order theory, priestley duality and semantic interpretation of logic
TRANSCRIPT
Order theory, Priestley duality and semanticinterpretation of logic
S.W. Gaaf
August 31, 2010
Bachelorthesis
Supervisor: Drs. Palmigiano
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KdV Instituut voor wiskunde
Faculteit der Natuurwetenschappen, Wiskunde en Informatica
Universiteit van Amsterdam
AbstractAfter an introduction to order theory we will establish a way of associatingany poset with a lattice endowed with extremely good properties. And wecan also establish the reverse correspondence. But we are interested in acorrespondence from posets to distributive lattices and back, which gets usto the same poset we started with. This we can do but only for specialdistributive lattices.From the finite case we will arrive at the general case, were we will seethat Boolean algebras are dual to Boolean spaces, and bounded distributivelattices are dual to Priestley spaces.Classical propositional logic is closely related to this. We will introduce theLindenbaum-Tarski algebra, which can be treated in the same way, obtainingvery nice results about maximal consistent theories.
DataTitle: Order theory, Priestley duality and semantic interpretation of logicAuthor: S.W. Gaaf, [email protected], 5619696Supervisor: Drs. PalmigianoDate: August 31, 2010
Korteweg de Vries Instituut voor WiskundeUniversiteit van AmsterdamPlantage Muidergracht 24, 1018 TV Amsterdamhttp://www.science.uva.nl/math
Contents
1 Introduction 3
2 Order 42.1 Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.1.1 Pre-orders and Partial orders . . . . . . . . . . . . . . . . . . . . . 42.1.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.1.3 Partial orders and strict orders . . . . . . . . . . . . . . . . . . . . 52.1.4 Order in propositional logic . . . . . . . . . . . . . . . . . . . . . . 52.1.5 Pre-orders and equivalence relations . . . . . . . . . . . . . . . . . . 62.1.6 Lexicographic order . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2.1 Covering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2.2 Construction of a diagram . . . . . . . . . . . . . . . . . . . . . . . 82.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.3 Duality Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.5 Down-sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.5.1 Subposets and down-sets . . . . . . . . . . . . . . . . . . . . . . . . 102.5.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.5.3 Properties of down-sets . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 Lattices 123.1 Upper bounds and Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.1.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.1.3 A canonical embedding . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.2 Properties of Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3 Filters and Ideals and their logical significance . . . . . . . . . . . . . . . . 15
3.3.1 Filters and Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.3.2 The logical significance of filters and ideals . . . . . . . . . . . . . . 163.3.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
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4 Boolean Algebra 184.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
4.1.1 Examples of Boolean algebras . . . . . . . . . . . . . . . . . . . . . 194.1.2 The Lindenbaum algebra as Boolean algebra . . . . . . . . . . . . . 19
4.2 The finite intersection property . . . . . . . . . . . . . . . . . . . . . . . . 204.2.1 The fip and the interpretation in propositional logic . . . . . . . . . 20
5 Finite representation 225.1 Join-irreducible elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.2 Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6 Maximality Principles 246.1 Prime ideals and filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246.2 Maximal ideals and ultrafilters . . . . . . . . . . . . . . . . . . . . . . . . . 26
7 Representation 297.1 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297.2 Stone’s representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
7.2.1 Boolean space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307.3 Priestley . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
8 Populaire samenvatting 35
2
Chapter 1
Introduction
This thesis starts with an introduction to order-theory. Sets can be ordered in differentways. Different ordering will give us different properties, which will be showed in examplesin the first chapter, Chapter 2.In Chapter 3 we will introduce lattices, which are partial ordered sets with very nice prop-erties. Then we will establish a way of associating any poset with a lattice endowed withextremely good properties. And we can also establish the reverse correspondence, in atrivial way, because every distributive lattice is also a poset. But we are interested in acorrespondence from posets to distributive lattices and back, which gets us to the sameposet we started with. This we can do but only for special distributive lattices.After having introduced join irreducibles in Chapter 5 we will be able to prove that everyfinite Boolean algebra is isomorphic to P(X) for some finite set X. And after that eventhat every Boolean algebra is isomorphic to a subalgebra of P(X). But it would be niceto have an isomorphism not only for the finite case, but for Boolean algebras in general.Therefor we will make in the last chapter the set of prime ideals of any Boolean algebrainto a topology space in such a way that it is dual to a Boolean algebra.At the end we will see that Boolean algebras are dual to Boolean spaces, and boundeddistributive lattices are dual to Priestley spaces. For the Boolean algebras we will use theStone representation theorem, and for the bounded distributive lattices there is Priestley’srepresentation theorem.
Also for the classical propositional logic these are all very nice results. Almost directly thelogic will be implemented in this thesis, starting with the consequence relation that canbe seen as an order relation. The Lindenbaum algebra can then be interpreted as a latticeand it is in fact a lattice. We will show a same kind of duality between the Lindenbaumalgebra (which we will first manipulate to make it into a Boolean algebra) and a canonicalmodel whose points are maximal consistent theories.
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Chapter 2
Order
He is older than that girl. Flour is an ingredient of bread and of pasta. Amsterdam isthe city with the geatest number of inhabitants of The Netherlands. In our daily lives wecompare a lot of things. We form groups and order their elements. A group of people canbe ordered by length, by age or by weight. But we could also order some cities by surfaceor by number of inhabitants. We see that there are many different things to order, andalso the way of ordering can differ. We could order dishes by ingredients or by calories,and clothes by size or by color, and there are also various ways to order families. Andfurther we have linear orders, for example time and lists.The following chapter outlines the mathematical formalization of this idea, and how wecan applicate this in logic, where order for example models the consequence relation.
2.1 Order
2.1.1 Pre-orders and Partial orders
Definition 2.1. Let P be a set. An order (or partial order) on P is a binary relation ≤on P such that, for all x, y, z ∈ P ,1. x ≤ x (reflexivity)2. x ≤ y and y ≤ x imply x = y (antisymmetry)3. x ≤ y and y ≤ z imply x ≤ z (transitivity)
A set P equipped with an order relation ≤ is said to be an ordered set or a partiallyordered set (poset). If we want to specify the order relation we write 〈P,≤〉.On any set = is an order, the discrete order. A quasi-order is a relation ≤ on a setP that is reflexive and transitive but not antisymmetric. This relation is also called apre-order.
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2.1.2 Examples
• We can order sets by inclusion. The inclusion relation is typically used to order thepowerset P(X), the set of all subsets of X. For A,B ∈ P(X) we say A ≤ B if andonly if A ⊆ B, and we write 〈P(X),⊆〉.
• The set N0 can be ordered with its usual order, i.e. m ≤ n ⇔ m = n or m < n.This gives us the poset 〈N0;≤〉, where for all m,n ∈ N0 either m ≤ n or n ≤ m. Ingeneral, an ordered set P is called a totally ordered set or a chain if for all p, q ∈ Peither p ≤ q or q ≤ p (if any two elements of P are comparable). The poset 〈N0;≤〉is a chain. A set X in which x | y (x 6≤ y and y 6≤ x) for every x, y ∈ X is called ananti-chain.
• The set N0 can also be ordered in a different way. Define the order 4 as x 4 y ifand only if x divides y (that is ∃k ∈ N0 such that kx = y). The obtained poset is〈N0;4〉.
2.1.3 Partial orders and strict orders
Example 2.2. Let P be a set on which a binary relation < is defined such that, for allx, y, z ∈ P ,
a) x < x is false (antireflexive)
b) x < y and y < z imply x < z
We can prove that if ≤ is defined by x ≤ y ⇐⇒ (x < y or x = y), then ≤ is an orderon P . We have to check the three conditions for an order:
1. The relation ≤ is reflexive, because ∀x ∈ P holds x = x, hence x ≤ x
2. Then x ≤ y and y ≤ x imply x = y, because if not both are equalities, we have x < xwhich is false. The relation is antisymmetric.
3. Transitivity can be shown as follows: Let x ≤ y and y ≤ z. Then we obtain one ofthe four following (in-)equalities: x < y < z, x < y = z, x = y < z or x = y = z.This implies x < z or x = z and hence x ≤ z.
2.1.4 Order in propositional logic
Definition 2.3. Let Fm be any logical language. Then a consequence relation is a relation` ⊆ P(Fm)× Fm which satisfies the following conditions
1. if φ ∈ Γ, then Γ ` φ.
2. if Γ ` δ for every δ ∈ ∆ ⊆ Fm and ∆ ` φ, then Γ ` φ.
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3. if Γ ` φ and Γ ⊆ Γ′, then Γ′ ` φ.
Note that 1. and 2. imply 3.. Fix therefor Γ ` φ and Γ ⊆ Γ′. Then by 1., Γ′ ` δ for allδ ∈ Γ. Then because Γ ` φ and by 2. we obtain Γ′ ` φ.
If we now take the restriction of ` to Fm× Fm, where the first Fm is identified with thesubset {{φ} |φ ∈ Fm} ⊆ P(Fm), then this is a pre-order.Reflexivity is certainly the case, by 1. and φ ∈ {φ}. The consequence relation is notantisymmetric, because it can be the case that φ ` ψ and ψ ` φ but φ 6= ψ. Take forexample the two formulas ¬φ and φ→ ⊥. We know φ→ ⊥ ` ¬φ∨⊥, and ¬φ∨⊥ ` ¬φ,hence (by 2.) φ → ⊥ ` ¬φ, and vice versa ¬φ ` φ → ⊥. But these two formulas arenot the same. They are logically equivalent, i.e. they have the same truth values underany assignment, but they are not identical because they are different strings of symbols.Hence the consequence relation is not antisymmetric. But the relation is transitive, whichfollows directly from 2. If we take Γ = {ψ} and ∆ = {δ}, then if {ψ} ` δ and {δ} ` φ, weobtain {ψ} ` φ.The relation ` is reflexive and transitive but not antisymmetric and hence a pre-order.
