Math Problem Statement

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Solution

To prove that PP is a partial order, we need to check that the relation PP defined on A×BA \times B satisfies the three properties of a partial order:

  1. Reflexivity: (a1,b1)P(a1,b1)(a_1, b_1) \, P \, (a_1, b_1) for all (a1,b1)A×B(a_1, b_1) \in A \times B.
  2. Antisymmetry: If (a1,b1)P(a2,b2)(a_1, b_1) \, P \, (a_2, b_2) and (a2,b2)P(a1,b1)(a_2, b_2) \, P \, (a_1, b_1), then (a1,b1)=(a2,b2)(a_1, b_1) = (a_2, b_2).
  3. Transitivity: If (a1,b1)P(a2,b2)(a_1, b_1) \, P \, (a_2, b_2) and (a2,b2)P(a3,b3)(a_2, b_2) \, P \, (a_3, b_3), then (a1,b1)P(a3,b3)(a_1, b_1) \, P \, (a_3, b_3).

Step 1: Reflexivity

The definition of PP is: (a1,b1)P(a2,b2)    a1P1a2 and b1P2b2.(a_1, b_1) \, P \, (a_2, b_2) \iff a_1 \, P_1 \, a_2 \text{ and } b_1 \, P_2 \, b_2. Since P1P_1 and P2P_2 are partial orders on AA and BB respectively, they are reflexive. Thus, a1P1a1a_1 \, P_1 \, a_1 and b1P2b1b_1 \, P_2 \, b_1 for all a1Aa_1 \in A and b1Bb_1 \in B. Hence: (a1,b1)P(a1,b1),(a_1, b_1) \, P \, (a_1, b_1), showing reflexivity of PP.

Step 2: Antisymmetry

Suppose (a1,b1)P(a2,b2)(a_1, b_1) \, P \, (a_2, b_2) and (a2,b2)P(a1,b1)(a_2, b_2) \, P \, (a_1, b_1). By the definition of PP, this implies: a1P1a2 and b1P2b2,a_1 \, P_1 \, a_2 \text{ and } b_1 \, P_2 \, b_2, and a2P1a1 and b2P2b1.a_2 \, P_1 \, a_1 \text{ and } b_2 \, P_2 \, b_1. Since P1P_1 and P2P_2 are partial orders, they are antisymmetric. Therefore: a1=a2andb1=b2.a_1 = a_2 \quad \text{and} \quad b_1 = b_2. Hence: (a1,b1)=(a2,b2),(a_1, b_1) = (a_2, b_2), proving antisymmetry of PP.

Step 3: Transitivity

Suppose (a1,b1)P(a2,b2)(a_1, b_1) \, P \, (a_2, b_2) and (a2,b2)P(a3,b3)(a_2, b_2) \, P \, (a_3, b_3). By the definition of PP, this implies: a1P1a2 and b1P2b2,a_1 \, P_1 \, a_2 \text{ and } b_1 \, P_2 \, b_2, and a2P1a3 and b2P2b3.a_2 \, P_1 \, a_3 \text{ and } b_2 \, P_2 \, b_3. Since P1P_1 and P2P_2 are partial orders, they are transitive. Therefore: a1P1a3andb1P2b3.a_1 \, P_1 \, a_3 \quad \text{and} \quad b_1 \, P_2 \, b_3. Thus: (a1,b1)P(a3,b3),(a_1, b_1) \, P \, (a_3, b_3), proving transitivity of PP.

Conclusion

Since PP satisfies reflexivity, antisymmetry, and transitivity, it is a partial order on A×BA \times B.


Do you need further details or examples? Here are five related questions you might find useful:

  1. How does the direct product of partial orders differ from other types of relations?
  2. Can you provide examples of AA, BB, and P1P_1, P2P_2 to illustrate the construction of PP?
  3. How does this concept generalize to other types of relations (e.g., equivalence relations)?
  4. What is the significance of the direct product in lattice theory or set theory?
  5. How would PP behave if P1P_1 or P2P_2 were not partial orders?

Tip: Visualizing the direct product A×BA \times B as a grid or matrix can help in understanding the behavior of the relation ( P \

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Math Problem Analysis

Mathematical Concepts

Partial Orders
Relations
Set Theory

Formulas

(a1, b1) P (a2, b2) iff a1 P1 a2 and b1 P2 b2

Theorems

Reflexivity
Antisymmetry
Transitivity

Suitable Grade Level

Undergraduate