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The logic of subset spaces, topologic and the
local difference modality K 6=
Isabel Bevort
July 18, 2013
Bachelor Thesis in Mathematics
Supervisor: dr. Alexandru Baltag
Korteweg-De Vries Instituut voor Wiskunde
Faculteit der Natuurwetenschappen, Wiskunde en Informatica
Universiteit van Amsterdam
Abstract
In this thesis we will study the logic of subset spaces and the logical system topologic.
These are logics that are strong enough for elementary topological reasoning. The first
chapter will provide a quick introduction to basic modal logic on relational models with
syntax, semantics, some correspondence theory, the canonical model and a description
of the standard modal completeness proof. In the second chapter we will give the syntax
and semantics of the logic of subset spaces and look at the soundness of the base axioms.
Also, we will prove completeness of the base axioms for subset spaces in this second
chapter. The third and last chapter is dedicated to the local difference modality K 6= ,
with the appropriate syntax, semantics and the K 6= 2-axioms. We shall finish this thesis
with the open question of completeness of the K 6= 2-axioms on subset spaces, with an
explicit explanation as to why this remains an open question.
Title: The logic of subset spaces, topologic and the local difference modality K 6=
Authors: Isabel Bevort, [email protected], 5881420
Supervisor: dr. Alexandru Baltag
Second grader: Prof. dr. Dick de Jongh
Date: July 18, 2013
Korteweg-De Vries Instituut voor Wiskunde
Universiteit van Amsterdam
Science Park 904, 1098 XH Amsterdam
http://www.science.uva.nl/math
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Contents
Introduction
1. Modal Logic on Relational
1.1. Syntax & Semantics . .
1.2. Correspondence theory .
1.3. Canonical model . . . .
5
Models
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2. Logic of subset spaces & topologic
2.1. Syntax . . . . . . . . . . . . . . . . .
2.2. Semantics . . . . . . . . . . . . . . .
2.3. Axioms, inference rules & soundness
2.4. Completeness for subset spaces . . .
2.4.1. Properties of theories . . . . .
2.4.2. The proof . . . . . . . . . . .
2.4.3. Construction . . . . . . . . .
2.4.4. Details of the construction . .
2.5. The system topologic . . . . . . . . .
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3. The local difference modality K 6=
3.1. Elementary topological reasoning . . . . . .
3.2. The local difference modality K 6= . . . . . .
3.2.1. Syntax . . . . . . . . . . . . . . . . .
3.2.2. Semantics . . . . . . . . . . . . . . .
3.2.3. Axioms, inference rules & soundness
3.3. Topology with the K 6= -modality . . . . . . .
3.4. Open problem: completeness . . . . . . . . .
3.4.1. Properties of theories . . . . . . . . .
3.4.2. The proof . . . . . . . . . . . . . . .
3.4.3. Construction . . . . . . . . . . . . .
3.4.4. Details of the construction . . . . . .
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Popular summary — in Dutch
43
Bibliography
45
A. Notes on Topological reasoning and the logic of knowledge from Dabrowski,
Moss and Parikh [3]
46
4
Introduction
Half a year away from completing my bachelor in mathematics, I realized that the courses
in mathematical logic were the ones that attracted most of my attention and enthusiasm.
This is why I decided to ask dr. Alexandru Baltag from the Institute of Logic, Language
and Computation, who taught the course “Introduction to Modal Logic” in the first
semester of this schoolyear, to be the supervisor of my bachelorthesis. He accepted and
soon came with a subject that I was very excited to start working on, Modal Logics of
Space. After having read the chapters Modal Logics of Space and Topology and Epistemic
Logic from the Handbook of Spatial Logics [1], my attention was drawn to the latter,
and in particular the logic of subset spaces and the logical system of topologic.
This is an area of research that considers the connection between topology and modal
logic, in particular epistemic logic. This area of research is relatively young as it started
with the work of Alfred Tarski and J.C.C. McKinsey in the 1940’s, who reviewed the
connection between the modal logic S4 and topology. When stated in the modern truthconditional format, the modality 2 can be interpreted as various topological concepts,
such as the interior operator or the derivative.
In this thesis we shall mostly work in a modal language with two modalities, the 2
and K. In some way, the 2 can be interpreted as effort because it shrinks an open, and
K as knowledge, because it tells us exactly where the point is.
We will give an introduction to basic modal logic on relational models in the first
chapter, which will be a quick overview of some standard definitions and results.
The second chapter is dedicated to the logic of subset spaces and topologic. First we
will define our language and give the semantics on this language. Then we will define the
logic of subset spaces by giving the base axioms and inference rules. The soundness of the
axioms will also be proven. We will spend an entire section on the completeness proof of
the base axioms for subset spaces, which is an exciting adaptation of the standard modal
completeness proof. We have followed the proof given by Dabrowski, Moss and Parikh
given partly in [8] and fully in [3]. Lastly, we will define the logical system topologic,
which is an extension of the logic of subset spaces with the addition of two more axioms,
(Un) and (WD).
In the third and last chapter we will consider a new modality, the local difference
modality K 6= . Before defining a new language, we will give a connection between topology and the logic of subset spaces by expressing some topological notions in the language
of the subset space logic. This will make clear why the new modality is needed, since
the possibilities of expressing topological notions are very limited. The modality K 6=
will allow us to express difference between points, and hence enlarge our capability of
expression. We will give the accompanying axioms and inference rules and define a
system called the K 6= 2-system. After proving that the axioms are sound, we will show
5
what new topological notions can then be expressed. The completeness of the K 6= 2axioms will be given as an open problem. We will follow the pattern of the completeness
proof for the subset space logic, adapt it to this new logic, and show exactly at what
point the adapted proof fails.
A section with notes on the article by Dabrowski, Moss and Parikh [3] is added as an
appendix.
I have very much enjoyed working on this project and would like to conclude by
thanking my supervisor, dr. Alexandru Baltag, for his kind and enthusiastic guidance.
His patient assistance in helping me overcome the problems I encountered, made writing
this bachelor thesis a rewarding and thought-provoking experience.
6
1. Modal Logic on Relational
Models
In this chapter we will introduce basic modal logic on relational models. The definitions
and results given in this chapter can be found here [2]. The first definition we need is
the following:
Definition 1.1. A relational structure is a tuple F whose first component is a nonempty set W called the universe (or domain) of F, and whose remaining components
are relations on W . The elements of W are called states or points.
1.1. Syntax & Semantics
In this section we will define the syntax and semantics of basic modal logic on relational
models. Let us begin with the syntax.
Definition 1.2. The basic modal language is defined using a set of proposition letters Φ
whose elements are usually denoted by p, q, r, . . . , and a unary modal operator 2 (‘box’).
The well-formed formulas φ of the basic modal language are given by the rule:
φ ::= p | ⊥ | ¬φ | ψ ∨ ψ | 2φ,
where p ranges over elements of Φ. This definition means that a formula is either a
propositional letter, the propositional constant falsum (‘bottom’), a negated formula, a
disjunction of formulas, or a formula prefixed by 2.
Just as the familiar first-order existential and universal quantifiers are duals to each
other (in the sense that ∀xα ↔ ¬∃x¬α), we have the dual operator 3 (‘diamond’) for our
box which is defined as 3φ := ¬2¬φ. We also make use of the classical abbreviations
for conjunction, implication, bi-implication and the constant true (‘top’): φ ∧ ψ :=
¬(¬φ ∨ ¬ψ), φ → ψ = ¬φ ∨ ψ, φ ↔ ψ := (φ → ψ) ∧ (ψ → φ) and > := ¬ ⊥.
Definition 1.3. A modal similarity type is a pair T = (O, τ ) where O is a non-empty
set, and τ is a function τ : O → N. The elements of O are called modal operators; we
use 4 (‘triangle’), 40 , 41 , . . . , to denote elements of O. The function τ assigns to each
operator 4 ∈ O a finite arity, indicating the number of arguments 4 can be applied to.
The similarity type of the basic modal language is called T0 , where O0 = 2 and
τ (2) = 1.
We will now define frames, models and the satisfaction relation for the basic modal
language.
7
Definition 1.4. A frame for the basic modal language is a pair F = (W, R) such that
(i) W is a non-empty set.
(ii) R is a binary relation on W , so R ⊆ W × W .
That is, a frame for the basic modal language is simply a relational structure bearing a
single binary relation.
A model for the basic modal language is a pair M = (F, V ), where F is a frame for
the basic modal language, and V is a function called a valuation. It assigns to each
proposition letter p in Φ a subset V (p) of W . Formally, V is a map: Φ → P(W ), where
P(W ) denotes the powerset of W . Informally, we think of V (p) as the set of points in
our model where p is true. Given a model M = (F, V ), we say that M is based on the
frame F, or that F is the frame underlying M.
In the remaining part of this section we will learn how to interpret the basic modal
language in models by defining the following satisfaction relation |=M .
Definition 1.5. Suppose w is a point in a model M = (F, V ). Then we inductively
define the notion of a formula being satisfied (or true) in M at a state w as follows:
w
w
w
w
w
|=M p
|=M ⊥
|=M ¬φ
|=M φ ∨ ψ
|=M 2φ
iff
never,
iff
iff
iff
w ∈ V (p), where p ∈ Φ,
not w |=M φ,
w |=M φ or w |=M ψ,
for all v ∈ W such that Rvw we have v |=M φ.
It follows from this definition that w |=M 3φ if and only if there exists some v ∈ W
such that Rvw and v |=M φ.
Notice how 2 and 3 are actually local versions of the usual ∀-‘for all’ and ∃-‘there
exists’ quantifiers. The following example should help the reader to become familiar
with this semantics.
Example 1.6. Let M = (F, V ) be such that F = (W, R), W = {w, v, u}, R =
{(w, v), (v, w), (w, u), (v, u)} and V is such that V (p) = {w, v, u} and V (q) = {v}.
8
In this model, we have the following:
• w |= p
• w |= 2p since w points to both u and v, and v |= p and u |= p.
• w |= 3p but w 6|= 2p. Indeed, Rwv and v |= q, but Rwu as well and u 6|= q.
• w |= 2(p ∨ q). Indeed, for every state that w points to (u and v), either p is true
or q is true.
These are just examples of formulas true in this model.
1.2. Correspondence theory
In this section we will present some correspondence theory results that we will use later
on in the completeness proof of the subset space logic. We refer to [2, chapter 3] for the
proofs of these results.
First, let us give the needed axioms together with their traditional names:
(T) p → 3p
(4) 2p → 22p
(B) p → 23p
(5) 3p → 23p
Proposition 1.7. For any frame F = (W, R), the axiom (T) corresponds to reflexivity
of the relation R:
F |= T iff F |= ∀xRxx.
Proposition 1.8. For any frame F = (W, R), the axiom (4) corresponds to transitivity
of the relation R:
F |= 4 iff F |= ∀x∀y∀z(Rxy ∧ Ryz → Rxz).
Proposition 1.9. For any frame F = (W, R), the axiom (B) corresponds to symmetry
of the relation R:
F |= B iff F |= ∀x∀y(Rxy → Ryx).
Proposition 1.10. For any frame F = (W, R), the axiom (5) corresponds to euclideanness of the relation R:
F |= 5 iff F |= ∀x∀y∀z((Rxy ∧ Ryz) → Ryz).
9
1.3. Canonical model
In this section we will define the modal canonical model in the basic language and give
the steps of the standard completeness proof. We will not prove any of the results, and
again refer to [2, section 4.2] for the strongly interested reader. First we need to define
modal logics and strong completeness.
Definition 1.11. (Modal Logics) A modal logic Λ is a set of modal formulas that
contains all propositional tautologies and is closed under modus ponens (that is, if φ ∈ Λ
and φ → ψ ∈ Λ then ψ ∈ Λ) and uniform substitution (that is, if φ belongs to Λ then so
do all of its substitution instances). If φ ∈ Λ we say that φ is a theorem of Λ and write
`Λ φ; if not, we write 6`Λ φ.
Definition 1.12. A logic Λ is strongly complete with respect to some class of structures
S if and only if every Λ-consistent set of formulas is satisfiable on some G ∈ S.
Definition 1.13. A modal logic is normal if it contains Kripke’s axiom and the axiom
for the duality of 2 and 3:
(K)
2(p → q) → (2p → 2q),
(Dual) 3p ↔ ¬2¬p,
and is closed under generalization (that is, if `Λ φ then `Λ 2φ).
Canonical models are introduced in order to prove completeness. Given a normal logic
Λ, its strong completeness is proven with respect to some class of structures by showing
that every Λ-consistent set of formulas can be satisfied in some suitable model. The
suitable models are built out of maximal consistent sets of formulas, and in particular,
canonical models are built.
