If R and S are relations on a set A, then prove that
(i) R and S are symmetric ⇒ R ∩ S and R ∪ S are symmetric
(ii) R is reflexive and S is any relation ⇒ R ∪ S is reflexive.
(i) R and S are symmetric relations on the set A.
$\Rightarrow R \subset A \times A$ and $S \subset A \times A$
$\Rightarrow R \cap S \subset A \times A$
Thus, $R \cap S$ is a relation on $A$.
Let $a, b \in A$ such that $(a, b) \in R \cap S$. Then,
$(a, b) \in R \cap S$
$\Rightarrow(a, b) \in R$ and $(a, b) \in S$
$\Rightarrow(b, a) \in R$ and $(b, a) \in S$ [Since $R$ and $S$ are symmetric]
$\Rightarrow(b, a) \in R \cap S$
Thus,
$(a, b) \in R \cap S$
$\Rightarrow(b, a) \in R \cap S$ for all $a, b \in A$
So, $R \cap S$ is symmetric on A.
Also,
Let $a, b \in A$ such that $(a, b) \in R \cup S$
$\Rightarrow(a, b) \in R$ or $(a, b) \in S$
$\Rightarrow(b, a) \in R$ or $(b, a) \in S$ [Since $R$ and $S$ are symmetric]
$\Rightarrow(b, a) \in R \cup S$
So, $R \cup S$ is symmetric on $A$.
(ii) R is reflexive and S is any relation.
Suppose $a \in A$. Then,
$(a, a) \in R$ [Since $R$ is reflexive]
$\Rightarrow(a, a) \in R \cup S$
$\Rightarrow R \cup S$ is reflexive on $A$
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