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Chapter 1: Problem 12

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

Short Answer

Expert verified
The collection \(\mathcal{F}(\mathbb{N})\) of all finite subsets of \(\mathbb{N}\) is countable. This is proven using mathematical induction, where we show that the set containing substrings of length 1 (base case) is countable, and further conclude that for any given \(k\), if a k-element subset of \(\mathbb{N}\) is countable, then a (k+1)-element subset (inductive step) is also countable.

Step by step solution

01

Introduction of required notation

Let's denote the collection or sets of all finite subsets of \(\mathbb{N}\) by \(\mathcal{F}(\mathbb{N})\). The goal is to find a sequence or order which can list all sets within \(\mathcal{F}(\mathbb{N})\). Hence, we have to establish a bijection between \(\mathcal{F}(\mathbb{N})\) and \(\mathbb{N}\).
02

Base Case: P(1)

The base case is the set of all subsets of size 1, denoted by \(P(1)\). The set of all finite subsets of the natural number set \(\mathbb{N}\) containing only 1 element is countable because there is a bijection between this set and the set of natural numbers itself. This is intuitively obvious, as any single-element set can be associated uniquely with that element itself.
03

Inductive assumption: P(k)

Assume that for some arbitrary natural number \(k\), the set \(P(k)\) of all finite subsets of the natural number set \(\mathbb{N}\) containing exactly \(k\) elements is countable.
04

Induction step: P(k+1)

We are to prove that the set \(P(k+1)\) of all finite subsets of \(\mathbb{N}\) containing exactly \(k+1\) elements is countable. However, each of these \(k+1\) element subsets can be seen as adding a single additional element to all the k-element subsets. By the inductive assumption, we know that the set of all k-element subsets is countable. Adding one element to each of these countable subset still results in a countable set. Therefore, P(k+1) is countable, and by mathematical induction, it follows that every set in \(\mathcal{F}(\mathbb{N})\) is countable.

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