Multivariable Advanced Calculus

10.6. JORDAN SEPARATION THEOREM 281because y ∈ K a component of R n \ C. Therefore, letting H 1 denote those sets of Hwhich contain some set of L, 10.19 is of the form1 = ∑d ( f, K, H ) ( )d f −1 , H, y .H∈H 1and it is still a finite sum because ( the terms ) in the sum are 0 for all but finitely manyH ∈ H 1 . I want to expand d f −1 , H, y as a sum of the form∑L∈L Hdusing Lemma 10.5.5. Therefore, I must verify( )f −1 , L, yy /∈ f −1 ( H \ ∪L H)but ( this is just ) Claim 2. By Lemma 10.5.5, I can write the above sum in place ofd f −1 , H, y . Therefore,1 = ∑H∈H 1d ( f, K, H ) d( )f −1 , H, y = ∑d ( f, K, H ) ∑ ( )d f −1 , L, yH∈H 1 L∈L Hwhere there are only finitely many H which give a nonzero ( term)and for each of these,there are only finitely many L in L H which yield d f −1 , L, y ≠ 0. Now the aboveequals= ∑ ∑d ( f, K, H ) ( )d f −1 , L, y . (10.21)L∈L HBy definition,H∈H 1d ( f, K, H ) = d ( f, K, x )where x is any point of H. In particular d ( f, K, H ) = d ( f, K, L ) for any L ∈ L H .Therefore, the above reduces to= ∑ d ( f, K, L ) ( )d f −1 , L, y(10.22)L∈LHere is why. There are finitely many H ∈ H 1 for which the term in the double sum of10.21 is not zero, say H 1 , · · · , H m . Then the above sum in 10.22 equalsm∑k=1∑L∈L Hkd ( f, K, L ) d( )f −1 , L, y +∑L\∪ m k=1 L H kd ( f, K, L ) d( )f −1 , L, yThe second sum equals 0 because those L are contained in some H ∈ H for which0 = d ( f, K, H ) ( )d f −1 , H, y = d ( f, K, H ) ∑ ( )f −1 , L, y= ∑L∈L Hd ( f, K, L ) dTherefore, the sum in 10.22 reduces tom∑k=1∑( )f −1 , L, y .L∈L Hkd ( f, K, L ) dL∈L Hd( )f −1 , L, y