Spatial logic of tangled closure operators and modal mu-calculus

Abstract There has been renewed interest in recent years in McKinsey and Tarski's interpretation of modal logic in topological spaces and their proof that S4 is the logic of any separable dense-in-itself metric space. Here we extend this work to the modal mu-calculus and to a logic of tangled closure operators that was developed by Fernandez-Duque after these two languages had been shown by Dawar and Otto to have the same expressive power over finite transitive Kripke models. We prove that this equivalence remains true over topological spaces. We extend the McKinsey–Tarski topological ‘dissection lemma’. We also take advantage of the fact (proved by us elsewhere) that various tangled closure logics with and without the universal modality ∀ have the finite model property in Kripke semantics. These results are used to construct a representation map (also called a d-p-morphism) from any dense-in-itself metric space X onto any finite connected locally connected serial transitive Kripke frame. This yields completeness theorems over X for a number of languages: (i) the modal mu-calculus with the closure operator ◇; (ii) ◇ and the tangled closure operators 〈 t 〉 (in fact 〈 t 〉 can express ◇); (iii) ◇ , ∀ ; (iv) ◇ , ∀ , 〈 t 〉 ; (v) the derivative operator 〈 d 〉 ; (vi) 〈 d 〉 and the associated tangled closure operators 〈 d t 〉 ; (vii) 〈 d 〉 , ∀ ; (viii) 〈 d 〉 , ∀ , 〈 d t 〉 . Soundness also holds, if: (a) for languages with ∀, X is connected; (b) for languages with 〈 d 〉 , X validates the well-known axiom G 1 . For countable languages without ∀, we prove strong completeness. We also show that in the presence of ∀, strong completeness fails if X is compact and locally connected.