2.1.5 Pre-orders and equivalence relations
Let 〈X,≤〉 be a pre-order. Define the relation ≡ ⊆ X ×X as follows:
x ≡ y ⇐⇒ x ≤ y and y ≤ x.
The relation≡ is an equivalence relation. Then consider the quotient setX/≡ := {[x] |x ∈ X},where [x] = {y ∈ X |x ≡ y}. Now define the following relation over X/≡
[x] ≤≡ [y] ⇐⇒ ∃x′ ∈ [x],∃y′ ∈ [y] : x′ ≤ y′.
The relation ≤≡ is a partial order.
We can applicate this to logic. Take therefor 〈Fm,`〉 as a pre-order, where Fm is anylogical language. The equivalence relation will be defined as
φ ≡ ψ ⇐⇒ φ ` ψ and ψ ` φ (notation φ a` ψ).
This is the interderivabiltiy relation, a proof theoretic relation, which tells us that φ logi-cally implies ψ, and vice versa. The quotient set then becomes Fm/≡ := {[φ] |φ ∈ Fm},where [φ] = {ψ ∈ Fm |φ ≡ ψ}. We define also the following relation over Fm/≡
[φ] ≤≡ [ψ] ⇐⇒ ∃φ′ ∈ [φ],∃ψ′ ∈ [ψ] : φ′ ` ψ′.
This relation is (as in the general case) a partial order.We obtained a set the of equivalence classes equipped with the order relation ≤≡. This isthe partial ordered set 〈Fm≡,`≡〉. The pre-ordered set 〈Fm,`〉 we started with, has nowturned into a partial ordered set, introducing equivalence relations.As we promised in the introduction of this chapter, a first connection already emergesbetween order and the consequence relation. In the next chapter the poset 〈Fm≡,`≡〉 willbecome the Lindenbaum-Tarski algebra under the condition that ≡ is a congruence.
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2.1.6 Lexicographic order
Let P and Q be two posets. We can define different orders on the Cartesian product P×Q.The canonical product order is defined compontent-wise, (p1, q1) ≤× (p2, q2) ⇔ p1 ≤Pp2 and q1 ≤Q q2.
Proposition 2.4. The product order is the greatest order relation on P ×Q which makesboth projections order preserving.
Proof. Let ≤′ ⊆ P × Q be an order relation such that both π1 : P × Q −→ P and π2 :P×Q −→ Q are order preserving. Let us show that if (p, q) ≤′ (p′, q′) then (p, q) ≤× (p′, q′).We need to show that p ≤P p′ and q ≤Q q′. Since π1 is order preserving, (p, q) ≤′ (p′, q′)implies π1(p, q) ≤P π1(p′, q′) which is equal to p ≤P p′. Likewise we can show that q ≤Q q′.This is enough to show that ≤′ ⊆ ≤×.
Another order on P ×Q is the lexicographic order, defined by (p1, q1) ≤ (p2, q2) if p1 < p2
or (p1 = p2 and q1 ≤ q2).The lexicographic order is different from the canonical order, and it is normally not so nicelybehaved. The only aspect in which the lexicographic order scores better than the canonicalproduct order is that if P and Q be chains, then also P ×Q, ordered lexicographically, isa chain. To show this take (p1, q1) and (p2, q2) arbitrarily. Sure is that either p1 ≤ p2 orp2 ≤ p1. If p1 < p2 we have (p1, q1) ≤ (p2, q2), and also if p2 < p1 we are done. Assumenow that p1 = p2. Then again we have p2 ≤ q2 or q2 ≤ p2 and the lexicographic order givesus (p1, q1) ≤ (p2, q2) or (p2, q2) ≤ (p1, q1) respectivily. This means that any two elementsof P ×Q are comparable, hence P ×Q is a chain in the lexicographic order.
2.2 Diagrams
We can represent an ordered set in a diagram. Therefore we need to introduce the termcovering.
2.2.1 Covering
Definition 2.5. Let P be an ordered set, and let x, y, z ∈ P . We say y covers x if x < yand x ≤ z < y implies z = x. We use the notation x ≺ y.
This covering relation is irreflexive and intransitive: Let P be an ordered set. Becausex < x is false for all x ∈ P , and thus x ⊀ x for all x ∈ P , the covering relation ≺ is anirreflexive relation. A relation ◦ is intransitive if x ◦ y and y ◦ z imply that x ◦ z is false,for all x, y, z ∈ P . We see that ≺ is intransitive because for all x, y, z ∈ P , if x ≺ y andy ≺ z then x < y < z, which implies that x ≤ y < z where y 6= x and hence x ⊀ z.
In case P is a finite ordered set, the following holds for every x, y ∈ P such that x 6= y:
x < y iff x ≺ x1 ≺ · · · ≺ xn ≺ y for some x1, . . . xn ∈ P.
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Assume x < y. If there is no element x1 ∈ P such that x ≤ x1 < y and x1 6= x we arefinished and conclude x ≺ y. Otherwise we continue and ask if there is an elements x2
such that x ≤ x2 < x1 and x2 6= x, and if there is an element x3 such that x1 ≤ x3 < y andx3 6= x1. In the case there is such an element, continue the same procedure. Otherwise wehave for example x ≺ x1. Once finished (this will allways be the case, because P is finite)we have found the elements that will form the covering chain.Suppose for the other direction that x ≺ x1 ≺ · · · ≺ xn ≺ y for some x1, . . . , xn ∈ P . Thismeans x < x1 < . . . < xn < y and we can conclude that x < y.
2.2.2 Construction of a diagram
To explain how to draw a diagram we use the following example. Let P = {a, b, c, d, e}be a set ordered by a < b, a < c, a < e, b < e, c < e, d < c and d < e. We construct adiagram for P linking every element of P to a point in the Euclidean plane R2. And everycovering pair x ≺ y will be drawn as a line segment between the two points. Line segmentsmay not accidentally cross points. Further, every element x has to be lower than y (thatis, the second coordinate has to be smaller) whenever y covers x. A few possible diagramaof P will now be:re
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2.2.3 Examples
In the figure on the next page you find examples of diagrams of different posets. Theposet 〈P({1, 2, 3}),⊆〉 and a chain of 4 elements (notation 4 = {1, 2, 3, 4}) and the chainN0 both ordered by ≤. The last diagram is a subset of the poset 〈N0,4〉, namely Q :=〈{1, 2, 4, 5, 6, 12, 20, 30, 60};4〉.
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2.3 Duality Principle
Definition 2.6. Let P = 〈X,≤〉 be a poset. The dual of P , notation P ∂, is the poset〈X,≥〉 where x ≥ y in P ∂ iff x ≤ y in P .
2.4 Maps
Definition 2.7. Let P and Q be ordered sets and φ a map from P to Q.
1. We call φ an order-preserving map if p ≤P q implies φ(p) ≤Q φ(q), for all p, q ∈ P .
2. We call φ an order-embedding if p ≤P q if and only if φ(p) ≤Q φ(q), for all p, q ∈ P .
3. We call φ an (order-)isomorphism if φ is onto and an order-embedding.
Example 2.8. An order-embedding is by definition always an injective order-preservingmap. It is now an interesting question which map is an order-preserving map but not anorder-embedding. Define P = {a, b, c} such that a < c and b < c. Define the map φ as inthe figure below.
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is not enough to characterize substructures in the context of posets and order-preservingmaps, as is the case with injective homomorphisms in algebras, injective continuous mapsin topological spaces and injective functions in sets.
2.5 Down-sets
2.5.1 Subposets and down-sets
In an ordered set we can indicate subsets, which themselves are again ordered sets, becausethey are naturally endowed with the inherited order.
Definition 2.9. Given a poset P = 〈X,≤〉. We call Q = 〈Y,≤〉 a subposet of P if Y ⊆ Xand if for every x, y ∈ Y , x ≤ y in Q iff x ≤ y in P .
Take for example the poset P = 〈N,≤〉. The set of all even numbers {2, 4, 6, · · · } equippedwith the same order-relation is a subposet of P . An other example of a subposet is adown-set.
Definition 2.10. A subset S of a poset P is a down-set iff s ∈ S, x ∈ P and x ≤ s implyx ∈ S. For arbitrary S ⊆ P and y ∈ P we define:
↓S = {x ∈ P | (∃s ∈ S)x ≤ x} and ↓y = {x ∈ P | x ≤ y}
These downsets are generated by S and y respectively. A down-set of the form ↓y (gen-erated by one element of P ) is called a principal down-set. The set of all down-sets of aposet X is denoted by O(X). Dually we can define up-sets, notation ↑S and ↑y.
2.5.2 Examples
• In the figure below we see the diagram of a poset P the diagram for the set of alldown-sets of P . Not all the down-sets in O(P ) are principal downsets. The downset{a, c} is a principal downset, because it is equal to ↓c. But the down-set {a, c, d} isgenerated by the set {c, d} and hence not a principal down-set.
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• For an antichain X the following equality holds: O(X) = P(X). The principal down-sets all contain only one element. Because O(X) ⊆ P(X), and because now everyelement of X is a down-set, and also every union of elements of X is a down-set, wehave P(X) ⊆ O(X), and hence O(X) = P(X)
2.5.3 Properties of down-sets
Lemma 2.11. Let P be an ordered set, and x, y ∈ P . Then the following are equivalent:
1. x ≤ y
2. ↓x ⊆ ↓y
3. (∀Q ∈ O(P ))y ∈ Q⇒ x ∈ Q
Proof. 1.⇒ 2. Assume x ≤ y. Then if x ∈ ↓x, thus x ∈ {a ∈ P | a ≤ x}. Because x ≤ ythen also x ∈ {a ∈ P | a ≤ y} = ↓y. Hence for all x ∈ ↓x we have x ∈ ↓y, whichimplies ↓x ⊆ ↓y.
2.⇒ 3. Assume ↓x ⊆ ↓y. Take a Q ∈ O(P ). Then if y ∈ Q we have ↓x ⊆ ↓y ⊆ Q, whichimplies x ∈ Q.
