# Companion to Real Analysis by John M. Erdman PDF

By John M. Erdman

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Definition. A lattice L is distributive if a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) and a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) . 14. Example. If S is a nonempty set, then P(S) (partially ordered by inclusion) is a distributive lattice. Of course, not every lattice is distributive. 15. Example. On the set S = {l, a, b, c, m} define a partial order ≤ so that l < a < m, l < b < m, l < c < m, and no two of the elements in the set {a, b, c} are comparable. Then (S, ≤) is a lattice but is not distributive. (Proof : a ∧ (b ∨ c) = a while (a ∧ b) ∨ (a ∧ c) = l.

3. Proposition. Let A and B be nonempty subsets of a partially ordered set S with B ⊆ A. If both A and B have least upper bounds, then sup B ≤ sup A. Also, if both A and B have greatest lower bounds, then inf B ≥ inf A. 4. Definition. Let ≤ be a partial ordering on a set S and A ⊆ S. An element m ∈ A is a maximal element of A if there is no element of A strictly larger than m; that is, if a = m whenever a ∈ A and a ≥ m. Similarly, m ∈ A is a minimal element of A if a = m whenever a ∈ A and a ≤ m.

10. Definition. A function f : S → T between preordered sets is order preserving if f (x) ≤ f (y) in T whenever x ≤ y in S. It is order reversing if f (x) ≥ f (y) in T whenever x ≤ y in S. We say that f is an order isomorphism if it is bijective and x ≤ y in S if and only if f (x) ≤ f (y) in T . 11. Exercise. Give an example to show that a bijective order preserving map between partially ordered sets need not be an order isomorphism. It is helpful to have a mental picture of objects with which one is working.