Introduction
Modal logic can be given semantics over topological spaces. In this setting, the modality 3 can be interpreted in more than one way. The first and most obvious way is as closure. Writing JϕK for the set of points (in a topological model) at which a formula ϕ is true, J3ϕK is defined to be the closure of JϕK, so that 3ϕ holds at a point x if and only if every open neighbourhood of x contains a point y satisfying ϕ.
In a seminal result, McKinsey and Tarski [27] proved that the logic of any given separable 1 dense-in-itself metric space in this semantics is S4: it can be axiomatised by the basic modal Hilbert system K augmented by the two axioms 2ϕ → ϕ (T) and 2ϕ → 22ϕ (4).
Motivated perhaps by the current wide interest in spatial logic, a wish to present simpler proofs in 'modern language', growing awareness of the work of particular groups such as Esakia's and Shehtman's, or involvement in new settings such as dynamic topology, interest in McKinsey and Tarski's result has revived in recent years. A number of new proofs of it have appeared, some for specific spaces or embodying other variants [30,4,1,31,39,24,17]. Very recently, strong completeness (every countably infinite S4-consistent set of modal formulas is satisfiable in every dense-in-itself metric space) was established by Kremer [20].
In this paper, we seek to extend McKinsey and Tarski's theorem to more powerful languages. We will extend the modal syntax in two separate ways: first, to the mu-calculus, which adds least and greatest fixed points to the basic modal language, and second, by adding an infinite sequence of new modalities 3 n of arity n (n ≥ 1) introduced in the context of Kripke semantics by Dawar and Otto [7]. The semantics of 3 n is given by the mu-calculus formula 3 n (ϕ 1 , . . . , ϕ n for a new atom q not occurring in ϕ 1 , . . . , ϕ n . The order and multiplicity of arguments of 3 n is immaterial, so we will abbreviate 3 n (γ 1 , . . . , γ n ) to t {γ 1 , . . . , γ n }. Fernández-Duque used this to give the modalities topological semantics, dubbed them tangled closure modalities (this is why we use the notation t ), and studied them in [9][10][11][12]. Dawar and Otto [7] showed that, somewhat surprisingly, the mu-calculus and the tangled modalities have exactly the same expressive power over finite Kripke models with transitive frames. We will prove that this remains true over topological spaces. So the tangled closure modalities offer a viable alternative to the mu-calculus in both these settings . We go on to determine the logic of an arbitrary dense-in-itself metric space X in these languages. We will show that in the mu-calculus, the logic of X is axiomatised by a system called S4μ comprising Kozen's basic system for the mu-calculus augmented by the S4 axioms, and the tangled logic of X is axiomatised by a system called S4t similar to one in [10]. We will establish strong completeness for countable sets of formulas.
We will also consider the extension of the tangled language with the universal modality, '∀'. (Earlier work on the universal modality in topological spaces includes [36,25].) This language can express connectedness: there is a formula C valid in precisely the connected spaces. Adding this and some standard machinery for ∀ to the system S4t gives a system called 'S4t.UC'. We will show that every S4t.UC-consistent formula is satisfiable in every dense-in-itself metric space. Thus, the logic of an arbitrary connected dense-in-itself metric space is S4t. UC. We also show that strong completeness fails in general, even for the modal language plus the universal modality.
A second and more powerful spatial interpretation of 3 is as the derivative operator. Following tradition, when considering this interpretation we will generally write the modal box and diamond as [d] and d . In this interpretation, J d ϕK is defined to be the set of strict limit points of JϕK: so d ϕ holds at a point x precisely when every open neighbourhood of x contains a point y = x satisfying ϕ. The original closure diamond is expressible by the derivative operator: 3ϕ is equivalent in any topological model to ϕ ∨ d ϕ, and 2ϕ to ϕ ∧ [d]ϕ. So in passing to d , we have not reduced the power of the language.
Already in [27, Appendix I], McKinsey and Tarski discussed the derivative operator and asked a number of questions about it. It has since been studied by, among others, Esakia and his Tbilisi group ( [8,3], plus many other publications), Shehtman [34,38], Lucero-Bryan [25], and Kudinov-Shehtman [23], section 3 of which contains a survey of results.
In the derivative semantics, determining the logic of a given dense-in-itself metric space is not a simple matter, for the logic can vary with the space. As McKinsey and Tarski observed, d (x ∧ d ¬x) ∨ (¬x ∧ d x) ↔ d x ∧ d ¬x is valid in R 2 but not in R ( [34] attributes this observation to Kuratowski (1922)). This formula is valid in the same topological spaces as the formula G 1 , where for each integer n ≥ 1, Here, p 0 , . . . , p n are pairwise distinct atoms, and for i = 0, . . . , n, The formulas G n were introduced by Shehtman [34, p. 43]. A sufficient (but not necessary) condition for G n to be valid in a space is that every open neighbourhood of an arbitrary point x contains an open neighbourhood N of x such that N \ {x} can be partitioned into at most n non-empty open sets (cf. [34, lemma 2, p. 3]). So, for example, G 1 is valid in R 2 , and G 2 in R. See Remark 8.6 below for further discussion . We now recall some relevant results on [d]-logics. R1. In [34], Shehtman proved that the logic of every separable zero-dimensional dense-in-itself metric space (such as Q and the Cantor space) is just KD4, axiomatised by the basic system K together with the axioms d (D) and [d]p → [d][d]p (4). This is the smallest possible logic of a dense-in-itself metric space in the derivative semantics. R2. [34] also proved that the logic of R n for finite n ≥ 2 is KD4G 1 , axiomatised by KD4 plus G 1 . In fact, rather more is shown: see Remark 8.6 below. R3. The logic of R was shown by Shehtman [38] and Lucero-Bryan [25] to be KD4G 2 , and KD4G 2 .UC if ∀ is added. R4. In [16], R1 and R3 were extended to tangled closure modalities and the separability assumption in R1 was eliminated. R5. [5] proved that there are continuum-many logics of subspaces of the rationals in the language with [d]. R6. It is plain that G 1 G 2 G 3 · · · , so the logics KD4G 1 ⊇ KD4G 2 ⊇ · · · form a decreasing chain, and by [25, corollary 3.11], its intersection is KD4.
Shehtman [34, problem 1] asked if KD4G 1 is the largest possible logic of a dense-in-itself metric space in the derivative semantics. In this paper, we answer Shehtman's question affirmatively: every KD4G 1 -consistent formula of the language with d is satisfiable in every dense-in-itself metric space. Thus, the logic of every dense-in-itself metric space that validates G 1 is exactly KD4G 1 . This strengthens R2 above. We also establish strong completeness for such spaces.
Adding the tangled closure operators, we prove similarly that the logic of every dense-in-itself metric space that validates G 1 is axiomatised by KD4G 1 t (including the tangle axioms). We also prove strong completeness.
Further adding the universal modality, we show similarly that KD4G 1 t.UC (and KD4G 1 .UC if the tangle closure operators are dropped) axiomatises the logic of every connected dense-in-itself metric space that validates G 1 . Strong completeness fails in general, as a consequence of the proof that it already fails for the weaker language with 2 and ∀.
The reader can find a summary of our results in Table 2 in section 10. Our proof works in a fairly familiar way, similar in spirit to McKinsey and Tarski's original argument in [27]. There are two main steps. 1. We establish Tarski's 'dissection lemma' [41, satz 3.10], [27, theorem 3.5] and a variant of it. 2. These topological results are used to construct a map from an arbitrary dense-in-itself metric space onto any finite connected KD4G 1 Kripke frame, preserving the required formulas.
Step 2, together with results from [14] and the mu-calculus canon establishing the finite model property for the various logics in Kripke semantics, proves completeness for all the languages, which is then lifted by a separate argument to strong completeness for languages without ∀.
It can be seen that our results concern the logic of each individual space within a large class of spaces (the dense-in-themselves metric spaces), rather than the logic of a large class of spaces, or of particular spaces such as R. This is as in [27]. We do not assume separability, we consider languages that have not previously been much studied in the topological setting, and we obtain some results on strong completeness, a matter that has only recently been investigated in this arena.

