Let S be the set of all sets and let R = {(A, B) : A ⊂ B)}, i.e., A is a proper subset of B. Show that R is (i)transitive (ii) not reflexive

Home » Let S be the set of all sets and let R = {(A, B) : A ⊂ B)}, i.e., A is a proper subset of B. Show that R is (i)transitive (ii) not reflexive

Let S be the set of all sets and let R = {(A, B) : A ⊂ B)}, i.e., A is a proper subset of B. Show that R is
(i)transitive
(ii) not reflexive

Let S be the set of all sets and let R = {(A, B) : A ⊂ B)}, i.e., A is a proper subset of B. Show that R is
(i)transitive
(ii) not reflexive

Solution:

(i) Suppose $A, B, C \in S$ such that $(A, B) \in R$ and $(B, C) \in R$
$\Rightarrow \mathrm{A}$ is a proper subset of $\mathrm{B}$ and $\mathrm{B}$ is a proper subset of $\mathrm{C}$
Therefore, A is a proper subset of $\mathrm{C}$
$\Rightarrow(\mathrm{A}, \mathrm{C}) \in \mathrm{R}$
As a result, $\mathrm{R}$ is transitive.
(ii) Suppose $\mathrm{R}=\{(\mathrm{A}, \mathrm{B}): \mathrm{A} \subset \mathrm{B})\}$ , i.e., $\mathrm{A}$ is a proper subset of $\mathrm{B}$ So now,
Any set is a subset of itself, but not a proper subset. $\Rightarrow(\mathrm{A}, \mathrm{A}) \notin \mathrm{R}$
As a result, $R$ is not reflexive.