Question 1: Equivalence Relations - Part A
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Understanding Equivalence Relations
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Today, we'll discuss equivalence relations. Can anyone tell me what three properties an equivalence relation must satisfy?
Is it reflexivity, symmetry, and transitivity?
Exactly! We often refer to these properties as the 'RST' properties - remember that acronym? R for Reflexive, S for Symmetric, and T for Transitive. Now, can anyone give me an example of a reflexive relation?
An example could be 'a is related to a' for any element.
Great! Now, how does symmetry come into play?
If a is related to b, then b should also relate back to a.
Exactly! Now, let's move to transitivity. If a is related to b and b is related to c, what must hold?
Then a must be related to c.
Perfect! So, we have our definitions well understood. Proceeding further, let's look at unions of equivalence relations.
Union of Equivalence Relations
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When we take the union of two equivalence relations, do we always get another equivalence relation? Let's explore.
Well, if both relations are reflexive, wouldn't their union also be reflexive?
Exactly! The union maintains reflexivity. Now, what about symmetry?
It should still be symmetric because if we have (a, b), we'll also have (b, a) in one of the two relations.
You're all spot on! But what about transitivity? Can anyone explain?
Just because we have (a, b) and (b, c) might not mean we have (a, c) in the union.
Right! This is why the union might fail to be an equivalence relation. A counterexample would be insightful here.
Let’s say R1 contains (a, b) and R2 contains (b, c). Without (a, c), it fails transitivity!
Exactly, excellent observation! So, unions retain reflexivity and symmetry but not necessarily transitivity.
Intersection of Equivalence Relations
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Now, let’s switch gears and discuss intersections of equivalence relations. How do you think their properties hold up?
It sounds like they should still be equivalence relations.
Precisely! Let’s show this step-by-step. First, how do we prove reflexivity for the intersection?
If both R1 and R2 are reflexive, then (a, a) will be in both.
Exactly! Now, what about symmetry?
If (a, b) is in the intersection, it means it’s in both relations, so (b, a) is also there.
Spot on! Now for transitivity, how do we prove that?
If we have (a, b) and (b, c) in the intersection, both pairs must come from R1 and R2, so (a, c) holds.
Correct! Thus, the intersection indeed is an equivalence relation. Good work, everyone!
Understanding Examples
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Let’s solidify your understanding by going through examples. Can anyone suggest a pair of equivalence relations over {1, 2, 3}?
How about R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} and R2 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}?
Great choice! What’s their union?
That would be {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}.
Is this union transitive?
No, it doesn’t include (1, 3) despite having (1, 2) and (2, 3).
Exactly! Now, let's check their intersection. What do we find?
The intersection is {(1, 1), (2, 2), (3, 3)} which is also reflexive, symmetric, and transitive.
Well done! This reinforces our findings about unions and intersections of equivalence relations.
Introduction & Overview
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Quick Overview
Standard
In this section, we analyze two equivalence relations defined on a set and determine whether their union forms an equivalence relation, establishing that the union is reflexive and symmetric but not necessarily transitive. The section also demonstrates that the intersection of two equivalence relations is always an equivalence relation, proving reflexivity, symmetry, and transitivity.
Detailed
Detailed Summary
In this section of the tutorial, we investigate equivalence relations on a set X. An equivalence relation must satisfy three properties: reflexivity, symmetry, and transitivity. We demonstrate this by examining two equivalence relations, R1 and R2, defined on a set X = {a, b, c}.
Union of Equivalence Relations
We first consider the union of two equivalence relations, R1 ∪ R2. While we show that this union is always reflexive (since it retains the reflexive pairs from both relations) and symmetric (as pairs in either relation will provide their symmetric pairs), it fails to be transitive in general. We illustrate this through a counterexample:
- Let R1 = {(a, a), (b, b), (c, c), (a, b), (b, a)} and R2 = {(a, a), (b, b), (c, c), (b, c), (c, b)}.
- The union R1 ∪ R2 includes the pairs of both relations but lacks the transitive pair (a, c).
Therefore, the union of two equivalence relations may not always yield an equivalence relation, especially since it does not guarantee transitivity.
Intersection of Equivalence Relations
In contrast, we prove that the intersection of two equivalence relations, R1 ∩ R2, on the same set X is indeed an equivalence relation. Each of the properties is verified:
- Reflexivity: Both relations retain reflexive pairs, ensuring they are found in the intersection.
- Symmetry: Through taking a pair present in the intersection and applying the symmetry property from both relations, we see they also occur in the intersection.
- Transitivity: By choosing pairs from the intersection, we can confirm that they lead to another pair also found in the intersection, confirming transitivity holds.
In summary, this section deeply delves into the properties of unions and intersections of equivalence relations and their implications, illustrating both successful and unsuccessful outcomes in maintaining the equivalence relation properties.
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Introduction to Equivalence Relations
Chapter 1 of 5
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This question you are given two equivalence relations on a non-empty set X. You are asked to prove or disprove whether R1 ∪ R2 is an equivalence relation or not? So, it turns out at a union of two equivalence relations need not be an equivalence relation.
Detailed Explanation
In this section, we are introduced to the idea of equivalence relations and their union. An equivalence relation is a relation that is reflexive, symmetric, and transitive. In this specific problem, we are examining two equivalence relations, denoted as R1 and R2, defined over a set X. The task is to determine whether the union of these two relations, written as R1 ∪ R2, also forms an equivalence relation. Interestingly, it has been concluded that this is not necessarily the case, meaning the union may lack one or more properties of equivalence relations.
