Question

Let \( f: [0,1] \to [0,1] \) be defined as follows:
\[ f(x) = \begin{cases} x, & \text{if } x \in \mathbb{Q} \cap [0,1] \\ x + \frac{2}{3}, & \text{if } x \in \mathbb{Q}^c \cap (0, \frac{1}{3}) \\ x - \frac{1}{3}, & \text{if } x \in \mathbb{Q}^c \cap (\frac{1}{3}, 1] \end{cases} \] Which of the following statements is (are) TRUE?

• A. \( f \) is one-to-one and onto
• B. \( f \) is not one-to-one but onto
• C. \( f \) is continuous on \( \mathbb{Q} \cap [0,1] \)
• D. \( f \) is discontinuous everywhere on \( [0,1] \)

Hint:

A function defined piecewise with different values on rationals and irrationals will be discontinuous at every point due to the density of rationals in any interval.

Answer 1

The Correct Option is A, D

Step 1: Understanding the function.
The function \( f(x) \) is piecewise defined for rational and irrational numbers within the interval \( [0,1] \). Since the rationals are dense in the real numbers and the function takes different values for rational and irrational numbers, the function will be discontinuous at every point in \( [0,1] \).
Step 2: Analyzing the options.
- (A) \( f \) is one-to-one and onto: This is true because the function takes distinct values for rational and irrational numbers, so it is injective (one-to-one). Additionally, the function is surjective because every value in the interval \( [0,1] \) is covered by the piecewise definitions for rationals and irrationals.
- (B) \( f \) is not one-to-one but onto: This is false because \( f \) is one-to-one and onto.
- (C) \( f \) is continuous on \( \mathbb{Q} \cap [0,1] \): This is false, as \( f \) is discontinuous everywhere on \( [0,1] \).
- (D) \( f \) is discontinuous everywhere on \( [0,1] \): This is correct, since the function is discontinuous at every point in the interval.
Step 3: Conclusion.
The correct answers are (A) and (D).