Prop. Every ordering is isomorphic to an ordering of sets under ⊆
Stmt. Let (L, ≤) be a preorder. For any element ℓ ∈ L let leq(ℓ) = { k ∈ L : k ≤ ℓ } be the set of all things below it. Let leq[L] = { leq(ℓ) : ℓ ∈ L }. Then (L, ≤) ≅ (leq[L], ⊆).
Proof. Need to show m ≤ n if and only if leq(m) ⊆ leq(n).
(⇒) Let m ≤ n. WTS leq(m) ⊆ leq(n). Let ℓ ∈ leq(m). Then ℓ ≤ m. By transitivity ℓ ≤ n. Then ℓ ∈ leq(n). This holds for all ℓ and so leq(m) ⊆ leq(n).
(⇐) Let leq(m) ⊆ leq(n). By reflexivity m ≤ m so m ∈ leq(m); then by subset m ∈ leq(n), so m ≤ n.
Cor.
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Every (partial order|linear order|semilattice|lattice) is isomorphic to a (partial order|linear order|semilattice|lattice) of sets under ⊆
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Lattice meet/join operations can be thought of as exactly set union/intersection, because they always are (up to isomorphism)
