Question

Everywhere Discontinuous Give a convincing argument that the following function is not continuous at any real number. $$f(x)=\left\\{\begin{array}{ll}{1,} & {\text { if } x \text { is rational }} \\\ {0,} & {\text { if } x \text { is irrational }}\end{array}\right.$$

Short answer

The function \( f(x) \) is not continuous at any real number. The function values jump between 0 and 1 at every point depending on whether the x-value is rational or irrational, not meeting the requirement of continuity that \( \lim_{x \to a} f(x) = f(a) \) for any real number \( a \).

Step-by-step solution

Understand the Definition of Continuity

The function \( f(x) \) is said to be continuous at a point \( a \) if \( \lim_ {x \to a} f(x) = f(a) \). This means that the value of the function at \( a \) is equal to the limit of the function as \( x \) approaches \( a \). This condition must hold for every point in the domain for the function to be called continuous everywhere.

Applying the Definition to the Function's Terms

Consider a number \( x \) such that \( x \) is irrational, and call it \( x_{0} \). Now take a sequence of rational numbers \( x_{n} \) that converge to \( x_{0} \). The limit of these numbers as \( n \) approaches to infinity \( \lim_ {n \to \infty} x _{n} \) is \( x_{0} \). Now consider the function values at these rational number sequences \( f(x _{n}) \). For each \( x_{n} \), \( f(x_{n}) \) equals 1. So, \( \lim_ {n \to \infty} f(x _{n}) = 1 \). However, \( f(x_{0}) = 0 \). Hence \( \lim_ {x \to x_{0} } f(x) \neq f(x_{0}) \), violating the definition of continuity at \( x_{0} \).

Extending the Argument to All Real Numbers

From this argument, it can be seen that the function \( f(x) \) is not continuous at any irrational number \( x_{0} \). The same procedure can be applied with a sequence of irrational numbers converging to any rational number, to show that \( f(x) \) is discontinuous at all rational numbers as well.