Question

Let \(f(x)=0\) if \(x\) is irrational and \(f(p / q)=1 / q\) if \(p\) and \(q\) are positive integers with no common factors. Show that \(f\) is discontinuous at every rational and continuous at every irrational on \((0, \infty)\).

Short answer

Short Answer: Yes, the function \(f(x)\) is continuous at all irrational numbers and discontinuous at all rational numbers in the interval \((0, \infty)\).

Step-by-step solution

Function definition and given information

First, let's recall the definition of the function \(f(x)\): - \(f(x) = 0\) if \(x\) is irrational; - \(f(p/q) = 1/q\) if \(p\) and \(q\) are positive integers with no common factors (it means \(p/q\) is rational and reduced to lowest terms). We have to show that: - \(f\) is continuous at every irrational number in \((0, \infty)\); - \(f\) is discontinuous at every rational number in \((0, \infty)\).

Continuity for irrational numbers

Let's try to show the continuity of the function for irrational numbers. Consider \(x_0\) to be an irrational number on \((0, \infty)\). The definition of continuity states that a function \(f\) is continuous at \(x_0\) if and only if: \(\lim \limits_{x \to x_0} f(x) = f(x_0)\) Since \(x_0\) is irrational, we already know \(f(x_0) = 0\). Now we need to show that the limit also equals 0 as x approaches \(x_0\). For any \(\epsilon > 0\), we have to find a \(\delta > 0\) such that whenever \(0 < |x - x_0| < \delta\), we get \(|f(x) - f(x_0)| < \epsilon\). To do that, we consider a neighborhood around \(x_0\). Choose \(\delta = \min \left(1, \frac{\epsilon}{2} \right)\). If \(x\) is irrational and \(0 < |x - x_0| < \delta\), then \(f(x) = 0\). In this case, \(|f(x) - f(x_0)| = |0 - 0| = 0 < \epsilon\). If \(x = \frac{p}{q}\) is rational and \(0 < |x - x_0| < \delta\), then \(f(x) = \frac{1}{q}\). Since \(|x - x_0| < \delta \leq \frac{\epsilon}{2}\), we have $$|f(x) - f(x_0)| = \left|\frac{1}{q} - 0 \right| \leq \frac{1}{q} < \frac{\epsilon}{2} < \epsilon$$ In both cases, we have shown that \(|f(x) - f(x_0)| < \epsilon\). Hence, \(\lim \limits_{x \to x_0} f(x) = f(x_0)\) for irrational \(x_0\), and \(f\) is continuous at every irrational number on \((0, \infty)\).

Discontinuity for rational numbers

Let's now examine the case for rational numbers. Suppose \(x_0 = p_0/q_0\) is a rational number on \((0, \infty)\). For the function to be discontinuous, the limit as \(x\) approaches \(x_0\) should not equal \(f(x_0)\). Given that \(x_0 = p_0/q_0\), we have \(f(x_0) = 1/q_0\). Now consider the sequence \(x_n = x_0 + \frac{\sqrt{2}}{n}\). This sequence converges to \(x_0\) as \(n\) goes to infinity, and \(x_n\) is irrational for every \(n\). By the definition of the limit, we have: $$\lim \limits_{n \to \infty} f(x_n) = \lim \limits_{n \to \infty} 0 = 0.$$ But \(f(x_0) = 1/q_0 \neq 0\). Since the function's value and the limit do not match, \(f\) is discontinuous at every rational number on \((0, \infty)\). In conclusion, we have shown that \(f\) is continuous at every irrational number and discontinuous at every rational number on the interval \((0, \infty)\).

Key concepts

Discontinuity at Rational Numbers

Understanding the concept of discontinuity at rational numbers requires a closer look at what happens to a function at these specific points. In the given exercise, the function defined as
\[f(x) = \begin{cases} 0 & \text{if } x \text{ is irrational} \frac{1}{q} & \text{if } x = \frac{p}{q} \text{ is rational}\end{cases} \] illustrates this concept vividly.

For rational numbers, the function assigns values based on the denominator when the fraction is in its lowest terms. To show discontinuity, it must be proven that there's a disruption in the behavior of the function at these points—meaning the function's limit does not equal its value at that point. A rational number can be embedded within a dense sea of irrational numbers, and since the function value for irrationals is consistently zero, approaching a rational number from either side through irrationals will yield a limit of zero, which will not match the non-zero value assigned to the rational number itself. This fundamental notion underpins why, for our function, there is a guaranteed discontinuity at every rational point in the interval \((0, \: \: \infty)\).

Continuity at Irrational Numbers

In real analysis, establishing continuity at irrational numbers can be intriguing, specifically because these points cannot be expressed as ratios of integers and are sprinkle irregularly throughout the number line. In the exercise, we are asked to show that the function \[f(x) = 0 \text{ if } x \text{ is irrational}\] is continuous at any irrational point.

The crux of proving this continuity lies in the limit definition of continuity, which asserts that a function is continuous at a point if the limit as \(x\) approaches the point equals the function's value at that point. Since the function assigns the same value (zero) to all irrational numbers, as we approach any such point through other irrational numbers or even rational numbers with large denominators, the function's output remains infinitesimally close to zero. This consistent behavior irrespective of the approach path ensures that the limit coincides with the function's value at irrational numbers, thus satisfying the condition for continuity.

Limit Definition of Continuity

The concept of continuity is pivotal in real analysis and is rigorously defined using limits. The limit definition of continuity states that a function \(f\) is continuous at a point \(x_0\) if \[\lim\limits_{x \to x_0} f(x) = f(x_0).\]This means as we get arbitrarily close to \(x_0\), the function's values converge to \(f(x_0)\).

Applying the definition

In practice, this involves checking if, for any given distance \(\epsilon > 0\), we can find a proximity \(\delta > 0\) such that any \(x\) within this \(\delta\)-neighborhood of \(x_0\) will result in a function value within \(\epsilon\) of \(f(x_0)\).

The beauty of this definition is its applicability to all types of functions and contexts, be they rational or irrational domains, as seen in our function example. It allows us to methodically prove continuity or find points of discontinuity, which are so critical in understanding the behavior of functions.

Sequence Convergence

The notion of sequence convergence is integral when dealing with functions and their behavior at specific points. A sequence \((x_n)\) is said to converge to a limit \(L\) if the sequence terms can be made as close as desired to \(L\) by going far enough out in the sequence. Mathematically, for any \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\), \(|x_n - L| < \epsilon\).

When assessing whether a function is discontinuous at a point, as in our textbook exercise, we deliberately choose a sequence of irrational numbers that converges to a rational point. By showing the limit of the function along this sequence is different from the function's value at the rational point, we demonstrate discontinuity. This employment of converging sequences is a powerful technique in real analysis, serving as a bridge between the discrete nature of sequences and the continuous nature of function limits.