Discussion
Let \(S = \{(x, \sin\frac{1}{x}) : 0 \lt x \le 1\}\) and \(T = S \cup \{(0,0)\}\). Under the usual metric on \(\mathbb{R}^{2}\),
Answer: D
Neither \(S\) nor \(T\) is closed.
As \(x \to 0^{+}\), \(\sin\frac{1}{x}\) takes every value in \([-1, 1]\) infinitely often. So every point \((0, b)\) with \(-1 \le b \le 1\) is a limit point of \(S\), for instance \((0, 1)\) as the limit of the points with \(x = \frac{1}{\pi/2 + 2k\pi}\).
None of these points lies in \(S\), so \(S\) is not closed. \(T\) adds only the origin and still misses \((0, 1)\), so \(T\) is not closed either. Its closure is \(S\) together with the whole segment \(\{0\} \times [-1, 1]\).
No comment is present. Be the first to comment.
Loading…
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