Definition 1.14. A set of formulas Γ is maximal Λ-consistent if Γ is Λ-consistent, and
any set of formulas properly containing Γ is Λ-consistent. If Γ is maximal Λ-consistent
we say it is a Λ-msc.
Proposition 1.15. (Properties of maximal consistent sets) If Λ is a logic and Γ
is a Λ-msc then:
(i) Γ is closed under modus ponens;
(ii) Λ ⊆ Γ;
(iii) for all formulas φ: φ ∈ Γ or ¬φ ∈ Γ;
(iv) for all formulas φ, ψ: φ ∨ ψ ∈ Γ iff φ ∈ Γ or ψ ∈ Γ.
The following lemma is to ensure that there are ‘enough’ maximal Λ-consistent sets.
This is important because they are the building blocks of this construction.
Lemma 1.16. (Lindenbaum’s Lemma) If U is a Λ-consistent set of formulas then
there is a Λ-msc U + such that U ⊆ U + .
10
Now we can define the canonical model.
Definition 1.17. The canonical model MΛ for a normal modal logic Λ in the basic
language is the triple (W Λ , RΛ , V Λ ) where:
(i) W Λ is the set of all Λ-mscs;
(ii) RΛ is the binary relation on W Λ defined by RΛ wu if for all formulas ψ, ψ ∈ u
implies 3ψ ∈ w. RΛ is called the canonical relation;
(iii) V Λ is the valuation defined by V Λ (p) = {w ∈ W Λ : p ∈ w}. V Λ is called the
canonical (ornatural ) valuation.
The pair F Λ = (W Λ , RΛ ) is called the canonical frame for Λ.
The following lemma ensures that there exist enough ‘coherently related’ mcss.
Lemma 1.18. (Existence Lemma) For any normal modal logic Λ and any state
w ∈ W Λ , if 3φ ∈ w then there is a state v ∈ W Λ such that RΛ wv and φ ∈ v.
Lemma 1.19. (Truth Lemma) For any normal modal logic Λ and any formula φ,
w |=MΛ φ if and only if φ ∈ w.
Theorem 1.20. (Canonical Model Theorem) Any normal modal logic is strongly
complete with respect to its canonical model.
11
2. Logic of subset spaces & topologic
In this chapter we will stuy the logic of subset spaces and the logical system topologic.
We will begin with the syntax and semantics, then define the logic of subset spaces and
prove completeness of the base axioms for subset spaces. We will finish this chapter by
having a look at topologic.
2.1. Syntax
Definition 2.1. A subset space is a pair X = (x, O) where X is a set and O is a set
of subsets of X. We assume that X ∈ O for convenience, though this is not really
necessary. The space X is closed under intersection, respectively union, if the following
holds: if S, T ∈ O, then S ∩ T ∈ O, respectively S ∪ T ∈ O.
Definition 2.2. Let A be an arbitrary countable set of atomic formulas. Define L as
the smallest set containing each A ∈ A that is closed under the following rules:
• if φ, ψ ∈ L, then also φ ∧ ψ ∈ L and ¬φ ∈ L,
• if φ ∈ L, then Kφ ∈ L and 2φ ∈ L.
Definition 2.3. Let X be a subset space. Thinking of X as a Kripke frame, we can
give the semantics of L as an interpretation of the atomic formulas. By this we mean a
map α : A → P(X). The pair (X , α) is a model, denoted by M.
2.2. Semantics
Definition 2.4. For p ∈ X and p ∈ u ∈ O, we define the satisfaction relation |=M on
(X × O) × L by recursion on φ:
p, u |=M
p, u |=M
p, u |=M
p, u |=M
p, u |=M
A
φ∧ψ
¬φ
Kφ
2φ
iff p ∈ α(A)
iff p, u |=M φ and p, u |=M ψ
iff p, u 6|=M φ
if for all q ∈ u, q, u |=M φ
if for all v ∈ O such that p ∈ v ⊆ u, p, v |=M φ
12
As usual, the dual of K is L, with Lφ ←→ ¬K¬φ, and the dual of 2 is 3, with
3φ ←→ ¬2¬φ. The satisfaction relation for L and 3 is defined dually as follows:
p, u |=M Lφ if there exists some q ∈ u such that q, u |=M φ
p, u |=M 2φ if there exists v ∈ O such that p ∈ v ⊆ u, p, v |=M φ.
This semantics can be represented visually, as is shown in the following picture.
Figure 2.1.: This is a visual representation: p, u |= Kφ iff for every q ∈ u, q, u |= φ, and
p, u |= 2φ iff for every v ⊆ u s.t. p ∈ v, p, v |= φ.
When M is clear from the context, we write p, u |= φ.
Let T ⊆ L. We write T |= φ if for all models M, all p ∈ X and all u ∈ O, the following
holds: if p, u |= ψ for each ψ ∈ T , then p, u |= φ.
Formulas of special interest
Given a model M and a formula φ,
• φ is local in M if for all p, u, v we have p, u |= φ iff p, v |= π. In other words, the
truth of φ depends only on the point.
• φ is local if it is local for all points in M.
• φ is persistent in M if φ → 2φ is valid in M.
• φ is persistent if φ → 2φ is persistent in every M.
Note that ever formula of the form K2φ is persistent. Also, if v ⊆ u, then every
persistent formula satisfied by p, u is also satisfied by p, v.
13
These formulas are of special interest because their truth is dependent only on the
points and not of the opens around them. Their interpretation in this language is similar
to usual modal interpretation. It also implies that if φ ∈ A, it can be interpreted as a
set of points.
2.3. Axioms, inference rules & soundness
The following are the base axioms.
• Propositional tautologies
• (A → 2A) ∧ (¬A → 2¬A), for A ∈ A;
• K2φ → 2Kφ, this is the Cross axiom;
• Kφ → (φ ∧ KKφ), this axiom combines reflexivity and transitivity for K;
•
– φ → KLφ, Brouwer’s axiom for symmetry. This is used in [8, 6, 5];
– Lφ → KLφ, the axiom for euclideanness. This is used in [1, 3];
For K to be S5-like, there needs an axiom for symmetry. This axiom is found in
two versions throughout the literature, as the Brouwer’s axiom or the axiom for
euclideanness. However, they are equivalent in the presence of the reflexivity and
transitivity axioms.
Indeed: suppose we have Lφ → KLφ as an axiom. The reflexivity axiom is
equivalent to φ → Lφ, so with the Eucliddeanness axiom Lφ → KLφ we have
φ → KLφ. Suppose now that we have Brouwer’s axiom. Use uniform substitution
with φ = Lφ, so we get Lφ → KLLφ. Now K-generalization on the transitivity
axiom gives K(LLφ → Lφ) and by the K-axiom and modus ponens it follows that
KLLφ → KLφ. Thus Lφ → KLφ.
• 2φ → (φ ∧ 22φ), this axiom combines reflexivity and transitivity for 2;
• 2(φ → ψ) → (2φ → 2ψ), this is Kripke’s axiom for the 2 modality;
• K(φ → ψ) → (Kφ → Kψ), this is the Kripke’s axiom for the K modality.
We add the following rules of inference: modus ponens, K-necessitation and 2-necessitation.
Definition 2.5. The logic of subset spaces is the logic described by the above axioms
and rules of inference.
We will now check the soundness of these axioms for subset spaces.
14
(A → 2A) ∧ (¬A → 2¬A), for A ∈ A
This axiom was added because the semantics for atomic formulas ignores the observation
whether p, u |= A depends only on p. It is an axiom that we need to include for atomic
preservation. Moss and Parikh put it in the following words: “Our framework is based
on the intuition that the interpretations of atomic formulas are “absolute” in this sense.”
Obviously this axiom is sound.
K2φ → 2Kφ
Let’s check the soundness of the Cross axiom. To do so, fix a subset space X , let p ∈ X
and u ∈ O and suppose p, u |= K2φ.
Then for all p0 ∈ u, we have p0 , u |= 2φ. So for all u0 ⊆ u such that p0 ∈ u0 ⊆ u we have
p0 , u0 |= φ. To show: p, u |= 2Kφ.
So let u00 be arbitrary such that u00 ⊆ u and p ∈ u00 ⊆ u. Now we want: p, u00 |= Kφ. So
let p00 ∈ u00 be arbitrary, we want: p00 , u00 |= φ. So let us use the premise. Let p0 := p00
and u0 := u00 . It follows that p00 , u00 |= φ, thus we are done. Note that the converse is not
valid.
(a) The premise
(b) the conclusion
Figure 2.2.: Visual representation of the Cross axiom
Kφ → (φ ∧ KKφ)
Fix a subset space X , let p ∈ X and u ∈ O and suppose p, u |= Kφ. Then for all q ∈ u
we have q, u |= φ. So obviously p, u |= φ. We need to show that p, u |= KKφ, i.e., for
all q ∈ u, q, u |= Kφ, i.e. for all q ∈ u and for all r ∈ u : r, u |= φ, which is obviously
true. So the axiom for reflexivity and transitivity of K is sound.
15
φ → KLφ
We will show the soundness of Brouwer’s axiom. So fix a subset space X , let p ∈ X and
u ∈ O and suppose p, u |= φ. We want to show that p, u |= KLφ, i.e., that for all q ∈ u
there exists a r ∈ u such that r, u |= φ, which is true since p, u |= φ. So the Brouwer’s
axiom is sound.
2φ → (φ ∧ 22φ)
Fix a subset space X , let p ∈ X and u ∈ O and suppose p, u |= 2φ. Then for all
v ⊆ u we have p, v |= φ. So clearly p, u |= φ as well, which shows the soundness of
the reflexivity of 2. We still need to show that p, u |= 22φ, i.e., for all v ⊆ u we have
p, v |= 2φ. So let v ⊆ u, and let w ⊆ v. Then also w ⊆ u, so p, w |= φ. So p, v |= 2φ.
So the axiom for reflexivity and transitivity of 2 is sound.
2(φ → ψ) → (2φ → 2ψ)
Fix a subset space X , let p ∈ X and u ∈ O. Suppose p, u |= 2(φ → ψ) and p, u |= 2φ.
We want to show that p, u |= 2ψ. So let v ⊆ u, then p, v |= φ → ψ, and p, v |= φ. By
modus ponens we have p, v |= ψ, thus p, u |= 2ψ. So the (K)-axiom for 2 is sound.
K(φ → ψ) → (Kφ → Kψ)
Fix a subset space X , let p ∈ X and u ∈ O. Suppose p, u |= K(φ → ψ) and p, u |= Kφ.
We want to show that p, u |= Kψ. So let q ∈ u, then q, u |= φ → ψ, and q, u |= φ. By
modus ponens we have q, u |= ψ, thus p, u |= Kψ. So the (K)-axiom for K is sound.
Note how these axioms resemble the axioms given in section 1.2. The axiom with the
2-modality that combines reflexivity and transitivity is actually a combination of (4)
and (T) (which is equivalent to 2p → p). The axiom with the K-modality that combines reflexivity and transitivity is actually just the same as for 2, but with a different
modality, namely the K-modality, which is also a local version of the ∀-quantifier. We
have already called the axiom for the symmetry for 2 “Brouwer’s axiom”, and with
good reason: it is just the (B) of section 1.2. The axiom for euclideanness is, similarly,
the (5) of this section.
Now that we have shown soundness for all of the base axioms, the following lemma
holds.
Lemma 2.6. (Soundness) If T ⊆ L and T ` φ, then T |= φ.
16
2.4. Completeness for subset spaces
In this section we will prove completeness of the base axioms for subset spaces.
2.4.1. Properties of theories
The proof uses maximal consistent subsets of L, they are called m-theories.
Fix a language L, and let T H be the set of m-theories in L. Use U, V, . . . to denote
m-theories. In order to prove that the proof system is complete, we need to show that
for every m-theory T , there exists a subset space model X = (X, O, α), a point p ∈ X,
and a subset u ∈ O such that p, u |=X T .
First we need to define relations on m-theories:
3
L
Definition 2.7. Define the relations −→ and −→ on m-theories by:
L
U −→ V iff if φ ∈ V , then Lφ ∈ U , and
3
U −→ V iff if φ ∈ V , then 3φ ∈ V .
Because of the maximal consistency of m-theories there are equivalent definitions, such
as:
L
U −→ V if when Kφ ∈ U , then φ ∈ V , and
3
U −→ V if when 2φ ∈ U , then φ ∈ V .
L3
L
3
3L
Further, define U −→ V if for some W , U −→ W −→ V , and define U −→ V if for
3
L
some W , U −→ W −→ V .
L
3
Proposition 2.8. Concerning the relations −→ and −→:
L
1. −→ is an equivalence relation.
