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Chapter 1: Problem 6

Exhibit a bijection between \(\mathbb{N}\) and a proper subset of itself.

Short Answer

Expert verified
The required bijection is the function \(f: \mathbb{N} \rightarrow \mathbb{N}\setminus\{0\}\) defined by \(f(n) = 2n\). This function is both injective and surjective, hence it establishes a one-to-one correspondence between \(\mathbb{N}\) and its proper subset \(\mathbb{N}\setminus\{0\}\).

Step by step solution

01

Define the function

Let's define a function \(f: \mathbb{N} \rightarrow \mathbb{N}\setminus\{0\} \) where \(f(n) = 2n\) . This function takes every natural number and maps it to its double, which will be an even number.
02

Prove the function is injective

A function is injective if different inputs give different outputs. To show that \(f\) is injective, let's assume \(f(a) = f(b)\) for some \(a, b \in \mathbb{N}\) . From the definition of \(f\) , we have \(2a = 2b\) , which implies \(a = b\) . Therefore, \(f\) is injective.
03

Prove the function is surjective

A function is surjective if every element in the codomain has a preimage in the domain. The codomain here is the set of even natural numbers. Let's pick an arbitrary element \(c\) in the codomain. As \(c\) is even, there exists a natural number \(b = \frac{c}{2}\) s.t. \(f(b) = 2b = c\) . Thus, \(f\) is surjective.

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Most popular questions from this chapter

Prove that the collection \(\mathcal{F}(\mathbb{N})\) of all finite subsets of \(\mathbb{N}\) is countable.

Show that if \(f: A \rightarrow B\) and \(G, H\) are subsets of \(B\), then \(f^{-1}(G \cup H)=f^{-1}(G) \cup f^{-1}(H)\) and \(f^{-1}(G \cap H)=f^{-1}(G) \cap f^{-1}(H)\).

Prove that a set \(T_{1}\) is denumerable if and only if there is a bijection from \(T_{1}\) onto a denumerable set \(T_{2}\).

Prove that \(n<2^{n}\) for all \(n \in \mathbb{N}\)

Prove that \(n^{3}+(n+1)^{3}+(n+2)^{3}\) is divisible by 9 for all \(n \in \mathbb{N}\).
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