Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Let X be a non-empty set and let S be the collection of all subsets of X. Let R be a relation in S defined by
$R=\left\{ \left( A,B \right):A\subset B \right\}$
Show that R is transitive but is neither reflexive nor symmetric.

seo-qna
Last updated date: 16th Apr 2024
Total views: 411.6k
Views today: 7.11k
MVSAT 2024
Answer
VerifiedVerified
411.6k+ views
Hint: Recall the definitions of reflexive, symmetric and transitive relations. Using definitions prove that the given relation is transitive, but not symmetric and reflexive.

Complete step-by-step answer:
Reflexive relation: A relation R on a set “A” is said to be reflexive if $\forall a\in A$ we have $aRa$.
Symmetric relation: A relation R on a set “A” is said to be symmetric if $aRb\Rightarrow bRa$
Transitive relation: A relation R on a set “A” is said to be transitive if $aRb,bRc\Rightarrow aRc$.
Before solving the question, we need to understand the difference between a subset and a proper subset of a set.
Subset of a set: A set X is said to be a subset of a set Y if every element of X is also in Y. The set X is a proper subset if there exists at least one element in Y, which is not in X. If X is a subset of Y, it is denoted as $X\subseteq Y$. If X is a proper subset of Y it is denoted as $X\subset Y$.
Reflexivity: Since $\forall A\in S$we have $a\in A\Rightarrow a\in A$. Hence $A\subseteq A$. But since no element of one set is absent in another, we have $A\not\subset A$. Hence $\left( A,A \right)\notin R$ and hence the relation is not reflexive.
Symmetricity: Since if $A\subset B\Rightarrow \exists x\in B$ such that $x\notin A$, we have $B\not\subset A$ and hence if $\left( A,B \right)\in R\Rightarrow \left( B,A \right)\notin R$ and hence the relation is not symmetric.
Transitivity: We have if $\left( A,B \right)\in R,\left( B,C \right)\in R\Rightarrow A\subset B,B\subset C$
Hence, we have $\forall x\in A,x\in B$ and $\forall y\in B,y\in C$.
Hence, we have $\forall x\in A,x\in C$
Hence, we have $A\subseteq C$.
Also since A is a proper subset of B $\exists x\in B$ such that $x\notin A$
Hence, we have $\exists x\in C$ such that $x\notin A$
Hence, we have $A\subset C$
Hence A is related to C, and hence the relation is transitive.

Note: [1] Students usually make a mistake while proving reflexivity of a relation. In reflexivity, we need all the elements of a to be related with themselves, and even if a single element is found such that it is not related with itself, then the relation is not reflexive.