Examples & Analogies
Think of equivalence relations like a group of friends who are connected by a unique set of common interests (like hobbies). If Friend A is connected to Friend B (R1) and Friend B is connected to Friend C (R2), you would expect them to all know each other (R1 ∪ R2). However, if Friend A doesn't share any interests with Friend C directly, their connection might not exist in the combined interests, just like how R1 ∪ R2 might not be an equivalence relation.
Counterexample of Non-transitivity
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And this is demonstrated by this counter example. So, I consider my set X={a,b,c}. And let me have equivalence relations R1 and R2. So, R1 = {(a,a),(b,b),(c,c),(a,b),(b,a)} and R2 = {(a,a),(b,b),(c,c),(b,c),(c,b)}.
Detailed Explanation
Here, a counterexample is introduced using a specific set X, which contains the elements {a, b, c}. Two equivalence relations, R1 and R2, are specified. R1 consists of the pairs emphasizing reflexivity and symmetry (a is related to b, and b is related to a). Similarly, R2 also captures these properties. This specific construction allows us to explore the union of R1 and R2 to see if it retains the properties of equivalence relations.
Examples & Analogies
Imagine you have three friends: Alice (a), Bob (b), and Charlie (c). If Alice thinks of Bob as a friend (R1), sensing a connection, and Bob thinks of Charlie the same way (R2), you might think Alice and Charlie are connected. However, if Alice doesn’t know Charlie directly, they don’t share a friendship (R1 ∪ R2), illustrating that knowing one friend does not always mean knowing their friends.
Proving Reflexivity and Symmetry
Chapter 3 of 5
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Now, if you take the union of these two relations, R1 ∪ R2, you will have these ordered pairs. And it is easy to verify that R1 ∪ R2 is a reflexive relation over the set X, because you have {(a,a),(b,b),(c,c)} present in the relation. The union is also symmetric because if you have any (a,b) in the union present and ordered pair (b,a) is also present in the union.
Detailed Explanation
In this chunk, after forming the union of relations R1 and R2, we can easily examine the properties of reflexivity and symmetry. Reflexivity is confirmed as all elements {a, b, c} have themselves as pairs in the relation, meaning every element relates to itself. The symmetry condition is also satisfied because for every connection (a, b), there’s a corresponding (b, a) present, maintaining the symmetric property expected from equivalence relations.
Examples & Analogies
Think of it like a mutual friendship network. If Alice is friends with Bob, and Bob is friends with Alice, their relationship is both mutual and connective. This reflects reflexivity because everyone believes they are at least their own friend, and symmetry shows that if one person is friends with another, the reverse is also true.
Failure of Transitivity
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But it is easy to see that union here is not a transitive relation, specifically you have (a,b) and an ordered pair (b,c) present in R1 ∪ R2, but (a,c) is not present in the union and hence the transitivity properties violated.
Detailed Explanation
In examining the union of R1 and R2, we find that while reflexivity and symmetry are satisfied, transitivity fails. If we have (a, b) from R1 and (b, c) from R2, we would expect (a, c) to also be included for transitivity to hold. However, since (a, c) is missing from the union, we see that the union does not maintain the transitive property required for an equivalence relation.
Examples & Analogies
Imagine organized friendships again: if Alice likes Bob and Bob likes Charlie, one might assume Alice also likes Charlie. However, if Alice doesn’t know Charlie at all, this assumption crumbles, mirroring how missing connections ruin the logical flow—like missing links in a chain.
Conclusion on Union of Equivalence Relations
Chapter 5 of 5
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So, in general the union of two equivalence relations need not be transitive. We can prove that they will be always reflexive and symmetric. If I take the union it will be always reflexive and symmetric, but the union need not be a transitive relation.
Detailed Explanation
To conclude this section, it’s highlighted that while unions of equivalence relations will consistently exhibit reflexivity and symmetry, they do not guarantee transitivity. Therefore, while examining unions of equivalence relations, one must be prepared to check these properties individually to establish if the resulting relation is indeed an equivalence relation.
Examples & Analogies
This is like asserting that while many friendships are either beneficial (reflexive) or mutually acknowledged (symmetric), a lack of common connections can make certain relationships weak (failing transitivity), underscoring the need for personal connections to truly form a united front.
Key Concepts
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Equivalence Relation: A relation that satisfies reflexivity, symmetry, and transitivity.
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Reflexivity: Each element is related to itself.
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Symmetry: If a is related to b, b is related to a.
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Transitivity: If a is related to b and b is related to c, then a must be related to c.
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Union of Relations: All pairs from either relation.
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Intersection of Relations: Only the pairs that are in both relations.
Examples & Applications
Example of Equivalence Relation: R = {(a, a), (b, b), (a, b), (b, a)} is an equivalence relation as it holds reflexivity, symmetry, and transitivity.
Counterexample for Union: For R1 = {(1, 1), (2, 2)} and R2 = {(1, 1), (3, 3)}, their union lacks the transitive pair (1, 3).
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Rhymes
Three properties in the mix, RST is the fix, reflexive for the self, symmetric for the rest, and transitive's the test.
Stories
Imagine a club where each member relates to themselves (reflexivity), greets others in return (symmetry), and if A likes B and B likes C, then A must like C (transitivity).
Memory Tools
Remember RST for Equivalence: R for Reflexivity, S for Symmetry, T for Transitivity.
Acronyms
RST - Reflexive, Symmetric, Transitive.
Flash Cards
Glossary
- Equivalence Relation
A relation that is reflexive, symmetric, and transitive.
- Reflexivity
A condition where every element is related to itself.
- Symmetry
A condition where if one element is related to another, then the second is related to the first.
- Transitivity
A condition where if one element is related to a second, and the second to a third, then the first is related to the third.
- Union
The set containing all pairs that are in either of the relations.
- Intersection
The set containing all pairs that are present in both relations.
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