Question

Let $X_{n}=\left\{z=x+i y:|z|^{2} \leq \frac{1}{n}\right\}$ for all integers $n \geq 1$. Then, $\overset{\infty}{\underset{n=1}{\cap}}$ is

• A. A singleton set
• B. Not a finite set
• C. An empty set
• D. A finite set with more than one elements

Answer 1

The Correct Option is A

Given, $ X_{n} =\left\{z=x+i Y:|z|^{2} \leq \frac{1}{n}\right\} $ $=\left\{x^{2}+y^{2} \leq \frac{1}{n}\right\} $ $\therefore X_{1}=\left\{x^{2}+y^{2} \leq 1\right\}$ $X_{2}=\left\{x^{2}+y^{2} \leq \frac{1}{2}\right\}$ $X_{3}=\left\{x^{2}+y^{2} \leq \frac{1}{3}\right\}$

$X_{\infty}=\left\{x^{2}+y \leq 0\right\}$ $ \therefore \overset{\infty}{\underset{n=1}{\cap}} X_{n} =X_{1} \cap X_{2} \cap X_{3} \cap \ldots \cap X_{\infty} $ $ =\left\{x^{2}+y^{2}=0\right\} $ Hence, $\overset{\infty}{\underset{n=1}{\cap}} X_{n}$ is a singleton set.