Use the well-ordering theorem to show that, if ∅ ⊂ A ⊆ Z and A is bounded
superiorly (that is, there exists c ∈ Z such that for all a ∈ A a ≤ c),
then A has a maximum element.
To use the well-ordering theorem to prove that A has a maximum element, we need to show that A is non-empty and if there is no maximum element, then A is not bounded superiorly. Here's how you can approach this proof:
1. Assume that A is non-empty. Since ∅ ⊂ A, there must exist an element x ∈ A.
2. Now, assume that there is no maximum element in A. This means that for every element a ∈ A, there exists an element b ∈ A such that a < b.
3. Consider the set B = {n ∈ Z | n ≥ x and n ≤ c}, where c is the upper bound of A. This set B is non-empty because it contains at least one element, x, and it is bounded superiorly by c.
4. By the well-ordering theorem, B must have a minimum element, say m. Since m is the minimum element of B, it must be the case that m ≤ n for all n ∈ B.
5. Now, let's consider m - 1. If m - 1 ≥ x, then m - 1 would be an element of B, and since m is the minimum element of B, we have m ≤ m - 1, which is a contradiction. Therefore, m - 1 < x.
6. Since m - 1 < x and x is an element of A, we have m - 1 < x ≤ c, which implies m - 1 ≤ c. But this contradicts the fact that m is the minimum element of B, as m - 1 is an element of B. Hence, our assumption that there is no maximum element in A must be incorrect.
7. Therefore, A must have a maximum element, which proves the desired result.
In this proof, we first assume that A is non-empty and has no maximum element, and then use the well-ordering theorem to derive a contradiction. This contradiction leads us to conclude that A must have a maximum element.