3.⇒ 1. Assume (∀Q ∈ O(P ))y ∈ Q⇒ x ∈ Q. Suppose x > y. Take Q = ↓y, thus y ∈ Q.But because x > y, x /∈ ↓y hence x /∈ Q, which is a contradiction. We can concludethat x ≤ y.
Lemma 2.12. Let P be an ordered set and O(P ) the set of all downsets of P . The mapφ : P → O(P ) is an order-embedding.
Proof. Define the map φ by sending an element of P to its downset: x → ↓x. Furtherwe take the inclusion order on O(P ). To check if φ is an order-embedding we first takex, y ∈ P such that x ≤ y. Then x ↓= {z ∈ P |z ≤ x} and ↓y = {z ∈ P |z ≤ y}. Whilex ≤ y we know that z ≤ x ≤ y holds for all z ∈ ↓x and hence for all z ∈ ↓x we have z ∈ ↓y.So ↓x ⊆ ↓y and hence φ is an order-preserving map.For the other direction, assume ↓x ⊆ ↓y. Then we know that p ∈ ↓y for all p ∈ ↓x. Hencefor all p ≤ x holds p ≤ y. Since x ∈ ↓x because x ≤ x holds x ∈ ↓y and hence x ≤ y.Because for all p, q ∈ P , p ≤ q if and only if φ(x) ⊆ φ(q), the map φ is an order-embedding.
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Chapter 3
Lattices
3.1 Upper bounds and Lattices
3.1.1 Definitions
Definition 3.1. Let P be an ordered set and let S ⊆ P . An element x ∈ P is an upperbound of S if s ≤ x for all s ∈ S. A lower bound is defined dually. For the set of all upperbounds of S and for the set of all lower bounds of S we write
Su = {x ∈ P | (∀s ∈ S) s ≤ x} and Sl = {x ∈ P | (∀s ∈ S)x ≤ s}.
Since ≤ is transitive, Su is always an up-set, and Sl a down-set. If Su has a least element,then it is called the least upperbound of S, also known as the supremum of S. Duallythe greatest element of Sl, if it exists, is called the greatest lowerbound, or the infimum ofS. Instead of writing supS (and inf S) we introduce the terms ’join’ and ’meet’, and wewrite
∨S and
∧S. This way we get, in case S contains two elements, sup{x, y} = x ∨ y
and inf{x, y} = x ∧ y.
Definition 3.2. The ordered set P is called a lattice if x ∨ y and x ∧ y exist for any twoelements x, y ∈ P . An ordered set is called a complete lattice if for every subset S ⊆ Pboth
∨S and
∧S exist. A bounded lattice is a lattice that has a top and a bottom, also
called 0 and 1.
If the join of two elements does not exist, it means that they have no least common upperbound, or that there is no common upper bound at all.
3.1.2 Examples
• The poset 〈P(X);⊆〉 is a lattice for every set X. For every A,B ∈ P(X) both A∨Band A ∧B exist. In this case they correspond to A ∪B and A ∩B.
• Every chain is a lattice. Let X be a chain. For every two elements x, y ∈ X eitherx ≤ y or y ≤ x. Then, assuming that x ≤ y, x ∧ y = x and x ∨ y = y. Hence forevery two elements the join and meet exist, which makes the chain a lattice.
12
• There are also posets that are not lattices. Consider the setQ = {1, 2, 4, 5, 6, 12, 20, 30, 60},equipped with the order 4 that is defined as m 4 n if m divides n, as we have seenin the previous chapter. When we draw the diagram of Q, we can easily see Q is nota lattice.
r60@@@
���r12
@@@
r20�
��
@@@
r30���r4
@@@
r6
r5r
2@@@r
1Q
r60@@@r20 r30 {2, 5}u = {20, 30, 60}
The join of 2 and 5 doesn’t exist, because there is no least common upper bound,and hence the poset is not a lattice.
3.1.3 A canonical embedding
Lemma 3.3. The set O(P ) of all downsets of P is a complete lattice.
Proof. To prove this it suffices to show that for every collection D of downsets also⋃D
and⋂D are downsets. Because then for every collection of downsets the join and the meet
exist.Take a collection of downsets D ⊆ O(P ). We need to proof that
⋂D is a downset. Let
y ∈⋂D and assume x ≤ y ∈
⋂D for an arbitrary x in P . While y ∈
⋂D we know that
y ∈ C for all downsets C in the collection D. But while C being a downset x ≤ y ∈ Cimplies that x ∈ C for all C, which gives us x ∈
⋂D. Hence
⋂D is a downset. We can
conclude that the meet of every collection of downsets is a downset.For the join the proof is almost the same. Let y ∈
⋃D and assume x ≤ y ∈
⋃D for
an arbitrary x in P . While y ∈⋃D we know that y ∈ C for at least one downset C1
in the collection D. But then x ≤ y ∈ C1 implies that x ∈ C1, which gives us x ∈⋃D.
Hence⋂D is a downset. We can conclude that the join of every collection of downsets is
a downset.Now for all D ⊆ O(P ) the join and meet exist, and hence O(P ) is a complete lattice.
We established a way of associating any poset with a lattice endowed with extremely goodproperties. But can we also establish the reverse correspondence? Yes, in a trivial way,every distributive lattice is also a poset. But we are interested in a correspondence fromposets to distributive lattices and back, which gets us to the same poset we started with.Can we do this? Yes we can but only for special distributive lattices. Therefor we needmore theory, so we will get back to this after having introduced join irreducibles.
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3.2 Properties of Lattices
Every chain is a lattice, hence N, Z, Q and R are lattices under their usual order. Butthey are not complete lattices, because the top is missing.
Theorem 3.4. Let L be a lattice. Then ∨ and ∧ satisfy, for all a, b, c ∈ L,
(L1) (a ∨ b) ∨ c = a ∨ (b ∨ c) (associativity)
(L1)∂ (a ∧ b) ∧ c = a ∧ (b ∧ c)
(L2) a ∨ b = b ∨ a (commutativity)
(L2)∂ a ∧ b = b ∧ a
(L3) a ∨ a = a (idempotency)
(L3)∂ a ∧ a = a
(L4) a ∨ (a ∧ b) = a (absorption)
(L4)∂ a ∧ (a ∨ b) = a
Proof. The duality enables us to prove only (L1)-(L4).For (L3) we know
{a, a}u = {x ∈ L | a ≤ x and a ≤ x}= {x ∈ L | a ≤ x}
Because ≤ is reflexive (a ≤ a) the least upperbound is a, hence a ∨ a = a.Then for (L2) the following equality holds:
{a, b}u = {x ∈ L | a ≤ x and b ≤ x}= {x ∈ L | b ≤ x and a ≤ x}= {b, a}u
Since the set of upperbounds of a and b is equal to the set of upperbounds of b and a, alsothe least upperbound will be one and the same, hence a ∨ b = b ∨ a.Then (L4) is easily done, when we see that a ∧ b ≤ a. Knowing this, we can concludedirectly that a ∨ (a ∧ b) = a.For (L1) we need to show that (a∨ b)∨ c = sup{a, b, c}. Because then it follows with (L2)that (a ∨ b) ∨ c = a ∨ (b ∨ c). We want to show that {a ∨ b, c}u = {a, b, c}u. We see thefollowing
d ∈ {a ∨ b, c}u ⇐⇒ a ∨ b ≤ d and c ≤ d⇐⇒ d ∈ {a, b}u and c ≤ d⇐⇒ a ≤ d and b ≤ d and c ≤ d⇐⇒ d ∈ {a, b, c}u.
Thus sup{a ∨ b, c} = sup{a, b, c} = sup{a, b ∨ c}, and hence (a ∨ b) ∨ c = a ∨ (b ∨ c).
14
Theorem 3.5. Let 〈L;∨;∧〉 be a non-empty set equipped with two binary operations whichsatisfy (L1)-(L4) and (L1)∂-(L4)∂. Then
1. For all a, b ∈ L we have a ∨ b = b if and only if a ∧ b = a.
2. Define ≤ on L by a ≤ b if a ∨ b = b. Then ≤ is an order-relation.
3. With ≤ defined as in (2) 〉L;≤ 〈 is a lattice in which the original operations agreewith the induces operations, that is, ∀a, b ∈ L : a ∨ b = supa, banda ∧ b = infa, b.
3.3 Filters and Ideals and their logical significance
3.3.1 Filters and Ideals
Definition 3.6. Let L be a lattice. Then a non-empty set J ⊆ L is an ideal if
1. a ∨ b ∈ J for all a, b ∈ J and
2. a ∈ L, b ∈ J and a ≤ b imply a ∈ J .
In words we can describe an ideal as a non-empty down-set closed under join. Further wecall the dual of an ideal, that is a non-empty up-set closed under meet a filter. If the ideal(filter) does not coincide with L we call it a proper ideal (filter). A principal filter is a filtergenerated by an element x ∈ L, i.e. it is the collection {y ∈ L |x ≤ y} for some x ∈ L. Inthe same way {y ∈ L | y ≤ x} is a principal ideal generated by x. A principal ideal (filter)generated by an element a ∈ L, is the down-set ↓a (up-set ↑a).If a lattice L has a maximum element 1, then every filter in L contains it. Dually if L hasa minimum element 0, then every ideal in L contains it.
Example 3.7. Let L and K be bounded lattices, and f : L→ K a {0, 1}-homomorphism.We can easily show that f−1(0) is an ideal of L and f−1(1) is an filter of L.We know that f(0) = 0 and f(1) = 1 because f is bounded, and because of the homomor-phism we have f(a ∨ b) = f(a) ∨ f(b) and f(a ∧ b) = f(a) ∧ f(b) for all a, b ∈ L.Now we wonder if a ∨ b ∈ f−1(0) for all a, b ∈ f−1(0). There holds
a, b ∈ f−1(0)⇔ f(a) = 0 and f(b) = 0⇔ f(a ∨ b) = f(a) ∨ f(b) = 0 ∨ 0 = 0
and hence a ∨ b ∈ f−1(0). Now it is left to show that f−1(0) is a downset, thus we takea ∈ L and b ∈ f−1(0) such that a ≤ b. There holds
f(b) = 0 and a ∨ b = b⇔ f(a ∨ b) = f(b) = 0⇔ f(a) ∨ f(b) = f(a) ∨ 0 = 0⇔ f(a) = 0
and hence a ∈ f−1(0), which shows that f−1(0) is in fact an ideal. The proof of f−1(1)being a filter is dual to this proof.
