11.3. HOMOGENEITY AND QUASIDISKS
I
•mI J ·
FIGURE 11.5
D
155
Next let (3j and (3j be the components of 8D \ { Wj, wj} and choose (ij E (3j and
(2j E (3j so that
where
D. = B ( Z1j + Z2j. )
J 2 'rJ '
See Figure 11.4. Then
((1j, (2j) n 8D = 0
and Wj and wj are in different components of 8D \ { ( 1 j, ( 2 j }. In addition
If we replace z 1 j and z 2 j by ( 1 j and (2j, and Wj by wj if necessary, we obtain the
sequence of open segments (z 1 j, z 2 j) described above.
By passing to a subsequence of the triples (z1j, Wj, z2j) we can divide the rest
of the proof for Theorem 11.3.1 into the following two cases.
1 ° For each j the open half-disk Hj and the segment ( z 1 j, z 2 j ) lie in D.
2° Every segment (z 1 j, z 2 j) lies in the complement D*.
Proof for case 1
Fix a E D and let wj be a point in 'Yj such that
2 lwj - z2jl 2 lwj - z2jl.
Let Hj be the open half-disk described above, set Zj = ~ ( z 1 j + z 2 j), and let mj be
the midpoint of the segment joining Zj to the midpoint of the circular boundary of
Hj. See Figure 11.5 for an illustration.
Next choose fj E QC(K) such that