离散习题讲解(10)

她来听我的演唱会......

Q1:

对于这两种闭包来说,你能够做的只有细心,别无他法。

Let $ R $ be the relation on the set $ {0, 1, 2, 3} $ containing the ordered pairs $ (0, 1), (1, 1), (1, 2), (2, 0), (2, 2), $ and $ (3, 0) $. Find the:

  • a) Reflexive closure of $ R $.
  • b) Symmetric closure of $ R $.

A1:

a) Reflexive closure of $ R $:

  • Definition: The reflexive closure of a relation $ R $ on a set $ S $ is the relation $ R' $ that contains all the pairs of $ R $ and also includes the pairs $ (x, x) $ for all elements $ x S $, ensuring that every element is related to itself.

    • Given set $ {0, 1, 2, 3} $, we need to ensure that each element in this set is related to itself. So we add the pairs $ (0, 0), (1, 1), (2, 2), (3, 3) $ to the relation $ R $.

    • **Reflexive closure of $ R \(**:\) {(0, 0), (0, 1), (1, 1), (1, 2), (2, 0), (2, 2), (3, 0), (3, 3)} $

b) Symmetric closure of $ R $:

  • Definition: The symmetric closure of a relation $ R $ is the relation $ R' $ that contains all the pairs in $ R $ and also includes the reverse of each pair in $ R $. Specifically, for every pair $ (a, b) R $, we must also include $ (b, a) $ in $ R' $.

    • Given the relation $ R = {(0, 1), (1, 1), (1, 2), (2, 0), (2, 2), (3, 0)} $, we add the symmetric pairs:
      • For $ (0, 1) $, we add $ (1, 0) $.
      • For $ (1, 2) $, we add $ (2, 1) $.
      • For $ (2, 0) $, we add $ (0, 2) $.
      • For $ (3, 0) $, we add $ (0, 3) $.
    • **Symmetric closure of $ R \(**:\) {(0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (3, 0)} $

Explanation:

  • Reflexive closure: Ensures every element is related to itself by adding the necessary pairs $ (x, x) $ for all elements in the set.

  • Symmetric closure: Adds the reverse of each existing pair to ensure that if $ (a, b) $ exists in the relation, $ (b, a) $ is also included.


Summary of Answers:

  • a) Reflexive closure of $ R \(:\) {(0, 0), (0, 1), (1, 1), (1, 2), (2, 0), (2, 2), (3, 0), (3, 3)} $

  • b) Symmetric closure of $ R \(:\) {(0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (3, 0)} $