Question

The set $\left( x:x\ne x \right)$ may be equal to
A. \left\\{ 0 \right\\}
B. \left\\{ 1 \right\\}
C. \left\\{ 3 \right\\}
D. \left\\{ {} \right\\}

Answer 1

We first try to define the difference between the null set and \left\\{ 0 \right\\}. Then we try to figure out the elements of the set $\left( x:x\ne x \right)$. We place the set in its respective option based on the number of elements it has and solves the problem.

Complete step-by-step solution:
First, we need to define the difference between two sets \left\\{ 0 \right\\} and \left\\{ {} \right\\}.
In the case of \left\\{ {} \right\\}, it means the null set. There is no element in the set and if A=\left\\{ {} \right\\} then $n\left( A \right)=0$. We also define this null set as $\phi$.
On the other hand we have \left\\{ 0 \right\\} which means this set has an element which is 0 and if B=\left\\{ 0 \right\\} then $n\left( B \right)=1$.
Now we check the given set $\left( x:x\ne x \right)$. It is irrelevant to the domain of x. Whatever be the value of x it will always be equal to itself. So, there exists no such x for which the set $\left( x:x\ne x \right)$ exists.
Number of elements in that set is 0. So, the set is null. The correct option is D.

Note: Having a second bracket always defines a set. But if we get such brackets inside another bracket then the first one defines the elements of the second one. For example: if we take A=\left\\{ 2,\left\\{ 5,6 \right\\},8 \right\\}. It means 5 and 6 are not elements of A rather \left\\{ 5,6 \right\\} is an element of A.