Union and Intersection of Relations
📺 Video Explanation
📝 Question
If \( R \) and \( S \) are relations on a set \( A \), prove:
(i) \( R \) and \( S \) are symmetric \( \iff \) \( R \cap S \) and \( R \cup S \) are symmetric
(ii) \( R \) is reflexive and \( S \) is any relation \( \iff \) \( R \cup S \) is reflexive
✅ Solution
🔹 Part (i): Symmetric
👉 Forward Direction
Assume \( R \) and \( S \) are symmetric.
For Intersection:
\[ (a,b) \in R \cap S \Rightarrow (a,b) \in R \text{ and } (a,b) \in S \]
\[ (b,a) \in R \text{ and } (b,a) \in S \]
\[ (b,a) \in R \cap S \]
✔ Hence, \( R \cap S \) is symmetric.
For Union:
\[ (a,b) \in R \cup S \Rightarrow (a,b) \in R \text{ or } (a,b) \in S \]
\[ (b,a) \in R \text{ or } (b,a) \in S \]
\[ (b,a) \in R \cup S \]
✔ Hence, \( R \cup S \) is symmetric.
👉 Converse Direction
Assume \( R \cap S \) and \( R \cup S \) are symmetric.
Since \( R \subseteq R \cup S \), if \( (a,b) \in R \), \[ (b,a) \in R \cup S \]
This ensures symmetry of \( R \). Similarly for \( S \).
✔ Hence, \( R \) and \( S \) are symmetric.
🔹 Part (ii): Reflexive
👉 Forward Direction
Assume \( R \) is reflexive.
\[ (a,a) \in R \quad \forall a \in A \]
Since \( R \subseteq R \cup S \), \[ (a,a) \in R \cup S \]
✔ Hence, \( R \cup S \) is reflexive.
👉 Converse Direction
Assume \( R \cup S \) is reflexive.
\[ (a,a) \in R \cup S \Rightarrow (a,a) \in R \text{ or } (a,a) \in S \]
Since \( S \) is arbitrary, for all \( a \), \[ (a,a) \in R \]
✔ Hence, \( R \) is reflexive.
🎯 Final Answer
✔ Symmetric property preserved under union and intersection
✔ Reflexivity preserved under union with a reflexive relation
🚀 Exam Insight
- Union → OR condition
- Intersection → AND condition
- Symmetry survives both
- Reflexivity depends on presence of all (a,a)