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Re: [obm-l] RE: [obm-l] Questão de Análise
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. Mostrar =
>que F(Uni=E3o
>> X_i) =3D Interse=E7=E3o F(X_i) e tamb=E9m F(Interse=E7=E3o X_i) =3D =
>Uni=E3o F(X_i).
>>=20
>> Uma fun=E7=E3o que satisfaz essas condi=E7=F5es =E9 F(X) =3D =
>Complementar X.
>>=20
>
>
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