15
3.3.2 The logical significance of filters and ideals
Given a logic 〈Fm,`〉. Suppose that the language has ∨ and ∧ with the properties above.A theory for the logic is a subset Σ ⊆ Fm such that ∀φ ∈ Fm (Σ ` φ =⇒ φ ∈ Σ). Wesay that Σ is closed under the consequence relation. Now for all Σ ⊆ Fm if Σ is a theory,then {[φ] |φ ∈ Σ} is a filter of the lattice 〈Fm/a`,≤a`〉, i.e. the Lindenbaum-Tarskialgebra.Conversely if F is a filter of the Lindenbaum-Tarski algebra, then Σ = ∪{[φ] | [φ] ∈ F} is atheory of 〈Fm,`〉. Proof: Let φ such that Σ ` φ, so by finitarity there is a Σ′ = {γ1, . . . , γn}such that Σ ` φ. Now take γ =
∧γi. Then by the ’filter condition’ [γ] ∈ F ⇐⇒
[γ1 ∧ . . . ∧ γn] = [γ1] ∧ . . . ∧ [γn] ∈ F and γ ` φ imply F 3 [γ] ≤ [φ]. Hence [φ] ∈ F , i.e.φ ∈ Σ.Filters and ideals are important for logic. Filters for example model logical theories.In propositional logic the meet-symbol ∧ is interpreted as ’and’. In proves it is used asfollows:
• φ ∧ ψ ` φ and φ ∧ ψ ` ψ.
• {φ, ψ} ` φ ∧ ψ, we would like to refer to it as the ’filter condition’.
Then for the join-symbol ∨ we have the interpretation ’or’ which is used in propositionallogic as follows:
• φ ` φ ∨ ψ and ψ ` φ ∨ ψ.
• If P ` φ ∨ ψ, then either P ` φ or P ` ψ, for a theory P . One of the two has to bechosen, and we could refer to this property as the ’prime ideal condition’.
If T is a theory (a set of formulas closed under logical consequence) and T ` a, then thereis a finite set of theorems b1, . . . , bn ∈ T such that
∧ni=1 bi ` a. But this means that a ∈ T ,
because e theory is closed under logical consequences. In general if Γ is a theory such thatΓ ` φ, then Γ′ ` φ for all Γ′ ⊇ Γ.We can see theories as filters, because it has the two properties for being a filter:
• If a, b ∈ T , thus T ` a ∧ b (see filter condition above), then also a ∧ b ∈ T because Tis closed under logical consequences.
• If a is a theorem, b ∈ T and b ` a, then follows, again by logical consequence, thata ∈ T .
If filters are theories, then we can think of ideals as cotheories.
3.3.3 Examples
• The lattice L itself is both a filter and an ideal.
16
• If 0 is the minimum and 1 is the maximum of L, then {0} is an ideal and {1} a filterof L.
• All the filters and ideals of a chain are principal.
• Let us search for a non-principal filter. In the finite-cofinite algebra the subset F :={Y ∈ FC(X) |Y is cofinite } of all cofinite subsets of X forms a non-principal filter.To check if it is a filter let Y1, Y2 ∈ F . Then X\Y1 and X\Y2 are finite and thusX\Y1 ∪ X\Y2 = X\(Y1 ∩ Y2) is finite. Hence Y1 ∩ Y2 is cofinite and (Y1 ∩ Y2) ∈ F .Now assume Y1 ∈ FC(X), Y2 ∈ F and Y2 ⊆ Y1. It is clear Y1 contains more elementsthan (or the same amount of elements as) Y2, because Y2 is included in Y1. Thenthen complement of Y1 contains less elements than the complement of Y2 which isfinite. We may conclude Y1 is cofinite, and hence Y1 ∈ F .We showed F is a filter. It is a non-principal filter because there is not one element inFC(X) that generates F . Assume there is such an element A ∈ FC(X). Then F ={V ∈ FC(X) |A ⊆ V }. This element A has to be cofinite, because A ∈ F . We writeA := X\{a1, . . . , an}, where n ∈ N. But then the cofinite set B := X\{a1, . . . , an, b}cannot be a member of F , because A 6⊆ B. This is a contradiction, whence F containsall cofinite subsetes of X.We have found a non-principal filter. And this set of cofinite subsets is in fact amaximal proper filter: an ultrafilter. Because if we add another element, then it willbe a finite subset (all the cofinite subsets are already taken). But this finite subset isthe complement of some element in F . Then the empty set will thus be in F . Thisimplies that all subsets of FC belong to F , because the empty set is a subset of everyother subset. We may conclude that F is a maximal proper filter, because addingone element will give us the whole set.
17
Chapter 4
Boolean Algebra
4.1 Definitions and examples
Definition 4.1. Let L be a lattice. We call L
1. distributive if it satisfies the distributive law:
(∀a, b, c ∈ L) a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c).
2. modular if it satisfies the modular law:
(∀a, b, c ∈ L) a ≥ c =⇒ a ∧ (b ∨ c) = (a ∧ b) ∨ c.
Example 4.2. 1. Every powerset lattice P(X) is distributive, because in this case thejoin and meet are the union and intersection.
Definition 4.3. Let L be a lattice with 0 and 1. For a ∈ L, we say b ∈ L is a complementof a if a ∧ b = 0 and a ∨ b = 1. If a has a unique complement, we denote this complementby a′.
Definition 4.4. A lattice L is called a Boolean lattice if
1. L is distributive,
2. L has 0 and 1,
3. each a ∈ L has a (necessarily unique) complement a′ ∈ L.
We see that a Boolean lattice is a special kind of distributive lattice. The necessity of theuniqueness of the complement of every element a ∈ L can be shown as follows. Let S bea distributive lattice in which every element has at least one complement. Assume x ∈ L,and let y and z both be complements of x. Then x ∧ y = 0, x ∧ z = 0, x ∨ y = 1 andx∨ z = 1. This means that y = y ∨ 0 = y ∨ (x∧ z) = (y ∨x)∧ (y ∨ z) = 1∧ (y ∨ z) = y ∨ z,which implies z ≤ y. In the same way z = y ∨ z, which implies y ≤ z. Hence y = z. Eacha ∈ L has a necessarily unique complement.
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Definition 4.5. A Boolean algebra is a structure 〈B;∨,∧,′ , 0, 1〉 such that
1. 〈B;∨,∧〉 is a distributive lattice,
2. a ∨ 0 = a and a ∧ 1 = a for all a ∈ B,
3. a ∨ a′ = 1 and a ∧ a′ = 0 for all a ∈ B.
In the following chapters we will prove that every finite Boolean algebra is isomorphic toP(X) for some finite set X. And after that even that every Boolean algebra is isomorphicto a subalgebra of P(X).
4.1.1 Examples of Boolean algebras
• The finite cofinite subsets of N is a Boolean algebra, but not isomorphic to anypowerset algebra, because of cardinality reason. The collection of all the cofinitesubsets is a non principal ultrafilter.
• P(X) is a Boolean algebra for every X. This algebra is join-generated by its atoms(see next chapter).
• The algebra of formula’s (as strings of symbols) is not a Boolean algebra, becausethe consequence relation restricted to formulas is not antisymmetric as we have seenin section 2.1.4.
• The Lindenbaum algebra (see next section) of classical propositional logic is a Booleanalgebra. As we will see in the next chapter it is atomless.
4.1.2 The Lindenbaum algebra as Boolean algebra
To be a Boolean lattice LA� needs to have a top and bottom and also join and meet haveto be defined as also the complement for every element in LA.Define top and bottom as follows:
1 = [φ] where φ is a tautology 0 = [¬φ] where φ is a tautology
Further define
[φ] ∨ [ψ] = [φ ∨ ψ][φ] ∧ [ψ] = [φ ∧ ψ]
[φ]′ = [¬φ]
LA has become a Boolean lattice, but we will go on and make a Boolean algebra of it.It is easy to see that 〈LA�;≤〉 is a lattice with join and meet given by ∨ and ∧. Then〈LA�;∨,∧〉 is also distributive, and finally there holds [φ]∨ [φ]′ = 1 and [φ]∧ [φ]′ = 0. TheBoolean algebra 〈LA�;∨,∧,′ , 0, 1〉 we obtained is also known as the Lindenbaum algebra.
19
4.2 The finite intersection property
A subset A of a Boolean algebra has the finite intersection property (from now on called‘fip’) if the infimum of any finite subset of A is not equal to 0. In this section we willdiscover that we can extend precisely these subsets to proper filters, and further how toextend a proper filter to an ultrafilter. All these steps can also be done in propositionallogic almost in the exact same way.
4.2.1 The fip and the interpretation in propositional logic
Let B be a Boolean algebra and let A be a set of elements of B. If A has the fip, then forany element x ∈ B either A ∪ {x} or A ∪ {x′} has the fip.To show this assume neither one of them has the fip. Then
∧(A ∪ {x}) = 0 and
∧(A ∪
{x′}) = 0. We can write this in a different way, having in mind A does have the fip:(∧A) ∧ x = a ∧ x = 0 and (
∧A) ∧ x′ = a ∧ x′ = 0, where a 6= 0. Then follows
0 = (a ∧ x) ∨ (a ∧ x′) = a ∧ (x ∨ x′) = a ∧ 1 = a,
which is a contradiction. It cannot be the case that neither one of them has the fip.The proof of the fact that, if one of the two has the fip, then the other cannot have thefip, will not be showed here.The following lemma shows that any subset that has the fip can be extended to a filter.Let therefor B be a Boolean algebra, and define
A0 = {x ∈ B | for some a ∈ A, a ≤ x} and Ac = {inf(X) |X is a finite subset ofA}
Lemma 4.6. For any subset A of a Boolean algebra, the set (Ac)0 is a filter. Any filtercontaining A contains (Ac)0. (Ac)0 is proper if and only if A has the fip.