3
2. −→ is a reflexive and transitive relation.
L
3. If Lφ ∈ T , then there is some U such that φ ∈ U and T −→ U .
3
4. If 3φ ∈ T , then there is some U such that φ ∈ U and T −→ U .
Proof. The following proofs are all simple consequences of the S4-ness of 3 and the
S5-ness of L. Note that 3. and 4. are equivalents of the Existence Lemma in standard
modal completeness proofs as in section 1.3.
L
1. To show : −→ is an equivalence relation.
• Reflexive. To show : if φ ∈ U , then Lφ ∈ U . Suppose φ ∈ U . By the
reflexivity axiom (equivalent to φ → Lφ) and the maximal consistency of U
L
it follows that Lφ ∈ U . So −→ is a reflexive relation.
L
L
• Symmetric. Suppose U −→ V . To show : V −→ U , i.e. if φ ∈ U then
Lφ ∈ V . Suppose φ ∈ U and Lφ ∈
/ V . Since V is a maximal consistent set,
L
¬L ∈ V , so K¬φ ∈ V . Then from U −→ V it follows that LK¬φ ∈ U . So by
17
the symmetry axiom φ → KLφ (equivalent to LKφ → φ) we have ¬φ ∈ U .
L
This is in contradiction with the consistency of U, so Lφ ∈ V . So −→ is
symmetric.
L
L
L
• Transitive. Suppose U −→ V −→ W . To show : U −→ W , i.e. if φ ∈ W
then Lφ ∈ U . Suppose φ ∈ W , then Lφ ∈ V and LLφ ∈ U . Thus by the
L
transitivity axiom (equivalent to LLφ → Lφ) Lφ ∈ U . So U −→ W , which
L
means −→ is transitive.
3
2. To show : −→ is reflexive and transitive.
• Reflexive. To show : if φ ∈ U then 3φ ∈ U . So suppose φ ∈ U . By the
reflexivity axiom (equivalent to φ → 3φ) and the maximal consistency of U
3
it follows that 3φ ∈ U . So −→ is reflexive.
3
3
3
• Transitive. Suppose U −→ V −→ W . To show : U −→ W , i.e. if φ ∈ W
then 3φ ∈ U . So suppose φ ∈ W . Then 3φ ∈ V and 33φ ∈ U . By the
transitivity axiom (equivalent to 33φ → 3φ) it follows that 3φ ∈ U . Thus
3
−→ is transitive.
3. Suppose Lφ ∈ T , and let A = {φ} ∪ {ψ : Kψ ∈ T }. We want to show that A
is consistent, so suppose towards a contradiction that it is not. Then there exist
ψ1 , . . . , ψn ∈ A such that Kψ1 , . . . , Kψn ∈ T and ψ1 ∧· · ·∧ψn → ¬φ. The maximal
theory T is closed under conjunction, so Kψ1 ∧ · · · ∧ Kψn ∈ T . We are working
in a normal modal logic since we have Kripke’s axiom for K, the dual of K and
K-necessitation [2, p. 33], so ` (Kψ1 ∧ . . . Kψn ) ↔ K(ψ1 ∧ · · · ∧ ψn ). Therefore
K(ψ1 ∧ · · · ∧ ψn ) ∈ T . Now apply necessitation to ψ1 ∧ · · · ∧ ψn → ¬φ to obtain
K(ψ1 ∧ · · · ∧ ψn → ¬φ) ∈ T . Followed by the use of Kripke’s axiom and modus
ponens we get K(ψ1 ∧ · · · ∧ ψn ) → K¬φ ∈ T , which by modus ponens again gives
K¬φ ∈ T . This is in contradiction to T being consistent since Lφ ∈ T . So A
is consistent, and by Lindenbaum’s lemma [2, p. 197] there exists a maximally
L
consistent U ⊇ A. It is easy to see that φ ∈ U and T −→ U .
4. Suppose 3φ ∈ T , and let A = {φ} ∪ {ψ : 2ψ ∈ T }. We want to show that A
is consistent, so suppose towards a contradiction that it is not. Then there exist
ψ1 , . . . , ψn ∈ A such that 2ψ1 , . . . , 2ψn ∈ T and ψ1 ∧ · · · ∧ ψn → ¬φ. The maximal
theory T is closed under conjunction, so 2ψ1 ∧ · · · ∧ 2ψn ∈ T . We are working
in a normal modal logic since we have Kripke’s axiom for 2, the dual of 2 and
2-necessitation [2, p. 33], so ` (2ψ1 ∧ . . . 2ψn ) ↔ 2(ψ1 ∧ · · · ∧ ψn ). Therefore
2(ψ1 ∧ · · · ∧ ψn ) ∈ T . Now apply necessitation to ψ1 ∧ · · · ∧ ψn → ¬φ to obtain
2(ψ1 ∧ · · · ∧ ψn → ¬φ) ∈ T . Followed by the use of Kripke’s axiom and modus
ponens we get 2(ψ1 ∧ · · · ∧ ψn ) → 2¬φ ∈ T , which by modus ponens again gives
2¬φ ∈ T . This is in contradiction to T being consistent since 3φ ∈ T . So A
is consistent, and by Lindenbaum’s lemma [2, p. 197] there exists a maximally
3
consistent U ⊇ A. It is easy to see that φ ∈ U and T −→ U .
18
3L
L3
Proposition 2.9. If U −→ V , then U −→ V . In other words, if U and V are m3
L
theories, and W is such that U −→ W −→ V , then there exists some T such that
L
3
U −→ T −→ V .
Proof. This proof makes use of the Cross Axiom. Suppose we have U , V and W such
3
L
that U −→ W −→ V . Let S := {3φ : φ ∈ V } ∪ {ψ : Kψ ∈ U }. Suppose towards a
contradiction that S is not consistent. Then there exists a finite subset S0 of S which is
inconsistent. Write
S0 = {3φ1 , 3φ2 , . . . , 3φn } ∪ {ψ1 , ψ2 , . . . , ψm },
where for each i and j, φi belongs to V and Kψj belongs to U . Let φ := φ1 ∧ · · · ∧ φn and
similarly ψ := ψ1 ∧ · · · ∧ ψm . Since V is closed under conjunction, φ ∈ V . And because
Kψ is equivalent to a conjunction K(ψ1 ∧ . . . ψm ) from U, we have Kψ ∈ U . Finally
3φ → 3φi for all i, and ψ → ψj for all j. Since S0 is inconsistent, ` 3φ → ¬ψ. By the
(K)-axiom and modus ponens, it follows that ` L3φ → L¬ψ, so this sentence belongs
to U . Also, φ ∈ V , so Lφ ∈ W and 3Lφ ∈ U . By the Cross axiom, L3φ ∈ U . By
modus ponens on ` L3φ → L¬ψ, it follows that L¬ψ ∈ U . But this is in contradiction
with U being consistent since Kψ ∈ U . So S is consistent. By Lindenbaum’s lemma
L
3
[2, p. 197] there exists T ⊇ S maximally consistent. By construction U −→ T −→ V ,
which proves the proposition.
Dabrowski, Moss and Parikh put the meaning of this proposition in such a nice wording
that we would like to cite them here: “This proposition is the embodiment of the Cross
Axiom scheme in the realm of m-theories.” [3, p. 85]
The next proposition is a generalization and will be used in the proof of completeness.
3
3
3
L
Proposition 2.10. Suppose that T1 −→ T2 −→ . . . −→ Tn , and suppose Tn −→ U .
3
3
3
L
Then there are U1 −→ U2 −→ . . . −→ Un such that 4Un = U , and for all i, Ti −→ Ui .
T1
L
U1
3
3
/
/ T2
3
/
...
3
L
/ Tn
U2
3
/
...
3
/
L
Un
3
L
Proof. Given T1 , T2 , . . . , Tn and U as above, let Un = U . Since Tn−1 −→ Tn −→ Un , we
L
3
get Un−1 from proposition 2.9 such that Tn−1 −→ Un−1 −→ Un . We continue backwards
in this way in order to get Un−2 , . . . , U1 .
This concludes this subsection, as we now know sufficiently about m-theories in order
to be able to prove completeness.
19
2.4.2. The proof
In this section we will give the proof of completeness. It contains the strategy of construction and the Truth Lemma, from which completeness will follow. We will follow
the proof of Dabrowski, Moss and Parikh [3] and give additional explanations along the
way.
Theorem 2.11. (Completeness for subset spaces) The base axioms are complete for
interpretation in subset spaces. That is, if T |= φ, then T ` φ.
Strategy
Like standard modal completeness proofs, the first idea in proving completeness is to
consider the canonical model of the logic. This is the set T H of m-theories with the
3
L
relations −→ and −→ as defined in subsection 2.4.1. It would be best if we could define
a family O of subsets of T H in order to obtain a subset space. Then we would hope
to show that every theory T is the theory of some pair p, u from that model, indeed we
might hope that P is T itself.
Unfortunately, this idea does not seem to work. The problem is that we do not know
any way to define a subset space structure on T H which leads to completeness. For this
reason, we do not approach completeness via the canonical model.
Our strategy is to build a space X of “abstract” points. The opens will also be given
in an abstract way, via a poset P and an order-reversing map i : P → P ∗ (X), where
P ∗ (X) is the set of non-empty subsets of X. The points are abstract since they are
not theories. But with each x and each p so that x ∈ i(p) there will be a “target”
m-theory t(x, p). The goal of the construction is to arrange that in the overall model,
th(x, i(p)) = t(x, p).
Even though we are not using the canonical model, the completeness proof reminds us
of the standard modal completeness proofs that do use the canonical modal of a logic.
We shall have a Truth Lemma and existence lemmas (proposition 2.8) as described in
section 1.3. By keeping this in mind when reading the proof, the reader can have a
better intuitive notion of what is happening.
We only need to prove that every m-theory T has a model in order to prove completeness. Let T be an m-theory. We build
(1) A set X containing a designated element x0 .
(2) A poset hP, 6i with least element ⊥.
(3) A function i : P → P ∗ (X) such that p 6 q iff i(q) ⊇ i(p), and i(⊥) = X. So i is an
order-reversing homomorphism from hP, 6, ⊥i to hP ∗ (X), ⊇, Xi.
(4) A partial function t : X × P → T H such that t(x, p) is defined iff x ∈ i(p). The
following properties need to be ensured for all p ∈ P, x ∈ i(p) and φ:
20
L
(a1) If y ∈ i(p), then t(x, p) −→ t(y, p).
(a2) If Lφ ∈ t(x, p), then for some y ∈ i(p), φ ∈ t(y, p).
3
(b1) If q > p, then t(x, p) −→ t(x, q).
(b2) If 3φ ∈ t(x, p), then for some q > p, φ ∈ t(x, q).
(c) t(x0 , ⊥) = T , where T is the m-theory that we fixed and want to make a model
for.
Suppose we have X, P, i and t, and that they satisfy these properties. Then we
consider the subset space model
X = hx, {i(p) : p ∈ P}, αi
where the valuation is as follows α(A) = {x : A ∈ t(x, ⊥)}.
Truth Lemma
The Truth Lemma is key in proving completeness for subset spaces.
Lemma 2.12. (The Truth Lemma). Assume conditions (1)-(4) for X,P,i and t. Then
for all x ∈ X and all p ∈ P such that x ∈ i(p),
thX (x, i(p)) = t(x, p).
Proof. The atomic case holds by definition of X .
Indeed, suppose A ∈ thX (x, i(p)), i.e. x, i(p) |= A. This means that x ∈ α(A), and from
the definition of the valuation α, this is equivalent to A ∈ t(x, ⊥). So what we want to
show is the following:
A ∈ t(x, ⊥) ⇐⇒ A ∈ t(x, p).
So suppose A ∈ t(x, ⊥). By the axiom for atomic preservation 2A ∈ t(x, ⊥), since
t(x, ⊥) ∈ T H. Since ⊥ is the least element of the poset P, we have p >⊥. By property
3
3
(b1), this implies t(x, ⊥) −→ t(x, p). Now we use the equivalent definition of −→ to
conclude that A ∈ t(x, p).
3
Suppose now that A ∈ t(x, p). As before, we have t(x, ⊥) −→ t(x, p), so 3A ∈ t(x, ⊥).
By the axiom for atomic preservation 3A → A, we have A ∈ t(x, ⊥).
The induction steps for Boolean connectives are consequences of the fact that the sets
in T H are m-theories.
So let us do the induction step for L. Suppose that Lφ ∈ thX (x, i(p)), i.e. that
x, i(p) |= Lφ. Then there is some y ∈ i(p) such that y, i(p) |= φ, i.e. φ ∈ thX (y, i(p)).
