Question

Let \( S \) be the set of all limit points of the set \[ \left\{ \frac{n}{\sqrt{2}} + \frac{\sqrt{2}}{n} : n \in \mathbb{N} \right\}. \] Let \( \mathbb{Q}_+ \) be the set of all positive rational numbers. Then

• A. \( \mathbb{Q}_+ \subset S \)
• B. \( S \subset \mathbb{Q}_+ \)
• C. \( S \cap (\mathbb{R} \setminus \mathbb{Q}_+) \neq \emptyset \)
• D. \( S \cap \mathbb{Q}_+ \neq \emptyset \)

Hint:

In sequences involving irrational limit points, be sure to identify whether the set of limit points intersects with rationals or irrationals. This often leads to clearer conclusions about the set's structure.

Answer 1

The Correct Option is B

Step 1: Understanding the set and limit points.
The set given is \( \left\{ \frac{n}{\sqrt{2}} + \frac{\sqrt{2}}{n} : n \in \mathbb{N} \right\} \), where each term involves \( n \), and as \( n \) increases, the terms approach \( \sqrt{2} \). This suggests that \( S \), the set of limit points, will include \( \sqrt{2} \), which is irrational.

Step 2: Analyzing the options.
- (A) \( \mathbb{Q}_+ \subset S \): This is incorrect because the limit points are irrational, and \( \mathbb{Q}_+ \) consists of positive rationals. Thus, \( \mathbb{Q}_+ \) is not a subset of \( S \).
- (B) \( S \subset \mathbb{Q}_+ \): This is incorrect, as the limit points include \( \sqrt{2} \), which is irrational, so \( S \) is not a subset of \( \mathbb{Q}_+ \).
- (C) \( S \cap (\mathbb{R} \setminus \mathbb{Q}_+) \neq \emptyset \): This is correct because \( S \) contains the irrational number \( \sqrt{2} \), meaning it intersects with real numbers that are not positive rationals.
- (D) \( S \cap \mathbb{Q}_+ \neq \emptyset \): This is incorrect because the limit points of the given sequence are irrational, and thus there is no intersection with \( \mathbb{Q}_+ \).

Step 3: Conclusion.
The correct answer is (C) \( S \cap (\mathbb{R} \setminus \mathbb{Q}_+) \neq \emptyset \), as the limit points of the sequence include irrational numbers like \( \sqrt{2} \).