1648 CHAPTER 51. FUNDAMENTALS OF COMPLEX ANALYSIS

of the four edges of S j and at this point, there is at least one edge of some Sl which alsocontains this point. As just discussed, this shared edge can be deleted without changing

∑k, j n(

z,γ jk

). Delete the edges of the Sk which intersect K but not the endpoints of these

edges. That is, delete the open edges. When this is done, delete all isolated points. Let theresulting oriented curves be denoted by {γk}

mk=1 . Note that you might have γ∗k = γ∗l . The

construction is illustrated in the following picture.

K1

K2

Ω

Then as explained above, ∑mk=1 n(p,γk) = 1. It remains to prove the claim about the

closed curves.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) hasthe property there are the same number of routes to and from that vertex. When an openedge whose closure contains a point of K is deleted, every vertex either remains unchangedas to the number of routes to and from that vertex or it loses both a route away and a routeto. Thus the property of having the same number of routes to and from each vertex ispreserved by deleting these open edges. The isolated points which result lose all routesto and from. It follows that upon removing the isolated points you can begin at any ofthe remaining vertices and follow the routes leading out from this and successive verticesaccording to orientation and eventually return to that end. Otherwise, there would be avertex which would have only one route leading to it which does not happen. Now if youhave 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 thisway till all routes are used exactly once, resulting in closed oriented curves, Γk. Then

∑k n(Γk, p) = ∑ j n(

γ j, p)= 1.

In case p ∈ K is on some line of the tiling, it is not on any of the Γk because Γ∗k ∩K = /0and so the continuity of z→ n(Γk,z) yields the desired result in this case also. This provesthe lemma.