The interpretation in propositional logic is very interesting. We saw in the previous chapterthat theories can be seen as filters. The lemma above shows that a set of theorems canbe extended to a consistent theory if and only if the set of theorems does not cause acontradiction. The next question is then if we could extend a set of theorems to a maximalconsistent theory. For the answer we need to prove the ’Ultrafilter Theorem’, but first weconsider the following lemma.
Lemma 4.7. If F is a filter in a Boolean algebra B, F is an ultrafilter if and only if foreach x ∈ B either x ∈ F of x′ ∈ F , but not both.
In propositional logic we can interpretate this lemma now in terms of maximal consistenttheories. An ultrafilter is a maximal proper filter, it has no proper extension. Everyextension of an ultrafilter would contain the element 0, which makes the extension equalto the Boolean algebra itself.A maximal consistent theory is an ultrafilter (in the appropriate context, see 3.3.2, becauseextending the maximal consistent theory would give us an inconsistent theory. A maximal
20
consistent theory always contains either φ or ¬φ, for all φ in the set of theorems. In fact,extending this theory by adding a theorem leads to a contradiction: φ ∧ ¬φ, and thus toan inconsistent theory.But let us come back to our question. The following theorem, the ’Ultrafilter Theorem’, isthe last piece of the puzzle. Adding this theorem, we will be able to construct a maximalconsistent theory out of a set of (consistent) theorems.
Theorem 4.8. (The Ultrafilter Theorem) Every filter in a Boolean algebra can beextended to an ultrafilter.
We will get to this theorem in Chapter 6.
21
Chapter 5
Finite representation
5.1 Join-irreducible elements
Definition 5.1. Let L be a lattice. An element x ∈ L is join-irreducible if
1. x 6= 0
2. x = a ∨ b implies x = a or x = b for all a, b ∈ L.
We use the notation J (L) to indicate the set of all join-irreducible elements of L. In thesame way we write M(L) for the set of meet-irreducible elements of L, which are defineddually.Let P be an ordered set, and let Q ⊆ P . Then Q is called join-dense if for every elementa ∈ P there is a subset A of Q such that a =
∨pA.
Example 5.2. For every two elements x and y in a chain we have either x ≤ y or y ≤ x.This implies that either x ∨ y = x or x ∨ y = y and hence every element in a chain isjoin-irreducible.In a diagram it is easy to distinguish the join-irreducible elements, these are namely thepoints with only one lower cover. Assume for example that point x has two lower coversa and b. This means x = a ∨ b while a < x and b < x. A point with two lower covers cantherefore never be a join-irreducible element.At last we take the lattice P(X). The join-irreducible elements of P(X) are the singletons{x} for x ∈ X, because those elements have only one lower cover, which is 0.
Picture of a chain, where join-irreducible elements cover each other.Picture of a Boolean algebra (P({1, 2, 3})) where the join-irreducible elements form ananti-chain.
Lemma 5.3. Let L be a lattice with least element 0. Then
1. 0 ≺ x implies x ∈ J (L)
2. if B is a Boolean lattice, x ∈ J (L) implies 0 ≺ x.
22
Proof. For the first part suppose by way of contradiction that 0 ≺ x and x /∈ J (L). Thusx = a ∨ b and a < x and b < x. Now 0 ≺ x implies a = b = 0 and hence x = 0 which is acontradiction (because 0 ≺ x).For the other part of the proof assume L is a Boolean lattice and x ∈ J (L). Then x = a∨bimplies x = a or x = b. Suppose that 0 ≤ y < x. To prove that 0 ≺ x we need to showthat y = 0. The fact that y ∨ y′ = 1 and the distributivity of the Boolean lattice gives us
x = x ∨ y = (x ∨ y) ∧ (y ∨ y′) = y ∨ (x ∧ y′).Now x = y ∨ (x ∧ y′) and x < y imply x = x ∧ y′. Because y < x ≤ y′ we obtainy = x ∧ y ≤ y′ ∧ y = 0. Hence y = 0 and we proved that 0 ≺ x.
5.2 Atoms
In the previous example of P(L) we have seen that the join-irriducible elements wereexactly the elements that cover 0.
Definition 5.4. Let L be a lattice with least element 0. Then a ∈ L is called an atom if0 ≺ a.
Lemma 5.5. This will be lemma 5.4 of the book. (We need it for the next proof.)
The set atoms of L is denoted by A(L). The lattice L is called atomic if for every x ∈ L(where x 6= 0) there is an a ∈ A(L) such that a ≤ x.
Theorem 5.6. Let B be a finite Boolean algebra. Then the map
η : a 7−→ {x ∈ A(B) | x ≤ a}is an isomorphism of B onto P(X) where X = A(B), with the inverse of η given byη−1(S) =
∨S for S ∈ P(X).
Proof. We will first show that η maps B onto P(X). Clearly ∅ = η(0). Let S = {a1, . . . , ak}be a set of atoms, thus S ∈ P(X), and define a =
∨S.
We claim S = η(a). Because for all i, ai is an atom and ai ≤ a = a1 ∨ . . .∨ ak, there holdsS ⊆ η(a). For the other inclusion let x be an atom such that x ≤ a = a1∨. . .∨ak. For each iwe have 0 ≤ x∧ai ≤ x. Assume that x∧ai = 0 for all i. Then because x ≤ a and because ofdistributivity we have x = x∧a = x∧(a1∨. . .∨ak) = (x∧a1)∨. . .∨(x∧ak) = 0∨. . .∨0 = 0.But x is an atom, and covers 0, hence x > 0. Contradiction! There must be at least onej such that x ∧ aj = x. But both x and aj are atoms, whence x = aj. For our arbitrarilychosen x ∈ η(a) we know that x ∈ S, and hence η(a) ⊆ S. We can conclude that S = η(a),and thus that η maps B onto P(X).
To show that η is an isomorphism let a, b ∈ B. If a ≤ b we easily see that η(a) ⊆ η(b)because { x ∈ A (B) | x ≤ a} ⊆ { x ∈ A (B) | x ≤ b}.Assume η(a) ⊆ η(b). By lemma 5.4 we have a =
∨{ x ∈ A (B) | x ≤ a} =
∨η(a),
and in the same way b =∨η(b). Because η(a) ⊆ η(b) we obtain a =
∨η(a) ≤
∨η(b) = b.
The map η is an isomorphism of B onto P(X).
23
Chapter 6
Maximality Principles
6.1 Prime ideals and filters
Definition 6.1. Let L be a lattice. A non-empty subset J of L is called an ideal if
1. a, b ∈ J implies a ∨ b ∈ J and
2. a ∈ L, b ∈ J , and a ≤ b imply a ∈ J .
We call J a proper ideal if J 6= L. A proper ideal J of L is said to be prime if a, b ∈ Land a ∧ b ∈ J imply a ∈ J or b ∈ J . The set of prime ideals of L is denoted by Ip(L).Dually defined we have a filter, a proper filter and a prime filter. We denote the set ofprime filters of L by Fp(L).
Lemma 6.2. A subset J of a lattice L is a prime ideal iff L\J is a prime filter.
Proof. Let L be a lattice and J a prime ideal. Then the following three properties hold:
• a, b ∈ J implies a ∨ b ∈ J ,
• a ∈ L, b ∈ J and a ≤ b imply a ∈ J ,
• a, b ∈ L and a ∧ b ∈ J imply a ∈ J or b ∈ J .
To prove that L\J is a prime filter we need to prove the three dual properties. For thefirst assume a, b ∈ L\J . If a ∧ b ∈ J then a ∈ J or b ∈ J because of the third property ofJ , which is a contradiction. Hence a, b ∈ L\J implies a ∧ b ∈ L\J .Then assume a ∈ L, b ∈ L\J and a ≥ b. If a ∈ J , the second property of J tells us thata ∈ J , b ∈ L and b ≤ a imply b ∈ J , which is again a contradiction. Hence a ∈ L, b ∈ L\Jand a ≥ b imply a ∈ L\J .At last assume that a, b ∈ L and a∨ b ∈ L\J . If both a and b are elements of J we have acontradiction, because the first property of J claims that if a, b ∈ J also a ∨ b ∈ J , whichis not the case. Hence a, b ∈ L and a ∨ b ∈ L\J imply a ∈ L\J or b ∈ L\J .The three dual properties hold for L\J and thus L\J is a prime filter. The proof of theconverse is obtained dually.
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Lemma 6.3. Let L be a distributive lattice and let x ∈ L with x 6= 0 in case L has a zero.Then te following are equivalent:
1. x is join-irreducible
2. if a, b ∈ L and a ≤ b then x ≤ a or x ≤ b.
Proof. 1.⇒ 2. Assume x ∈ J (L). Thus for x = a ∨ b we have x = a or x = b. Supposex ≤ a ∨ b. Then by distributivity of L and because x ≤ a ∨ b, x = x ∧ (a ∨ b) =(x∧ a)∨ (x∧ b). And the join-irreducibility of x implies x = a∧ b or x = x∧ b, hencex ≤ a or x ≤ b. We conclude that if a, b ∈ L and a ≤ b then x ≤ a or x ≤ b.
2.⇒ 1. Assume now that if a, b ∈ L and a ≤ b then x ≤ a or x ≤ b. Suppose x = c ∨ d.We want to show that x = c or x = d. Because x = c∨ d we can write also x ≤ c∨ d.This implies that x ≤ c or x ≤ d. But x = c ∨ d implies c ≤ x and d ≤ x. Weobtained x ≤ c ≤ x or x ≤ d ≤ x, hence either x = c or x = d. Which means that xis join-irreducible.
Lemma 6.4. Let L be a lattice satisfying (DCC). Suppose a, b ∈ L and a 6≤ b. Then thereexists x ∈ J (L) such that x ≤ a and x 6≤ b.