Basic definitions
In this section, we lay out the main definitions, notation, and some basic results.

Notation for sets and binary relations
Let X, Y, Z be sets. We let ℘(X) denote the power set (set of all subsets) of X. We write X \ Y for {x ∈ X : x / ∈ Y }. Note that (X ∩ Y ) \ Z = X ∩ (Y \ Z), so we may omit the parentheses in such expressions. For a partial function f : X → Y , we let dom f denote the domain of f , and rng f its range.
A binary relation on a set W is a subset of W × W . Let R be a binary relation on W . We write any of R(w 1 , w 2 ), Rw 1 w 2 , w 1 Rw 2 to denote that (w 1 , w 2 ) ∈ R. We say that R is reflexive if R(w, w) for all w ∈ W , and transitive if R(w 1 , w 2 ) and R(w 2 , w 3 ) imply R(w 1 , w 3 ). We write R * for the reflexive transitive closure of R: the smallest reflexive transitive binary relation that contains R. We also write The notation is loosely motivated by the traditional use of • for a reflexive world and • for an irreflexive world in diagrams of frames in modal logic. For w ∈ W , we say that w is reflexive if Rww, and irreflexive otherwise. We let R(w) denote the set {w ∈ W : R(w, w )}, sometimes called the set of R-successors or R-alternatives of w. For W ⊆ W , we write R W for the binary relation R ∩ (W × W ) on W .
We write Z for the set of integers, Q for the set of rational numbers, R for the set of real numbers, and ω for the first infinite ordinal. A set will be said to be countable if its cardinality is at most ω.