L
By the induction hypothesis φ ∈ t(y, p). By condition (a1), t(x, p) −→ t(y, p). Therefore
Lφ ∈ t(x, p). Now suppose Lφ ∈ t(x, p). Then by property (a2) there is some y ∈ i(p)
such that φ ∈ t(y, p). By the induction hypothesis we have φ ∈ thX (y, i(p)), i.e. y, i(p) |=
φ. Thus x, i(p) |= Lφ, i.e. Lφ ∈ thX (x, i(p)).
21
Next we do the induction step for 3. First suppose that 3φ ∈ thX (x, i(p)), i.e.
x, i(p) |= 3φ. Then there is some i(q) ⊆ i(p) such that x ∈ i(q) and x, i(q) |= φ. By the
induction hypothesis we have φ ∈ t(x, q). By property (3), p 6 q, so by property (b1) it
3
follows that t(x, p) −→ t(x, q). So 3φ ∈ t(x, p). Now suppose that 3φ ∈ t(x, p). Then
by (b2) there is some q > p, φ ∈ t(x, q). From the induction hypothesis it follows that
φ ∈ thX (x, q), i.e. that x, i(q) |= φ. But i(q) ⊆ i(p) by property (3), so x, i(p) |= 3φ,
i.e. φ ∈ thX (x, p).
This proves the Truth Lemma.
Now we have almost proven completeness. All we need to do is note that our m-theory
T has a model: by the Truth Lemma and property (4c) we have
T = t(x0 , ⊥) = thX (x0 , i(⊥)) = thX (x0 , X),
thus the base axioms are complete for subset space models.
2.4.3. Construction
In this section we will build X, P, i and t by recursion. That is, we build approximations
Xn , Pn , in and tn satisfying certain local and global properties. Fix two objects x0 and
⊥. The following local properties will be satisfied by the construction.
(L1) Xn is a finite set containing x0 .
(L2) Pn is a finite poset with ⊥ as minimum, and with the property that for each p ∈ Pn ,
the lower set of p, {q ∈ Pn : q 6 p}, is linearly ordered.
(L3) The map in : Pn → P ∗∗ (Xn ), where P ∗∗ (X) is the collection of subsets of Xn
with at least two elements, has the property that p 6 q iff in (q) ⊆ in (p). Also,
in (⊥) = Xn .
(L4) The function tn : Xn ×Pn → T H is partial with the property that tn (x, p) is defined
iff x ∈ in (p). Furthermore, we assume the following properties for all x ∈ Xn and
p ∈ Pn :
L
(a) If x, y ∈ in (p), then tn (x, p) −→ tn (y, p).
3
(b) If x ∈ in (q) and q > p, then tn (x, p) −→ tn (x, q).
(c) t0 (x0 , ⊥) = T .
These conditions are required for specific reasons. In (L2), the requirement that the
lower sets of points is linear is essential to our construction. By maintaining this property
throughout the construction, we can use proposition 2.9 to add points to the model. The
condition in (L3) that each in (p) has at least two elements is not really necessary, but
it leads to a simplification of the overall construction.
22
There will also be global properties satisfied by the construction:
(G1) Xn ⊆ Xn+1 .
(G2) Pn+1 is an end extension of Pn . This means that Pn is a suborder of Pn+1 and if
p ∈ Pn+1 , q ∈ Pn and p 6 q, then p ∈ Pn .
(G3) For all p ∈ Pn+1 , in+1 (p) ∩ Xn = in (p).
(G4) The restriction of tn+1 to Xn × Pn is tn .
Finally, our construction has to satisfy some overall requirements:
(R4a) If Lφ ∈ tn (x, p), then for some m > n, there is some y ∈ im (p) such that
φ ∈ tm (y, p).
(R4b) If 3φ ∈ tn (x, p), then for some m > n, there is some q > p in Pm such that
φ ∈ tm (x, q).
Suppose we build Xn , Pn , in and tn in such a way they satisfy the (L), (G) and (R)
requirements.
S
Definition 2.13. Let S
X = n∈N Xn , and let P be the limit of the posets Pn . Let i
be defined by i(p) = n>m in+1 (p), where m is the least number such that p ∈ Pm .
Finally, define t(x, p) = tn (x, p) where n is any number such that tn (x, p) is defined. The
construction has arranged that tn (x, p) = tn+1 (x, p) whenever the latter is defined.
Proposition 2.14. Suppose we build Xn , Pn , in and tn in accordance with the (L), (G)
and (R) requirements. Then X, P, i and t as defined above satisfy conditions (1)-(4)
above.
Proof. It is clear that (1) holds as we have constructed X in this way. In the same way
we have constructed P to have ⊥ as its least element so (2) also holds. To check (3),
first note that x ∈ i(p) if and only for some n, x ∈ in (p). Now suppose p > q and let
x ∈ i(p). Then there is an n such that x ∈ in (p). By (L3) in (p) ⊆ in (q). So x ∈ in (q),
i.e. x ∈ i(q). Thus i(p) ⊆ i(q). On the other hand, if p 6> q, then let n be such that
both p, q ∈ Xn . By (G2), p 6> q in Xn , so by (L3), in (p) 6⊆ in (q). This means there is
some x such that x ∈ in (p) \ in (q). By (G3), in (q) = Xn ∩ i(q), so since x ∈ in (p) ⊆ Xn
we see that x ∈
/ i(q). Hence i(p) 6⊆ i(q). Furthermore, it is easy to see that i(⊥) = X.
This completes the verification of (3).
The verifications of (4) are consequences of the overall (R) requirements and conditions
(G4) and (L4a).
23
2.4.4. Details of the construction
In this section we will go deeper into the construction and give the details.
At the beginning we fix a map
v : ω −→ {L, 3} × ω × ω
with the property that if v(n) = (x, m, k) then m < n, and for all (x, m, k) there is some
n > m such that v(n) = (x, m, k). Note that we have used a suggestive notation with
{L, 3}, it could have been {1, 2} just as well.
We define by recursion on n a tuple hXn , Pn , in , tn , αn , βn i. The last two, αn and βn ,
are functions whose purpose is to ensure all of the (R) requirements are met in an orderly
fashion in countably many steps:
αn : ω → {(x, p, φ) ∈ Xn × Pn × L : Lφ ∈ tn (x, p)}
βn : ω → {(x, p, φ) ∈ Xn × Pn × L : 3φ ∈ tn (x, p)}
Let X0 be a two-element set {x0 , x1 }, let P0 be the trivial poset {⊥}, let i0 (⊥) = X
and t0 (x0 , ⊥) = T , where T is the m-theory that we started with. All of these (L)
requirements are satisfied for n = 0. We also fix functions α0 and β0 , but this is just to
make sure that they are not arbitrary.
Now suppose we are at stage n+1 of the construction. There are two cases, depending
on the value of v(n + 1).
Case 1: v(n + 1) = (L, m, k) for some m and k. Consider αm (k), and let it be equal
to the triple (x, p, φ). This means that Lφ ∈ tm (x, p).
Let y be a new point not in Xn . Let Xn+1 := Xn ∪ {y}. Let Pn+1 := Pn . Let
(
in (q) ∪ {y} if q 6 p,
in+1 (q) =
in (q)
otherwise.
Note that conditions (L1), (L2), (G1),(G2) and (G3) are trivial. Indeed, Xn+1 is still
finite (L1) and Xn ⊆ Xn+1 (G1), and Pn+1 = Pn (L2,G2). For (G3), let q ∈ Pn+1 . Then
if q 6 p, in+1 (q) = in (q) ∪ {y} so in+1 (q) ∩ Xn = in (q) since y ∈ Xn+1 \ Xn . Otherwise
in+1 (q) = in (q) so obviously in+1 (q) ∩ Xn = in (q). Also, in+1 (⊥) = Xn+1 .
Next we will check condition (L3), i.e. that q 6 r if and only if in+1 (r) ⊆ in+1 (q).
For this we look at three cases. First, if neither q 6 p nor r 6 p, then in+1 (q) = in (q)
and in+1 (r) = in (r). So we are done, since (L3) holds for n. Second, if both q and r are
below p, then in+1 (q) = in (q) ∪ {y} and in+1 (r) = in (r) ∪ {y}. So in+1 (r) ⊆ in+1 (q) if and
only if in (r) ⊆ in (q), which, again, is equivalent to q 6 r. For the last case, suppose that
q 6 p but r 66 p. Then also r 66 q. And since y ∈
/ in+1 (r), we have in+1 (q) 6⊆ in+1 (r).
Now, since in+1 (q) = in (q) ∪ {y}, we again have that in+1 (r) ⊆ in+1 (q) if and only if
in (r) ⊆ in (q). Now we have completed the verification of (L3).
Next we wish to define tn+1 . For this, we stipulate that (G4) holds. Remember that
tn+1 is a map from Xn+1 × Pn+1 to T H, so we only need to define tn+1 (y, q) for q 6 p,
24
since y is the only point in Xn+1 that is not in Xn , and when q 66 p we have in (q) ⊂ in (p),
so no changes are made.
We use condition (L2) to see that {q : q 6 p} is a finite, linearly ordered set. So write
this set as q1 6 q2 6 · · · 6 qN = p. Then i(p) ⊆ · · · ⊆ i(q2 ) ⊆ i(q1 ) by (L3), so for all i,
x ∈ i(qi ). Thus we can write
3
3
3
t(x, q1 ) −→ t(x, q2 ) −→ . . . −→ t(x, qN ).
Note that we have used the notation t(x, qi ) here instead of tn+1 (x, qi ). This is because
of condition (G4) combined with the definition of t: the restriction of tn+1 to Xn × Pn
is tn , and t(z, r) = tn (z, r), where n is any number such that tn (z, r) is defined.
L
Now let U be such that t(x, qN ) −→ U and φ ∈ U . This exists by 3. of proposition
2.8. To be able to apply this proposition we need to realize that Lφ ∈ t(x, qN ). We
began with the triple αm (k) which told us that Lφ ∈ tm (x, p). We know that p = qN
so Lφ ∈ tm (x, qN ). And of course by the definition of t again t(x, qN ) = tm (x, qn ). So
Lφ ∈ t(x, qN ) and we can use the proposition.
To give a visual representation:
t(x, q1 )
3
/
t(x, q2 )
3
/
...
3
/
t(x, qN )
L
U
L
By proposition 3.12 there exist m-theories U1 , U2 , . . . , UN such that U = UN , t(x, qi ) −→
3
3
3
Ui for all i, and U1 −→ U2 −→ . . . −→ UN . Let t(y, qi ) := Ui for 1 6 i 6 N . So we get:
t(x, q1 )
3
/
L
t(x, q2 )
t(y, q1 )
3
/
3
/
...
3
/
L
t(x, qN )
t(y, q2 )
3
/
...
3
/
L
t(y, qN )
and this definition of tn+1 ensures (L4b).
Next we need to check whether (L4a) holds for this definition of tn+1 . So suppose
a, b ∈ in+1 (q). We might as well assume that a = y, and hence that q 6 p. If b = y as well,
L
L
then we have tn+1 (a, q) −→ tn+1 (b, q) by the reflexivity of −→. So assume now that b 6= y,
i.e. that b ∈ i( q). From q 6 p it follows that in+1 (q) ⊇ in+1 (p); indeed we have shown
above that (L3) holds for n+1. Also, x ∈ im (p) and m < n+1 by the property of v, so by
(G3) it follows that x ∈ in+1 (p), thus x ∈ in+1 (q). Now tn+1 (x, q) is defined. Since both
x and b are not equal to y, we know that tn+1 (x, q) = tn (x, q) and tn+1 (b, q) = tn (b, q).
L
Also, (L4a) holds for n, so tn+1 (x, q) −→ tn+1 (b, q). By construction of tn+1 , and by (G4),
L
L
tn+1 (x, q) −→ tn+1 (y, q). So by symmetry and transitivity, tn+1 (y, q) −→ tn+1 (b, q). This
completes the verification of (L4a).
Case 2: v(n + 1) = (3, m, k) for some m and k. Now consider βm (k), and let it be
equal to the triple (x, p, φ). This means that 3φ ∈ tm (x, p).
25
Let y be a new point not in Xn , and let q be a new point not in Pn . Let Xn+1 =
Xn ∪ {y}. Let Pn+1 = Pn ∪ {q} and extend the partial order so that for al r ∈ Pn , r < q
in Pn+1 if and only if r 6 q. Then the new point q is not below any element of Pn , and
its lower set is a chain which means it is linearly ordered. We now have (L1) and (G1)
since Xn+1 is still finite and Xn ⊆ Xn+1 . And we have (L2) and (G2) since Pn+1 is still
finite and Pn is a suborder of Pn+1 such that if p ∈ Pn+1 , q ∈ Pn and p 6 q, then p ∈ Pn .
