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Chapter 1: Problem 82

Show that the Dirichlet function \(f(x)=\left\\{\begin{array}{ll}0, & \text { if } x \text { is rational } \\\ 1, & \text { if } x \text { is irrational }\end{array}\right.\) is not continuous at any real number.

Short Answer

Expert verified
The Dirichlet function is not continuous at any real number, because the limit \(\lim_{x\to c} f(x)\) does not exist at any point \(c\) in its domain.

Step by step solution

01

Understand the Definition of Continuity

A function \(f(x)\) is said to be continuous at a point \(c\) if \(\lim_{x\to c} f(x) = f(c)\). This means that the values of the function get closer and closer to \(f(c)\) as \(x\) approaches \(c\). Let's consider an arbitrary real number \(c\).
02

Consider c as a Rational Number

If \(c\) is rational, according to the given function \(f(c) = 0\). To obtain the left hand limit and right hand limit of \(f(x)\) as \(x\to c\), we need to consider rational and irrational numbers in the vicinity of \(c\), and these numbers are dense. So for a rational \(c\), the function always switches between 0 and 1 as \(x\to c\) and doesn't approach \(f(c) = 0\). Therefore \(\lim_{x\to c} f(x)\) does not exist.
03

Consider c as an Irrational Number

Similarly, if \(c\) is irrational, according to the given function \(f(c) = 1\). However, similar to the case of rational \(c\), the function also switches between 0 and 1 as \(x\to c\) because of the dense nature of rational and irrational numbers around \(c\), and doesn't approach \(f(c) = 1\). Therefore \(\lim_{x\to c} f(x)\) does not exist.
04

The Function is Not Continuous Anywhere

Since the limit \(\lim_{x\to c} f(x)\) does not exist for both cases (whether \(c\) is rational or irrational), the function \(f(x)\) is not continuous at any real number \(c\), since the definition of continuity, \(\lim_{x\to c} f(x) = f(c)\), is not satisfied anywhere.

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