Question

Which of the following function is discontinuous at every point of \(\mathbb{R}\)?

• A. \( f(x) = \begin{cases} 1, & \text{if } x \text{ is rational}
  -1, & \text{if } x \text{ is irrational} \end{cases} \)
• B. \( f(x) = \begin{cases} x, & \text{if } x \text{ is rational}
  0, & \text{if } x \text{ is irrational} \end{cases} \)
• C. \( f(x) = \begin{cases} x, & \text{if } x \text{ is rational}
  2x, & \text{if } x \text{ is irrational} \end{cases} \)
• D. \( f(x) = \begin{cases} x, & \text{if } x \text{ is rational}
  -x, & \text{if } x \text{ is irrational} \end{cases} \)

Hint:

Functions defined piecewise on rationals/irrationals are continuous only at points where the defining rules give the same value. To find points of continuity for \(f(x) = \begin{cases} g(x) & x \in \mathbb{Q}
h(x) & x \notin \mathbb{Q} \end{cases}\), solve \(g(x)=h(x)\). If there are no solutions, the function is discontinuous everywhere (assuming g and h are continuous).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A function \(f\) is continuous at a point \(c\) if \( \lim_{x \to c} f(x) = f(c) \). For functions defined piecewise on the rational and irrational numbers, the limit \( \lim_{x \to c} f(x) \) exists only if the values of both pieces approach the same number as \( x \to c \). The function is discontinuous everywhere if this limit fails to exist at every point \( c \in \mathbb{R} \). This happens because both rational and irrational numbers are dense in \(\mathbb{R}\), meaning any interval contains points of both types.

Step 2: Detailed Explanation:

A. Dirichlet Function variation: \( f(x) = 1 \) for rational \(x\), and \( f(x) = -1 \) for irrational \(x\). Let \(c\) be any real number. In any neighborhood of \(c\), there are both rational numbers (where \(f(x)=1\)) and irrational numbers (where \(f(x)=-1\)). This means that as \(x\) approaches \(c\), the function values oscillate between 1 and -1 and do not approach a single limit. Therefore, \( \lim_{x \to c} f(x) \) does not exist for any \(c\). The function is discontinuous everywhere.
B. Thomae-like function: \( f(x) = x \) for rational \(x\), and \( f(x) = 0 \) for irrational \(x\). The function can only be continuous at points \(c\) where the two pieces meet, i.e., where \( c = 0 \). At \(c=0\), \( \lim_{x \to 0} f(x) = 0 \) and \(f(0)=0\). Thus, this function is continuous at \(x=0\) and discontinuous everywhere else.
C. \( f(x) = x \) for rational \(x\), and \( f(x) = 2x \) for irrational \(x\): The function can only be continuous at points \(c\) where \( c=2c \), which implies \( c=0 \). At \(c=0\), \( \lim_{x \to 0} f(x) = 0 \) and \(f(0)=0\). Thus, this function is continuous at \(x=0\) and discontinuous everywhere else.
D. \( f(x) = x \) for rational \(x\), and \( f(x) = -x \) for irrational \(x\): The function can only be continuous at points \(c\) where \( c=-c \), which implies \( c=0 \). At \(c=0\), \( \lim_{x \to 0} f(x) = 0 \) and \(f(0)=0\). Thus, this function is continuous at \(x=0\) and discontinuous everywhere else.
Step 3: Final Answer:
Only the function in option (A) is discontinuous at every point of \(\mathbb{R}\).