C******a
发帖数: 115 | 1
Let A_1=A, B_1=B, A_{n+1}=g(B_n), B_{n+1}=f(A_n),
then A_1>A_2>A_3>...., B_1>B_2>B_3...., by induction.
A=(A_1\A_2)+(A_2\A_3)+....+the intersection of all A_n;
B=(B_1\B_2)+(B_2\B-3)+....+the intersection of all B_n;
("+" means disjoint union.)
f(A_{2n-1}\A_{2n})=B_{2n}\B_{2n+1};
g(B_{2n-1}\B_{2n})=A_{2n}\B_{2n+1}.
f(the intersection of all A_n)=the intersection of all B_n. | C******a
发帖数: 115 | 2 F当然是分段定义了。有些部分定义为f，而另一些部分的定义为g的逆。 | C******a
发帖数: 115 | |
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