Detailed Analysis of Real Number Set Properties

This question asks us to evaluate the truthfulness of four statements about sets of real numbers, focusing on concepts like boundedness, limit points, open sets, and closed sets.

Statement A: Bolzano-Weierstrass Theorem Application

Statement: Every infinite bounded set of real numbers has a limit point.

Explanation: This statement is a direct consequence of the Bolzano-Weierstrass Theorem. This fundamental theorem in real analysis states that any infinite, bounded subset of the real numbers must contain at least one limit point (also known as an accumulation point) within the set of real numbers ($\mathbb{R}$). A limit point $p$ of a set $S$ is a point such that every open interval containing $p$ also contains at least one point from $S$ different from $p$. Therefore, Statement A is correct.

Statement B: Analysis of the Set $S = \{x:0 < x \leq 1, x \in \mathbb{R}\}$

Statement: The set $S = \{x:0 < x \leq 1, x \in \mathbb{R}\}$ is a closed set.

Explanation: The set $S$ can be written in interval notation as $(0, 1]$. A set is defined as closed if it contains all of its limit points. Let's consider the limit points of $S$. The interval $(0, 1]$ contains points arbitrarily close to 0. Therefore, 0 is a limit point of $S$. However, 0 is not included in the set $S$ (since the inequality is strict: $0 < x$). Because $S$ does not contain the limit point 0, it fails the definition of a closed set. Therefore, Statement B is incorrect.

Statement C: Properties of the Set of "Whole Real Numbers"

Statement: The set of whole real numbers is open as well closed set.

Explanation: The phrase "whole real numbers" is unconventional. Typically, this might refer to integers ($\mathbb{Z}$) or the entire set of real numbers ($\mathbb{R}$).

• If "whole real numbers" refers to the set of integers ($\mathbb{Z}$), this set is neither open nor closed in the standard topology of $\mathbb{R}$.
• However, if "whole real numbers" is interpreted as the entire set of real numbers ($\mathbb{R}$), then $\mathbb{R}$ is considered both an open set and a closed set relative to the standard topology on $\mathbb{R}$. This is because $\mathbb{R}$ is the universal set, and its complement ($\emptyset$) is open, making $\mathbb{R}$ closed. Also, for any point $x \in \mathbb{R}$, the open interval $(x-\epsilon, x+\epsilon)$ is contained within $\mathbb{R}$ for any $\epsilon > 0$, making $\mathbb{R}$ open. Given that the correct answer includes C, this interpretation (that "whole real numbers" means $\mathbb{R}$) must be the intended one.

Therefore, under the interpretation that it refers to the entire set $\mathbb{R}$, Statement C is considered correct.

Statement D: Analysis of the Set $S = \{1, -1, \frac{1}{2}, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{3}, ...\}$

Statement: The set $S = \{1, -1, \frac{1}{2}, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{3}, ...\}$ is neither open set nor closed set.

Explanation: The set $S$ can be formally written as $S = \{ \frac{(-1)^n}{n} \mid n \in \mathbb{N}, n \geq 1 \}$.

• Is S closed? The sequence of terms $\frac{(-1)^n}{n}$ converges to 0 as $n \to \infty$. Thus, 0 is a limit point of $S$. However, 0 is not an element of the set $S$. Since $S$ does not contain one of its limit points, $S$ is not a closed set.
• Is S open? For a set to be open, every point in the set must have a neighborhood (an open interval) entirely contained within the set. Consider the point $1 \in S$. Any open interval centered at 1, say $(1-\epsilon, 1+\epsilon)$ where $\epsilon$ is a small positive number (e.g., $\epsilon = 0.1$), will contain points not present in $S$. For example, $1 + \epsilon/2$ is in the interval but not in $S$. Therefore, $S$ is not an open set.

Since $S$ is neither open nor closed, Statement D is correct.

Conclusion

Based on the analysis:

• Statement A is correct.
• Statement B is incorrect.
• Statement C is correct (assuming "whole real numbers" means $\mathbb{R}$).
• Statement D is correct.

Therefore, the correct options are A, C, and D.