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Today we'll explore the union of two equivalence relations. Can anyone tell me what an equivalence relation is?
An equivalence relation is a relation that is reflexive, symmetric, and transitive.
Correct! Now, here's a key point: If you have two equivalence relations, their union is always reflexive and symmetric. But is it always transitive?
No, right? In some cases, it might not be transitive.
Exactly! For instance, let's consider the relations R1 and R2 over the set X = {a, b, c}...
We can observe through a counterexample that R1 ∪ R2 can fail to have the transitivity property.
So the union can create issues with transitivity but not with reflexivity or symmetry?
Indeed! Remembering the acronym 'RST' for Reflexive, Symmetric, but not necessarily Transitive can help you.
To sum up, while the union of two equivalence relations is always reflexive and symmetric, it may not be transitive.
Now, let's shift gears and discuss the intersection of two equivalence relations. How does this behavior differ from their union?
I think the intersection is also an equivalence relation, isn't it?
You're spot on! Let's prove this. Both R1 and R2 are reflexive, so what about R1 ∩ R2?
R1 ∩ R2 will definitely also be reflexive because each element will relate to itself in both R1 and R2.
Right! What about symmetry?
If (a, b) is in the intersection, then it must also be in both R1 and R2. And since those are symmetric, (b, a) must also be in both of them.
Correct! And how about transitivity?
If (a, b) and (b, c) are in the intersection, they must also be in both R1 and R2. So, (a, c) must also be there.
Exactly! Thus, R1 ∩ R2 is an equivalence relation. Remember the phrase 'PIE' - for Properties Include Equivalence.
Let's delve into the composition of two equivalence relations now. What do we need to show if we're asked whether R1 ∪ R2 equals R1 ◦ R2?
We need to show that the composition must also hold the same properties as the union, right?
Yes, and we demonstrate the sufficiency of transitivity. Let's provide a general proof by breaking it down into cases.
What are the various cases we consider for (a, b) and (b, c) in the union?
Great question! We need to check all configurations. If both pairs come from R1, they hold. If they come from R2 or they are mixed, we can show continuity.
So we are utilizing transitivity from the individual relations to validate for the union!
Exactly! And understanding this leads to deeper insights into connectivity in mathematical relations!
In summary, the conditions under which the composition and the union equate leads to notable implications regarding equivalence relations.
Demonstrates that the union is not guaranteed to be transitive.
Always results in an equivalence relation.
Links the understanding of the connection between unions and composite relations.
Example of equivalence relations R1 = {(a,a), (b,b), (c,c), (a,b), (b,a)} and R2 = {(a,a), (b,b), (c,c), (b,c), (c,b)} showing their union is not transitive.
Example illustrating that the intersection of R1 and R2 yields another equivalence relation, confirming reflexivity, symmetry, and transitivity.
Reflexive, Symmetric too, Transitive will get a view. In any union we may lose, Transitivity, that's the dues.
Once in a math land, two relations met. Their union was grand, but failed the transitive pet. They shook hands with an intersection, forming a trio's perfection!
Remember 'RST': Reflexive, Symmetric, but beware when Transitive does not see!
Use 'PIE' to recall: Properties Include Equivalence for intersections that stand tall.
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Properties of Equivalence Relations
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Union of Equivalence Relation
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Intersection of Equivalence Relation
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A relation that is reflexive, symmetric, and transitive.
A relation R on a set S is reflexive if every element is related to itself.
A relation R is symmetric if for all a, b in S, if aRb, then bRa.
A relation R is transitive if for all a, b, c in S, if aRb and bRc, then aRc.
The union of two sets contains all elements that are in either set.
The intersection of two sets contains all elements that are common to both sets.
The composition of two relations is defined when a relation of the first and a relation of the second act together.
In the first part of Tutorial 4, various properties of equivalence relations are explored, specifically focusing on union and intersection operations.
This section elaborates on the characteristics of two equivalence relations over a non-empty set, discussing the conditions under which the union is not necessarily an equivalence relation while the intersection is. It further delves into implications regarding compositions.
In this section, we delve into the properties of equivalence relations, specifically examining operations involving two equivalence relations over a non-empty set. The union of two equivalence relations is analyzed, showcasing a counterexample where the union fails to satisfy transitivity, even though it remains reflexive and symmetric. The interpretation here is that while both individual equivalence relations exhibit reflexivity and symmetry, their union does not guarantee transitivity.
In contrast, the intersection of the same two equivalence relations is proven to always remain an equivalence relation since it continues to satisfy all three fundamental properties: reflexivity, symmetry, and transitivity. The implications of the union and intersection of equivalence relations lead us into a conjoint understanding of the conditions under which one may maintain equivalence in relation to compositions, guided by the equality of the composite and the union of equivalence relations.