Proof. Suppose a, b ∈ L and a 6≤ b. Let S = {x ∈ L |x ≤ a and x 6≤ b}. Because a ∈ S, Sis not-empty. By lemma 2.39 (dual) the subset S ⊆ L has a minimal element. We need toproof that this minimal element, say x, is in J (L).Assume x 6∈ J (L). Then x = c ∨ d and c < x and d < x. Because x is a minimal elementof S, c and d are not in S. Thus c ≤ b and d ≤ b. Hence x = c ∨ d ≤ b, which impliesx ≤ b. But x ∈ S, so we have reached a contradiction. Hence x ∈ J (L) and x ≤ a andx 6≤ b.
Lemma 6.5. Let L be a finite distributive lattice and let a ∈ L. Then the map x 7−→ L\ ↑xis an order-isomorphism of J (L) onto Ip(L), that maps {x ∈ J (L) |x ≤ a} onto {I ∈Ip(L) | a 6∈ I}.
Proof. We first prove that ↑x is a prime filter. For a, b ∈↑ x holds a ≥ x and b ≥ x whichimplies a ∧ b ≥ x. Hence a ∧ b ∈ ↑x.Then suppose a ∈ L, b ∈ ↑x and a ≥ b. This means that b ≥ x and a ≥ b ≥ x and thusa↑x.At last suppose a, b ∈ L and a ∨ b ∈↑ x. Because a ∨ b ≥ x, lemma 5.1 gives us a ∈↑ x orb ∈↑ x.We conclude that ↑x is a prime filter, because it has the three properties it needs to have.In a finite lattice every ideal/filter is principal, i.e. of the form ↑x or ↓x which is the up-or downset generated by a. As we have seen in lemma 6.1 above, L\↑x is a prime ideal.Onto because, take I ∈ {I ∈ Ip(L) | a 6∈ I} = {L\↑x | a 6∈ L\↑x and x ∈ J (L)}. For everyideal L\↑x there is a x ∈ J (L), because we take the complement of the filter generated by
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that x.Lemma 1.30 helps us to say dually that x ≤ y iff ↑y ⊆ ↑x which gives us the order-embedding. Hence the map is an order-isomorphism.
Lemma 6.6. Let L be a finite distributive lattice and let a 6≤ b in L. Then there existsI ∈ Ip(L) such that a 6∈ I and b ∈ I.
Proof. Let L be a finite distributive lattice. Because L is finite the (DDC) holds. Supposenow that a 6≤ b. Then by lemma 2.45 there is a x ∈ J (L) such that x ≤ a and x 6≤ b.By lemma 10.8 the map x 7−→ L\↑x is an order-isomorphism of J (L) onto Ip(L), thatmaps {x ∈ J (L) |x ≤ a} onto {I ∈ Ip(L) | a 6∈ I}. The x ∈ J (L) is mapped onto L\↑x,where a 6∈ L\↑x. We want to know now whether b ∈ L\↑x or not. Suppose b 6∈ L\↑x,then b ∈ ↑x, which means that x ≤ b. Contradiction! Hence b ∈ L\↑x. We have found theI ∈ Ip(L) such that a 6∈ I and b ∈ I.
6.2 Maximal ideals and ultrafilters
Definition 6.7. Let L be a lattice and let I ⊆ L be an ideal. We call I a maximal ideal if,for every ideal J ⊆ L which contains I properly (thus I ( J), we have J = L. A maximalfilter, also kown as an ultrafilter, is dually defined.
Theorem 6.8. Let L be a distributive lattice with 1. Then every maximal ideal in L isprime. (Dually in a distributive lattice with 0, every ultrafilter is a prime filter.)
Proof. Let L be a ditributive lattice with 1. Let I be a maximal ideal in L. An ideal I isprime if a, b ∈ L and a ∧ b ∈ I imply a ∈ I or b ∈ I.Suppose a ∧ b ∈ I and without loss of generality suppose that a 6∈ I. We need to provethat b ∈ I. Define Ia = ↓{a ∨ c | c ∈ I}. This is an ideal because
• let a ∨ x1 and a ∨ x2 be elements of Ia, thus x1, x2 ∈ I. Then, because x1 ∨ x2 ∈ I,we have (a ∨ x1) ∨ (a ∨ x2) = a ∨ (x1 ∨ x2) ∈ Ia.
• if a∨x1 ∈ L, a∨x2 ∈ Ia and a∨x1 ≤ a∨x2, then x1 ≤ x2 hence x1 ∈ I. We concludethat a ∨ x1 ∈ Ia.
Let now c1 ∈ I. Then a∨ c1 ≥ c1 implies c1 ∈ Ia, because Ia is a downset. Hence is I ⊆ Ia.We started with I being a maximal ideal, which implies that Ia = L. Because 1 ∈ Ia,there is a d ∈ I such that 1 = a ∨ d. We get d ∨ (a ∧ b) ∈ I, and by distributivity(d ∨ a) ∧ (d ∨ b) ∈ I ⇔ 1 ∧ (d ∨ b) ∈ I ⇔ d ∨ b ∈ I. Because d ∨ b ≥ b we obtain b ∈ I,which we wanted to prove. Every maximal ideal in L is prime.
Theorem 6.9. Let B be a Boolean lattice and let I be a proper ideal in B. Then thefollowing are equivalent:
1. I is a maximal ideal
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2. I is a prime ideal
3. for all a ∈ B holds: a ∈ I ⇐⇒ a′ 6∈ I.
Proof. We will prove 1.⇒ 2.⇒ 3.⇒ 1. to obtain the equivalence.
1.⇒ 2. Assume I is a maximal ideal. To prove that I is also a prime ideal we need toshow that for a ∧ b = x ∈ I either a ∈ I or b ∈ I, or both. Suppose x = a ∧ b ∈ Iand assume a 6∈ I. Required is now that b ∈ I.Define Ia = ↓{a ∨ c | c ∈ I}. For all c ∈ I we have c ≤ a ∨ c and hence c ∈ Ia,which implies I ⊆ Ia. But because I is a maximal ideal, and 0 ∈ I we obtaina ∨ 0 = a ∈ Ia ⇒ I ⊂6= Ia ⇒ Ia = B. Now 1 ∈ Ia, thus there is a d ∈ B such thata ∨ d = 1. But then
(a ∧ b) ∨ d = (a ∨ d) ∧ (b ∨ d) ∈ I =⇒ b ∨ d ∈ I =⇒ b ≤ b ∨ d ∈ I =⇒ b ∈ I.
The maximal ideal is also a prime ideal.
2.⇒ 3. Assume I is a prime ideal. Because a ∧ a′ = 0 ∈ I and I is a prime ideal eithera ∈ I or a′ ∈ I. Assume without loss of generality that a ∈ I. Suppose a′ ∈ Iaswell. Then also a ∨ a′ = 1 ∈ I which is a contradiction because I is proper. Hencea ∈ I ⇐⇒ a′ 6∈ I.
3.⇒ 1. Assume for all a ∈ B it is the case that a ∈ I ⇐⇒ a′ 6∈ I. Let J be an ideal suchthat I ( J and fix a ∈ J\I. Because a 6∈ I, we have a′ ∈ I. But then 1 = a∨ a′ ∈ J ,which implies J = B, and hence I is maximal.
Let M = (X, . . . , V ) be a model and Tw = {φ |w �V φ} be a theory. The valuation V isa function that maps a proposition letter to the subset of X where this letter is true. Thetheory is thus a maximal consistent theory, because for every theorem φ either φ ∈ Tw or¬φ ∈ Tw.Every consistent theory is the intersection of all the maximal consistent theories extendingit. If Γ 6` φ, then there is a maximal consistent theory such that Γ ⊆ Γ′ and φ 6∈ Γ′. HenceΓ = ∩{Γ′ |Γ′ is a maximal consistent theory and Γ ⊆ Γ′}.
Theorem 6.10. (The ultrafilter theorem) Every filter in a Boolean algebra can be extendedto an ultrafilter.
Proof. Let F be a filter in a Boolean algebraB. Define F := {Fi ⊆ B | Fi is a filter and F ⊆Fi}. Clearly F is not-empty, because F ∈ F . Order F by inclusion, and we obtain theposet 〈F ;⊆〉.We claim now that every chain in 〈F ;⊆〉 has an upper bound. To show this let S = {Si ∈F | i ∈ I} be a chain in F . Define S :=
⋃i∈I Si. If x, y ∈ S, then there are i, j ∈ I such
that x ∈ Si and y ∈ Sj. Without loss of generality we may assume Si ⊆ Sj, because S is
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a chain. Thus x ∈ Si ⊆ Sj, imply x ∈ Sj. Because Sj is a filter also x ∧ y ∈ Sj ⊆ S. Nowsuppose z ∈ B and x ≤ z. Then z ∈ Si ⊆ Sj ⊆ S, hence z ∈ S. Finally, for all i ∈ I, wehave 0 6∈ Si, and hence 0 6∈ S. We can conclude now that S is a proper filter containingF , which means S ∈ S. Hence S is an upper bound for S in F .Every chain in 〈F ;⊆〉 has an upper bound. By Zorn’s Lemma we can conclude that Fhas a maximal element. This maximal element is a filter G that contains F and for everyfilter Fi such that if Fi ⊇ G, holds Fi ⊆ G (because F ⊆ Fi, hence Fi = G and G is aultrafilter.
If Fm is a language over a denumerable set of variables, then |Fm| = ℵ0 and we have analternative proof of the ultrafilter theorem for the Boolean algebra 〈Fm/≡,≤≡ 〈 whichdoes not need Zorn’s Lemma (or any other property equivalent to AC):This is done in the so called Henkin construction. Since |Fm| = ℵ0, we can enumerate theformulas of Fm, Fm = {γ0, γ1, . . . , γi, . . . | i ∈ N}. Let T be our consistent theory. Definea chain Γ0 ⊆ Γ1 . . . ⊆ Γn ⊆ . . . for n ∈ N by induction as follows:Γ0 := T . Let n ≥ 0. Suppose we have defined Γn. In order to define Γn+1 consider γn and
Γn+1 :={
Γn∪{γn} if Γn∪{γn}6`⊥Γn∪{¬γn} if Γn∪{γn}`⊥
Let Γ :=⋃n∈N. It now rests us to show that Γ is a maximal consistent theory. This is
obvious, because for every formula γ in our language is decided whether it is contained inΓ or not, and if not then ¬γ is contained in Γ. Adding a formula will lead to absurdity,and hence Γ is a maximal consistent theory.
