Question

Let $X_n=\{z=x+iy:{\left|z\right|}^2\le \frac{1}{n}\}$ for all integers $n\ge 1$. Then, $\bigcap _{n=1}^{\infty }{X}_n$ is

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

Answer 1

Answer: (A)

A singleton set

Given, $X_n=\{z=x+iy:{\left|z\right|}^2\le \frac{1}{n}\}$
$=\{x^2+y^2\le \frac{1}{n}\}$
$\therefore X_1=\{x^2+y^2\le 1\}$
$X_2=\{x^2+y^2\le \frac{1}{2}\}$
$X_3=\{x^2+y^2\le \frac{1}{3}\}$
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$X_{\infty }=\{x^2+y^2\le 0\}$
$\therefore {\bigcap }_{n=1}^{\infty }{X}_n=X_1\cap X_2\cap X_3\cap \cdots \cap X_{\infty }$
$=\{x^2+y^2=0\}$
Hence, ${\bigcap }_{n=1}^{\infty }{X}_n$ is a singleton set.