Let X_{n} = {z = x + iy : |z|^{2} leq frac{1} | vedclass

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(A) Given, $X_{n} = \{z = x + iy : |z|^{2} \leq \frac{1}{n}\}$.This represents a disk centered at the origin with radius $\frac{1}{\sqrt{n}}$.For $n =

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a singleton set

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(A) Given, $X_{n} = \{z = x + iy : |z|^{2} \leq \frac{1}{n}\}$.This represents a disk centered at the origin with radius $\frac{1}{\sqrt{n}}$.For $n = 1$, $X_{1} = \{x^{2} + y^{2} \leq 1\}$.For $n = 2$, $X_{2} = \{x^{2} + y^{2} \leq \frac{1}{2}\}$.As $n 	o \infty$, the radius $\frac{1}{\sqrt{n}} 	o 0$.The intersection of all such sets $X_{n}$ for $n \geq 1$ is the set of points that satisfy $x^{2} + y^{2} \leq \frac{1}{n}$ for all $n \geq 1$.This implies $x^{2} + y^{2} \leq 0$.Since $x^{2} + y^{2}$ cannot be negative, the only solution is $x^{2} + y^{2} = 0$, which means $x = 0$ and $y = 0$.Thus, $\bigcap_{n=1}^{\infty} X_{n} = \{0 + 0i\} = \{0\}$.Therefore, the intersection is a singleton set.

Let $X_{n} = \{z = x + iy : |z|^{2} \leq \frac{1}{n}\}$ for all integers $n \geq 1$. Then, $\bigcap_{n=1}^{\infty} X_{n}$ is

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