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Magazine Chapter 10: Problem 13
Let \(A=\\{a, b\\}\). Describe all partial order relations on \(A\).
Short Answer
Expert verified
The partial order relations on the set \(A = \{a, b\}\) are as follows:
1. \(R_1 = \{(a, a), (b, b)\}\)
2. \(R_2 = \{(a, a), (a, b), (b, b)\}\)
3. \(R_3 = \{(a, a), (b, a), (b, b)\}\)
Step by step solution
01
Define the set and relations
We are given the set A containing two elements, 'a' and 'b'. Mathematically, this set can be represented as \(A = \{a, b\}\). We need to define the partial order relations on this set, which can be written as \(R \subseteq A \times A\).
02
Determine the possibilities of relations
We need to find all potential relations that can be considered as a partial order. First, recall that a partial order relation on a set must satisfy three properties: reflexivity, antisymmetry, and transitivity.
- Reflexivity means that for every element 'x' in A, the relation also contains (x, x), i.e., (a, a) and (b, b) must be present in a reflexive partial order.
- Antisymmetry means that if (a, b) is in the relation and (b, a) is in relation as well, then a = b.
- Transitivity means that if (a, b) and (b, c) are in the relation, then (a, c) should also be in the relation.
03
List the possible relations
We will list all the possible relations on A that contain (a, a) and (b, b), as these elements are necessary for reflexivity. We will then check which of these relations satisfy all the properties of a partial order.
1. \(R_1 = \{(a, a), (b, b)\}\)
2. \(R_2 = \{(a, a), (a, b), (b, b)\}\)
3. \(R_3 = \{(a, a), (b, a), (b, b)\}\)
4. \(R_4 = \{(a, a), (a, b), (b, a), (b, b)\}\)
04
Analyze the relations for antisymmetry and transitivity
Now that we have all the possible relations, we must check if they satisfy antisymmetry and transitivity.
1. \(R_1\) is reflexive by definition. It does not have any (a, b) and (b, a) elements in it, so it's trivially antisymmetric. There is no need to check transitivity since there are no different elements related. This means R1 is a partial order.
2. \(R_2\) contains (a, a) and (b, b), so it's reflexive. Since (a, b) is present but (b, a) is absent, it's antisymmetric. Transitivity is satisfied because (a, b) is in R2, but there is no (b, c) present for the same b and c elements. So, R2 is also a partial order.
3. Analyzing \(R_3\) similarly, we realize that it is also a partial order.
4. \(R_4\) is reflexive, but it fails antisymmetry because both (a, b) and (b, a) are present. Therefore, R4 is not a partial order.
05
Write down the partial order relations
Finally, we write down all the relations that are partial orders on the set A:
1. \(R_1 = \{(a, a), (b, b)\}\)
2. \(R_2 = \{(a, a), (a, b), (b, b)\}\)
3. \(R_3 = \{(a, a), (b, a), (b, b)\}\)
These are all the partial order relations on A.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Reflexivity
The concept of reflexivity in the context of partial orders is crucial.
It ensures that every element of a set relates to itself in the relation. For the given set \(A = \{a, b\}\), reflexivity requires the pairs \((a, a)\) and \((b, b)\) to be part of any relation that claims to be reflexive.
This property is one of the simplest to check, as one only needs to confirm that these pairs are included.
- If you find these pairs, the relation satisfies reflexivity.- In the example, every listed relation \(R_1, R_2, R_3,\) and \(R_4\) considers these pairs, indicating initial reflexive steps.Remember, reflexivity does not concern itself with relationships between different elements, only the elements with themselves.
It ensures that every element of a set relates to itself in the relation. For the given set \(A = \{a, b\}\), reflexivity requires the pairs \((a, a)\) and \((b, b)\) to be part of any relation that claims to be reflexive.
This property is one of the simplest to check, as one only needs to confirm that these pairs are included.
- If you find these pairs, the relation satisfies reflexivity.- In the example, every listed relation \(R_1, R_2, R_3,\) and \(R_4\) considers these pairs, indicating initial reflexive steps.Remember, reflexivity does not concern itself with relationships between different elements, only the elements with themselves.
Antisymmetry
Antisymmetry can be a bit trickier to grasp, but it's all about the unique directionality of relationships.
For a relation to be antisymmetric, if \((a, b)\) and \((b, a)\) are both present, then it must imply that \(a = b\).
In other words, no two distinct elements can mutually relate in both directions unless they are the same element.
- In \(R_2\), we see \((a, b)\) is present, but \((b, a)\) is absent. This satisfies antisymmetry.- However, \(R_4\) includes both \((a, b)\) and \((b, a)\), which violates antisymmetry unless \(a = b\), which is not the case here.Thus, antisymmetry helps to maintain order, ensuring that cyclic relationships (where different elements mutually hold each other) don't occur without equality.
For a relation to be antisymmetric, if \((a, b)\) and \((b, a)\) are both present, then it must imply that \(a = b\).
In other words, no two distinct elements can mutually relate in both directions unless they are the same element.
- In \(R_2\), we see \((a, b)\) is present, but \((b, a)\) is absent. This satisfies antisymmetry.- However, \(R_4\) includes both \((a, b)\) and \((b, a)\), which violates antisymmetry unless \(a = b\), which is not the case here.Thus, antisymmetry helps to maintain order, ensuring that cyclic relationships (where different elements mutually hold each other) don't occur without equality.
Transitivity
Transitivity links elements together through common connections. If you think of it as a chain or bridge, it makes sure the route is complete.
When you have pairs like \((a, b)\) and \((b, c)\) in your relation, transitivity demands that \((a, c)\) must also be part of the relation.
- In the solutions, we are often dealing with a relatively small set. For \(R_2\), adding \((a, b)\) means checking for another span like \((b, c)\), and ensuring \((a, c)\) completes the chain if needed.- In \(R_3\), where \((b, a)\) exists, transitivity doesn't demand any further links, as no additional parts needed to complete bridges.Transitivity ensures that longer connections don't end abruptly, making pathways through the relation logically consistent and linked.
When you have pairs like \((a, b)\) and \((b, c)\) in your relation, transitivity demands that \((a, c)\) must also be part of the relation.
- In the solutions, we are often dealing with a relatively small set. For \(R_2\), adding \((a, b)\) means checking for another span like \((b, c)\), and ensuring \((a, c)\) completes the chain if needed.- In \(R_3\), where \((b, a)\) exists, transitivity doesn't demand any further links, as no additional parts needed to complete bridges.Transitivity ensures that longer connections don't end abruptly, making pathways through the relation logically consistent and linked.
