Exercise 5.1.6

If there is a mapping of B onto A , then 2 | A | 2 | B | . [Hint: Given g mapping B onto A , let f ( X ) = g 1 [ X ] , for all X A .]

Answers

Proof. Suppose that f is a mapping from B onto A . We shall construct an injective F : 2 A 2 B so that

2 | A | = | 2 | | A | = | 2 A | | 2 B | = | 2 | | B | = 2 | B | .

So for any g 2 A let F ( g ) = h where h 2 B is defined by

h ( b ) = g ( f ( b ) )

for b B . To show that F is injective consider any g 1 , g 2 2 A where g 1 g 2 . Then there is an a A such that g 1 ( a ) g 2 ( a ) . Since f : B A is onto there is a b B such that f ( b ) = a . Thus we have

F ( g 1 ) ( b ) = g 1 ( f ( b ) ) = g 1 ( a ) g 2 ( a ) = g 2 ( f ( b ) ) = F ( g 2 ) ( b )

so that F ( g 1 ) F ( g 2 ) . Thus F is injective. □

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2024-07-15 11:42
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