Question

Let S be any set and P (S) be its power set, We define a relation R on P(S) by ARB to mean $A\subseteq B;\forall A,B \in P(S)$. Then R is
A. Equivalence relation
B. Not an equivalence but partial order relation
C. Both equivalence and partial order relation
D. None of these

Answer 1

Hint: If and only if a relation R on a set A is reflexive, symmetric, and transitive, then it qualifies as an equivalence relation. So check the relation for each case. If any condition holds then the relation is not an equivalence relation.



Formula used:R is said to be reflective, if $(x,x) \in R$, for every $x \in set A$
R is said to be symmetric, if $(x,y) \in R$, then $(y, x) \in R$
R is said to be transitive, if $(x, y) \in R$ and $(y,z)\in R$, then $(x, z) \in R$

Complete step by step solution: S be any set and P (S) be its power set, such that R on P(S) by ARB to mean $A\subseteq B;\forall A,B \in P(S)$.
The features of relation R that we notice are as follows:
(i) Reflexive Relation:
$A \subseteq A$
ie, $ARA, \forall A \in P(S)$
Hence, R is reflexive.
(ii) Symmetric Relation:
$A \subseteq B$
$B \subseteq A$
Therefore ARB BRA.
Hence, R is not symmetric.
Since ARB and BRA
$\Rightarrow A \subseteq B and B \subseteq A$
$\Rightarrow A=B $
Therefore, R is anti-symmetric.
(iii)Transitive Relation:
ARB and BRC
$\Rightarrow A \subseteq B and B \subseteq C$
$\Rightarrow A \subseteq C$
$\Rightarrow ARC$
therefore R is a transitive relation.
Hence, R is a partially ordered relation but not an equivalence relation.


Thus, Option (B) is correct.

Note: The collection of all subsets of a set S is known as the power set, usually P(S). You must demonstrate transitivity, symmetry, and reflexivity to establish an equivalence relationship.