Oi Artur, > >Oi Duda! >Se X_1,... e X_n estao em P(A), entao cada X_i esta contido s em Uniao = >X_i. >Pelas condicoes dadas, segue-se que F(Uniao X_i) estah contido em cada = >um >dos F(X_i). Logo, F(Uniao X_i) estah contido em Interseccao F(X_i). Alem >disto, temos que Interseccao F(X_i) esta contido em cada um dos F(x_i), = >o >que acarreta que cada F(F(X_i)) =3D X_i esteja contido em F(Interseccao >F(X_i)). Prosseguindo, temos que Uniao X_i esta contido em F(Interseccao >F(X_i), o que implica que F(F(Interseccao F(X_i)) =3D Interseccao F(X_i) >esteja contido em F(Uniao X_i), Assim concluimos que F(Uni=E3o >X_i) =3D Interse=E7=E3o F(X_i) --Ufa! .Interessante observar que isto eh = >valido >mesmo para subcolecoes nao numeraveis de P(A).=20 > >Agora, temos que Interseccao X_i estah contido em cada X_i, de modo que = >cada >F(X_i) estah contido em F(Interseccao X_i). Logo, Uniao F(X_i) esta = >contido >em F(Interseccao X_i). Alem disto, cada F(X_i) estah contido em Uniao >F(x_i), de modo que F(Uniao F(X_i)) esta contido em cada um dos = >F(F(X_i)) =3D >X_i. Segue-se que F(Uniao F(X_i)) esta contido em Interseccao (X_i), do = >que >concluimos que F(Interseccao X_i) esta contido em F(F(Uniao F(X_i))) =3D = >Uniao >F(X_i). E assim, segue-se que F(Interse=E7=E3o X_i) =3D Uni=E3o F(X_i), = >completando >a prova. Verificamos de novo que isto eh valido mesmo para subcolecoes = >nao >numeraveis de P(A). > >Das condicoes dadas segue-se que F eh bijetora. Sendo 0 o conjunto = >vazio, >temos para todo X de P(A) que 0 estah contido em X e que, portanto, F(X) >esta contido em F(0). Mas como F eh bijetora, para algum X temos F(X) = >=3D A, >de modo que F(0) =3D A. Logo, F(A) =3D F(F(0)) =3D 0. Isto nao prova, = >mas >desconfio que F eh a funcao complemento.
Acho que nao e' sempre assim nao: seja f uma involucao de A e F(X)={f(x), x no complementar de A}. Entao F satisfaz as condicoes do enunciado. Abracos, Gugu > >Um abraco! >Artur=20 > >> Ol=E1 Pessoal! >>=20 >> Estou resolvendo o livro do Elon de An=E1lise e h=E1 um exerc=EDcio = >que n=E3o >> estou >> conseguindo resolver. >>=20 >> Seja A um conjunto e P(A) o conjunto das partes de A. Considere uma = >fun=E7=E3o >> f:P(A)->P(A) que satisfaz as propriedades: se X est=E1 contido em Y = >(ambos >> de >> P(A)) ent=E3o F(Y) est=E1 contido em F(X); e F(F(X)) =3D X. 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