1. Tardigrade
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  3. Mathematics
  4. Let Xn= z=x+i y:|z|2 ≤ (1/n) for all integers n ≥ 1. Then, overset∞ undersetn=1∩ is

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Q. Let $X_{n}=\left\{z=x+i y:|z|^{2} \leq \frac{1}{n}\right\}$ for all integers $n \geq 1$. Then, $\overset{\infty}{\underset{n=1}{\cap}}$ is

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Solution:

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\} $
$\therefore X_{1}=\left\{x^{2}+y^{2} \leq 1\right\}$
$X_{2}=\left\{x^{2}+y^{2} \leq \frac{1}{2}\right\}$
$X_{3}=\left\{x^{2}+y^{2} \leq \frac{1}{3}\right\}$
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$X_{\infty}=\left\{x^{2}+y \leq 0\right\}$
$ \therefore \overset{\infty}{\underset{n=1}{\cap}} X_{n} =X_{1} \cap X_{2} \cap X_{3} \cap \ldots \cap X_{\infty} $
$ =\left\{x^{2}+y^{2}=0\right\} $
Hence, $\overset{\infty}{\underset{n=1}{\cap}} X_{n}$ is a singleton set.