Let $S$ be the set of all sets and let $R=\{(A, B): A \subset B)\}, i . e ., A$ is a proper subset of $B$. Show that $R$ is
(i) transitive
(ii) not reflexive
(iii) not symmetric.
Let $R=\{(A, B): A \subset B)\}$, i.e., $A$ is a proper subset of $B$, be a relation defined on $S .$
Now,
Any set is a subset of itself, but not a proper subset.
$\Rightarrow(A, A) \notin R \forall A \in S$
$\Rightarrow \mathrm{R}$ is not reflexive.
Let $(A, B) \in R \forall A, B \in S$
$\Rightarrow A$ is a proper subset of $B$
$\Rightarrow$ all elements of $A$ are in $B$, but $B$ contains at least one element that is not in $A$.
$\Rightarrow B$ cannot be a proper subset of $A$
$\Rightarrow(B, A) \notin R$
For e.g., if $B=\{1,2,5\}$ then $A=\{1,5\}$ is a proper subset of $B$. we observe that $B$ is not a proper subset of $A$.
$\Rightarrow \mathrm{R}$ is not symmetric
Let $(A, B) \in R$ and $(B, C) \in R \forall A, B, C \in S$
$\Rightarrow A$ is a proper subset of $B$ and $B$ is a proper subset of $C$
$\Rightarrow \mathrm{A}$ is a proper subset of $\mathrm{C}$
$\Rightarrow(A, C) \in R$
For e.g. , if $B=\{1,2,5\}$ then $A=\{1,5\}$ is a proper subset of $B$.
And if $C=\{1,2,5,7\}$ then $B=\{1,2,5\}$ is a proper subset of $C$.
We observe that $A=\{1,5\}$ is a proper subset of $C$ also.
$\Rightarrow \mathrm{R}$ is transitive.
Thus, $R$ is transitive but not reflexive and not symmetric.
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