Chapter 21: Problem 7
Let \(<\) be an order on a set \(S\). Prove that \(\alpha<\beta\) implies \(\beta \nless \alpha\) for all \(\alpha, \beta \in S\).
Short Answer
Expert verified
\( \alpha < \beta \) implies \( \beta \nless \alpha \) due to the antisymmetry of strict orders.
Step by step solution
01
Understand the Definitions
First, recall the definition of a strict order. An order "<" on a set \(S\) is a strict total order if for any two elements \(\alpha, \beta \in S\), exactly one of the following is true: \(\alpha < \beta\), \(\beta < \alpha\), or \(\alpha = \beta\). This property ensures the comparability and the antisymmetry of the relationship.
02
Define the Problem's Contrapositive
The statement to prove is: \( \alpha < \beta \) implies \( \beta less \alpha \). We can express this using a contrapositive approach. The contrapositive of the implication is: If \( \beta < \alpha \) or \( \beta = \alpha \), then \( \alpha less \beta \). Proving the contrapositive is logically equivalent to proving the original statement.
03
Consider the Properties of an Order
In a strictly ordered set, the relationship is such that if \( \alpha < \beta \), then by definition \( \beta < \alpha \) cannot be true because each element can only relate to another element in one unique way regarding the order relationship. This follows from the antisymmetry in strict ordering.
04
Arrive at the Conclusion
Since \( \alpha < \beta \) inherently means that \( \beta < \alpha \) cannot occur due to the definition of a strict order, we can conclude \( \beta less \alpha \). Thus, the original claim that \( \alpha < \beta \) implies \( \beta less \alpha \) is proven.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Antisymmetry
Antisymmetry is a fundamental concept in the study of ordered sets. It defines a crucial aspect of how elements compare to one another within a set. In a set with antisymmetric order, if you have two elements, say \( \alpha \) and \( \beta \), and both \( \alpha < \beta \) and \( \beta < \alpha \) hold true, this situation indicates a contradiction. Thus, for antisymmetry to hold, at least one must change. In simpler terms, if an element \( \alpha \) is less than another element \( \beta \), then it can never be true that \( \beta \) is less than \( \alpha \) at the same time.
- Antisymmetry helps to prevent circular comparisons within a set.
- It ensures a clear, distinct hierarchical relationship between elements.
Total Order
A total order is another essential concept in understanding ordered sets. In a total order, every pair of elements is comparable. This means for any two elements \( \alpha \) and \( \beta \) in the set \( S \), you can always say whether \( \alpha < \beta \), \( \beta < \alpha \), or \( \alpha = \beta \). This comprehensive ability to compare all elements makes total order very structured.
- Total order implies there are no gaps or ambiguities between any elements in the set.
- Each pair of elements can be ordered in precisely one way, enhancing predictability and consistency.
Comparability
Comparability is an intuitive yet vital aspect of ordered sets, closely tied to the concepts of antisymmetry and total order. When we say elements are comparable, we mean that you can always determine the order between any two elements \( \alpha \) and \( \beta \) in your set \( S \).
- This comparability ensures that no relationships are left undefined.
- It allows implementations like sorting algorithms to function continuously without encountering undefined scenarios.