Question

Let $X_n=\left\{z=x+iy:|z{|}^2\le \frac{1}{n}\right\}$ for all integers $n\ge 1$. Then, $\stackrel{\infty }{\underset{n=1}{\cap }}$ 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)

Given, $X_n=\left\{z=x+iY:|z{|}^2\le \frac{1}{n}\right\}$
$=\left\{x^2+y^2\le \frac{1}{n}\right\}$
$\therefore X_1=\left\{x^2+y^2\le 1\right\}$
$X_2=\left\{x^2+y^2\le \frac{1}{2}\right\}$
$X_3=\left\{x^2+y^2\le \frac{1}{3}\right\}$

$X_{\infty }=\left\{x^2+y\le 0\right\}$
$\therefore \stackrel{\infty }{\underset{n=1}{\cap }}{X}_n=X_1\cap X_2\cap X_3\cap \dots \cap X_{\infty }$
$=\left\{x^2+y^2=0\right\}$
Hence, $\stackrel{\infty }{\underset{n=1}{\cap }}{X}_n$ is a singleton set.