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Chapter 7
Representation
We have seen that a finite Boolean algebra is isomorphic to a powerset algebra. Butit would be nice to have an isomorphism not only for the finite case, but for Booleanalgebras in general. We will make therefor the set of the prime ideals of any Booleanalgebra into a topology space in such a way that it is homeomorphic to a Boolean algebra.At the end we will see that Boolean algebras are dual to Boolean spaces, and boundeddistributive lattices are dual to Priestley spaces. For the Boolean algebras we will use theStone representation theorem, and for the bounded distributive lattices there is Priestley’srepresentation theorem.Also for the logic this is a very nice result. We will give a same order-homeomorphismbetween the Lindenbaum algebra (which we will first manipulate to make it into a Booleanalgebra) and a canonical model whose points are maximal consistent theories.
7.1 Topology
For the Priestley space and the Boolean space we need a bit of topology. This sectiondescribes some basics of topology. There are also two lemmas that are needed for someproofs in this chapter.
A topology on X is a family of subsets of X closed under finite intersections, arbitraryunions and which contains ∅ and X. A topological space is a set X together with a topol-ogy on X, notation (X; T ).Let B be a family of subsets of X, containing ∅ and X, and which is closed under finiteintersections. Then B is a basis for the topology T which contains all arbitrary unions ofsets in B.Let S be a family of subsets of X, containing ∅ and X. We can form B by taking all thefinite intersections of sets in S. Then B is again a basis for the topology containing allarbitrary unions of sets in B. We call S a subbasis for T .Let (X; T ) and (X ′; T ′) be topological spaces and f : X −→ X ′ a map. Then f is contin-uous iff U ⊆ X ′ open implies f−1(U) ⊆ X open. And f is said to be a homeomorphism
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if f is bijective and if both f and f−1 are continuous.A topological space (X; T ) is said to be Hausdorff if for all x, y ∈ X such that x 6= ythere are U, V ⊆ X open such that x ∈ U , y ∈ V and U ∩ V = ∅.A topological space (X; T ) is compact if for every open cover U of X there is a finitesubcover U ⊆ U of X.
Lemma 7.1. Let (X; T ) be a compact Hausdorff space. A subset Y of X is compact iff Yis closed.
Proof. Assume Y ⊆ X is a compact subset of X. Fix a z ∈ X\Y . Then for all y ∈ Y wehave y 6= x and because X is Hausdorff there are open Uy and Vy such that z ∈ Uy andy ∈ Vy. Define Y := {Vy | y ∈ Y, z 6∈ Vy}. This is an open cover of Y . We assumed Y iscompact, thus there are y1, . . . , yn ∈ Y such that Y ⊆
⋃ni=1 Vyi . Define then Uz :=
⋂ni=1 Uyi
(note this is an open set). Then z ∈ Uz and Uz ∩ Y = ∅. For all z ∈ X\Y we can find aUz such that Uz ∩ Y = ∅. We conclude X\Y =
⋃z∈X\Y Uz is open, and hence Y is closed.
Assume then that Y is a closed subset of X. Let U = {Ui}i∈I be an open cover of Y .Because Y is closed, X\Y is open, and hence U ∪X\Y is an open cover of X. We knowthat X is compact, thus there is a finite subcover, U1 ∪ . . . ∪ Un ∪X\Y , of X. But thenY ⊆ U1 ∪ · · · ∪ Un. We found a finite subcover for an arbitrary open cover of Y , hence Yis compact.
Lemma 7.2. Alexander’s Subbasis LemmaLet (X; T ) be a topological space and S be a subbasis for T . Then X is compact if everyopen cover of X by members of S has a finite subcover.
7.2 Stone’s representation
7.2.1 Boolean space
The Stone’s representation theorem will give us a order-homeomorphism between Booleanalgebras and Boolean spaces. First we will define these terms:
Definition 7.3. A topological space (X; T ) is called a totally disconnected space if for allx, y ∈ X such that x 6= y there exists a clopen (open and closed) subset U ⊆ X such thatx ∈ U and y 6∈ U .
Definition 7.4. A topological space is called a Boolean space if it is both compact andtotally disconnected.
Now we have to ask ourselves which set X to choose to achieve the right representation.It turns out to be the prime ideal space of the Boolean algebra, 〈Ip(B); T 〉, that gives us
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exact the right properties for the order-homeomorphism. The topology on Ip(B) will bedefined as follows:
T := {U ⊆ Ip(B) |U is a union of Xa for a collection of a ∈ B}
where Xa := {I ∈ Ip(B) | a 6∈ I}. These sets Xa of prime ideals of B are clopen sets (wewill show in this section that the clopen subsets of X are exactly the sets Xa for a ∈ B).This is a nice choise, because in a topological space the family of clopen subsets form aBoolean algebra. (have I showed this yet? )To show that 〈Ip(B); T 〉 is in fact a Boolean space we will first show that the space iscompact.
Proposition 7.5. Let B be a Boolean algebra. Then the prime ideal space 〈Ip(B); T 〉 iscompact.
Proof. We have to show that for every open cover of X := Ip(B) there exist finitely manymembers of this open cover whose union is X. Let U be an open cover of X. The collectionB := {Xa | a ∈ B} (where Xa = {I ∈ Ip(B) | a 6∈ I}) forms a basis for the topology T . Allunions of Xa are open sets, and without loss of generality we may assume that U ⊆ B. Wewrite U = {Xa | a ∈ A} where A ⊆ B. Let J be the smallest ideal containing A, that isJ = {b ∈ B | b ≤ a1 ∨ . . . ∨ an for some a1, . . . , an ∈ A}. Suppose J is not proper. Then1 ∈ J , and 1 = a1 ∨ . . . ∨ an for some a1, . . . , an ∈ A. We obtain
Ip(B) = X = X1 = Xa1∨...∨an = {I ∈ Ip(B) | a1 ∨ . . . ∨ an 6∈ I}= {I ∈ Ip(B) | a1 6∈ I or . . . or an 6∈ I}= {I ∈ Ip(B) | a1 6∈ I} ∪ . . . ∪ {I ∈ Ip(B) | an 6∈ I}=
⋃i = 1, nXai
Hence we found a finite subcover {Xa1 , . . . , Xan} of U that covers X.Suppose that J is proper. Then (BPI) confirms us the existence of a prime ideal I suchthat J ⊆ I. But then A ⊆ I ∈ Ip(B) = X. Also a ∈ I implies I 6∈ {I ∈ Ip(B) | a 6∈ I} forall a ∈ A. Hence I 6∈ Xa for all a ∈ A. This means that U doesn’t cover X. Contradiction.We conclude that J has to be not proper which implies that for every open cover of Xthere exists a finite subcover. Hence the prime ideal space {Ip(B); T } is compact.
Now we have proved the compactness, we have to prove that {Ip(B); T } is totally dis-connected. We therefor need the following proposition, which also shows that the clopensubsets of Ip(B) are exactly the sets Xa for a ∈ B, as we promised above.
Proposition 7.6. Let X := Ip(B) and let 〈X; T 〉 be the prime ideal space of the Booleanalgebra B. Then the clopen subsets of X are exactly the sets Xa for a ∈ B. Further, givendistinct points x, y ∈ X, there exists a clopen subset V of X such that x ∈ V and y 6∈ V .
Proof. The sets Xa form the basis for the topology and hence are open sets. Then X\Xa =I(B)\{I ∈ Ip(B) | a 6∈ I}. Theorem 10.12 tells us that a 6∈ I iff a′ ∈ I. Thus X\Xa ={I ∈ Ip(B) | a′ 6∈ I} = Xa′ ∈ B. Because Xa′ is open, all sets Xa are closed and open:
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clopen.Let I1 and I2 be two distinct prime ideals (i.e. points of X). Without loss of generalitywe have a ∈ I1\I2. Then I2 ∈ Xa but I1 6∈ Xa. We found our clopen V := Xa such thatx := I2 ∈ V and y := I1 6∈ V . Hence {X; T } is Hausdorffs (???).Let now U be a clopen subset of X. Then U :=
⋃a ∈ A for some A ⊆ B. But U is also
closed, and by A7 compact. Hence there is a finite A1 ⊆ A such that U =⋃a ∈ A1 =
Xa1∨...∨an = X∨A1 . The clopen subsets of X are exactly the sets Xa for a ∈ B.
Lemma 7.7. Let (X; T ) be a Boolean space.
1. Let Y be a closed subset of X and x 6∈ Y . Then there exists a clopen set V such thatY ⊆ V and x 6∈ V .
2. Let Y and Z be closed disjoint subsets of X. Then there exists a clopen set U suchthat Y ⊆ U and Z ∩ U = ∅.
Proof. 1. Let Y be a closed subset of X and x 6∈ Y . Then proposition ... (11.3) tells us thatfor every y ∈ Y there exists a clopen subset Vy of X such that y ∈ Vy and x 6∈ Vy. Thus⋃y∈Y Vy ⊇ Y is an open cover of Y and x 6∈
⋃y∈Y Vy. Because Y is closed, lemma ... implies
the compactness of Y , and thus there is a finite subcover of Y , namely V ′ :=⋃ni=1 Vi ⊆ Y .
