Discussion
Let \(C[0,1] = \{f : [0,1] \to \mathbb{R} : f \text{ is continuous}\}\).
Consider the metric space \((C[0,1], d_{\infty})\), where \(d_{\infty}(f,g) = \sup\{|f(x)-g(x)| : x \in [0,1]\}\) for \(f, g \in C[0,1]\).
Let \(f_{0}(x) = 0\) for all \(x \in [0,1]\) and \(X = \{f \in (C[0,1], d_{\infty}) : d_{\infty}(f_{0}, f) \ge \frac{1}{2}\}\).
Let \(f_{1}, f_{2} \in C[0,1]\) be defined by \(f_{1}(x) = x\) and \(f_{2}(x) = 1-x\) for all \(x \in [0,1]\).
Consider the following statements:
P: \(f_{1}\) is in the interior of X.
Q: \(f_{2}\) is in the interior of X.
Which of the following statements is correct?
Answer: D
Both P and Q are TRUE.
X is the set of continuous functions whose supremum norm is at least 1/2. Both \(f_{1}(x) = x\) and \(f_{2}(x) = 1-x\) have norm 1 (at x = 1 and x = 0 respectively).
Take any g with \(d_{\infty}(g, f_{1}) \lt \tfrac{1}{2}\). Then \(|g(1)| \gt 1 - \tfrac{1}{2} = \tfrac{1}{2}\), so \(\|g\|_{\infty} \gt \tfrac{1}{2}\) and g is in X. The open ball of radius 1/2 around \(f_{1}\) lies inside X, so \(f_{1}\) is an interior point. The same argument at x = 0 shows \(f_{2}\) is interior.
A boundary point of X would need norm exactly 1/2; neither function is near that, so neither is on the boundary.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development