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Math 475 Homework #3 March 1, 2010 Section 4.6

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Math 475 Homework #3 March 1, 2010 Section 4.6 Student: Yu Cheng (Jade) Math 475 Homework #3 March 1, 2010 Section 4.6 Exercise 36-a Let ͒ be a set of ͢ elements. How many different relations on ͒ are there? Answer: On set ͒ with ͢ elements, we have the following facts. ) Number of two different element pairs ƳͦƷ Number of relations on two different elements ) ʚ͕, ͖ʛ ∈ ͌, ʚ͖, ͕ʛ ∈ ͌ 2 Ɛ ƳͦƷ Number of relations including the reflexive ones ) ʚ͕, ͕ʛ ∈ ͌ 2 Ɛ ƳͦƷ ƍ ͢ ġ Number of ways to select these relations to form a relation on ͒ 2ͦƐƳvƷͮ) ͦƐ)! ġ ͮ) ʚ ʛ v 2ͦƐƳvƷͮ) Ɣ 2ʚ)ͯͦʛ!Ɛͦ Ɣ 2) )ͯͥ ͮ) Ɣ 2) . b. How many of these relations are reflexive? Answer: We still have ) number of relations on element pairs to choose from, but we have to 2 Ɛ ƳͦƷ ƍ ͢ include the reflexive one, ʚ͕, ͕ʛ ∈ ͌. There are ͢ relations of this kind. Therefore there are ͦƐ)! ġ ġ ʚ ʛ 2ƳͦƐƳvƷͮ)Ʒͯ) Ɣ 2ͦƐƳvƷ Ɣ 2ʚ)ͯͦʛ!Ɛͦ Ɣ 2) )ͯͥ . c. How many of these relations are symmetric? Answer: To select only the symmetric relations on set ͒ with ͢ elements, we have the following facts. ) Number of symmetric relation pairs between two elements ƳͦƷ Number of relations including the reflexive ones ) ʚ͕, ͕ʛ ∈ ͌ ƳͦƷ ƍ ͢ ġ Number of ways to select these relations to form a relation on ͒ 2ƳvƷͮ) )! )ʚ)ͯͥʛ )ʚ)ͮͥʛ ġ ͮ) ͮ) 2ƳvƷͮ) Ɣ 2ʚ)ͯͦʛ!Ɛͦ Ɣ 2 ͦ Ɣ 2 ͦ . d. How many of these relations are anti-symmetric? Answer: To select only the anti-symmetric relations on set ͒ with ͢ elements, we have the following facts. ) Number of anti-symmetric relations between two elements ƳͦƷ Number of relations including the reflexive ones ) ʚ͕, ͕ʛ ∈ ͌ ƳͦƷ ƍ ͢ ġ Number of ways to select these relations to form a relation on ͒ 2ƳvƷͮ) )! )ʚ)ͯͥʛ )ʚ)ͮͥʛ ġ ͮ) ͮ) 2ƳvƷͮ) Ɣ 2ʚ)ͯͦʛ!