Kripke frames
A (Kripke) frame is a pair F = (W, R), where W is a non-empty set of 'worlds' (sometimes referred to as the domain of F), and R is a binary relation on W . We attribute properties to a frame by the usual extrapolation from the frame's components. So, we say that F is finite if W is finite, reflexive if R is reflexive, serial if R(w) = ∅ for every w ∈ W , and transitive if R is transitive. Two frames are said to be disjoint if their respective sets of worlds are disjoint. And so on.
A root of F is an element w ∈ W such that W = R * (w). Roots of a frame may not exist, nor be unique when they do. We say that F is rooted if it has a root. At the other end, an element w ∈ W is said to be R-maximal if R • (w) = ∅. Such an element has no 'proper' R-successors, of which it is not itself an R-successor.
If F is transitive, a cluster in F is an equivalence class of the equivalence relation R • ∪ {(w, w) : w ∈ W } on W . A cluster consists either of a single irreflexive world, in which case we say it is degenerate, or a non-empty set of reflexive worlds, in which case we say it is nondegenerate. For example, if w is R-maximal then R * (w) is a cluster.
A subframe of F is a frame of the form F = (W , R W ), for non-empty W ⊆ W . It is simply a substructure of F in the usual model-theoretic sense. We call F the subframe of F based on W . We say that F is a proper subframe of F if W = W . We say that F is a generated or inner subframe of F if R(w) ⊆ W for every w ∈ W -equivalently, R W = R ∩ (W × W ). For w ∈ W , we write: • F(w) for the subframe (R(w), R R(w)) of F based on R(w), • F * (w) for the subframe (R * (w), R R * (w)) of F generated by w.
For an integer n ≥ 1, we say that F is connected if it is not the union of two pairwise disjoint generated subframes (recall that subframes are non-empty), and locally connected if for each w ∈ W , the subframe F(w) is connected. Note that F is connected iff the equivalence relation (R ∪ R −1 ) * on W has a single equivalence class (i.e., it is the global relation W × W ). Every rooted frame is connected.

Topological spaces
We will assume some familiarity with topology, but we take a little time to reprise the main concepts and notation. A topological space is a pair (X, τ ), where X is a set and τ ⊆ ℘(X) satisfies: 1. if S ⊆ τ then S ∈ τ , 2. if S ⊆ τ is finite then S ∈ τ , on the understanding that ∅ = X. We use the signs int, cl, d to denote the interior, closure, and derivative operators, respectively. So for S ⊆ X, That is, closure and d are additive and interior is multiplicative. It follows that each of these three operators is monotonic: We follow standard practice and identify (notationally) the space (X, τ ) with X. The reader should note that we do allow empty topological spaces, where X = ∅. This is particularly useful when dealing with subspaces.
A subspace of X is a topological space of the form (Y, {O ∩ Y : O ∈ τ }), for (possibly empty) Y ⊆ X. It is a subset of X, made into a topological space by endowing it with what is called the subspace topology. It is said to be an open subspace if Y is an open subset of X. As with X, we identify (notationally) the subspace with its underlying set, Y . We write int Y , cl Y for the operations of interior and closure in the subspace Y . It can be checked that for every We will be considering various properties that a topological space X may have. We leave most of them for later, but we mention now that X is said to be dense in itself if no singleton subset is open, connected if it is not the union of two disjoint non-empty open sets, and separable if it has a countable subset D with X = cl D. We say that X is T1 if every singleton subset {x} is closed, and T D if the derivative d {x} of every singleton is closed, which is equivalent to requiring , is strictly weaker than T1.

Metric spaces
A metric space is a pair (X, d), where X is a set and d : X × X → R is a 'distance function' (having nothing to do with the operator d above) satisfying, for all x, y, z ∈ X, We assume some experience of working with this definition, in particular with the triangle inequality. Examples of metric spaces abound and include the real numbers R with the standard distance function d(x, y) = |x − y|, R n with Pythagorean distance, etc. As usual, we often identify (notationally) (X, d) with X.
Let (X, d) be a metric space, and x ∈ X. For non-empty S ⊆ X, define We leave d( We frequently regard a metric space (X, d) equally as a topological space (X, τ d ). So, we will say that a metric space has a given topological property (such as being dense in itself) if the associated topological space has the property. As an example, it can be checked that every metric space is It is plainly a metric space, and the topological space (Y, τ d Y ×Y ) is a subspace of (X, τ d ).

