Intersection of Equivalence Relations | If R is Equivalence Relation then R^(−𝟏) is also Equivalence

Summary

In this lecture, Dr. Anuj Kumar explores the concept of equivalence relations, characterized by reflexivity, symmetry, and transitivity. He discusses two important theorems related to these relations: the intersection and union of equivalence relations also form equivalence relations under certain conditions. The lecture delves into proofs for these theorems, using examples to illustrate how these properties hold true. The discussion also touches on the inverse of a relation and its properties, providing a comprehensive understanding of equivalence relations in set theory.

Takeaways

• 🎶 The speaker is Dr. Anuj Kumar, an Assistant Professor of Mathematics.
• 📚 The previous video discussed equivalence relations and important examples.
• 📖 An equivalence relation satisfies three properties: reflexive, symmetric, and transitive.
• 🔄 The first theorem discussed proves that the intersection of two equivalence relations is also an equivalence relation.
• 🤔 The second part of the video addresses whether the union of two equivalence relations can also be an equivalence relation, with examples provided.
• ✅ The reflexive property of the intersection is explained using common elements in both relations.
• ↔️ The symmetric property of the intersection is proven using pairs of elements belonging to both relations.
• 🔗 The transitive property is discussed by showing how pairs of elements in both relations maintain transitivity in their intersection.
• ⚖️ The video concludes that while the intersection of equivalence relations holds all three properties, the union does not always satisfy them.
• 🌀 A final example shows that the inverse of an equivalence relation retains all properties—reflexive, symmetric, and transitive.

Q & A

What is an equivalence relation?

An equivalence relation is a relation defined over a set that satisfies three properties: reflexivity, symmetry, and transitivity.

What does it mean for a relation to be reflexive?

A relation is reflexive if every element in the set is related to itself.

Can you explain the symmetry property of a relation?

A relation is symmetric if whenever an element 'a' is related to an element 'b', then 'b' is also related to 'a'.

How is the transitivity property defined?

A relation is transitive if whenever an element 'a' is related to an element 'b', and 'b' is related to an element 'c', then 'a' is also related to 'c'.

What is the significance of the theorem stating that the intersection of two equivalence relations is also an equivalence relation?

This theorem is significant as it assures that if you have two equivalence relations, the properties of equivalence are preserved in their intersection, which is also an equivalence relation.

What is the condition for the union of two relations to be an equivalence relation?

The union of two relations might not necessarily be an equivalence relation unless specific conditions are met that ensure it satisfies reflexivity, symmetry, and transitivity.

Can you provide an example to illustrate the intersection of two equivalence relations?

Sure, consider two equivalence relations R1 and R2 over a set. Their intersection, R1 ∩ R2, will include all elements that are related in both R1 and R2, and this intersection will also be an equivalence relation.

How do you prove that the intersection of two equivalence relations is reflexive?

To prove reflexivity for the intersection, take any element 'x' in the set. Since R1 and R2 are both reflexive, 'x' is related to 'x' in both R1 and R2. Therefore, 'x' is also related to 'x' in R1 ∩ R2.

What is the process to prove that the intersection of two equivalence relations is symmetric?

To prove symmetry, consider any elements 'x' and 'y' such that 'x' is related to 'y' in R1 ∩ R2. Since both R1 and R2 are symmetric, 'y' is related to 'x' in both R1 and R2, which means 'y' is related to 'x' in R1 ∩ R2 as well.

Explain how to demonstrate that the intersection of two equivalence relations is transitive.

To demonstrate transitivity, assume 'x' is related to 'y' and 'y' is related to 'z' in R1 ∩ R2. Since both R1 and R2 are transitive, 'x' is related to 'z' in both R1 and R2, implying 'x' is related to 'z' in R1 ∩ R2.

What does the script suggest about the union of equivalence relations?

The script suggests that the union of two equivalence relations might not always be an equivalence relation unless it satisfies all three properties of equivalence relations.

Can you provide an example from the script where the union of two relations does not form an equivalence relation?

Yes, the script provides an example where the union of a reflexive, symmetric relation (R) and its complement (S) does not hold the transitivity property, hence their union does not form an equivalence relation.