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: Core Idea:
This question assesses understanding of fundamental point-set topology concepts on the real number line, specifically focusing on limit points, closed sets, and open sets.

Step 2: Detailed Analysis:
Let's evaluate each statement:
A. Every infinite bounded set of real numbers has a limit point.
This statement reflects the Bolzano-Weierstrass Theorem, a foundational 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 includes all 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. Thus, 0 is a limit point of S. However, \( 0 otin S \). As S does not contain its limit point 0, it is not a closed set. This statement is false.
C. The set of all real numbers is open and closed.
"All real numbers" refers to the set of real numbers, \( \mathbb{R} \). In the standard topology of \( \mathbb{R} \), only \( \mathbb{R} \) itself and the empty set \( \emptyset \) 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 all its points have a neighborhood entirely contained within the set. For instance, at the point 1, any open interval \( (1-\epsilon, 1+\epsilon) \) includes points (like \(1-\epsilon/2\)) that are not in S. Therefore, 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. Also, the sequence \( y_n = -1/n \) is in S, with a limit of 0. Thus, 0 is a limit point of S. But, 0 is not in S. Since S doesn't contain the limit point 0, it is not closed. - Consequently, 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).