Let in+1 (q) = {x, y}, and for r ∈ Xn , let
(
in (r) ∪ {y} if r 6 p,
in+1 (r) =
in (r)
otherwise.
Now we wish to check property (L3). First note that p 6 q, so we wish to show that
in+1 (q) ⊆ in+1 (p). But from p 6 p it follows that in+1 (p) = in (p) ∪ {y}. Since (L3) holds
for n, in (p) contains at least two points, so it contains a point x0 6= x. Also y 6= x0 , thus
in+1 (q) = {x, y} is a proper subset of in+1 (p).
Second, if r 6 q in Xn+1 then either r = q or r < q. In the first case we are done; in
the second case we have r 6 p in+1 (q) ⊂ in+1 (r) for the same reason as above, so we are
done as well.
Now if r 66 q, then r 66 p so in+1 (r) = in (r) hence y ∈
/ in+1 (r). So in this case,
in+1 (q) 6⊆ in+1 (r). This verifies most of (L3), and the remainder of the verification is as
in Case 1 above.
So all we need to do now is define tn+1 . As in subcase 1b we stipulate that (G4) holds
for n. For all r 6 p in Pn , let tn+1 (y, r) = tn+1 (x, r) and for all z ∈ ?? tn+1 (z, r) =
tn+1 (r, x). In addition, let tn+1 (y, q) = tn+1 (x, q) = U , where U is any m-theory so that
3
tn (x, p) −→ U and φ ∈ U , which exists by proposition 3.10. The fact that both (L4a)
and (L4b) hold for n implies easily that they hold for n + 1 as well.
This concludes the definition of Xn+1 , Pn+1 , in+1 and tn+1 for both case 1 and case
2. Now fix enumerations αn+1 and βn+1 as above to complete the construction in both
cases.
The construction is now complete. At each step, the (L) and (G) requirements have
been satisfied. It remains to check the (R) requirements on the overall construction. So
let us check (R4a). First suppose that Lφ ∈ tn (x, p). Let k be such that αn (k) = (x, p, φ).
Let N be such that v(N ) = (L, n, k). Then at stage N we ensure that there is some
y ∈ XN such that φ ∈ tn (y, p). This verifies condition (R4a).
This completes the completeness proof of the base axioms for subset spaces.
26
2.5. The system topologic
In this section we will define the topologic system, which is an extension of the logic
of subset spaces. It brings us closer to topology since it ensures closure of O under
intersection and union.
Definition 2.15. A directed space is one where for every p, u, v with p ∈ u and p ∈ v
there is a w ∈ O such that p ∈ w and w ⊆ (u ∩ v). An intersection space is one where
we can take w = u ∩ v. A union space is a space that is closed under union: for any
u, v ∈ O: u ∪ v ∈ O.
Definition 2.16. The system whose axioms are the base axioms together with the
Weak-Directedness axiom and the Union axiom will be called topologic. The idea is that
topologic will be strong enough to support elementary topological reasoning. [1, p. 313]
Definition 2.17. The two following axioms, the Weak-Directedness axiom and the
Union axiom, ensure closure under union and intersection.
(WD) 32φ → 23φ
(Un) 3φ ∧ L3ψ → 3 (3φ ∧ L3ψ ∧ K3L(φ ∨ ψ))
However, (WD) does not lead to a complete axiomatisation of the valid sentences on
intersection spaces. This is discussed here [2, p. 316].
Proposition 2.18. The Weak-Directedness Axiom is sound for directed spaces.
Proof. Fix a directed space X = (X, O) and let x ∈ X and u ∈ O. Suppose that
x, u |= 32φ. Then there is a u0 ⊆ u such that x, u0 |= 2φ. Thus for all u00 ⊆ u0 ,
x, u00 |= φ.
Now let v ⊆ u such that x ∈ v. Let w ∈ O be such that x ∈ w ⊆ u0 ∩ v. Since
w ⊆ u0 , x, w |= φ. Also w ⊆ v, hence x, v |= 3φ. We chose v to be arbitrary, thus
x, u |= 23φ.
(a) The premise
(b) the conclusion
Figure 2.3.: Visual representation of the (WD) axiom
27
Proposition 2.19. The Union Axioms are sound on union spaces.
(Un)
3φ ∧ L3ψ → 3 (3φ ∧ L3ψ ∧ K3L(φ ∨ ψ))
Proof. Fix a union space X = (X, O), let x ∈ X and u ∈ O. Suppose x, u |= 3φ ∧ L3φ.
Then:
- There exists v such that x ∈ v ⊆ u and x, v |= φ,
- there exists x0 ∈ u such that x0 , u |= 3φ,
and there exists v 0 ⊆ u such that x ∈ v 0 and x0 , v 0 |= ψ.
To show: x, u |= 3 (3φ ∧ L3ψ ∧ K3L(φ ∨ ψ)).
Figure 2.4.: Visual representation of the (Un) axiom
- Let w := v ∪ v 0 , then w ⊆ u.
- Now v ⊆ w and x, v |= φ so x, w |= 3φ,
- v 0 ⊆ w and x0 ∈ v 0 ⊆ w, so from x0 , v 0 |= ψ follows x, w |= L3ψ.
- Let y ∈ w be arbitrary. Then either y ∈ v or y ∈ v 0 . If y ∈ v, then since
x ∈ v and x, v |= φ we have y, v |= Lφ, thus y, v |= L(φ ∨ ψ). If y ∈ v 0 , then
since x0 ∈ v 0 and x0 , v 0 |= ψ we have y, v 0 |= Lψ, thus y, v 0 |= L(φ ∨ ψ). So
y, w |= 3L(φ ∨ ψ). Since y was arbitrary it follows that x, w |= K3L(φ ∨ ψ).
So x, w |= 3φ ∧ L3ψ ∧ K3L(φ ∨ ψ).
Thus x, u |= 3 (3φ ∧ L3ψ ∧ K3L(φ ∨ ψ)).
28
The following completeness result was proven by Georgatos [6].
Theorem 2.20. (Georgatos 1993 [6]) The topologic axioms are complete for topological
spaces, indeed for complete lattice spaces.
There are different versions of the (Un) axiom to be found in the literature: the version
above is to be found in [1, p. 312] and in [3, p. 79], whereas the following version is used
by Moss and Parikh in [8, p. 105]:
(Un) L3(φ ∧ Kχ) ∧ L3(ψ ∧ Kχ) → 3(L3φ ∧ L3ψ ∧ K3χ)
and Georgatos uses the following for his completeness proof in [6, p. 7]:
(Un) 3(Kφ ∧ ψ) ∧ L3(Kφ ∧ χ) → 3(L3ψ ∧ 3ψ ∧ L3χ)
All these versions look very similar, and there is a reason for this not to matter.
Georgatos notes that any formula, sound in the class of subset spaces closed under finite
union and intersection, which implies the formula
3(Kφ ∧ ψ) ∧ L3(Kφ ∧ χ) → 3(Kφ ∧ ψ ∧ Lχ)
where 3φ → 2φ, 3ψ → ψ and χ → 2χ are theorems, can replace the Union axiom in
the completeness proof of topologic.
An equivalent form of the Weak-Directedness axiom appeared here [8, p. 103]:
32φ ∧ 32φ → 32(φ ∧ ψ).
With this we conclude the chapter on the logic of subset spaces and topologic.
29
3. The local difference modality K 6=
This chapter is dedicated to the local difference modality K 6= . Previous work has been
done on combining topology with a difference modality. Andrey Kudinov has studied
propositional modal logic with two modal operators 2 and [6=] [7], and has shown that
some important topological properties are expressible in this language. He notes that
the [6=] modality and its interpretation in Kripke frames have been studied more deeply
by Maarten de Rijke [4].
We are interested in doing some topological reasoning, and found the need for a
difference modality. However, in our system, the truth of a formula depends on a point
and an open around that point, which is why the local difference modality K 6= will the
result of “localizing” the standard difference modality by restricting to the current open.
This is how our work differs from previous work.
Before introducing the new modality K 6= , let us do some elementary topological reasoning in the language L of the previous chapter.
3.1. Elementary topological reasoning
Let X = (X, O) be a subset frame such that O is a topology on X. Now we can express
certain topological notions such as open, closed and dense sets. The above definitions
are actually motivated by topology, as the following observations will show.
Claim. The set α(A) is open if and only if the formula A → 3KA is valid in the model.
Proof. “⇐” Suppose p, u |= A → 3KA. We want to show that A is an open set. So let
p ∈ A (this is equivalent to p, u |= A. Then by modus ponens p, u |= 3KA, i.e., there
exists a v ⊆ u such that p ∈ v ⊆ u and p, v |= KA. So there exists v ⊆ u such that for
all q ∈ v : q ∈ A (i.e., q, v |= A). But this means that p ∈ A implies that there exists an
open v such that p ∈ V ⊆ A, i.e., p ∈ Int(A. So A ⊆ Int(A), and of course Int(A) ⊆ A,
so A = Int(A), thus A is an open set.
“⇒” Suppose α(A) is open. Let p ∈ X and u ∈ O be arbitrary such that p ∈ u ∈ O.
Suppose p, u |= A, i.e. that p ∈ α(A). Now consider α(A) ∩ u, this is an open set
since O is a topology on X, and obviously α(A) ∩ u ⊆ u. Also, p ∈ α(A) ∩ u, and
p, α(A) ∩ u |= KA since clearly α(A) ∩ u ⊆ α(A). Thus p, u |= 3KA.
30
Claim. The set α(A) is closed if and only if the formula 2LA → A is valid in the model.
Proof. We use the duality of 2 and 3 in this proof.
A is closed ⇔ X\A is open
⇔ ¬A → 3K¬A is valid
⇔ ¬3K¬A → A is valid
⇔ 2LA → A is valid
Claim. The set α(A) is dense iff the formula 2LA is valid, and it is nowhere dense if
the formula 2LK¬A is valid.
Proof. The set α(A) is dense if and only if for every point x ∈ X, every open neighborhood around x has a nonempty intersection with α(A). This is exactly what 2LA being
valid means.
Claim. A point x belongs to the boundary of α(A) if p, X |= 2(LA ∧ L¬A).
Proof. A point x belongs to the boundary of α(A) if and only if every open neighborhood
around x has a nonempty intersection with both α(A) and X \ α(A). This is exactly
what p, X |= 2(LA ∧ L¬A) means.
Next we considered expressing the topological derivative:
Definition 3.1. The derivative of A ⊆ X is defined as the set of limit points of A.
It is denoted by d(A). A point x is called a limit point of A ⊆ X is for each open
neighborhood U of x, the set A ∩ (U \ {x}) is nonempty.
However, before we can begin doing so, we have to realize that we have no ability
to express difference in our language, which is necessary for expressing things like the
derivative.
Consider therefore the following language.
3.2. The local difference modality K 6=
We will now introduce a new modality that will enable us to express difference between
two points.
3.2.1. Syntax
Definition 3.2. Define a subset space as in definition 2.1. Let A be an arbitrary
countable set of atomic formulas. Define L0 as the smallest set containing each A ∈ A
that is closed under the following rules:
31
• if φ, ψ ∈ L0 , then also φ ∧ ψ ∈ L0 and ¬φ ∈ L0 ,
• if φ ∈ L0 , then K 6= φ ∈ L0 and 2φ ∈ L0 ,
Definition 3.3. Let X be a subset space. Thinking of X as a Kripke frame, we can
give the semantics of L0 as an interpretation of the atomic formulas. By this we mean a
map α : A → P(X). The pair (X , α) is a model, denoted by M.
3.2.2. Semantics
Definition 3.4. For p ∈ X and p ∈ u ∈ O, we define the satisfaction relation |=M on
(X × O) × L0 by recursion:
p, u |=M
p, u |=M
p, u |=M
p, u |=M
p, u |=M
A
φ∧ψ
¬φ
K 6= φ
2φ
iff p ∈ α(A)
iff p, u |=M φ and p, u |=M ψ
iff p, u 6|=M φ
if for all q ∈ u such that p 6= q, q, u |=M φ
if for all v ∈ P such that p ∈ v ⊆ u, p, v |=M φ
Note that the only difference with the semantics in section 2.2 lies in the satisfaction
relation for formulas of the form K 6= φ.
As usual, we use the abbreviation L6= φ ↔ ¬K 6= ¬φ. This logic actually contains the
subset space logic since we can define K from it as follows:
def
Kφ = φ ∧ K 6= φ,
and equivalently:
def
Lφ = φ ∨ L6= φ.
