The next theorem deals with the existence of a cycle with nice properties. Basically, you go around the compact subset of an open set with suitable contours while staying in the open set. The method involves the following simple concept. If a cycle Γ consists of oriented curves

Also, for convenience define

Definition 16.5.1 A tiling of ℝ^{2} = ℂ is the union of infinitely many equally spaced vertical and horizontal lines. You can think of the small squares which result as tiles. To tile the plane or ℝ^{2} = ℂ means to consider such a union of horizontal and vertical lines. It is like graph paper. See the picture below for a representation of part of a tiling of ℂ.
Theorem 16.5.2 Let K_{1},K_{2},

so if p is in some K_{k},

each Γ_{j} being the union of oriented simple closed curves, while for all z

Also, if p ∈ Γ_{j}^{∗}, then for i≠j,n
Proof: Consider z → dist
Let S_{j} denote the set of all the closed squares in this tiling which have nonempty intersection with K_{j}.Thus, ∪S_{j} ⊇ K_{j}, all the squares of S_{j} are contained in Ω, and none have nonempty intersection with any K_{k} for k≠j. First suppose p is a point of K_{j} which is in the interior of one of these squares in the tiling. Denote by ∂Q_{k} the boundary of Q_{k} one of the squares in S_{j}, oriented in the counter clockwise direction and Q_{m} denote the square of S_{j} which contains the point, p in its interior. Thus n
If the two squares of S_{j} have a common edge, then this edge is oriented in opposite directions and so this edge the two squares have in common can be deleted without changing ∑ _{j,k}n
From the construction, if any of the γ_{k}^{j∗} contains a point of K_{j} then this point is on one of the four edges of some Q_{s} ∈ S_{j} and at this point, there is at least one edge of some Q_{r} ∈ S_{j},r≠s which also contains this point. As just discussed, this shared edge can be deleted without changing ∑ _{k,j}n
Then as explained above, if p is not on any of the lines in the original tiling but in some K_{j} then ∑ _{k=1}^{m}n
Each orientation on an edge corresponds to a direction of motion over that edge. Call such a motion over the edge a route. Initially, every vertex, (corner of a square in S_{j}) has the property there are the same number of routes to and from that vertex. When an open edge whose closure contains a point of K is deleted, every vertex either remains unchanged as to the number of routes to and from that vertex or it loses both a route away and a route to. Thus the property of having the same number of routes to and from each vertex is preserved by deleting these open edges. The isolated points which result lose all routes to and from. It follows that upon removing the isolated points you can begin at any of the remaining vertices and follow the routes leading out from this and successive vertices according to orientation and eventually return to that end. Otherwise, there would be a vertex which would have only one route leading to it which does not happen. Now if you have used all the routes out of this vertex, pick another vertex and do the same process. Otherwise, pick an unused route out of the vertex and follow it to return. Continue this way till all routes are used exactly once, resulting in closed oriented curves, Γ_{k} for S_{j}. Then if p ∈ K_{j},

because, as discussed above, the sum of the winding numbers started out as 1 and remains unchanged. If p ∈ K_{k},k≠j, then ∑ _{k}n
In case p ∈ K_{j} is on some line of the original tiling, it is not on any of the Γ_{k} because Γ_{k}^{∗}∩ K_{j} = ∅ and so the continuity of z → n
Now note that K_{k} and Γ_{k}^{∗},Γ_{j}^{∗},K_{j},j≠k are disjoint compact sets in Ω. Thus the same process just used will provide a cycle

Also n
Corollary 16.5.3 In the context of Theorem 16.5.2, there exist continuous, closed, bounded cycles

so if p is in some K_{k},

each

Also, if p ∈