Question

Let \( f : [0,1] \to [0,1] \) be a non-constant continuous function such that \( f \circ f = f. \) Define \[ E_f = \{x \in [0,1] : f(x) = x\}. \] Then

• A. \(E_f\) is neither open nor closed.
• B. \(E_f\) is an interval.
• C. \(E_f\) is empty.
• D. \(E_f\) need not be an interval.

Hint:

If a continuous map satisfies \(f(f(x)) = f(x)\), its image (and hence the set of fixed points) must be connected and closed — i.e., an interval.

Answer 1

The Correct Option is B

Step 1: Property of \(f\).
The equation \(f \circ f = f\) implies that \(f\) is an idempotent continuous map, so its image equals its set of fixed points: \[ \text{Im}(f) = E_f. \]
Step 2: Continuity of the image.
Since \(f\) is continuous and \([0,1]\) is compact and connected, \(\text{Im}(f)\) is also compact and connected — hence an interval.
Step 3: Conclusion.
Thus, \(E_f\) is an interval.