Question

Prove the following. (a) If ≤ is an order relation on P, then < is an antireflexive (that is for all p∈P we have p≮p ) and transitive relation. (b) If < is antireflexive and transitive, then ≤:=<∪= is an order relation.

Answer 1

An order relation ≤ implies that the relation < is antireflexive and transitive. Conversely, if < is antireflexive and transitive, then ≤ defined as < ∪ = is an order relation. These properties are fundamental to the understanding of relational structures in set theory.

Step-by-step explanation:

The subject of the question involves proving properties related to order relations in set theory, a fundamental branch of mathematics.

Proof (a)

To prove that if ≤ is an order relation on a set P, then < is antireflexive and transitive:

1. Antireflexivity means for all elements p in P, it is not the case that p < p. Since ≤ is an order relation, it is reflexive, which means p ≤ p is true for all p in P. The symbol < is defined such that p < q means p ≤ q and p ≠ q. Since p = p, we cannot have p < p, hence < is antireflexive.
2. To prove transitivity, we assume that p < q and q < r for some elements p, q, r in P. By definition of <, we have p ≤ q and q ≤ r. Since ≤ is transitive, we then derive p ≤ r. However, since p ≠ q and q ≠ r, it also means p ≠ r. Therefore, p < r and the relation < is transitive.

Proof (b)

To prove that if < is antireflexive and transitive, then ≤ defined as < ∪ = is an order relation:

1. Since < is antireflexive, when we combine it with the equality =, which is reflexive, the resulting relation ≤ is reflexive, as for all p, p = p implies p ≤ p.
2. As < is transitive and = is also transitive (trivially, since if p = q and q = r, then p = r), the union of the two, ≤, is also transitive.
3. Lastly, ≤ is antisymmetric because if p ≤ q and q ≤ p, then either p < q or p = q, and similarly for q < p or q = p. In both cases, it can only be that p = q for ≤ to hold, thus it is antisymmetric.

Answer 2

(a) If ≤ is an order relation on P, then < is an antireflexive and transitive relation. (b) If < is antireflexive and transitive, then ≤:=<∪= is an order relation.

Step-by-step explanation:

(a) To prove that < is antireflexive, consider any element p in P. Since p≤p (as ≤ is an order relation), it implies p cannot be less than itself, making < antireflexive. Now, to show transitivity, let p₁, p₂, and p₃ be elements in P such that p₁ < p₂ and p₂ < p₃. This implies p₁ ≤ p₂ and p₂ ≤ p₃. By the transitivity of ≤, we have p₁ ≤ p₃, and thus, p₁ < p₃, proving the transitivity of <.

(b) Assume < is antireflexive and transitive. The relation ≤ is defined as ≤ := < ∪ =. To show that ≤ is an order relation, we need to prove three properties: reflexivity, antisymmetry, and transitivity. Referring to the definition of ≤, it includes the equality relation (=).

Since < is antireflexive, and ≤ includes =, it automatically satisfies reflexivity. Antisymmetry is inherited from <, as it is antireflexive. Finally, transitivity of ≤ is guaranteed by the transitivity of <, as well as the fact that = is transitive. Therefore, ≤ is an order relation.