Externalities, Nonconvexities, and Fixed Points

Now we have a contradiction: First, because n(K(z 0 )) is a minimal essential setfor N(K(z 0 )) in S K(z 0 ) and because n(K(z 0 )) ⊂ £ W 1 ∪ W 2¤ , there exists a positivenumber δ ∗ > 0 such that for all E ∈ B hw ∗ ×w ∗ (δ ∗ ,K(z 0 )) ∩ S K(z 0 ),N(E) ∩ £ W 1 ∪ W 2¤ 6= ∅. (69)But because δ > 0 can be chosen arbitrarily, choosing δ = δ∗ 3, we have by (68) andthe 3M property, the existence of a Ky Fan setE ∈ B hw ∗ ×w ∗ (3 δ∗ 3 ,K(z0 )) ∩ S K(z 0 ) = B hw ∗ ×w ∗ (δ ∗ ,K(z 0 )) ∩ S K(z 0 )such thathN(E) ∩ W 1 ∪ W 2i = ∅.Thus we must conclude that F ∗ (z) ⊆ N (z) for all z ∈ Z.(3) Finally, we will show that F ∗ (z) ∈ M N (z) (C w ∗ f(X)) for all z ∈ Z. Becauseη(z) ⊂ F ∗ (z) for all z ∈ Z and η ∈ [N ] ρZ -w ∗, we know that F ∗ (z) ∈ E N (z) (C w ∗ f(X))for all z ∈ Z.Let z 0 be any point in Z and Let C 0 ∈ C w ∗ f(X) be a proper subset of F ∗ (z 0 ) andconsider the nonempty set, F ∗ (z 0 )\C 0 . Because F ∗ (z 0 ) is the unique subcontinuumirreducible about η(z 0 ),η(z 0 ) ∩ [F ∗ (z 0 )\C 0 ] 6= ∅.Let x 0 ∈ η(z 0 )∩[F ∗ (z 0 )\C 0 ] and let B ε 0(x 0 ) be an open ball of radius ε 0 > 0 about x 0such that B ε 0(x 0 ) ∩ C 0 = ∅. Therefore, η(z 0 )\B ε 0(x 0 ) ⊂ η(z 0 ) and η(z 0 )\B ε 0(x 0 ) ∈P w ∗ f(X). Becauseη(z 0 ) ∈ M N (z 0 )(P w ∗ f(X))there exists some eε > 0 such that for all n, thereexistsz n ∈ B 1 (z 0 ) such thatnBut for some n 0 ,N (z n ) ∩ B w ∗(eε, η(z 0 )\B ε 0(x 0 )) = ∅. (70)N (z 0 ) ∩ B w ∗(eε, η(z 0 )) 6= ∅ for all z 0 ∈ B 1n 0 (z0 ). (71)Because B ε 0(x 0 ) ∩ C 0 = ∅, given (70) and (71), we must conclude that for all n, thereexists z n ∈ B 1 (z 0 ) such thatnwhereas, for some n 00 ,N (z n ) ∩ B w ∗(eε,C 0 )=∅,N (z 00 ) ∩ B w ∗(eε,F ∗ (z 0 )) 6= ∅ for all z 00 ∈ B 1 (z0 ).n 00Therefore, for any proper subset C 0 of F ∗ (z 0 ) in C w ∗ f(X), C 0 is not essential forN (z 0 ), implying therefore thatF ∗ (z) ∈ M N (z) (C w ∗ f(X)) for all z ∈ Z.39