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Challenge: Find a partial ordering with exactly one minimal element but where that element is not a minimum element.

Short Answer

Expert verified
Set \( \{a, b, c\} \) with relation \( \{(b, a), (c, a), (b, c)\} \) creates the desired partial ordering.

Step by step solution

01

Understand the Terms

To solve this exercise, we first need to understand the terms. A partial ordering is a binary relation that is reflexive, antisymmetric, and transitive. A minimal element in a set with respect to a partial order is an element that has no other element less than it in the set. A minimum element is a special type of minimal element that is less than or equal to every other element in the set.
02

Identify the Objective

Our goal is to construct a partial ordering with exactly one minimal element that is not a minimum element. This means we have to construct a set where one element has no other element less than it, but it is not the smallest under the ordering rule for all other elements.
03

Construct the Set

Let's consider a set \( S = \{ a, b, c \} \) and define a relation \( R \) such that \((b, a)\), \((c, a)\), and \((b, c)\) are in \( R \). This implies that \( b \) is related to \( a \), \( c \) is related to \( a \), and \( b \) is related to \( c \).
04

Validate Partial Order

Check the conditions for a partial order:- Reflexivity: Every element is related to itself, for example, \((a, a)\), \((b, b)\), \((c, c)\).- Antisymmetry: If \((x, y)\) and \((y, x)\), then \(x = y\). Here, no such pairs violate this condition.- Transitivity: If \((x, y)\) and \((y, z)\), then \((x, z)\). In this set, this holds true, e.g., \((b, c)\) and \((c, a)\) imply \((b, a)\). Thus, \(R\) is a partial order.
05

Identify Minimal and Minimum Elements

In the set with relation \(R\):- \(a\) is minimal because there are no elements \(x\) such that \((x, a)\) in \( R \).- However, \(a\) is not a minimum since \((b, a)\) and \((c, a)\) exist, meaning \(b\) and \(c\) are not greater than \(a\) for all possible pairs. Hence, \(b\) and \(c\) are comparable with \(a\) but not lesser for the entire set.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Minimal Element
In a set with a partial order, a minimal element is like a unique marker that stands at a certain position where no other elements come before it in the order. It is crucial to differentiate between minimal and minimum elements in order theory. A minimal element is one that does not have another distinct element preceding it in the same set. This means there are no elements that the minimal element is "less than" within that ordered structure. If you imagine elements sort of "competing" for being the smallest, a minimal element simply isn’t surpassed by any other single piece.

A common misconception is confusing minimal elements with global minimum elements. A minimum element, conversely, beats every other element in the set in being the smallest. In other words, it is related to every other element in such a way that it is lower or equal. This exercise cleverly illustrates this notion by presenting a scenario where an element is minimal but not a global minimum.

In our constructed set, the element \(a\) is minimal because no elements \((b\) or \(c)\) have the relation \((x, a)\) for any \(x\) in the set other than \(a\) itself. By mastering this, you will enhance your ability to analyze and identify these elements in varied ordered frameworks.
Transitive Relation
Transitive relations are a fundamental component of partial orders. A relation is transitive if whenever it holds between a first and a second element, and then the second and a third element, it also holds between the first and the third element. This is often written in a formal statement that if \(x, y\) is in the relation R and \(y, z\) is also in R, then \(x, z\) must be in R.

Let's apply this to our constructed example. Consider the relations given: \(b, c\) and \(c, a\). Since both are established in the relation R, transitivity dictates that \(b, a\) should also be in R. This chain-like behavior helps maintain a consistent creative framework, ensuring that all possible indirect relationships are directly considered within the set, maintaining orderliness.

Keep in mind that losing sight of the transitive property can lead to misinterpretations of an ordered set and undermine the soundness of its ordering. Recognizing transitive relations empowers you to visualize connections and design ordered systems that comprehensively adhere to logical principles.
Antisymmetry
Antisymmetry is another pillar of establishing proper partial orders. It's essential to review its meaning clearly: a relation R on a set is antisymmetric if, for any two elements \(x\) and \(y\), whenever both \(x, y\) and \(y, x\) are in R, then \(x = y\). Simply put, two different elements can’t mutually relate to one another in both directions without them actually being the same element.

In the context of our partial ordering example, none of the pairs involve a two-way street that contradicts antisymmetry. For instance, we can see pairs like \(b, a\) and \(c, a\) without any bidirectional counterparts such as \(a, b\) or \(a, c\). This asymmetrical nature allows the ordering to enforce distinct linear or branching hierarchies, typical of partial orders.

Understanding antisymmetry is crucial, as confusing or neglecting it could result in an incorrect framework that masquerades as a valid order. This concept is the bedrock upon which distinctions between different elements are maintained, aiding in ordering complex sets effectively.

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Most popular questions from this chapter

Let \(A=(1,2,3,4\\} .\) Find the transitive closure of the relation \(R\) defined on \(A\) as $$R=\\{(1,2),(2,1),(2,3),(3,4)\\}$$

Determine which of the following five relations defined on \(\mathbb{Z}\) are equivalence relations: (a) \(\\{(a, b) \in \mathbb{Z} \times \mathbb{Z}:(a>0\) and \(b>0)\) or \((a<0\) and \(b<0)\\}\) (b) \(\\{(a, b) \in \mathbb{Z} \times \mathbb{Z}:(a \geq 0\) and \(b>0\) ) or \((a<0\) and \(b \leq 0)\\}\) (c) \(\\{(a, b) \in \mathbb{Z} \times \mathbb{Z}:|a-b| \leq 10\\}\) (d) \(\\{(a, b) \in \mathbb{Z} \times \mathbb{Z}:(a \leq 0\) and \(b \geq 0)\) or \((a \leq 0\) and \(b \leq 0)\\}\) (e) \(\\{(a, b) \in \mathbb{Z} \times \mathbb{Z}:(a \geq 0\) and \(b \geq 0\) ) or \((a \leq 0\) and \(b \leq 0)\\}\)

Let \(R\) and \(S\) be equivalence relations on a set \(X\). (a) Show that \(R \cap S\) is an equivalence relation. (b) Show by example that \(R \cup S\) need not be an equivalence relation. (c) Show that \((R \cup S)^{*}\), the reflexive and transitive closure of \(R \cup S\), is the smallest equivalence relation containing both \(R\) and \(S\).

Let \(X=\\{0,1\\}\). Let \(B=\mathcal{P}(X \times X)\) be the set of all binary relations on \(X\). (a) List all the elements of \(B\). (b) Since elements of \(B\) are themselves relations, it makes sense to ask whether two of those relations are inverses of each other. Let $$\text { IslnerseOf }=\left\\{(R, S): R \in B \text { and } S \in B \text { and } R=S^{-1}\right\\}$$ List all elements of IslmerseOf. (c) Since IslmerseOf is a binary relation, it has an inverse. What is IsimerseOf \(^{-1}\) ? (d) What is Islmerse of o IslnverseOf?

Let \(M=\\{1,2, \ldots, 10] .\) Define a relation \(R\) on elements \(x, y \in M\) such that \((x, y) \in\) \(R\) if and only if there is a positive integer \(k\) such that \(x=k y\). Find the elements of \(R\).

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