit
enclosing $\mathcal{A}$ but containing $\mathcal{B}$ in its exterior.

Part (d): The chosen dual vertex $x^{\star}$ of $\mathcal{I}_{j+1}$ is the
endpoint of dual edge crossing $\mathcal{I}_{j+1}$. Set that edge (along with
its primal counterpart to closed). Additionally, set all primal edges with at
least one endpoint in $\mathsf{ext}(\mathcal{I}_{j+1})$ to closed (along with
the dual counterparts). Set all edges on $\mathcal{I}_{j}$ or with at least
one endpoint in $\mathsf{int}(\mathcal{I}_{j})$ to closed. Then, there is no
closed dual path between $x^{\star}$ and $\mathcal{I}_{j}$ in the original
environment if and only if, in the modified environment, the cluster of dual
closed edges containing those edges passing through $\mathcal{I}_{j}$ is
bounded. In the latter case, Lemma 5.6 implies that there is an open circuit
$\mathcal{C}$ surrounding $\mathcal{I}_{j}$ in the modified environment (which
i=s also open in the original environment). This open circuit is edge disjoint
from $\mathcal{I}_{j}$ and encloses $\mathcal{I}_{j}$ by Lemma 6.1. The point
$x^{\star}$ lies in $\mathsf{ext}(\mathcal{C})$ since, in the dual
environment, there is an infinite, closed dual path starting from
$x^{\star}$–this path cannot cross $\mathcal{C}$.

Similarly as in Part (b), Lemma 6.11 contradicts the construction of
$\mathcal{I}_{j+1}$ as the next innermost circuit.

Part (e): This is almost the same as Part (d). The only difference is that,
instead of setting edges on and inside $\mathcal{I}_{j}$ to closed, we set all
edges which connect vertices in $\mathcal{A}$ to closed. The rest of the proof
goes the same.

The proofs of Items (viii) and (ix) of Proposition 2.5 are postponed until the
end of the section.

###  7.3. Existence closed circuits hitting the closed edges (Proof of Items
(vii), and (xi))

Take $\gamma$ to be any geodesic from $\mathcal{A}$ to $\mathcal{B}$. For
every circuit $\mathcal{I}_{j}$, let $x$ and $y$ be the first and last
vertices of $\gamma$ on $\mathcal{I}_{j}$. Then, we modify the portion of
$\gamma$ between $x$ and $y$ to follow the open circuit $\mathcal{I}_{j}$ (in
either direction) if it does not already. We can similarly force $\gamma$ to
take no edges between the circuits $\mathcal{I}_{j}$ and $\mathcal{I}_{j+1}$
if these two are not vertex-disjoint. This is Item (vii). To prove Item (xi),
enumerate the closed edges along the path, in order starting from
$\mathcal{A}$ as $e_{1},\ldots,e_{V}$, and choose one of the edges $e_{i}$. We
continue to work on the event $\Omega_{\infty}$ and this time make use of a
deterministically chosen open circuit $\widehat{\mathcal{C}}$ that encloses
both $\mathcal{A}$ and $\mathcal{B}$. Modify the environment in the following
manner: set all edges with at least one endpoint in the exterior of
$\widehat{\mathcal{C}}$ to open. Set all edges along $\gamma$ to open, except
for the edge $e_{i}$. Set all edges connecting two vertices in $\mathcal{A}$
to open, as well as all edges connecting two vertices in $\mathcal{B}$. This
creates three open clusters of edges: the cluster $W_{\mathcal{A}}$ consisting
of all open edges with an open path to $\mathcal{A}$, the cluster
$W_{\mathcal{B}}$ of open edges with a path to $\mathcal{B}$, and the
unbounded cluster $W_{\widehat{\mathcal{C}}}$ of the circuit
$\widehat{\mathcal{C}}$. Since $\gamma$ is a geodesic, $W_{\mathcal{A}}\neq
W_{\mathcal{B}}$. Otherwise, there is an open path from $\mathcal{A}$ to
$\mathcal{B}$ in the modified environment. In the original environment, this
path may still take some of the closed edges $e_{j}$, but it avoids the closed
edge $e_{i}$, and thus it has a strictly lower passage time than the geodesic
$\gamma$, a contradiction.

