Externalities, Nonconvexities, and Fixed Points

is convex or empty.Let x ∈ U 1 , then because U 1 and U 2 are disjoint,{y ∈ D(E ∗ ):(y, x) /∈ E ∗ } = © y ∈ D(E ∗ ):(y,x) /∈ E 1ª ,a convex or empty set because E 1 ∈ S eE .Let x ∈ U 2 , then because U 1 and U 2 are disjoint,{y ∈ D(E ∗ ):(y, x) /∈ E ∗ } = © y ∈ D(E ∗ ):(y,x) /∈ E 2ª ,a convex or empty set because E 2 ∈ S e E .Let x ∈ D(E ∗ )\U 1 ∪ U 2 .Then{y ∈ D(E ∗ ):(y, x) /∈ E ∗ }= © y ∈ D(E ∗ ):(y, x) /∈ E 1ª ∩ © y ∈ D(E ∗ ):(y, x) /∈ E 2ª ,the later being the intersection of convex or empty sets. Therefore,is convex or empty.(2) E ∗ ∈ B hw ∗ ×w ∗ (3δ 0 ,E 0 ):Wehaveand by the triangle inequality,{y ∈ D(E ∗ ):(y, x) /∈ E ∗ }E ∗ =[E 1 ∩ (X × U 2 ) c ] ∪ [E 2 ∩ (X × U 1 ) c ] (93)h w ∗ ×w ∗(E1 ,E 2 ) ≤ h w ∗ ×w ∗(E1 ,E 0 )+h w ∗ ×w ∗(E2 ,E 0 ) < 2δ 0 ,andh w ∗ ×w ∗(E∗ ,E 0 ) ≤ h w ∗ ×w ∗(E∗ ,E 1 )+h w ∗ ×w ∗(E1 ,E 0 ).(94)We know already that h w ∗ ×w ∗(E1 ,E 0 ) < δ 0 .Considerh w ∗ ×w ∗(E∗ ,E 1 ).Wehaveh w ∗ ×w ∗(E∗ ,E 1 ):=max © e w ∗ ×w ∗(E∗ ,E 1 ),e w ∗ ×w ∗(E1 ,E ∗ ) ª .It is easy to check that,e w ∗ ×w ∗(E∗ ,E 1 )=sup (y,x)∈E ∗ ρ w ∗ ×w ∗((y, x),E1 )=sup (y,x)∈[E 2 ∩(X×U 1 ) c ] ρ w ∗ ×w ∗((y,x),E1 )≤ sup (y,x)∈E 2 ρ w ∗ ×w ∗((y, x),E1 )=e w ∗ ×w ∗(E2 ,E 1 ).To show that e w ∗ ×w ∗(E1 ,E ∗ ) ≤ e w ∗ ×w ∗(E1 ,E 2 ) observe thate w ∗ ×w ∗(E1 ,E ∗ )=sup (y,x)∈E 1 ρ w ∗ ×w ∗((y, x),E∗ )=sup (y,x)∈E 1 ρ w ∗ ×w ∗((y, x), [E1 \(X × U 2 )] ∪ [E 2 \(X × U 1 )]).55