Understanding the Modified Dirichlet Function's Continuity

The question asks us to analyze the continuity of a specific function, denoted as $f(x)$. Let's break down the function definition and the concept of continuity before evaluating the options.

Function Definition

The function $f(x)$ is defined as follows:

$ f(x) = \begin{cases} 0, & \text{if } x \text{ is irrational} \\ \frac{1}{n}, & \text{if } x \text{ is rational, } x = \frac{m}{n}, \text{ with gcd(m,n)=1 and } n>0 \end{cases} $

Here:

• For irrational numbers, the function value is always 0.
• For rational numbers expressed in their simplest form $\frac{m}{n}$ (where $m$ and $n$ have no common factors other than 1, and $n$ is positive), the function value is $\frac{1}{n}$.

What is Continuity?

A function $f$ is said to be continuous at a point $c$ if the following condition holds:

$ \lim_{x \to c} f(x) = f(c) $

In simpler terms, for a function to be continuous at a point, its value at that point must equal the limit of the function as $x$ approaches that point. This also implies that small changes in the input $x$ result in small changes in the output $f(x)$. We often use the epsilon-delta definition:

For every $\epsilon > 0$, there exists a $\delta > 0$ such that if $|x - c| < \delta$, then $|f(x) - f(c)| < \epsilon$.

Analyzing Continuity at Irrational Points

Let's consider a point $c$ that is an irrational number. According to the function definition, $f(c) = 0$.

To check for continuity at $c$, we need to see if $\lim_{x \to c} f(x) = 0$. We examine the values of $f(x)$ for $x$ near $c$. There are two cases for $x$ near $c$:

• Case 1: $x$ is irrational. If $x$ is irrational, then $f(x) = 0$.
• Case 2: $x$ is rational. If $x$ is rational, let $x = \frac{m}{n}$ in lowest terms. Then $f(x) = \frac{1}{n}$.

We need to show that as $x$ approaches $c$ (an irrational number), $f(x)$ approaches $0$.

Let's use the epsilon-delta definition. We want to show that for any $\epsilon > 0$, there exists a $\delta > 0$ such that if $|x - c| < \delta$, then $|f(x) - f(c)| < \epsilon$. Since $f(c)=0$, we need $|f(x)| < \epsilon$.

• If $x$ is irrational, $f(x) = 0$, so $|f(x)| = 0 < \epsilon$. This part is satisfied.
• If $x$ is rational ($x = \frac{m}{n}$, gcd(m,n)=1), $f(x) = \frac{1}{n}$. We need $\frac{1}{n} < \epsilon$, which is equivalent to $n > \frac{1}{\epsilon}$.

So, for a given $\epsilon > 0$, we choose $N = \lceil \frac{1}{\epsilon} \rceil$. We need to find a $\delta > 0$ such that any rational number $x = \frac{m}{n}$ (in lowest terms) within the interval $(c-\delta, c+\delta)$ has a denominator $n \ge N$.

It's a known property that the rational numbers are dense in the real numbers. This means that for any irrational number $c$ and any $\delta > 0$, the interval $(c-\delta, c+\delta)$ contains infinitely many rational numbers. Furthermore, within any interval $(a, b)$, there exist rational numbers with arbitrarily large denominators.

Therefore, we can choose $\delta > 0$ small enough such that the interval $(c-\delta, c+\delta)$ contains at least one rational number $x = \frac{m}{n}$ (in lowest terms) with $n \ge N$.

If $x$ is irrational and $|x-c| < \delta$, then $|f(x)-f(c)| = |0-0| = 0 < \epsilon$.

If $x$ is rational ($x = \frac{m}{n}$, gcd(m,n)=1) and $|x-c| < \delta$, we choose $\delta$ such that there is such an $x$ with $n \ge N$. Then $|f(x)-f(c)| = |\frac{1}{n}-0| = \frac{1}{n}$. Since $n \ge N$, we have $\frac{1}{n} \le \frac{1}{N} \le \epsilon$.

Thus, for any $\epsilon > 0$, we can find a suitable $\delta > 0$. This proves that $f$ is continuous at every irrational point $c$.

Analyzing Continuity at Rational Points

Now, let's consider a point $c$ that is a rational number. Let $c = \frac{p}{q}$ where $p$ and $q$ are integers, $q \neq 0$, and gcd(p,q)=1. Assume $q>0$. According to the function definition, $f(c) = f(\frac{p}{q}) = \frac{1}{q}$.

For $f$ to be continuous at $c$, we need $\lim_{x \to c} f(x) = f(c) = \frac{1}{q}$.

Let's consider a sequence of irrational numbers, $y_k$, such that $y_k \to c$ as $k \to \infty$. Since $y_k$ is irrational for all $k$, $f(y_k) = 0$ for all $k$.

The limit of the function along this sequence of irrational numbers is:

$ \lim_{k \to \infty} f(y_k) = \lim_{k \to \infty} 0 = 0 $

For continuity at $c$, this limit must equal $f(c)$. However, we found that $f(c) = \frac{1}{q}$. Since $q$ is a positive integer, $q \ge 1$. Therefore, $f(c) = \frac{1}{q}$ is always positive and less than or equal to 1.

Since $0 \neq \frac{1}{q}$ for any positive integer $q$, the limit from the irrational sequence does not match the function value at the rational point $c$.

Therefore, the function $f$ is discontinuous at every rational point.

Conclusion on Function Continuity

Based on the analysis:

• The function $f(x)$ is continuous at every irrational point.
• The function $f(x)$ is discontinuous at every rational point.

This means the function is not continuous everywhere, nor is it nowhere continuous. It is also not continuous at every rational point. The statement "$f$ is continuous at every irrational point" accurately describes the continuity property of this modified Dirichlet function.