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A short class with Sarah and her students. The other ways are under More.
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Let's discuss the union of two equivalence relations. If we have two equivalence relations, R1 and R2, on a set X, what can we say about their union?
I think both of them should remain reflexive, right?
Exactly! The union will always be reflexive if both R1 and R2 are reflexive. Can anyone tell me if the union is symmetric?
It should also be symmetric because if (a, b) is in R1 ∪ R2, then (b, a) must also be there.
Great! So union is reflexive and symmetric. Now, what about transitivity? Is R1 ∪ R2 guaranteed to be transitive?
I don't think so. I mean, we could have cases where (a, b) is in R1 and (b, c) is in R2, but (a, c) isn't in the union.
Right! It is essential to examine that. In fact, we can create a counterexample to see this clearly.
Could you show us a specific example?
Of course! Let's consider two equivalence relations defined on the set X = {a, b, c}. What if R1 = {(a, a), (b, b), (c, c), (a, b), (b, a)} and R2 = {(a, a), (b, b), (c, c), (b, c), (c, b)}. Can anyone calculate R1 ∪ R2?
That gives us {(a, a), (b, b), (c, c), (a, b), (b, a), (b, c), (c, b)}.
Correct! Now can someone check its transitivity by providing any two pairs that are there?
We have (a, b) and (b, c), but then (a, c) is missing.
Exactly! This proves the union is not always transitive. We conclude that the union of two equivalence relations is not necessarily an equivalence relation.
To summarize, while unions are reflexive and symmetric, they may not always be transitive.
Now, let’s look at the intersection of two equivalence relations, R1 & R2. Who can tell me if the intersection is also an equivalence relation?
I guess it should be an equivalence relation, but we need to prove it.
Correct! Let's start by proving that it's reflexive. If we know both R1 and R2 are reflexive, what follows?
That means both (a, a) pairs will be in R1 and R2, so they will definitely be in the intersection.
Great, so we established that R1 ∩ R2 is reflexive. What about symmetry?
If (a, b) is in R1 ∩ R2, then (a, b) must be in both R1 and R2, meaning (b, a) must also exist in both.
Exactly right! Now, what about transitivity?
If we have (a, b) and (b, c) in the intersection, since they are in both relations, we can conclude (a, c) must also be there!
Perfect! Since reflexivity, symmetry, and transitivity hold for the intersection, we conclude that it is also an equivalence relation.
To summarize, unlike unions, the intersection of two equivalence relations is always an equivalence relation.
Let’s move on to a significant point: the union of two equivalence relations can be an equivalence relation under certain conditions. Can anyone suggest what those conditions might be?
Is it if R1 composed with R2 is equal to their union?
Correct! The statement says that R1 ∪ R2 is an equivalence relation if and only if R1 ∘ R2 equals R1 ∪ R2. Who can help me understand the implications here?
We need to show both implications separately: If R1 ∘ R2 = R1 ∪ R2, then R1 ∪ R2 is an equivalence relation, and vice versa!
Exactly! Let’s first assume that R1 ∘ R2 = R1 ∪ R2. So, what do we focus on proving?
We only need to check if the union is transitive since we know it’s already reflexive and symmetric.
Right on target! Now, we can check cases based on the combinations of the pairs belonging to R1 and R2. Can someone outline these cases?
Case 1 includes both pairs in R1 and Case 2 includes both in R2. What do we know about R1?
There, transitivity holds due to R1 being an equivalence relation!
Excellent! Then let’s consider doing something similar for the other cases. What's the conclusion we can have for the direction where we assume union is an equivalence relation?
We can solidly conclude R1 ∘ R2 = R1 ∪ R2!
Exactly! We've discussed the key properties and implications surrounding the condition under which the union is an equivalence relation today.
The union of two equivalence relations is always reflexive and symmetric but not necessarily transitive.
The intersection of two equivalence relations is always an equivalence relation.
The union of two equivalence relations is itself an equivalence relation if and only if their composition is equal to their union.
Example of an equivalence relation: R defined on {1, 2, 3} where R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} is reflexive, symmetric, but to confirm transitivity, we check pairs.
Consider two relations over {a, b, c}: R1 = {(a, a), (b, b), (c, c), (a, b), (b, a)} and R2 = {(a, a), (b, b), (c, c), (b, c), (c, b)}. Their union R1 ∪ R2 would be reflective and symmetric but not transitive.
Reflexive means you return, symmetry and transpose your turn. Transitivity links A to C, without it, an equivalence won't be!
Imagine two friends holding hands — they both must return to each other (reflexivity), they can swap partners (symmetry), and if one helps the next, they should be linked too (transitivity).
Remember 'RST' for 'Reflexive, Symmetric, Transitive' — essential for equivalence relations!
Use 'RST' to recall the key properties of equivalence relations: Reflexive, Symmetric, Transitive.
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What are the properties of an equivalence relation?
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What is the intersection of two relations?
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The union of two equivalence relations is always...
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A relation that is reflexive, symmetric, and transitive.
The set of all ordered pairs that are in at least one of the relations.
The set of all ordered pairs that are in both relations.
A relation R on a set X is reflexive if for every element a in X, (a, a) is in R.
A relation R on a set X is symmetric if for all a, b in X, if (a, b) is in R, then (b, a) is also in R.
A relation R on a set X is transitive if for all a, b, c in X, if (a, b) is in R and (b, c) is in R, then (a, c) is also in R.
This section discusses equivalence relations, particularly focusing on unions and intersections, providing proofs, counterexamples, and exploring their properties.
This section examines equivalence relations on sets, detailing the properties of unions and intersections. While the union of two equivalence relations is not necessarily an equivalence relation, the intersection is always an equivalence relation. The discussion is supported by proofs and counterexamples, as well as a focus on the implications of these properties.
In this section, we explore equivalence relations defined on a non-empty set X, discussing their critical properties, particularly focusing on unions and intersections. An equivalence relation is classified by three properties: reflexivity, symmetry, and transitivity.
The section begins with a discussion on the union of two equivalence relations, denoted as R1 and R2. It is established that while both R1 and R2 are reflexive and symmetric, their union, R1 ∪ R2, does not necessarily maintain transitivity. A counterexample using a set X = {a, b, c} illustrates this point. Here, we see that although R1 and R2 are both reflexive and symmetric, R1 ∪ R2 fails to be transitive, demonstrating that the union of two equivalence relations does not always yield another equivalence relation.
Next, we explore the intersection of two equivalence relations, R1 ∩ R2. This operation consistently produces an equivalence relation since reflexivity, symmetry, and transitivity are preserved. The reasoning provided involves verifying these properties through proof methods.
Furthermore, the section addresses a condition for the union of two equivalence relations to be an equivalence relation itself: specifically stating that R1 ∪ R2 is an equivalence relation if and only if R1 ∘ R2 = R1 ∪ R2, with detailed proofs in both directions.
The exploration of equivalence relations concludes with the application of combinatorial properties and further implications within set theory, providing a foundation for understanding discrete structures in mathematics.