Question

(L,) complete lattice and f L >L an order-preserving function (i) Let a, bE L with a b. Define a, b]= {z € L | a b} Show that (a, b], 3) is a complete lattice (ii) Consider X = {x € L \ f(x) = x}. The Tarski-Knaster Theorem shows that there exists a E X such that for all r E X, a x. Let S C X. Since (L, is a complete lattice, S has a g.l.b in L. Show that a 3 s (iii By considering f restricted to [a, s], for s and a defined in (ii), show that S has a g.l.b. in X. Note that it is not necessarily the case that the g.1.b. of S in X is the same as the g.1.b. of S in L

please help with discrete math HW. please write clearly and dont make up an answer.............i will rate for the best answer thank you. will appreciate if you can answer all 5 parts.

Answer 1

Given that (L, lessthanorequalto) Lattice and F: L rightarrow L an order preserving function An order preserving Function We mean that if a, b elementof L such that a lessthanorequalto b then f(a) lessthanorequalto f(b) Let a, b elementof L with a lessthanorequalto b. Define [a, b] = {x elementof L: a lessthanorequalto) is a compLete lattice in order to prove that [a, b], lessthanorequalto) is partially ordered set and every subset of [a, b] has infamous and supreme since a lessthanorequalto a Forall a elementof [a, b] Rightarrow 'lessthanorequalto' is reflexive relation also a lessthanorequalto b & b lessthanorequalto a Rightarrow a = b Rightarrow 'lessthanorequalto' is anti symmetric (or) [a, b] Also We have a lessthanorequalto b & b lessthanorequalto c Rightarrow a lessthanorequalto c Rightarrow 'lessthanorequalto' is every element of [a, b] is combinable so ([a, b], lessthanorequalto) is totally ordered set so every subset of [a, b] has a chain
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