A finite set of clopen sets is clopen, and hence V ′ is a clopen set such that Y ⊆ V ∗ andx 6∈ V ′.2. Let Y and Z be closed disjoint subsets of X. The Boolean space (X; T ) is totallydisconnected which means that for x, y ∈ X such that x 6= y there is a clopen subset V ofX such that x ∈ V and y 6∈ V . We define Uy,z
1 as a clopen subset of X such that y ∈ Uy,z1
and z 6∈ Uy,z1 . Define then U1 := {Uy,z
1 | y ∈ Y }. This is an open cover of Y . Because Y isclosed and the Boolean space is Hausdorff, we know by the lemma that Y is compact. Wecan now find a finite subcover U z
1 :=⋃
1≤i≤n Uyi,z1 of Y such that z 6∈ U z
1 .After this we define U2 := {X\U z
1 | z ∈ Z} where X\U z1 6⊇ Y and z ∈ X\U z
1 holds for allz ∈ Z. This U2 is again an open cover, this time of Z and Y 6⊆ U z
2 . For the same reasonsalso Z is compact, and hence we can find a finite subcover U z
2 :=⋃
1≤i≤mX\Uzi1 of Z such
that Y 6⊆ U z2 .
Define finally U := X\U z2 = X\(
⋃1≤i≤mX\U
zi1 ) =
⋂1≤i≤m U
zi1 . Then Y ⊆ U and Z ∩U =
∅.
Lemma 7.8. Let Y be a Boolean space, let B be the algebra PT (Y ) of clopen subsets of Yand let X be the dual space of B. Then Y and X are homeomorphic.
Proof. Define ε : Y −→ X by ε(y) := {a ∈ B | y 6∈ a}. Then as we see below ε(y) is aprime ideal in B:
• Assume c, d ∈ ε(y). Then y 6∈ c ∪ d (= c ∨ d), whence y 6∈ c and y 6∈ d.
• If c ∈ B, d ∈ ε(y) and c ⊆ d, then y 6∈ d and thus y 6∈ c. This implies c ∈ ε(y).
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• Let I be an ideal such that ε(y) ( I. Then there is an element x ∈ I\ε(y), thusy ∈ x. Both x and x′ are clopen sets and y 6∈ x′, hence x′ ∈ ε(y) ⊂ I. Becausex, x′ ∈ I and because I is an ideal, x ∨ x′ ∈ I ⇒ I = X.
We now want to prove that ε is a continuous bijection, because then by the lemma ofsection Topology it follows that ε is a homeomorphism.Let us start with the injectivity. Let y 6= z in Y . Because Y is totally disconnected thereis a clopen subset a of Y such that y ∈ a and z 6∈ a. But then ε(y) 6= ε(z). Hence ε is aninjection.To show that ε is continuous we need to show that every ε−1(Xa) is clopen in Y for everya ∈ B. Because Xa = {I ∈ X | a 6∈ I}, we obtain
ε−1(Xa) = {y ∈ Y | ε(y) ∈ Xa}= {y ∈ Y | a 6∈ ε(y)}= {y ∈ Y | a 6∈ {b ∈ B | y 6∈ b}}= {y ∈ Y | a ∈ {b ∈ B | y ∈ b}}= {y ∈ Y | y ∈ a}= a.
This is obviously a clopen subset of Y , and hence ε is continuous.Finally, for surjectivity, assume there is a x ∈ X\ε(Y ). By Topology lemma ... we know,because ε is continuous, that ε(Y ) is compact, and hence ε(Y ) is a closed subset of X.By proposition ... and lemma ... above there is a clopen subset Xa of X such thatε(Y ) ⊆ Xa and x 6∈ Xa. But then x ∈ X\Xa = Xa′ and Xa′ ∩ ε(y) = ∅, which implies∅ = ε−1(Xa′) = a′. This is a contradiction and hence ε is a surjection.By lemma ... of the section Topology it follows now that ε is a homeomorphism.
7.3 Priestley
Theorem 7.9. Let L be a bounded distributive lattice. Then the prime ideal space 〈Ip(L); T 〉is compact.
Proof. We want to prove that for every open cover U of X := Ip(L) there is a finitesubcover of X. Alexander’s Subbasis Lemma helps us to prove only that for every opencover of X that consists of sets of a subbasis S there is a finite subcover of X.Let U := {Xb | b ∈ A0}∪ {X\Xc | c ∈ A1}, where A0, A1 ⊆ L. Let J be the ideal generatedby A0 (which is {0} if A0 = ∅) and let G be the filter generated by A1 (which is {1} ifA1 = ∅). Suppose J ∩G = ∅. With (DPI) we can find a I ∈ Ip(L) and F = L\I ∈ Fp(L)such that J ⊆ I and G ⊆ F . Thus G ∩ I = ∅. Now b ∈ I implies I 6∈ Xb for all b ∈ A0.Further c 6∈ I implies I ∈ Xc for all c ∈ A1 and hence I 6∈ X\Xc for all c ∈ A1. But thenU doesn’t cover Ip(L), which is a contradiction.We may conclude now that J ∩ G 6= ∅. Take an a ∈ J ∩ G, and assume A0 and A1
are both non-empty. Then because a ∈ J and a ∈ G there a are a1, . . . , an ∈ A0 suchthat a ≤ a1 ∨ . . . ∨ an and there are b1, . . . , bm ∈ A1 such that a ≤ b1 ∧ . . . ∧ bm. Hence
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b1 ∧ . . . ∧ bm ≤ a ≤ a1 ∨ . . . ∨ an. We obtain X = X1 = Xa1∨...∨an ∪ (X\Xb1∧...∧bm) =⋃ni=1Xan ∪ (X\
⋂mi=1Xbi) = Xa1 ∪ . . . ∪Xan ∪X\Xb1 ∪ . . . ∪X\Xbm . We have found the
finite subcover.
Definition 7.10. A Priestley space is a totally order-disconnected space that is compact.
Lemma 7.11. Let 〈X;≤, T 〉 be a Priestley space.
1. x ≤ y in X ⇐⇒ y ∈ U implies x ∈ U for every U ∈ OT (X).
2. • Let Y be a closed downset in X and let x 6∈ Y . Then there exists a clopendownset U such that Y ⊆ U and x 6∈ U .
• Let Y and Z be disjoint closed subsets of X such that Y is a down-set andZ is an up-set. Then there exists a clopen down-set U such that Y ⊆ U andZ ∩ U = ∅.
Proof. 1. In the Priestley space 〈X;≤, T 〉 there is a clopen downset U ∈ OT (X) such thatx ∈ U and y 6∈ U for all x, y ∈ X with x 6≥ y.Let x, y ∈ X and x ≤ y. Then for the case x = y we have for all U ∈ OT (X) that if y ∈ Uthen x ∈ U . For the case x < y (thus x 6≥ y) we have property of a Priestley space namelythat there is a U ∈ OT (X) such that x ∈ U and y 6∈ U . But if x < y and y ∈ U thenbecause U is a down-set also x ∈ U . Hence if y ∈ U then also x ∈ U , but the other way isnot always true.2. Let Y be a closed down-set in X and let x 6∈ Y . For each y ∈ Y holds x 6≤ y otherwisex ∈ Y . Because of the Priestley space there is a clopen down-set U ∈ OT (X) such thaty ∈ Uy and x 6∈ Uy. Thus {Uy | y ∈ Y } forms an open cover of Y . Because Y is closed, thelemma ... sais Y is compact. Hence there is a finite subcover {Uyi | y1, . . . , yn ∈ Y } of Y .By construction x 6∈
⋃ni=1 Uyi and Y ⊆
⋃ni=1 Uyi .
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Chapter 8
Populaire samenvatting
Stel je voor dat iemand een prijs wint en op reis mag met een paar familieleden. Maar welmet familieleden die op een speciale manier geselecteerd worden, namelijk als volgt: Webeginnen met de winnaar. Alle kinderen van de winnaar mogen mee. En vervolgens ookde kinderen van deze personen, enzovoort. Daarnaast mogen ook die mensen mee, die eenouder zijn van twee mensen uit de al geselecteerde mensen.
Concreter (zie ook het plaatje): stel je voor dat Piet de prijs wint. Clara en Joost, dekinderen van Piet, mogen dan mee op reis. Vervolgens mag ook de moeder van Clara enJoost mee op reis, omdat zij de ouder is van deze twee. Als Piet geen broers en zussen zouhebben, dan mogen de vader en moeder van Piet helaas niet mee, omdat zij allebei geenouder zijn van twee personen. We hebben nu met twee simpele eisen een manier gevondenom een aantal mensen uit te kiezen uit bijvoorbeeld een familie.
bOpaAAA
bOma���rPiet
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rMoeder@@@r
Joostr
Clara
Dit kunnen we nu heel algemeen maken en in willekeurige geordende verzamelingen toepassen.Een geordende verzameling is bijvoorbeeld de verzameling natuurlijke getallen, die geor-dend is met de natuurlijk ordening (zo geldt bijvoorbeeld 5 < 7).Vervolgens kunnen we overstappen naar de propositielogica en op een bepaalde manier deproposities ordenen, bijvoorbeeld door te bekijken welke premissen nodig zijn om bepaaldeconclusies te kunnen trekken. Uit de zin ”Het is maandag en het regent.” kunnen wenatuurlijk concluderen dat het regent. Op deze manier brengen we een ordening aan, doorte zeggen welke proposities afleidbaar zijn uit andere propositie.En nu kunnen we net zoals bij de familie ook binnen de propositielogica een specifieke
35
verzameling proposities aanwijzen die consistent is (niet tot een absurditeit leidt).
We kunnen dit steeds verder doorvoeren. Er zijn speciale geordende verzamelingen, dieheel mooie eigenschappen hebben. Daardoor kunnen we deze verzamelingen gebruiken omingewikkelde wiskundige structuren duidelijker te maken.
36
Bibliography
[1] B.A. Davey and H.A. Priestley, Introduction to Lattices and Order, 2nd edition, Came-bridge University Press, 2002.
[2] S. Abramsky, Domain theory in logical form, Annals of Pure and Applied Logic 51(1991) 2–6, 1988.
[3] J.L. Bell and A.B. Slomson, Models and Ultraproducts: an introduction, North Hol-land, Amsterdam 1969.
37