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

Answer: (A)

A singleton set

Answer 2

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.