# solHMK1 - Section 1.3#6 There are many examples For...

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Unformatted text preview: Section 1.3 #6) There are many examples. For instance f : N inj , → N given by f ( n ) = n + 1, and g : N inj , → N given by g ( n ) = n 2 . #7) T 2 is a denumerable set. There exists a bijection f : T 2 bij → N . If T 1 is denumerable, then there is a bijection g : T 1 bij → N . Thus, f- 1 ◦ g : T 1 → T 2 is a bijection. Conversely, if h : T 1 bij → T 2 is a bijection, then f ◦ h : T 1 → N is a bijection, and hence T 1 is denumerable. #9) Since S ⊆ S ∪ T and S is infinite, we know that S ∪ T is infinite. Thus, we only need to show it is countable. By Theorem 1.3.10, it is enough to find an injection h : S ∪ T inj , → N . Let f : S bij → N and g : T bij → N be bijections. Define ¯ f : S inj , → N by ¯ f ( s ) = 2 f ( s )( ∀ s ∈ S ), so that ¯ f is a bijection from S onto the positive even numbers. Define ¯ g : T inj , → N by ¯ g ( t ) = 2 g ( t )- 1( ∀ t ∈ T ), so that ¯ g is a bijection from T onto the positive odd numbers. Define h : S ∪ T → N as follows: h ( x ) = ¯ f ( x ) if x ∈ S ¯ g ( x ) if x ∈ T \ S To show that h is an injection, suppose that for some x 1 , x 2 ∈ S ∪ T and some c ∈ N , we have h ( x 1 ) = c = h ( x 2 ). If c is even, then we must have x 1 , x 2 ∈ S (elements not in S are mapped to odd numbers.) Hence, ¯ f ( x 1 ) = h ( x 1 ) = h ( x 2 ) = ¯ f ( x 2 ). Now x 1 = x 2 follows from the injectivity of ¯ f .....
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solHMK1 - Section 1.3#6 There are many examples For...

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