3.2.3. Axioms, inference rules & soundness
The following axioms will be called the K 6= 2-axioms:
• Propositional tautologies
• (A → 2A) ∧ (¬A → 2¬A);
• 2φ → (φ ∧ 22φ), this axiom combines reflexivity and transitivity, i.e. S4 for 2;
• 2(φ → ψ) → (2φ → 2ψ), this is the (K)-axiom for 2;
• φ ∧ K 6= φ → K 6= K 6= φ, this is the axiom for quasi-transitivity;
• φ → K 6= L6= φ, this is the axiom for symmetry;
• K 6= 2φ → 2K 6= φ, this is the Cross axiom;
• K 6= (φ → ψ) → (K 6= φ → K 6= ψ), this is the (K)-axiom for K 6= .
32
We add the following inference rules:
• 2-necessitation
• K 6= -necessitation
• modus ponens
Definition 3.5. The axioms and inference rules above describe the logical system we
will call the K 6= 2-system.
Let us check the soundness of these axioms for subset spaces. For the first three
axioms, soundness for subet spaces was shown in section 2.3, so we will only show
soundness for the remaining axioms.
φ ∧ K 6= φ → K 6= K 6= φ
Fix a subset space X , let p ∈ X and u ∈ O, and suppose p, u |= φ ∧ K 6= φ. Then p, u |= φ
and p, u |= K 6= φ. So let q ∈ u be such that q 6= p and q, u |= φ. Since p, u |= φ and
for all r ∈ u : r, u |= φ we have that q, u |= K 6= φ. Since q was chosen to be arbitrary,
p, u |= K 6= K 6= φ.
φ → K 6= L6= φ
Fix a subset space X , let p ∈ X and u ∈ O, and suppose p, u |= φ. If there is no point
in u other than p, we are done. So suppose there is and let q be any point in u such
that q 6= p. Since q 6= p and p, u |= φ, we have q, u |= L6= φ. So p, u |= K 6= L6= φ since q
was chosen to be arbitrary.
K 6= 2φ → 2K 6= φ
Fix a subset space X , let p ∈ X and u ∈ O, and suppose p, u |= K 6= 2φ. Then for
all p0 ∈ u with p0 6= p: p0 , u |= 2φ. So for all p0 ∈ u with p0 6= p and for all u0 ⊆ u :
p0 , u0 |= φ. To show: p, u |= 2K 6= φ.
So let u00 be arbitrary such that p ∈ u00 and u00 ⊆ u. And let p00 ∈ u00 be such that
00
p 6= p. Then p00 ∈ u since u00 ⊆ u. Let p0 := p00 and u0 := u00 , then since p00 6= p we can
use the premise and conclude that p00 , u00 |= φ. Thus p, u |= 2K 6= φ.
K 6= (φ → ψ) → (K 6= φ → K 6= ψ)
Fix a subset space X , let p ∈ X and u ∈ O. Suppose p, u |= K 6= (φ → ψ) and p, u |= K 6= φ.
We want to show that p, u |= K 6= ψ. So let q ∈ u be such that q 6= p, then q, u |= φ → ψ,
and q, u |= φ. By modus ponens we have q, u |= ψ, and since q was chosen to be arbitrary,
p, u |= K 6= ψ. So the (K)-axiom for K 6= is sound.
33
3.3. Topology with the K 6=-modality
Now that we have this new modality in the language there are some more topological
notions that we can express. So let X = (X, O) be a subset frame such that O is a
topology on X.
Claim. A point p belongs to the derivative of α(A) if and only if p, X |= 2L6= A.
Proof. Let p ∈ X. Then p ∈ d(α(A)), where d(α(A)) is the derivative of α(A), if and
only if p is a limit point of α(A),
i.e. for every open neighborhood U around p, α(A) ∩ (U \ {p}) 6= ∅,
i.e. for every open U around p, there exists a q ∈ α(A) ∩ U such that q 6= p,
i.e. for every open U around p, there exists a q such that q, U |= A and q 6= p,
i.e. for every open U around p, p, U |= L6= A,
i.e. p, X |= 2L6= A.
Definition 3.6. A subset X0 ⊆ X is called dense-in-itself if X0 ⊆ d(X0 )
Claim. The set α(A) is dense-in-itself if and only if the formula A → 2L6= A is valid in
the model.
Proof. By the previous claim, this is trivial.
Definition 3.7. A topological space X is called a T1 -space if for each pair of different
points there exists an open set containing exactly one of the points. Equivalently, X is
a T1 -space if and only if each singleton {x} is closed in X.
Claim. A topological space X is a T1 -space if and only if p, X |= (A ∧ K 6= ¬A) →
K(2LA → A). (Remember that Kφ is an abbreviation for φ ∧ K 6= φ, and dually, Lφ is
an abbreviation for φ ∨ L6= φ.)
Proof. We will show that p, X |= (A ∧ K 6= ¬A) → K(2LA → A) is equivalent to each
singleton being closed. The formula A∧K 6= ¬A describes the fact that α(A) is a singleton.
Indeed, p, X |= (A ∧ K 6= ¬A) means that p ∈ α(A), and every other point in X is not in
α(A), so that α(A) = {p}. The implication (A ∧ K 6= ¬A) → K(2LA → A) then ensures
this singleton is closed.
Notice that we did not express the notions T0 -space and Hausdorff-space. This is
something we still cannot do because we cannot do any separation arguments in this
language. The topological reasoning we can therefore do in this language remains very
much elementary.
3.4. Open problem: completeness
We do not claim completeness for the K 6= 2-axioms and will leave this as an open question
in this thesis. We have followed the pattern of the completeness proof of the base axioms
34
as above and found that at one crucial point there occurs a difficulty that we were not
able to overcome. We will explicitly describe this difficulty later on, when it occurs in
subsection 3.4.4 on the details of the construction. But first let us do the preliminaries
with theories.
3.4.1. Properties of theories
Let us consider maximal consistent subsets of L, they are called m-theories.
Fix a language L, and let T H be the set of m-theories in L. Use U, V etc., to denote
m-theories. In order to prove that the proof system is complete, we need to show that
for every m-theory T , there exists a subset space model X = (X, O, α), a point x ∈ X,
and a subset u ∈ O such that p, u |=X T .
First we need relations on m-theories:
3
L6=
Definition 3.8. Define the relations −→ and −→ on m-theories by:
L6=
U −→ V
3
U −→ V
Because of
tion:
L6=
U −→ V
3
U −→ V
iff if φ ∈ V , then L6= φ ∈ U , and
iff if φ ∈ V , then 3φ ∈ U .
the maximal consistency of m-theories,the following is an equivalent definiif when K 6= φ ∈ U , then φ ∈ V , and
if when 2φ ∈ U , then φ ∈ V .
L6= 3
L6=
3
3L6=
Further, define U −→ V if for some W , U −→ W −→ V , and define U −→ V if for
3
L6=
some W , U −→ W −→ V .
Definition 3.9. Let X be a set. A binary relation R ⊆ X × X ic called quasi-transitive
when the following holds:
if Rwv and Rvu, then either w = u or Rwu.
L6=
3
Proposition 3.10. Concerning the relations −→ and −→:
L6=
1. −→ is a symmetric and quasi-transitive relation.
3
2. −→ is a reflexive and transitive relation.
L6=
3. If L6= φ ∈ T , then there is some U such that φ ∈ U and T −→ U .
3
4. If 3φ ∈ T , then there is some U such that φ ∈ U and T −→ U .
Proof. We will prove 1. and 3., since 2. and 4. have been proven above in proposition 2.8
in subsection 2.4.1. Note that 3. and 4. are equivalents of the Existence Lemma in
standard modal completeness proofs [2, p. 198].
35
1.
L6=
L6=
• Symmetric. Suppose U −→ V . To show: V −→ U , i.e., if φ ∈ U then
L6= ∈ V . Let φ ∈ U and suppose L6= φ ∈
/ V . Then since V is maximally
L6=
consistent, ¬L6= φ ∈ V , so K 6= ¬φ ∈ V . Since U −→ V , L6= K 6= ¬φ ∈ U . It
now follows from the symmetry axiom (equivalent to L6= K 6= ¬φ → ¬φ) that
¬φ ∈ U , which is in contradiction to maximal consistency of U since φ ∈ U .
L6=
So L6= ∈ V . Thus −→ is symmetric.
L6=
L6=
• Quasi-transitive. Suppose U −→ V −→ W . To show: either U = W or
L6=
L6=
U −→ W . Suppose that U −→ W is not true. Then there is a θ such that
θ ∈ W and L6= θ ∈
/ U . Let φ ∈ U . By maximal consistency of U , ¬L6= θ ∈ U ,
6=
i.e. K ¬φ ∈ U . So φ∧K 6= ¬θ ∈ U . But then θ → φ ∈ U and K 6= (θ → φ) ∈ U
by the truth value of an implication. So (θ → φ) ∧ K 6= (θ → φ) ∈ U . Now we
can use the axiom for quasi-transitivity and obtain that K 6= K 6= (θ → φ) ∈ U
L6=
L6=
by modus ponens. Because U −→ V −→ W , it follows that θ → φ ∈ W ,
so by modus ponens, φ ∈ W . Thus U ⊆ W , but this implies U = W by
L6=
maximal consistency of both U and W . Thus −→ is quasi-transitive.
2. Suppose L6= φ ∈ T , and let A = {φ} ∪ {ψ : K 6= ψ ∈ T }. We want to show that A
is consistent, so suppose towards a contradiction that it is not. Then there exist
ψ1 , . . . , ψn ∈ A such that K 6= ψ1 , . . . , K 6= ψn ∈ T and ψ1 ∧ · · · ∧ ψn → ¬φ. The
maximal theory T is closed under conjunction, so K 6= ψ1 ∧ · · · ∧ K 6= ψn ∈ T . We are
working in a normal modal logic since we have Kripke’s axiom for K 6= , the dual L6=
of K 6= and K 6= -necessitation [2, p. 33], so ` (K 6= ψ1 ∧. . . K 6= ψn ) ↔ K 6= (ψ1 ∧· · ·∧ψn ).
Therefore K 6= (ψ1 ∧ · · · ∧ ψn ) ∈ T . Now apply necessitation to ψ1 ∧ · · · ∧ ψn → ¬φ
to obtain K 6= (ψ1 ∧ · · · ∧ ψn → ¬φ) ∈ T . Followed by the use of Kripke’s axiom
and modus ponens we get K 6= (ψ1 ∧ · · · ∧ ψn ) → K 6= ¬φ ∈ T , which by modus
ponens again gives K 6= ¬φ ∈ T . This is in contradiction to T being consistent since
L6= φ ∈ T . So A is consistent, and by Lindenbaum’s lemma [2, p. 197] there exists
L6=
a maximally consistent U ⊇ A. It is easy to see that φ ∈ U and T −→ U .
3L6=
L6= 3
Proposition 3.11. If U −→ V , then U −→ V . In other words, if U and V are m3
L6=
theories, and W is such that U −→ W −→ V , then there exists some T such that
L6=
3
U −→ T −→ V .
Proof. This proof makes use of the Cross Axiom for K 6= .
3
L6=
Suppose we have U , V and W such that U −→ W −→ V . Let S := {3φ : φ ∈
V } ∪ {ψ : ψ ∈ U }. Suppose towards a contradiction that S is not consistent. Then there
exists a finite subset S0 of S which is inconsistent. Write
S0 = {3φ1 , 3φ2 , . . . , 3φn } ∪ {ψ1 , ψ2 , . . . , ψm },
where for each i and j, φi belongs to V and K 6= ψj belongs to U . Let φ = φ1 ∧ · · · ∧ φn
and similarly define ψ = ψ1 ∧ · · · ∧ ψm . Since V is closed under conjunction, φ ∈ V . And
36
because K 6= ψ is equivalent to a conjunction K 6= (ψ1 ∧ . . . ψm ) from U, we have K 6= ψ ∈ U .
Finally 3φ → 3φi for all i, and ψ → ψj for all j. Since S0 is inconsistent, ` 3φ → ¬ψ.
By the (K)-axiom for K 6= and modus ponens, it follows that ` L6= 3φ → L6= ¬ψ, so this
sentence belongs to U . Also, φ ∈ V , so L6= φ ∈ W and 3L6= φ ∈ U . By the Cross axiom,
L6= 3φ ∈ U . By modus ponens on ` L6= 3φ → L6= ¬ψ, it follows that L6= ¬ψ ∈ U . But
this is in contradiction with U being consistent since K 6= ψ ∈ U . So S is consistent.
By Lindenbaum’s lemma [2, p. 197] there exists T ⊇ S maximally consistent. By
3
L6=
construction U −→ T −→ V , which proves the proposition.
