Question

Which of the following are correct:
A. Every infinite bounded set of real number has a limit point
B. The set \( S = \{x : 0<x \le 1, x \in \mathbb{R}\} \) is a closed set
C. The set of whole real numbers is open as well closed set
D. The set \( S = \{1, -1, \frac{1}{2}, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{3}, ... \} \) is neither open set nor closed set

• A. A, B and C Only
• B. A, C and D Only
• C. B, C and D Only
• D. D Only

Hint:

To test if a set is closed, find all its limit points. If every single limit point is already in the set, the set is closed. A common way to find limit points is to consider sequences within the set and find their limits.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question tests fundamental concepts of point-set topology on the real number line, including limit points, closed sets, and open sets.

Step 2: Detailed Explanation:
Let's analyze each statement:
A. Every infinite bounded set of real number has a limit point.
This is the statement of the Bolzano-Weierstrass Theorem, a fundamental result in real analysis. This statement is true.
B. The set \( S = \{x : 0<x \le 1, x \in \mathbb{R}\} \) is a closed set.
This set is the interval \( (0, 1] \). A set is closed if it contains all of its limit points. Consider the sequence \( x_n = 1/n \) for \( n \ge 2 \). All points of this sequence are in S. The limit of this sequence is 0. Therefore, 0 is a limit point of S. However, \( 0 \notin S \). Since S does not contain one of its limit points, it is not a closed set. This statement is false.
C. The set of whole real numbers is open as well closed set.
The phrase "whole real numbers" is interpreted to mean the entire set of real numbers, \( \mathbb{R} \). In the standard topology of \( \mathbb{R} \), the set \( \mathbb{R} \) itself and the empty set \( \emptyset \) are the only two sets that are both open and closed. Therefore, \( \mathbb{R} \) is both open and closed. This statement is true.
D. The set \( S = \{1, -1, \frac{1}{2}, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{3}, ... \} \):
- Is S open? A set is open if every point in it has a neighborhood entirely contained within the set. For any point in S, e.g., 1, any open interval \( (1-\epsilon, 1+\epsilon) \) contains points (like \(1-\epsilon/2\)) that are not in S. So, S is not open. - Is S closed? A set is closed if it contains all its limit points. The sequence \( x_n = 1/n \) is in S, and its limit is 0. The sequence \( y_n = -1/n \) is also in S, and its limit is also 0. Thus, 0 is a limit point of S. However, the number 0 itself is not an element of S. Since S does not contain this limit point, it is not closed. - Therefore, the set is neither open nor closed. This statement is true.
Step 3: Final Answer:
Statements A, C, and D are correct, while B is incorrect. This corresponds to option (B).