Question

If $X_n=\left\{z=x+iy:|z{|}^2\le \left(\frac{1}{n}\right)\right\}$ for all integers $n\ge 1$. Then $\underset{n=1}{\overset{\infty }{\cap }}{X}_n$ 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

The correct option is A

A singleton set

Explanation for the correct option.

$\begin{array}{rcl}{z}^2& =& x^2+i^2{y}^2\\ \Rightarrow z^2& =& x^2+y^2\\ & \Rightarrow & x^2+y^2\le \frac{1}{n}\end{array}$

So, for $n=\infty$, $x^2+y^2\le 0$

This will be true only for $x=0,y=0$.

So, $\underset{n=1}{\overset{\infty }{\cap }}{X}_n$ is a singleton set

Hence, option A is correct.