Now that we have established $W_{\mathcal{A}}\neq W_{\mathcal{B}}$, the dual
counterpart of Lemma 5.7 implies the existence of a closed circuit
$\mathcal{U}$ lying in $\mathsf{int}(\widehat{\mathcal{C}})$ that contains
$W_{\mathcal{A}}$ (and therefore also $\mathcal{A}$) in its interior and
$W_{\mathcal{B}}$ (therefore also $\mathcal{B}$) in its exterior, or vise
versa. The geodesic from $\mathcal{A}$ to $\mathcal{B}$ thus passes through
vertices in $\mathsf{ext}(\widehat{\mathcal{C}})$ and vertices in
$\mathsf{int}(\widehat{\mathcal{C}})$. The path between these must then cross
$\widehat{\mathcal{C}}$. Since all edges on the geodesic $\gamma$ except for
$e_{i}$ were set to open, the edge $e_{i}$ is the only edge on $\gamma$ whose
dual belongs to $\mathcal{U}$.

###  7.4. Separate argument in the Bernoulli case (Proof of Item (xi))

In the case of Bernoulli weights, the dual version of the construction of the
circuits
$\mathcal{I}_{1},\ldots,\mathcal{I}_{L},\mathcal{I}_{L+1},\ldots,\mathcal{I}_{P}$
allows us to successively construct edge-disjoint innermost closed circuits
surrounding $\mathcal{A}$, $\mathcal{U}_{1},\ldots,\mathcal{U}_{U}$ followed
by successive outermost closed circuits
$\mathcal{U}_{U+1},\ldots,\mathcal{U}_{V}$, surrounding $\mathcal{B}$. Since
the circuits are edge disjoint, any geodesic $\gamma$ must pass through each
circuit at least once. By a symmetric argument to the construction in Section
7.2, we may construct a primal path $\gamma$ that crosses each of the circuits
$\mathcal{U}_{1},\ldots,\mathcal{U}_{U},\mathcal{U}_{U+1},\ldots,\mathcal{U}_{V}$
exactly once. This path is therefore a geodesic.

By Lemma 6.9, for each pair of open and closed circuit in this sequence, they
either have disjoint interior, or one encloses the other. This gives a natural
ordering of the sequence $\mathcal{C}_{1},\ldots,\mathcal{C}_{K}$: we start
with the circuits that enclose $\mathcal{A}$, starting from the innermost,
then move to the circuits that enclose $\mathcal{B}$, starting from the
outermost, and this implies that, for some index $W\in\\{0,\ldots,K\\}$,
$\mathsf{int}(\mathcal{I}_{W})\cap\mathsf{int}(\mathcal{I}_{W+1})=\varnothing$.
Furthermore,

$\mathcal{A}\subseteq\mathsf{int}(\mathcal{C}_{1})\subseteq\cdots\subseteq\mathsf{int}(\mathcal{C}_{W}),\quad\text{and}\quad\mathsf{int}(\mathcal{C}_{W+1})\supseteq\cdots\mathsf{int}(\mathcal{C}_{K})\supseteq
B.$

###  7.5. Modification of $\gamma$ and $\zeta$ to be disjoint.

Recall that $\gamma$ is a primal path while $\zeta$ is a dual path. If
$\gamma$ and $\zeta$ meet at the midpoint of an open edge, then that edge must
lie along one of the circuits
$\mathcal{I}_{1},\ldots,\mathcal{I}_{L},\mathcal{I}_{L+1},\ldots,\mathcal{I}_{P}$,
and so $\gamma$ contains an edge $e$ that lies on this circuit. Call the
circuit at which the intersection occurs $\mathcal{C}$. There exists an open
path between the endpoints of $e$ that goes the opposite direction around
$\mathcal{C}$, thus avoiding $e$. Since $\zeta$ only crosses $\mathcal{C}$
once, by rerouting $\gamma$ in this manner, we have avoided intersection with
$\zeta$ at $e$ without incurring additional passage time.

If $\gamma$ and $\zeta$ intersect at the midpoint of a closed edge, then we
showed that this edge of $\gamma$ is the primal counterpart of an edge on a
closed circuit. We similarly reroute $\zeta$ to avoid this edge.

We make the observation here that, because the circuits
$\mathcal{I}_{1},\ldots,\mathcal{I}_{P}$ (as well as the dual circuits
containing the closed edges along $\gamma$) are not necessarily vertex-
disjoint, this procedure could have created paths $\gamma$ and/or $\zeta$ that
are not self-avoiding. If this is the case, we can simply create self-avoiding
paths by deleting some of the edges along the paths.