Ɛͦ Ɣ 2 ͦ Ɣ 2 ͦ . e. How many of these relations are reflexive and symmetric? Answer: To select only the reflexive and symmetric relations on set ͒ with ͢ elements, we have the following facts. ) Number of symmetric and reflexive relations between two elements ƳͦƷ ġ Number of ways to select these relations to form a relation on ͒ 2ƳvƷ )! ġ )ʚ)ͯͥʛ 2ƳvƷ Ɣ 2ʚ)ͯͦʛ!Ɛͦ Ɣ 2 ͦ . f. How many of these relations are reflexive and anti-symmetric? Answer: Subtracting the number of relations on ͒ that are reflexive from the number of relations on ͒ that are anti-symmetric would result the number of relations on ͒ that are reflexive and anti- symmetric. )! ġ )ʚ)ͯͥʛ 2ƳvƷ Ɣ 2ʚ)ͯͦʛ!Ɛͦ Ɣ 2 ͦ . Exercise 37 Let ͌′ and ͌" be two partial orders on a set ͒. Define a new relation ͌ on ͒ by ͬ͌ͭ if and only if both ͬ͌′ͭ and ͬ͌"ͭ hold. Prove that ͌ is also a partial order on ͒. ( ͌ is called the intersection of ͌′ and ͌".) Answer: According to the definition of partial order set, we need to prove three properties on the given relation ͌: reflexive, transitive, and anti-symmetric hold. Reflexivity: For all ͬ ∈ ͒, we have ʚͬ, ͬʛ ∈ ͌′, since ͌′ is a partial order on ͒. For all ͬ ∈ ͒, we have ʚͬ, ͬʛ ∈ ͌′′ , since ͌′′ is a partial order on ͒. ⇒ For all ͬ ∈ ͒, we have ʚͬ, ͬʛ ∈ ͌, since both ʚͬ, ͬʛ ∈ ͌′ and ʚͬ, ͬʛ ∈ ͌′′ hold. Therefore, ʚͬ, ͬʛ ∈ ͌ for all ͬ ∈ ͒. ͌ is a reflexive relation on ͒. Transitivity: Assume ʚͬ, ͭʛ ∈ ͌ and ʚͭ, ͮʛ ∈ ͌ ⇒ ʚͬ, ͭʛ ∈ ͌′, ʚͭ, ͮʛ ∈ ͌′ and ʚͬ, ͭʛ ∈ ͌′′ , ʚͭ, ͮʛ ∈ ͌′′ . ⇒ ʚͬ, ͮʛ ∈ ͌′, since ͌′ is a partial order; ʚͬ, ͮʛ ∈ ͌′′ , since ͌′′ is a partial order. ⇒ ʚͬ, ͮʛ ∈ ͌, since ʚͬ, ͮʛ ∈ ͌′ and ʚͬ, ͮʛ ∈ ͌′′ both hold. Therefore, ʚͬ, ͭʛ ∈ ͌ ʚͭ, ͮʛ ∈ ͌ is followed by ʚͬ, ͮʛ ∈ ͌. ͌ is a transitive relation on ͒. Anti-symmetry: Assume ʚͬ, ͭʛ ∈ ͌ and ʚͭ, ͬʛ ∈ ͌, where ͬ ƕ ͭ. ⇒ ʚͬ, ͭʛ ∈ ͌′, ʚͭ, ͬʛ ∈ ͌′ and ʚͬ, ͭʛ ∈ ͌′′ , ʚͭ, ͬʛ ∈ ͌′′ , where ͬ ƕ ͭ. ⇒ This is a conflict, since ͌′ and ͌′′ are anti-symmetric relations on ͒. Therefore, ʚͬ, ͭʛ ∈ ͌ is followed by ʚͭ, ͬʛ ∉ ͌. ͌ is an anti-symmetric relation on ͒. Exercise 39 Let ʚ̈́, ƙʛ be the partially ordered set with ̈́ Ɣ ʜ0,1ʝ and with 0 Ɨ 1. By identifying the subsets of a set ͒ of ͢ elements with the ͢-tuples of 0’s and 1’s, prove that the partially ordered set ʚ͒, ⊆ʛ can be identified with the n-fold direct product ʚ̈́, ƙʛ Ɛ ʚ̈́, ƙʛ Ɛ ⋯ Ɛ ʚ̈́, ƙʛ ʚ͢ ͚͕͗ͨͣͦͧ ʛ Answer: By identifying the subset relation of a set with a one element set and an empty set, we have, ʚ͌, ⊆ʛ is a partial order with ͌ Ɣ ƣ∅, ʜͬʝƧ, and with ∅ ⊆ ʜͬʝ. There are ͢ elements in set ͒, so we construct the following relations, ͌ͥ: Ƴƣ∅, ʜͬͥʝƧ, ⊆Ʒ ↔ ̈́ͥ: ʚʜ0, 1ʝ, ƙʛ ͌ͦ: Ƴƣ∅, ʜͬͦʝƧ, ⊆Ʒ ↔ ̈́ͦ: ʚʜ0, 1ʝ, ƙʛ : : : : : ͌): Ƴƣ∅, ʜͬ)ʝƧ, ⊆Ʒ ↔ ̈́): ʚʜ0, 1ʝ, ƙʛ ⇒ ͌+-*0/ Ɣ ʚ͌ͥ, ⊆ʛ Ɛ ʚ͌ͦ, ⊆ʛ Ɛ ⋯ Ɛ ʚ͌) ⊆ʛ ↔ ʚ̈́ͥ, ƙʛ Ɛ ʚ̈́ͦ, ƙʛ Ɛ ⋯ Ɛ ʚ̈́) ƙʛ ʚ̈́ͥ, ƙʛ Ɛ ʚ̈́ͦ, ƙʛ Ɛ ⋯ Ɛ ʚ̈́), ƙʛ defines a partial order set on ͢-tuples of 0’s and 1’s. So now we need to prove relation, ͌+-*0/ Ɣ ʚ͌ͥ, ⊆ʛ Ɛ ʚ͌ͦ, ⊆ʛ Ɛ ⋯ Ɛ ʚ͌), ⊆ʛ, is the same thing as the subset relation on ͒ Ɣ ʜͬͥ, ͬͦ, ⋯ ͬ)ʝ. Therefore, if ͒ͥ, and ͒ͦ are subsets of ͒, the goal is to prove ʚ͒ͥ, ͒ͦʛ ∈ ͌+-*0/ ., if and only ͒ͥ ⊆ ͒ͦ. If ͒ͥ, and ͒ͦ are two subsets of ͒, and ͒ͥ ⊆ ͒ͦ, then let ͒ͥ Ɣ ʜͬͥ, ͬͦ, ⋯ , ͬ(ʝ, and ͒ͦ Ɣ ʜͬͥ, ͬͦ, ⋯ , ͬ(, ͬ(ͮͥ, ⋯ , ͬ(ͮ&ʝ. We can then rewrite these two sets in the form of, Column #1 Column #2 Column #3 ͒ͥ: ʜͬͥʝ, ʜͬͦʝ, ⋯ , ʜͬ(ʝ, ∅(ͮͥ, ∅(ͮͦ, ⋯ , ∅(ͮ&, ∅(ͮ&ͮͥ , ⋯ , ∅) ͒ͦ: ʜͬͥʝ, ʜͬͦʝ, ⋯ , ʜͬ(ʝ, ʜͬ(ͮͥʝ, ʜͬ(ͮͦʝ ⋯ , ʜͬ(ͮ&ʝ, ∅(ͮ&ͮͥ , ⋯ , ∅) For column #1 ʜͬͥʝ ⊆ ʜͬͥʝ; ʜͬͦʝ ⊆ ʜͬͦʝ; ⋯ ; ʜͬ(ʝ ⊆ ʜͬ(ʝ , For column #2 ∅(ͮͥ ⊆ ʜͬ(ͮͥ ʝ; ∅(ͮͦ ⊆ ʜͬͦʝ; ⋯ ; ∅(ͮ& ⊆ ʜͬ(ͮ&ʝ , For column #3 ∅(ͮ&ͮͥ ⊆ ∅(ͮ&ͮͥ; ∅(ͮ&ͮͦ ⊆ ∅(ͮ&ͮͦ; ⋯ ; ∅) ⊆ ∅) ⇒ ʚ͒ͥ, ͒ͦʛ ∈ ͌+-*0/ according to the definition of relation product. If ʚ͒ͥ, ͒ͦʛ ∈ ͌+-*0/ then ͒ͥ and ͒ͦ have to satisfy the form above, therefore ͒ͥ ⊆ ͒ͦ . Therefore, we’ve shown that the partially ordered set ʚ͒, ⊆ʛ is identified by the subsets of a set ͒ of ͢ elements with the ͢-tuples of 0’s and 1’s. The following example illustrates this corresponding relationship between ͢-tuples of 0’s and 1’s with and subset relation. We have ͒ Ɣ ʜ͕, ͖, ͗ʝ, and 3-tuples of 0’s and 1’s. Exercise 40 Generalize Exercise 39 to the multiset of all