Question

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

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

Answer 1

The correct option is A A singleton set

$X_n=\{z=x+iy:{\left|z\right|}^2\le \frac{1}{n}\}n\ge 1$

${\bigcap }_{n=1}^{\infty }{X}_n$

${\left|z\right|}^2\le \frac{1}{n}$

$-\frac{1}{n}\le z\le \frac{1}{n}$

$n\ge 1$

So $X_n$ is singleton set.