###  7.6. Modifying the geodesic $\gamma$ between any two of its closed edges

We perform a final modification of he geodesic $\gamma$ to obain the precise
results we need. With the circuits
$\mathcal{I}_{1},\ldots,\mathcal{I}_{L},\mathcal{I}_{L+1},\ldots,\mathcal{I}_{P}$
and the geodesic $\gamma$ constructed thus far, for $i\in\\{0,\ldots,P\\}$,
let $e_{i,1},\ldots,e_{i,J_{i}}$ be an enumeration of the closed edges of the
geodesic between the $i$th and $i+1$st circuit, in order, starting from the
set $\mathcal{A}$. For each $e_{i,j}$, let $x_{i,j}^{1}$ be its first vertex
along the path and let $x_{i,j}^{2}$ be the second vertex. If
$x_{i,j}^{2}=x_{i,j+1}^{1}$, this means that $e_{i,j}$ and $e_{i,j+1}$ are
successive edges on the geodesic. In this case, we perform no modification, so
assume $x_{i,j}^{2}\neq x_{i,j+1}^{1}$. Let $\gamma_{i,j}$ be the portion of
$\gamma$ between the two vertices. We may replace $\gamma_{i,j}$ with any open
path between $x_{i,j}^{2}$ and $x_{i,j+1}^{1}$, and the modified path is still
a geodesic. Instead of vertices, we may also consider portions of the geodesic
to the next circuit or the next set. There are, in total, 9 possibilities of
open paths that could lie along the geodesic, but by careful use of
definitions, we will reduce the number of total cases.

  1. (1)

Path from $\mathcal{A}$ to $\mathcal{B}$ (if the set
$\mathcal{I}_{1},\ldots,\mathcal{I}_{L},\mathcal{I}_{L+1},\ldots,\mathcal{I}_{P}$
is empty) and there are no closed edges on $\gamma$.

  2. (2)

Path from $\mathcal{A}$ to $\mathcal{I}_{1}$ (if there are no closed edges on
$\gamma$ between $\mathcal{A}$ and $\mathcal{I}_{1}$).

  3. (3)

Path from $\mathcal{A}$ to $x_{0,1}^{1}$.

  4. (4)

Path from $\mathcal{I}_{P}$ to $\mathcal{B}$ (if there are no closed edges of
$\gamma$ from $\mathcal{I}_{P}$ to $\mathcal{B}$).

  5. (5)

Path from $\mathcal{I}_{i}$ to $\mathcal{I}_{i+1}$ (if $\mathcal{I}_{1}$ and
$\mathcal{I}_{i+1}$ are disjoint and there are no closed edges of $\gamma$
between the two circuits).

  6. (6)

Path from $\mathcal{I}_{i}$ to $x_{i,1}^{1}$.

  7. (7)

Path from $x_{P,J_{P}}^{2}$ to $\mathcal{B}$.

  8. (8)

Path from $x_{i,J_{i}}^{2}$ to $\mathcal{I}_{i+1}$.

  9. (9)

Path from $x_{i,j}^{2}$ to $x_{i,j+1}^{2}$.

When the path starts at $\mathcal{A}$ or $\mathcal{I}_{i}$, we say the first
end of the path is free. We may move the starting point to a different vertex
$\mathcal{A}$ or a different vertex of the open circuit $\mathcal{I}_{i}$ and
still have a geodesic. Similarly, when the path ends at $\mathcal{B}$ or
$\mathcal{I}_{i}$, we say the second end is free. Otherwise, we say the end is
fixed. This notation will greatly simplify the construction with fewer cases
to handle.

For our chosen pair of ends (either free or fixed), we let $\Pi$ be the set of
open paths between them. Our task is to choose an element $\pi\in\Pi$ in a
manner so that each edge $e$ will have a closed dual connection to $\zeta$.
For each $\pi\in\Pi$, we associate a Jordan curve $\mathcal{J}_{\pi}$ that
lives on the union of the primal and dual lattices. By a simple modification
of Lemma 6.8, the area of each such Jordan curve is an integer multiple of
$\frac{1}{4}$, so we choose a path from the set
$\operatorname*{arg\,min}_{\pi}J_{\pi}$ according to some deterministic total
ordering of such circuits. Then, we argue that our chosen path satisfies the
desired properties.

We construct the Jordan curve for four separate and exhaustive cases. the
following observation will be used several times: Any circuit that contains
$\mathcal{A}$ in its interior and $\mathcal{B}$ in its exterior (or vise
versa) must intersect the path $\zeta$ between $\mathcal{A}$ and
$\mathcal{B}$. Figure 14 illustrates the case when one end is free and one end
is fixed.

Figure 14. The case when one end is free and one end is fixed. Open edges are
shown in blue/thin and closed edges are shown in red/thick. The fixed end
crosses through a closed dual circuit with a closed primal edge. The path
$\zeta$ crosses one of the $\mathcal{I}_{i}$ with an open dual edge. We use
edges and half-edges to create a Jordan curve seen in the figure.

