Question

Let {X_n} = \left\\{ {z = x + iy:{{\left| z \right|}^2} \leqslant \dfrac{1}{n}} \right\\} for all integers $n \geqslant 1$. Then, $\bigcap\limits_{n = 1}^\infty {{X_n}}$ is:
A. A singleton set
B. Not a finite set
C. An empty set
D. A finite set with more than one element.

Answer 1

Here, in the question, we are given a set of complex numbers with a condition. And we are asked to find the intersection of${X_n}$, if the value of $n$ is taken from one to infinity. We will first simplify the given set and then put the values on $n$ one by one to get the corresponding values. And then at the end, we will find the intersection of all the values to get the desired answer.

Complete step by step solution:
Given: {X_n} = \left\\{ {z = x + iy:{{\left| z \right|}^2} \leqslant \dfrac{1}{n}} \right\\}
We already know that, the modulus of $z$, $\left| z \right|$ is given by $\sqrt {{x^2} + {y^2}}$, where $z = x + iy$.
Therefore, ${\left| z \right|^2} = {x^2} + {y^2}$
So, we can write ${X_n}$ as:
{X_n} = \left\\{ {{x^2} + {y^2} \leqslant \dfrac{1}{n}} \right\\}
If we put $n = 1$, we get,
{X_1} = \left\\{ {{x^2} + {y^2} \leqslant 1} \right\\}
If we put $n = 2$, we get,
{X_2} = \left\\{ {{x^2} + {y^2} \leqslant \dfrac{1}{2}} \right\\}
If we put $n = 3$, we get,
{X_3} = \left\\{ {{x^2} + {y^2} \leqslant \dfrac{1}{3}} \right\\}
And this goes on……..
If we put $n = \infty$, we get,
{X_\infty } = \left\\{ {{x^2} + {y^2} \leqslant 0} \right\\}
$\therefore \bigcap\limits_{n = 1}^\infty {{X_n}} = {X_1}\bigcap {{X_2}} \bigcap {{X_3}\bigcap . } \ldots \ldots \bigcap {{X_\infty }}$
= \left\\{ {{x^2} + {y^2} = 0} \right\\}
Set of $\bigcap\limits_{n = 1}^\infty {{X_n}}$ contains only one element i.e. $0$ and hence, it is a singleton set.

So, the correct answer is “Option A”.

Note: If a set $A$ has only one element, then set $A$ is called a singleton set. Like \left\\{ a \right\\} is a singleton set.
A set $A$ is not said to be a finite set if it does not contain a finite number of elements in it.
A set is called an Empty set if it does not contain any element. It is denoted by \left\\{ {} \right\\} or $\phi$.
A finite set having more than one element means a set which has a definite number of elements but at-least two elements such as $A =${North, West, South, East}.