Problem 92 - Single Variable Calculus

Question

The Dirichlet Function The function $$ f(x)=\left\\{\begin{array}{ll} 1 & \text { if } x \text { is rational } \\ 0 & \text { if } x \text { is irrational } \end{array}\right. $$ is called the Dirichlet function. For example, \(f\left(\frac{1}{2}\right)=1\), \(f\left(\frac{(20}{21}\right)=1, f(\sqrt{2})=0\), and \(f(-\pi)=0 .\) Show that for every a, \(\lim _{x \rightarrow a} f(x)\) does not exist.

Step-by-Step Solution

Answer

In conclusion, the limit of the Dirichlet function as \(x \rightarrow a\) does not exist for any 'a,' regardless of whether it is rational or irrational. This is because, in both cases, the difference between \(f(x)\) and \(f(a)\) is 1, so the values of \(f(x)\) will not get closer to \(f(a)\) as \(x\) approaches 'a,' no matter how small the interval is.

Step 1: Case 1: 'a' is rational

First, we will prove that the limit does not exist when 'a' is rational. Suppose that 'a' is rational. Then, given any positive number ε, we can pick an irrational number 'x' in the interval (a - ε, a + ε). Notice that the function f(x) will be 0 since x is irrational. Now, let's compare it to the function evaluated at 'a.' Since 'a' is rational, we have f(a) = 1. Therefore, the difference between f(x) and f(a) is 1, so no matter how small the interval is, the values of f(x) will not get closer to f(a) as x approaches 'a.' Consequently, the limit does not exist when 'a' is rational.

Step 2: Case 2: 'a' is irrational

Now, let's prove that the limit does not exist when 'a' is irrational. Suppose that 'a' is irrational. Then, given any positive number ε, we can pick a rational number 'x' in the interval (a - ε, a + ε). Notice that the function f(x) will be 1 since x is rational. Now, let's compare it to the function evaluated at 'a.' Since 'a' is irrational, we have f(a) = 0. Therefore, the difference between f(x) and f(a) is 1, so no matter how small the interval is, the values of f(x) will not get closer to f(a) as x approaches 'a.' Consequently, the limit does not exist when 'a' is irrational. In conclusion, the limit of the Dirichlet function as x approaches 'a' does not exist for any 'a,' regardless of whether it is rational or irrational.

Key Concepts

Limits of Functions

Limits of functions help us understand how a function behaves as it approaches a specific point. Imagine sailing toward an island from afar, even without reaching the island, one can see it's there. Likewise, limits help us "see" where a function is heading, even without actually reaching the point. The formal definition states that the limit of a function \( f(x) \) as \( x \) approaches \( a \) is \( L \), if for every number \( \varepsilon > 0 \), there is a corresponding \( \delta > 0 \) such that whenever \( 0 < |x - a| < \delta \), it follows that \( |f(x) - L| < \varepsilon \).

In simpler terms, this means that as \( x \) gets closer to \( a \), \( f(x) \) gets closer to \( L \).

• The Dirichlet function provides an interesting challenge for understanding limits because it behaves very differently when \( x \) is rational compared to when it is irrational.
• This unique behavior ensures that no matter how close we get to \( a \), the function values keep oscillating between 0 and 1.
• Hence, no singular value \( L \) can be approached, leading to the non-existence of limits for the Dirichlet function.

Continuity

Continuity in functions can be thought of as a smooth path without breaks, jumps, or gaps. If you can draw a function's graph without lifting your pen, the function is continuous. A function \( f(x) \) is continuous at a point \( a \) if \( \lim_{x \to a} f(x) = f(a) \). In simple terms, not only must the function reach a unique value at \( x = a \), but also approach that value smoothly from either side.

The Dirichlet function is a prime example of discontinuity everywhere.

• Its design inherently mandates a "jump" every time a rational point is followed by an irrational point or vice versa.
• The values of the function never settle into a steady approach toward a single point as \( x \) moves through any neighborhood of any real number \( a \).
• Therefore, for all inputs \( a \), regardless of whether they are rational or irrational, the function always remains discontinuous.

Rational and Irrational Numbers

Rational and irrational numbers together form the real number line, yet they are fundamentally different.
Rational numbers can be expressed as fractions of integers – for example, \( \frac{1}{2} \), \( \frac{3}{4} \), or \( -\frac{7}{5} \). Everything that can be represented neatly and precisely as a fraction falls under this category.
On the other hand, irrational numbers cannot be expressed as simple fractions. They go on forever without repeating. Examples include \( \sqrt{2} \), \( \pi \), and \( e \).

The interplay between these sets is crucial for the Dirichlet function. This function highlights the stark differences between rational and irrational numbers:

• If within any given small interval, however tiny, between any two numbers on the real number line, there exist both rational and irrational numbers.
• This unavoidable blending ensures that no matter how close one tries to narrow down to any point \( a \), the function's output toggles between 1 and 0.
• Thus, the behavior of the Dirichlet function is rooted directly in the distinct properties of rational and irrational numbers.