Case 1: Both ends are free For this case, we assume that if $\pi$ is a path
from $\mathcal{I}_{i}$ to $\mathcal{I}_{i+1}$, that the two circuits are
disjoint becuase, if not, Item (vii) states that the geodesic passes from
$\mathcal{I}_{i}$ to $\mathcal{I}_{i+1}$ without ever leaving the union of the
circuits. We also assert that the only vertices along $\pi$ that intersect
$\mathcal{I}_{i},\mathcal{I}_{i+1},A$, or $\mathcal{B}$, are the ends.
Otherwise, we can take a subset of the path that achieves the same purpose.
Traverse the path $\pi$ between its ends. The latter end connects either to an
open circuit or to $\mathcal{B}$. Follow either the open circuit (in any
direction) or a self-avoiding path of edges connecting vertices in
$\mathcal{B}$ until we reach $\zeta$. Then, we can connect the two paths with
half of a primal edge and then half of a dual edge. Follow $\zeta$ backwards
until we reach either $\mathcal{A}$ or the open circuit containing the first
end of $\pi$. Again, we connect this path onto the primal lattice with a half
dual edge and a half primal edge. From there, we follow the open circuit in
either direction, or a self-avoiding path of edges connecting vertices of
$\mathcal{A}$ until reaching the first end. A circuit has now been formed.
This is a well-defined circuit because $\zeta$ and $\gamma$ are disjoint, and
$\zeta$ can only intersect each of the open circuits once.

Case 2: One end is free. Without loss of generality, assume that the free end
comes second as we traverse between $\mathcal{A}$ and $\mathcal{B}$. Traverse
$\pi$ between its ends until meeting the free end. Connect this path to
$\zeta$ just as in Case 1. Let $\mathcal{U}$ be the closed circuit constructed
in Section 7.3 associated to the fixed end. Follow $\zeta$ backwards until
meeting $\mathcal{U}$. Then, follow $\mathcal{U}$ (in either direction) until
we can connect it to the first end of $\pi$ with a concatenation of half dual
and half primal edges.

Case 3: Both ends are fixed. We let $\mathcal{U}_{i},\mathcal{U}_{i+1}$ be the
associated closed circuits constructed in Section 7.3. For this case, we need
two subcases.

Subcase 3.1: $\mathcal{U}_{i}$ and $\mathcal{U}_{i+1}$ are (both edge and
vertex) disjoint. Traverse $\pi$ between its fixed edges until reaching
$\mathcal{U}_{i+1}$, and connect the paths just as in Case 2. Follow
$\mathcal{U}_{i+1}$ in either direction until meeting $\zeta$. Follow $\zeta$
backwards until meeting $\mathcal{U}_{i}$. Then, follow $\mathcal{U}_{i}$ in
either direction until meeting the first fixed end, connecting the paths in
the usual way.

Subcase 3.2: $\mathcal{U}_{i}$ and $\mathcal{U}_{i+1}$ share at least one
vertex. Traverse the path $\pi$ between its fixed ends. Connect to
$\mathcal{U}_{i+1}$ in the usual way and then follow this circuit in either
direction until we first meet $\mathcal{U}_{i}$. Follow $\mathcal{U}_{i}$ in
either direction until we can connect to the first fixed end with two half-
edges as usual. It is possible that $\mathcal{U}_{i}$ hits $\mathcal{U}_{i+1}$
again, but it will not intersect the portion of $\mathcal{U}_{i+1}$ that we
used as part of the Jordan curve since we started along $\mathcal{U}_{i}$ at
the first place that the two circuits met.

We emphasize that in the construction of the associated Jordan curves
$\mathcal{J}_{\pi}$, there were often several choices of path (traversing the
circuits in either direction, etc.) For each portion of the geodesic, we
modify by choosing $\pi$ from the set of open paths between the ends with the
smallest possible area of such a Jordan curve (breaking ties
deterministically), and let $\mathcal{J}_{\pi}$ the associated smallest Jordan
curve.

