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Chapter 4: Problem 22

Show that the set \(\mathbb{R}\) of all real numbers is non-enumerable.

Short Answer

Expert verified
By using Cantor's Diagonal argument, it can be shown that the set of real numbers \(\mathbb{R}\) is non-enumerable.

Step by step solution

01

Define Enumeration Requirement

All sets whose elements can be placed in one-to-one correspondence with the set of natural numbers are said to be enumerable. The objective here is to show the set \(\mathbb{R}\) of all real numbers is not like this.
02

Define a List by Supposition

Suppose \(\mathbb{R}\) could be enumerated. This would mean all real numbers could be ordered as a list \(r_1, r_2, r_3, ...\)
03

Define a Contradictory Real Number

Define a new real number \(r'\) such that it is different from the nth digit of \(r_n\) for every number \(n\). This will mean that \(r'\) is different from every number in the list of real numbers.
04

Conclude the Contradiction

The existence of \(r'\) contradicts the initial supposition that the list includes all real numbers. This shows that \(\mathbb{R}\) cannot be enumerable, so the set of all real numbers is indeed non-enumerable.

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Most popular questions from this chapter

Show that \(\mathbb{N}^{\omega}\), the set of infinite sequences of natural numbers, is non-enumerable by a reduction argument.

Show that if \(B \subseteq A\) and \(A\) is enumerable, so is \(B\). To do this, suppose there is a surjective function \(f: \mathbb{Z}^{+} \rightarrow A .\) Define a surjective function \(g: \mathbb{Z}^{+} \rightarrow B\) and prove that it is surjective. What happens if \(B=\varnothing ?\)

Show that \(\wp(\mathrm{N})\) is non-enumerable by a diagonal argument.

Let \(S\) be the set of all surjections from \(\mathrm{N}\) to the set \(\\{0,1\\},\) i.e., \(S\) consists of all surjections \(f: \mathbb{N} \rightarrow \mathbb{B}\). Show that \(S\) is non-enumerable.

Show that if there is an injective function \(g: B \rightarrow A,\) and \(B\) is non-enumerable, then so is \(A\). Do this by showing how you can use \(g\) to turn an enumeration of \(A\) into one of \(B\).
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