3
3
3
L6=
Proposition 3.12. Suppose that T1 −→ T2 −→ . . . −→ Tn , and suppose Tn −→ U .
3
3
3
L6=
Then there are U1 −→ U2 −→ . . . −→ Un such that Un = U , and for all i, Ti −→ Ui .
T1
L6=
U1
3
3
3
/
/ T2
/
...
3
L6=
/ Tn
3
U2
/
...
3
/
L6=
Un
3
L6=
Proof. Given T1 , T2 , . . . , Tn and U as above, let Un = U . Since Tn−1 −→ Tn −→ Un , we
L6=
3
get Un−1 from proposition 3.11 such that Tn−1 −→ Un−1 −→ Un . We continue backwards
in this way in order to get Un−2 , . . . , U2 , U1 .
3.4.2. The proof
We suspect that if we could prove completeness, the proof would be very similar to the
proof of completeness of the base axioms in section 2.4 above. In was only when checking
the details of the construction that we realized there was one part that did not work out
as we suspected. We will first give the proof, assuming that we can do the construction,
and describe the details of the construction with the difficulty in the next section.
Again, we would not work with the canonical model. Instead, we would want to build
a space X of “abstract” points.
Our strategy would be to build a space X of “abstract” points. The opens will also
be given in an abstract way, via a poset P and an order-reversing map i : P → P ∗ (X),
where P ∗ (X) is the set of non-empty subsets of X. The points are abstract since they
are not theories. But with each x and each p so that x ∈ i(p) there will be a “target”
m-theory t(x, p). The goal of the construction would be to arrange that in the overall
model, th(x, i(p)) = t(x, p).
We only need to prove that every m-theory T has a model in order to prove completeness. Remember that we do not claim that we can prove completeness, but assume in
this section that the construction goes well. Let T be an m-theory. We would build
1. A set X containing a designated element x0 .
2. A poset hP, ≤i with least element ⊥.
37
3. A function i : P → P ∗ (X) such that p 6 q iff i(q) ⊇ i(p), and i(⊥) = X. So i is
an order-reversing homomorphism from hP, 6, ⊥i to hP ∗ (X), ⊇, Xi.
4. A partial function t : X × P → T H such that t(x, p) is defined iff x ∈ i(p). The
following properties need to be ensured for all p ∈ P, x ∈ i(p) andφ:
L6=
(a1) If y ∈ i(p) and y 6= x, then t(x, p) −→ t(y, p).
(a2) If L6= φ ∈ t(x, p), then for some y ∈ i(p) such that y 6= x : φ ∈ t(y, p).
3
(b1) If q > p, then t(x, p) −→ t(x, q).
(b2) If 3φ ∈ t(x, p), then for some q > p, φ ∈ t(x, q).
(c) t(x0 , ⊥) = T , where T is the m-theory that we fixed and want to make a model
for.
Suppose we have X, P, i and t that satisfy these properties. Then we consider the
subset space model
X = hx, {i(p) : p ∈ P}, αi,
where the valuation is as follows α(A) = {x : A ∈ t(x, ⊥)}.
Next, consider the Truth Lemma, which is key in proving Completeness, assuming
that we can do the construction.
Lemma 3.13. (The Truth Lemma). Assume conditions (1)-(4) for X,P,i and t. Then
for all x ∈ X and all p ∈ P such that x ∈ i(p),
thX (x, i(p)) = t(x, p).
Proof. We will only show the induction step for L6= since the atomic case, boolean steps,
and induction step for 3 are exactly the same as in the proof of the Truth Lemma in
subsection 2.4.2
So let us do the induction step for L6= . Suppose that L6= φ ∈ thX (x, i(p)), i.e. that
x, i(p) |= L6= φ. Then there is some y ∈ i(p) such that y 6= x and y, i(p) |= φ. By the
L6=
induction hypothesis φ ∈ t(y, p). By condition (a1), t(x, p) −→ t(y, p). So L6= φ ∈ t(x, p).
Now suppose L6= φ ∈ t(x, p). Then by property (a2) there is some y ∈ i(p) such that
y 6= x and φ ∈ t(y, p). By the induction hypothesis we have φ ∈ thX (y, i(p)), i.e.
y, i(p) |= φ. Thus x, i(p) |= L6= φ, i.e. L6= φ ∈ thX (x, i(p)).
This proves the Truth Lemma.
Now that we have the Truth Lemma, we can use property (4c) to show that T had a
model, since T = t(x0 , ⊥) = thx (x0 , i(⊥)) = thx (x0 , X). If only we could construct X,
P, i and t, this would prove that the K 6= 2-axioms are complete for subset spaces.
38
3.4.3. Construction
We would want to build X, P, i and t by recursion. That is, we would build approximations Xn , Pn , in and tn satisfying certain local and global properties. Fix two objects x0
and ⊥. Assuming that we can do the construction, it would have to satisfy the following
local properties.
(L1) Xn is a finite set containing x0 .
(L2) Pn is a finite poset with ⊥ as minimum, and with the property that for each p ∈ Pn ,
the lower set of p, {q ∈ Pn : q 6 p}, is linearly ordered.
(L3) The map in : Pn → P ∗∗ (Xn ), where P ∗∗ (X) is the collection of subsets of Xn
with at least two elements, has the property that p 6 q iff in (q) ⊆ in (p). Also,
in (⊥) = Xn .
(L4) The function tn : Xn ×Pn → T H is partial with the property that tn (x, p) is defined
iff x ∈ i(p). Furthermore, we assume the following properties for all x ∈ Xn and
p ∈ Pn :
L6=
(a) If x, y ∈ in (p) and x 6= y, then tn (x, p) −→ tn (y, p).
3
(b) If x ∈ in (q) and q > p, then tn (x, p) −→ tn (x, q).
(c) t0 (x0 , ⊥) = T .
These conditions are required for specific reasons. In (L2), the requirement that the
lower sets of points is linear is essential to our construction. By maintaining this property
throughout the construction, we can use proposition 3.12 to add points to the model.
The condition in (L3) that each in (p) has at least two elements is not really necessary,
but it leads to a simplification of the overall construction.
Assuming that we can do the construction, it would have to satisfy the following global
properties:
(G1) Xn ⊆ Xn+1 .
(G2) Pn+1 is an end extension of Pn . This means that Pn is a suborder of Pn+1 and if
p ∈ Pn , q ∈ Pn and p 6 q, then p ∈ Pn .
(G3) For all p ∈ Pn+1 , in+1 (p) ∩ Xn = in (p).
(G4) The restriction of tn+1 to Xn × Pn is tn .
Finally, our construction would have to satisfy some overall requirements:
(R4a) If L6= φ ∈ tn (x, p), then for some m > n, there is some y ∈ im (p) such that y 6= x
and φ ∈ tm (y, p).
(R4b) If 3φ ∈ tn (x, p), then for some m > n, there is some q > p in Pm such that
φ ∈ tm (x, q).
39
Note that an adjustment in requirement (R4a) was made compared to requirement (R4a)
in subsection 2.4.3.
Now suppose we have managed to build Xn , Pn , in and tn in such a way that they
satisfy the (L), (G) and (R) requirements.
S
Definition 3.14. Let S
X = n∈N Xn , and let P be the limit of the posets Pn . Let i
be defined by i(p) = n>m in+1 (p), where m is the least number such that p ∈ Pm .
Finally, define t(x, p) = tn (x, p) where n is any number such that tn (x, p) is defined. The
construction has arranged that tn (x, p) = tn+1 (x, p) whenever the latter is defined.
Proposition 3.15. Suppose we build Xn , Pn , in and tn in accordance with the (L), (G)
and (R) requirements. Then X, P, i and t as defined above satisfy conditions (1)-(4)
above.
Proof. It is clear that (1) holds as we have constructed X in this way. In the same way
we have constructed P to have ⊥ as its least element so (2) also holds. To check (3),
first note that x ∈ i(p) if and only for some n, x ∈ in (p). Now suppose p > q and let
x ∈ i(p). Then there is an n such that x ∈ in (p). By (L3) in (p) ⊆ in (q). So x ∈ in (q),
i.e. x ∈ i(q). Thus i(p) ⊆ i(q). On the other hand, if p 6> q, then let n be such that
both p, q ∈ Xn . By (G2), p 6> q in Xn , so by (L3), in (p) 6⊆ in (q). This means there is
some x such that x ∈ in (p) \ in (q). By (G3), in (q) = Xn ∩ i(q), so since x ∈ in (p) ⊆ Xn
we see that x ∈
/ i(q). Hence i(p) 6⊆ i(q). Furthermore, it is easy to see that i(⊥) = X.
This completes the verification of (3).
The verification of (4) are consequences of the overall (R) requirements and conditions
(G4) and (L4c).
3.4.4. Details of the construction
In this section we will give the details of the construction and point out exactly where
it goes wrong, which is at the very end.
At the beginning we fix a map
v : ω −→ {L6= , 3} × ω × ω
with the property that if v(n) = (x, m, k) then m < n, and for all (x, m, k) there is some
n > m such that v(n) = (x, m, k).
Note that we have again used a suggestive notation with {L6= , 3}, it could have been
{1, 2} just as well.
We define by recursion on n a tuple hXn , Pn , in , tn , αn , βn i. The last two, αn and βn are
functions whose purpose is to ensure all of the (R) requirements are met in an orderly
fashion in countably many steps:
αn : ω → {(x, p, φ) ∈ Xn × Pn × L : L6= ∈ tn (x, p)}
βn : ω → {(x, p, φ) ∈ Xn × Pn × L : 3 ∈ tn (x, p)}
40
Let X0 be a two-element set {x0 , x1 }, let P0 be the trivial poset {⊥}, let i0 (⊥) = X
and t0 (x0 , ⊥) = T , where T is the m-theory that we started with. Now all of the (L)
requirements for n = 0 are met. We also fix functions α0 and β0 , but this is just to make
sure that they are now arbitrary.
Now suppose we are at stage n+1 of the construction. There are two cases, depending
on the value of v(n + 1). We will only consider the first case. The second case, where
v(n + 1) = (3, m, k), is exactly the same as in subsection 2.4.4. At the end of the first
case is where the problem occurs.
Case 1: v(n + 1) = (L6= , m, k) for some m and k. Now consider αm (k), and let it
be equal to the triple (x, p, φ). This means that L6= φ ∈ tm (x, p). The first thing we do
now is check whether there is already a witness for φ in Xn . In other words: is there a
y ∈ in (p) ⊆ Xn , different from x such that φ ∈ tn (y, p) ?
Subcase 1a: There is an y 0 ∈ in (p) such that y 0 6= x and φ ∈ tn (y 0 , p). Then let
Xn+1 := Xn , Pn+1 := Pn and tn+1 := tn . Note that then in+1 = in .
Subcase 1b: There is no y 0 ∈ in (p) that is different from x such that φ ∈ t(y 0 , p). So
let y be a new point not in Xn , so y 6= x. Let Xn+1 := Xn ∪ {y}. Let Pn+1 := Pn . Let
(
in (q) ∪ {y} if q 6 p,
in+1 (q) =
in (q)
otherwise.
Note that conditions (L1), (L2), (G1),(G2) and (G3) are trivial. Indeed, Xn+1 is still
finite (L1) and Xn ⊆ Xn+1 (G1), and Pn+1 = Pn (L2,G2). For (G3), let q ∈ Pn+1 . Then
if q 6 p, in+1 (q) = in (q) ∪ {y} so in+1 (q) ∩ Xn = in (q) since y ∈ Xn+1 \ Xn . Otherwise
in+1 (q) = in (q) so obviously in+1 (q) ∩ Xn = in (q). Also, in+1 (⊥) = Xn+1 .
Next we will check condition (L3), i.e. that q 6 r if and only if in+1 (r) ⊆ in+1 (q).
For this we look at three cases. First, if neither q 6 p nor r 6 p, then in+1 (q) = in (q)
and in+1 (r) = in (r). So we are done. Second, if both q and r are below p, then
in+1 (q) = in (q) ∪ {y} and in+1 (r) = in (r) ∪ {y}. So in+1 (r) ⊆ in+1 (q) if and only if
in (r) ⊆ in (q), which is equivalent to q 6 r. For the last case, suppose that q 6 p
but r 66 p. Then also r 66 q. And since y ∈
/ in+1 (r), in+1 (q) 6⊆ in+1 (r). Now, since
in+1 (q) = in (q) ∪ {y}, we again have that in+1 (r) ⊆ in+1 (q) if and only if in (r) ⊆ in (q).
Now we have completed the verification of (L3).
