If $$A_n = \left( -\frac{1}{n}, \frac{1}{n} \right), \quad n \in \mathbb{N}$$ (N is the set of positive integers) be a subset of the set of real numbers R, then the set $$\bigcap_{n=1}^{\infty} A_n$$ is​

Correct option is A

Solution:

$$\textbf{Given:} \\A_n = \left(-\frac{1}{n}, \frac{1}{n}\right), \quad n \in \mathbb{N} \\[8pt]\text{We are asked to compute:} \\\bigcap_{n=1}^{\infty} A_n \\[8pt]\textbf{Step 1: Understand the intervals} \\A_1 = (-1, 1), \quad A_2 = \left(-\frac{1}{2}, \frac{1}{2}\right), \quad A_3 = \left(-\frac{1}{3}, \frac{1}{3}\right), \dots \\[6pt]\text{Each } A_n \text{ is an open interval centered at } 0 \text{ and } A_{n+1} \subset A_n \\[8pt]\textbf{Step 2: Infinite Intersection} \\\text{If } x \ne 0, \text{ then } \exists n \in \mathbb{N} \text{ such that } |x| > \frac{1}{n} \Rightarrow x \notin A_n \\[6pt]\Rightarrow \text{Only } x = 0 \text{ lies in all intervals } A_n \\[6pt]\therefore \bigcap_{n=1}^{\infty} A_n = \{0\} \\[8pt]\textbf{Step 3: Nature of the Set} \\\{0\} \text{ is:} \\\quad \text{- Not open in } \mathbb{R} \\\quad \text{- Closed in } \mathbb{R} \\[8pt]\boxed{\text{Final Answer: (c) closed}}$$​