Given a non-empty set X, consider P(X) which is a set of all subsets of X. Define the relation R in P(X) as follows-
For subsets A, B in P(X), ARB if and only if $\mathrm A\subset\mathrm B$. Is R an equivalence relation on P(X)? Justify your answer.

Hint: We will solve this problem by using an example. We will form the set of subsets on that and form a relation and check if it is an equivalence relation or not. To prove a relation to be equivalent, it should be reflexive, symmetric and transitive.

Complete step-by-step answer:
Let X = {1, 2, 3} be a set of first three natural numbers.
The set of the subsets P(X) will contain all the subsets of X, including the empty set and the set X itself as-
$\mathrm P\left(\mathrm X\right)=\left\{\varnothing,\;1,\;2,\;3,\;\left\{1,\;2\right\},\;\left\{2,\;3\right\},\;\left\{1,\;3\right\},\;\left\{1,\;2,\;3\right\}\right\}$
Now, ARB is a relation from A to B such that A is a subset of B.
We will now check if the relation is reflexive, symmetric and transitive.
For A to be reflexive A should be an subset of A, which is true because every set is a subset of itself.
$\mathrm A\subset\mathrm A,\;\left(\mathrm A,\mathrm A\right)\in\mathrm R$
{1} is a subset of {1}.
For A to be symmetric, if A is a subset of B then B should also be a subset of A.
$\mathrm{If}\;\left(\mathrm A,\mathrm B\right)\in\mathrm R\;\mathrm{then}\;\left(\mathrm B,\mathrm A\right)\in\mathrm R$
But we know that-
$\mathrm A\subset\mathrm B\;\mathrm{but}\;\mathrm B\subset\mathrm A\;\mathrm{is}\;\mathrm{not}\;\mathrm{true}$
{1} is a subset of {1, 2, 3} but converse is not true.
So, R is not symmetric.
But for R to be an equivalence relation, it should be reflexive, symmetric and transitive. It is not symmetric so it cannot be equivalent.

Note: In such types of questions, one should have knowledge about set theory and know the meaning of each symbol. Also, it is not necessary to check all three conditions because if one of them is false then there is no need to check the others. One example of an equivalence relation is-
R = {(1, 1), (1, 2), (2, 1), (2, 2)} on the set A = {1, 2}
Here, (1, 1) and (2, 2) exist so R is reflexive.
Also, (1, 2) and (2, 1) both exist so R is symmetric.
(1, 2) and (2, 1) exist which implies that (1, 1) should also exist, which is true. Hence, R is transitive as well.