Find the value

Solution:

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.