combinations of the multiset ͒ Ɣ ʜͥ͢ ⋅ ͕ͥ, ͦ͢ ⋅ ͕ͦ, ⋯ , ͢( ⋅ ͕(ʝ. Answer: Let ʚ̈́, ƙʛ be the partially ordered set with ̈́ Ɣ ʜ0,1, ⋯ , ͢ʝ and with 0 Ɨ 1 Ɨ, ⋯ , Ɨ ͢. The partially ordered set ʚ͒, ⊆ʛ on a multiset ͒ Ɣ ʜͥ͢ ⋅ ͕ͥ, ͦ͢ ⋅ ͕ͦ, ⋯ , ͢( ⋅ ͕(ʝ, can be identified by a ͡-tuples of 0’s, 1’s, …, and n’s. ͌ͥ: Ƴƣ∅, ͥ͢ ⋅ ʜͬͥʝƧ, ⊆Ʒ ↔ ̈́ͥ: ʚʜ0, 1, ⋯ , ͥ͢ʝ, ƙʛ ͌ͦ: Ƴƣ∅, ͦ͢ ⋅ ʜͬͦʝƧ, ⊆Ʒ ↔ ̈́ͦ: ʚʜ0, 1, ⋯ , ͦ͢ʝ, ƙʛ : : : : : ͌(: Ƴƣ∅, ͢( ⋅ ʜͬ(ʝƧ, ⊆Ʒ ↔ ̈́(: ʚʜ0, 1, ⋯ , ͢(ʝ, ƙʛ ⇒ ͌+-*0/ Ɣ ʚ͌ͥ, ⊆ʛ Ɛ ʚ͌ͦ, ⊆ʛ Ɛ ⋯ Ɛ ʚ͌( ⊆ʛ ↔ ʚ̈́ͥ, ƙʛ Ɛ ʚ̈́ͦ, ƙʛ Ɛ ⋯ Ɛ ʚ̈́( ƙʛ The proof proceeds the same as the proof in Exercise 39. The following example illustrates this corresponding relationship between ͢-tuples of 0’s, 1’s, …, and n’s with and subset relation on a multiset. We have ͒ Ɣ ʜ͕, ͕, ͖, ͖, ͗ʝ, and 3-tuples of 0’s, 1’s, and 2’s. Exercise 42 Describe the cover relation for the partial order ⊆ on the collection Ěʚ͒ʛ of an subsets of a set ͒. Answer: The cover relation for the partial order ⊆ on set ͒ is the ⊂ relation. In other words, the transitive closure of ⊂ together with the reflexive relations on set ͒ makes the relation ⊆. Exercise 43 Let ͒ Ɣ ʜ͕,͖,͗,͘,͙,͚ʝ and let the relation ͌ on ͒ be defined by ͕͖͌ , ͖͌͗ , ͗͌͘ , ͕͙͌ , ͙͚͌ , ͚͌͘ . Verify that ͌ is the cover relation of a partially ordered set, and determine all the linear extensions of this partial order. Answer: The direct graph representation and the adjacent matrix representation of the given relation on set ͒ Ɣ ʜ͕,͖,͗,͘,͙,͚ʝ are shown below. ͕ ͖ ͗ ͘ ͙ ͚ ͕ 0 1 0 0 1 0 ͖ ˥0 0 1 0 0 0˨ ͗ ˦ ˩ ͇ Ɣ ˦0 0 0 1 0 0˩ ͘ ˦0 0 0 0 0 0˩ ͙ ˦0 0 0 0 0 1˩ ͚ ˧0 0 0 1 0 0˪ The transitive and reflexive closure of the given relation can be represented as the following direct graph. ͕ ͖ ͗ ͘ ͙ ͚ ͕ 11 1 1 1 1 ͖ ˥01 1 1 0 0˨ ͗ ˦ ˩ ͇ Ɣ ˦0 01 1 0 0˩ ͘ ˦0 0 01 0 0˩ ͙ ˦0 0 01 1 1˩ ͚ ˧0 0 0 1 0 1˪ By evaluating the graph, and the adjacency matrix, it’s clear that this relation is reflexive, transitive and anti-symmetric.
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