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Let $X_{n}=\left\{z=x+i y:|z|^{2} \leq \frac{1}{n}\right\}$ for all integers $n \geq 1 .$ Then, $\underset{n=1}{\cap} X_{n}$ is
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The correct answer is:
a singleton set
Given, $X_{n}=\left\{z=x+i y:|z|^{2} \leq \frac{1}{n}\right\}$
$=\left\{x^{2}+y^{2} \leq \frac{1}{n}\right\}$
$\therefore$
$X_{1}=\left\{x^{2}+y^{2} \leq 1\right\}$
$\begin{array}{c}
X_{2}=\left\{x^{2}+y^{2} \leq \frac{1}{2}\right\} \\
X_{3}=\left\{x^{2}+y^{2} \leq \frac{1}{3}\right\} \\
\ldots \\
\qquad \begin{aligned}
X_{\infty} &=\left\{x^{2}+y^2 \leq 0\right\} \\
\bigcap_{n=1}^{\infty} x_n &=X_{1} \cap X_{2} \cap X_{3} \cap \ldots \cap X_{\infty} \\
&=\left\{x^{2}+y^{2}=0\right\}
\end{aligned}
\end{array}$
Hence, $\underset{n=1}{\cap} X_{n}$ is a singleton set.
$=\left\{x^{2}+y^{2} \leq \frac{1}{n}\right\}$
$\therefore$
$X_{1}=\left\{x^{2}+y^{2} \leq 1\right\}$
$\begin{array}{c}
X_{2}=\left\{x^{2}+y^{2} \leq \frac{1}{2}\right\} \\
X_{3}=\left\{x^{2}+y^{2} \leq \frac{1}{3}\right\} \\
\ldots \\
\qquad \begin{aligned}
X_{\infty} &=\left\{x^{2}+y^2 \leq 0\right\} \\
\bigcap_{n=1}^{\infty} x_n &=X_{1} \cap X_{2} \cap X_{3} \cap \ldots \cap X_{\infty} \\
&=\left\{x^{2}+y^{2}=0\right\}
\end{aligned}
\end{array}$
Hence, $\underset{n=1}{\cap} X_{n}$ is a singleton set.
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