Chapter 2.3, Problem 64E

To determine

To explain: A convincing argument that the function $f\left(x\right)=\{\begin{array}{l}1,\text{if}x\text{is rational}\\ \text{0,}\text{if }x\text{ is irrational}\end{array}$ is not continuous at any real number.

Answer to Problem 64E

The function is not continuous at every real number as the limiting value at any point is not equal to the functional value.

Explanation of Solution

Given information: The function is $f\left(x\right)=\{\begin{array}{l}1,\text{if}x\text{is rational}\\ \text{0,}\text{if }x\text{ is irrational}\end{array}$.

A function is said to be continuous at any number $x=a$ if it is continuous in small open interval around it.

But the open interval around any number contains many irrational and rational numbers. So, the limiting value at the points is must not equal to the functional value at that point.

As the limiting value of function at irrational point is $1$ as it is near a rational number and the functional value at irrational number is equal to $0$ . So, the function is not continuous at every point.

Therefore, the function is not continuous at every real number as the limiting value at any point is not equal to the functional value.