Exercise 5.1.4

Prove that κ κ 2 κ κ .

Answers

Proof. Suppose that κ = | A | for a set A . Then we construct an injective F : A A 2 A × A so that

κ κ = | A A | | 2 A × A | = | 2 | | A × A | = 2 κ κ .

So for any f A A define F ( f ) = g where g 2 A × A is defined by

g ( a 1 , a 2 ) = { 0 f ( a 1 ) a 2 1 f ( a 1 ) = a 2

for ( a 1 , a 2 ) A × A . To show that F is injective consider any f , g A A where f g . Then there is an a A such that f ( a ) g ( a ) . Now let a 1 = f ( a ) and a 2 = g ( a ) so that

f ( a ) = a 1 a 2 = g ( a ) .

Since f ( a ) = a 1 it follows by definition that F ( f ) ( a , a 1 ) = 1 . Similarly since g ( a ) = a 2 a 1 it follows that F ( g ) ( a , a 1 ) = 0 . Hence we have

F ( f ) ( a , a 1 ) = 1 0 = F ( g ) ( a , a 1 )

so that clearly F ( f ) F ( g ) . Thus F is injective. □

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2024-07-15 11:42
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