Next we wish to define tn+1 , and it is when chekcing condition (L4a) that we will
realize there is a problem. First, we stipulate that (G4). Remember that tn+1 is a map
from Xn+1 × Pn+1 to T H, so we only need to define tn+1 (y, q) for q 6 p, since y is the
only point in Xn+1 that is not in Xn , and when q 66 p we have in (q) ⊂ in (p), so no
changes are made.
We use condition (L2) to see that {q : q 6 p} is a finite, linearly ordered set. So write
this set as q1 6 q2 6 · · · 6 qN = p. Then i(p) ⊆ · · · ⊆ i(q2 ) ⊆ i(q1 ) by (L3), so for all i,
x ∈ i(qi ). Thus we can write
3
3
3
t(x, q1 ) −→ t(x, q2 ) −→ . . . −→ t(x, qN ).
41
Note that we have used the notation t(x, qi ) here instead of tn+1 (x, qi ). This is because
of condition (G4) combined with the definition of t: the restriction of tn+1 to Xn × Pn
is tn , and t(z, r) = tn (z, r), where n is any number such that tn (z, r) is defined.
L6=
Now let U be such that t(x, qN ) −→ U and φ ∈ U . This exists by 3. of proposition
3.10. To be able to apply this proposition we need to realize that L6= φ ∈ t(x, qN ). We
began with the triple αm (k) which told us that L6= φ ∈ tm (x, p). We know that p = qN
so L6= φ ∈ tm (x, qN ). And of course by the definition of t again t(x, qN ) = tm (x, qn ). So
L6= φ ∈ t(x, qN ) and we can use the proposition.
To give a visual representation:
t(x, q1 )
3
/
t(x, q2 )
3
/
...
3
/
t(x, qN )
L6=
U
L6=
By proposition 3.12 there exist m-theories U1 , U2 , . . . , UN such that U = UN , t(x, qi ) −→
3
3
3
Ui for all i, and U1 −→ U2 −→ . . . −→ UN . Let t(y, qi ) := Ui for 1 6 i 6 N . So we get:
t(x, q1 )
L6=
t(y, q1 )
3
3
/
/
t(x, q2 )
3
/
...
3
/
L6=
t(x, qN )
t(y, q2 )
3
/
...
3
/
L6=
t(y, qN )
and this definition of tn+1 ensures (L4b).
Up to this point everything seems to work out. Next we would want to check whether
(L4a) holds for this definition of tn+1 , and here is where the problem occurs. Remember,
L6=
(L4a) for n + 1 is : “if x, y ∈ in+1 (p) and x 6= y, then tn+1 (x, p) −→ tn+1 (y, p)”. So
suppose a, b ∈ in+1 (q). We might as well assume that a = y, and hence that q 6 p. So
let q := qi for some i. To satisfy the premise we need to assume that b 6= y, i.e. that
b ∈ in (q). From qi 6 p it follows that in+1 (qi ) ⊇ in+1 (p). Also, x ∈ im (p), and since
m > n + 1 by the condition imposed on v, we have x ∈ in+1 (p) thus x ∈ in+1 (p). By
the definition of tn+1 we have tn+1 (x, q) = tn (x, q) and tn+1 (b, q) = tn (b, q) (since b 6= y),
L6=
and by the assumption that (L4a) holds for n, we get tn+1 (x, q) −→ tn+1 (b, q). By
L6=
construction, and by (G4), tn+1 (x, q) −→ tn+1 (y, q). By symmety and quasi-transitivity,
L6=
either tn+1 (y, q) = tn+1 (b, q) or tn+1 (y, q) −→ tn+1 (b, q).
This is the problem. What we would want to show is that tn+1 (y, q) = tn+1 (b, q) can’t
happen. We have not found a way to overcome this problem. We have tried to add an
axiom to the system or alter the conditions, but nothing we tried resulted in solving this
difficulty.
Therefore the completeness of the K 6= 2-axioms for subset spaces remains an open
problem. It is interesting to see that even though everything seems to work nicely
throughout the whole proof and construction, the problem arises in the very last step of
checking the conditions.
42
Popular summary — in Dutch
Het onderwerp van deze scriptie ligt tussen de modale logica en topologie in. Na een
kleine inleiding zullen we kort uitleggen wat topologie, een logica, en modale logica is.
Ook zullen we kort uitleggen wat we precies hebben gedaan.
In de wiskunde wordt het meest gebruikt gemaakt van de de wiskundige taal. Deze
taal bestaat
R bekende symbolen zoals 1, 2, 3, . . . , +, −, , ×, =, etc., en ook symbolen
Puit
als π, x, f, , . Formules in deze taal zien er bijvoorbeeld als volgt uit:
Z ∞
2
2
2
a +b =c ,
x3 dx, 3x + 9y 2 = 10.
0
Het meeste van wiskunde wordt in deze taal gedaan. Er is echter ook een tak van de
wiskunde die een aparte taal gebruikt, en deze tak wordt ‘wiskundige logica’ genoemd.
In deze scriptie alleen worden al twee nieuwe talen geı̈ntroduceerd, en het is ook zo dat
elk onderwerp in de wiskundige logica een eigen taal heeft. Er zijn dus heel veel logische
talen.
Het doel van deze scriptie was logica’s te bestuderen waarin elementaire topologische
redenaties in gedaan kunnen worden. Dit werd gedaan met behulp van modale logica.
Wat topologie eigenlijk is, kan in deze setting het beste beschreven worden door te zeggen
dat voor een wiskundige die topologie bestudeert, een koffiekop en een donut hetzelfde
zijn. Het is een tak van de wiskunde waarin de vorm van een object wordt bekeken.
Vooral van belang van deze vorm zijn de eigenschappen die behouden blijven door het
object te vervormen, dus te buigen, uit te rekken, of te verdraaien, maar zonder het te
scheuren of stukken aan elkaar te plakken.
We hebben in deze scriptie een logica bestudeerd die “the logic of subset spaces” heet
(logica van ruimten van deelverzamelingen). Ook hebben we een uitbreiding, “topologic”,
bekeken, en een aangepaste versie van de ‘logic of subset spaces’. Laten we eerst uitleggen
wat een logica is, en dan bekijken we daarna modale logica.
Een logica is in wezen niets anders dan een verzameling axioma’s met een paar ‘rekenregels’ zoals bijvoobeeld modus ponens. De axioma’s zijn formules waarvan we stellen
dat ze altijd waar zijn en de rekenregels vertellen ons hoe je met die formules moet
werken. Een voorbeeld van modus ponens gaat als volgt: bekijk de zin “als het regent,
dan wordt de grond nat”. Als “het regent” waar is, dan volgt uit voorgaande zin met
modus ponens dat “de grond is nat” ook waar moet zijn. In formules: als φ → ψ en φ
beiden waar zijn, dan kunnen we afleiden dat ψ ook waar is. De verzameling axioma’s
bevat in ieder geval de propositionele tautologieën, zoals p ∨ ¬p. Een tautologie is een
zin die altijd waar is. Dus als p voor de zin “het regent” staat, dan is het altijd waar
dat het òf regent, òf niet regent.
43
Een modale logica is een logica waaraan de concepten van mogelijkheid en noodzaak
zijn toegevoegd. Deze concepten worden uitgedrukt aan de hand van locale versies van
de ∀-‘voor alle’ en ∃-‘er bestaat’ kwantoren. In modale logica zijn uitspraken als “het is
noodzakelijk dat p waar is” of “het is mogelijk dat p waar is” mogelijk en toegestaan.
Een groot deel van deze scriptie is gewijd aan het bewijs van de volledigheid en
correctheid van de axioma’s van de ‘logic of subset spaces’. Deze concepten kunnen vrij
makkelijk uitgelegd worden. Correctheid betekent dat wanneer een formule bewezen kan
worden vanuit de axioma’s, deze ook waar moet zijn. Volledigheid is precies andersom:
wanneer een formule waar is, moet deze ook bewezen kunnen worden vanuit de axioma’s.
Met deze logica konden we elementaire topologische begrippen uitdrukken, en dit
hebben we ook gedaan. Onze mogelijkheid om dingen te zeggen was echter nog vrij
beperkt. We konden in deze taal bijvoorbeeld nog niet zeggen dat twee punten ongelijk
aan elkaar zijn. De oplossing was het maken van een nieuwe taal, met een nieuwe
verzameling axioma’s, en een verandering in de rekenregels. Deze logica lijkt sterk op de
‘logic of subset spaces’, met kleine aanpassingen in de taal, axioma’s en rekenregels. Dit
hebben we het K 6= 2-systeem genoemd. Met dit systeem kunnen al meer topologische
begrippen worden uitgedrukt. De correctheid van de K 6= 2-axioma’s is in deze scriptie
bewezen, de volledigheid geven we als een open probleem. We vermoeden dat het bewijs
sterk lijkt op het volledigheidsbewijs van de ‘logic of subset spaces’, echter in de laatste
stap blijkt er een, voor ons onoverkomelijk, probleem te zitten.
Om terug te keren naar wiskundige en logische talen, hebben we in deze scriptie in
feite niets anders gedaan dan logica’s bestuderen die topologische concepten vertalen
van de wiskundige taal naar een logische taal.
44
Bibliography
[1] Marco Aiello, Ian E Pratt-Hartmann, and Johan FAK van Benthem. Handbook of
spatial logics. Springer, 2007.
[2] Patrick Blackburn, Maarten De Rijke, and Yde Venema. Modal logic, volume 53.
Cambridge University Press, 2002.
[3] A Dabrowski, LS Moss, and R Parikh. Topological reasoning and the logic of knowledge. ANNALS OF PURE AND APPLIED LOGIC, 78(1-3):73–110, APR 4 1996.
[4] Maarten de Rijke. The modal logic of inequality. Journal of Symbolic Logic, pages
566–584, 1992.
[5] Konstantinos Georgatos. Knowledge theoretic properties of topological spaces.
In Knowledge Representation and Reasoning Under Uncertainty, pages 147–159.
Springer, 1994.
[6] Konstantinos Georgatos.
cs/0007038, 2000.
Modal logics for topological spaces.
arXiv preprint
[7] Andrey Kudinov. Topological modal logics with difference modality. Advances in
modal logic, 6:319–332, 2006.
[8] Lawrence S Moss and Rohit Parikh. Topological reasoning and the logic of knowledge:
preliminary report. In Proceedings of the 4th conference on Theoretical aspects of
reasoning about knowledge, pages 95–105. Morgan Kaufmann Publishers Inc., 1992.
45
A. Notes on Topological reasoning
and the logic of knowledge from
Dabrowski, Moss and Parikh [3]
For this thesis the completeness proof of the logic of subset spaces in the article Topological reasoning and the logic of knowledge by Dabrowski, Moss and Parikh [3] was
carefully read. In this appendix we wish to point out typos that were found. Some of
them were easily recognizable, others required some thinking and we think it is useful
for the reader that we give some notes on them.
First the easily recognizable typos and their supposed correction:
• on p. 86 the definition of α is given as follows:
“[...] where α(P ) = {x : A ∈ t(x, ⊥)}”.
We believe the authors meant the following:
“[...] where α(A) = {x : A ∈ t(x, ⊥)}”.
• on p. 90 in the second alinea, the arguments of the function tn+1 are exchanged:
“[...], let tn+1 (r, y) = tn+1 (r, x) and tn+1 (r, z) = tn+1 (r, x)”.
Since this function is given as a function from Xn × Pn to T H, it is quite clear
that this should be:
“[...], let tn+1 (y, r) = tn+1 (x, r) and tn+1 (z, r) = tn+1 (x, r)”.
Now let us have a look at the typos that required some thought before being recognized.
• on p. 89 in the first case, the statement “We check that q 6 r iff i(r) ⊆ i(q)”
tells us that we are checking that requirement (L3) is met. We think it would be
better to put “We check that q 6 r iff in+1 (r) ⊆ in+1 (q)”. This is of course but a
small correction since the definition of i makes sure that it does not make much
of a difference. However, this addition makes this step more precise and it would
help the reader to see what exactly is happening at this point of the construction.
46
• on p. 89 at the bottom of the page, while checking that the (L3) requirement in
the second case is met, we encountered the following: “For the same reason, if
r 6 q in Xn+1 , then in+1 (q) ⊂ in+1 ”. In our opinion the authors meant for this to
be “For the same reason, if r 6 q in Xn+1 , then in+1 (q) ⊆ in+1 ”. Namely, if r 6 q
in Xn+1 then either r = q or r < q. In the first case then in+1 (q) = in+1 , and in
the second case in+1 (q) ⊂ in+1 , so if r 6 q in Xn+1 then in+1 (q) ⊆ in+1 .
47