###  7.7. Proof of consistency and existence of the closed arms (Proof of
Items (viii) and (ix))

The consistency of the geodesics in Item (viii) follows because the
construction of the paths $\zeta$ and $\gamma$ between any pair of successive
open circuits depends only on the environment between those circuits. We turn
to Item (ix). Let $e\in\gamma$ be an open edge with
$e\notin\mathcal{I}_{1}\cup\cdots\cup\mathcal{I}_{L}\cup\mathcal{I}_{L+1}\cup\cdots\cup\mathcal{I}_{P}$.
Let $\pi$ be the associated portion of $\gamma$ between successive closed
edges or open circuits, as in Section 7.6, and let $\mathcal{J}_{\pi}$ be the
associated Jordan curve of smallest area. In each case of the construction,
$e$ is an edge along $\mathcal{J}_{\pi}$. Let $x_{1}^{\star},x_{2}^{\star}$ be
the endpoints of the associated dual edge $e^{\star}$. Since
$\mathcal{J}_{\pi}$ lives on the union of the primal and dual lattices, either
at least one of these points lies on $\mathcal{J}_{\pi}$ (in which case the
desired path is trivial), or Lemma 6.3 implies that one of these points lies
in $\mathsf{int}(\mathcal{J}_{\pi})$ and the other lies in
$\mathsf{ext}(\mathcal{J}_{\pi})$. Without loss of generality, assume that
$x_{1}^{\star}$ lies in $\mathsf{int}(\mathcal{J}_{\pi})$.

We perform a modification of the environment as follows. Set each primal edge
with at least one endpoint in $\mathsf{ext}(\mathcal{J}_{\pi})$ to closed
(along with the associated dual edges). Then, $x_{1}^{\star}$ is connected to
$\zeta$ by a closed dual path in the original environment if and only if the
closed dual cluster of $x_{1}^{\star}$ in the modified environment is
unbounded. If, by way of contradiction, this is not the case, then in the
modified environment, there is an open circuit $\mathcal{C}$ containing
$x^{\star}$ in its interior. The open circuit must lie entirely on or in the
interior of $\mathcal{J}_{\pi}$ because all edges outside were set to closed.
In other words,
$\mathsf{int}(\mathcal{C})\subseteq\mathsf{int}(\mathcal{J}_{\pi})$. Hence,
both $\mathcal{J}_{\pi}$ and $\mathcal{C}$ contain $x_{1}^{\star}$ in their
interior and $x_{2}^{\star}$ in their exterior, so
$e\in\mathcal{J}_{\pi}\cap\mathcal{C}$. Let $W$ be the connected component of
$\mathcal{J}_{\pi}\cap\mathcal{C}$ containing $e$. Then, modifying $\pi$ by
replacing $W$ with $\mathcal{C}\setminus W$ results in a new path between the
two ends (possibly changing the starting and/or end point in the case of free
ends). Let $\pi^{\prime}$ be the modified path, and let
$\mathcal{J}_{\pi^{\prime}}$ be the Jordan curve obtained by replacing $\pi$
with $\pi^{\prime}$. It is straightforward to see that
$\mathcal{J}_{\pi^{\prime}}$ is a Jordan curve that can be obtained from
$\pi^{\prime}$ in the setup of Section 7.6. We show that the area of
$\mathcal{J}_{\pi^{\prime}}$ is smaller than the area of $\mathcal{J}_{\pi}$,
thereby obtaining a contradiction. The path $\pi$ cannot be the same as
$\pi^{\prime}$: otherwise $\mathcal{C}=W$. But $W$ is a subset of the self-
avoiding path $\pi$ and can not contain a circuit. By Lemma 6.8, it is
sufficient to show that
$\mathsf{int}(\mathcal{J}_{\pi^{\prime}})\subseteq\mathsf{int}(\mathcal{J}_{\pi})$.

To prove this, take $x\in\mathsf{int}(\mathcal{J}_{\pi^{\prime}})$. We already
established
$\mathsf{int}(\mathcal{C})\subseteq\mathsf{int}(\mathcal{J}_{\pi})$, so we may
assume $x\in\mathsf{ext}(\mathcal{C})$. Let $\theta$ be any infinite path
starting from $x$. We show that $\theta$ must cross $\mathcal{J}_{\pi}$.
Suppose not. The path $\theta$ must cross $\mathcal{J}_{\pi^{\prime}}$, so the
crossing must be along $\mathcal{C}\setminus\pi$. Since
$x\in\mathsf{ext}(\mathcal{C})$, Lemma 6.4 implies that when this crossing
occurs, $\theta$ crosses into $\mathsf{int}(\mathcal{C})$. But since
$\mathsf{int}(\mathcal{C})\subseteq\mathsf{int}(\mathcal{J}_{\pi})$, $\theta$
must cross $\mathcal{J}_{\pi}$, completing the proof. ∎

##  8\. Acknowledgments

We thank Michael Damron for suggesting this problem, many helpful discussions,
and invaluable input. Part of this work was conducted at the International
Centre for Theoretical Sciences (ICTS), Bengaluru, India during the program
“First-passage percolation and related models” in July 2022 (code:
ICTS/fpp-2022/7). We thank ICTS for the privilege of attending this meeting
and for its hospitality.

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