Fixed points
Let X be a set and f : ℘(X) → ℘(X) be a map. We say that f is monotonic if f (S) ⊆ f (S ) whenever S ⊆ S ⊆ X. By a well known theorem of Knaster and Tarski [42], actually formulated for complete lattices, every monotonic f : ℘(X) → ℘(X) has a least and a greatest fixed point -there is a unique ⊆-minimal subset L ⊆ X such that f (L) = L, and a unique ⊆-maximal G ⊆ X such that f (G) = G. We write L = LFP(f ) and G = GFP(f ).
There is a useful way to 'compute' these fixed points. A subset S ⊆ X is said to be a pre-fixed point of f if f (S) ⊆ S, and a post-fixed point if f (S) ⊇ S. Now, the Knaster-Tarski theorem [42] states that LFP(f ) is the intersection of all pre-fixed points of f , and dually for GFP(f ): Least fixed points are used in the semantics of the mu-calculus, coming up next.

Languages
We assume some familiarity with modal languages and the mu-calculus. We fix an infinite set Var of propositional variables, or atoms. We will be considering various logical languages. The biggest of them is denoted by L μ t dt 2[d]∀ , which is a set of formulas defined as follows: is a formula, 3. if ϕ, ψ are formulas then so are ¬ϕ, (ϕ ∧ ψ), 2ϕ, [d]ϕ, and ∀ϕ, 4. if Δ is a non-empty finite set of formulas then t Δ and dt Δ are formulas, 5. if q ∈ Var and ϕ is a formula that is positive in q (that is, every free occurrence of q as an atomic subformula of ϕ is in the scope of an even number of negations in ϕ; free means 'not in the scope of any μq in ϕ'), then μqϕ is a formula, in which all occurrences of q are bound. Bound atoms arise only in this way.
For formulas ϕ, ψ, and q ∈ Var, the expression ϕ(ψ/q) denotes the result of replacing every free occurrence of q in ϕ by ψ, where the result is well formed -that is, all of its subformulas of the form μpθ are such that θ is positive in p. We leave ϕ(ψ/q) undefined if the result is not well formed. For example, if ϕ = μp q then ϕ(¬p/q) is undefined, since μp ¬p is not well formed. We use standard abbreviations: being true at w in M. The definition is by induction on ϕ, as follows: 6. The truth condition for [d]ϕ is exactly the same as for 2ϕ. 2 . . ∈ W with R(w n , w n+1 ) for each n < ω and such that for each δ ∈ Δ there are infinitely many n < ω with M, w n |= δ. 9. The truth condition for dt Δ is exactly the same as for t Δ. 10 Since ϕ is positive in q, it can be shown that f is monotonic, so it has a least fixed point, LFP(f ) (see section 2.5). We define M, w |= μqϕ iff w ∈ LFP(f ).
In the notation of the last clause, it can be checked that M, w |= νqϕ iff w ∈ GFP(f ). For a set Γ of formulas, we write M, w |= Γ if M, w |= γ for every γ ∈ Γ. A word on the semantics of t and dt . Let us temporarily write ϕ ≡ ψ to mean that M, w |= ϕ ↔ ψ for every transitive Kripke model M = (W, R, h) and every w ∈ W . Then it can be checked that for every non-empty finite set Δ of formulas, and w ∈ W , we have M, w |= ϕ iff M , w |= ϕ.

Topological semantics
Given a topological space X, an assignment into X is simply a map h : Var → ℘(X). A topological model is a pair (X, h), where X is a topological space and h an assignment into X. We will also be considering topological models where Var is replaced by some other set of atoms. Details will be given later. As with Kripke models, we attribute a topological property to a topological model if the underlying topological space has the property.
For every topological model (X, h) and every point x ∈ X, we define (X, h), x |= ϕ, for a L μ t dt 2[d]∀ -formula ϕ, by induction on ϕ: We do not require ϕ to hold at x itself. 7. (X, h), x |= ∀ϕ iff (X, h), y |= ϕ for every y ∈ X. 8. For a non-empty finite set Δ of formulas for which we have inductively defined semantics, write JδK = {x ∈ X : (X, h), x |= δ}, for each δ ∈ Δ. Then define: 9. Suppose that ϕ is positive in q and (inductively) that JϕK h = {x ∈ X : (X, h), x |= ϕ} is well defined, for every assignment h into X. Define a map f : where h[S/q] is defined as in Kripke semantics. Again, f is monotonic, and we define ( at which δ is true (and with z = y in the case of dt ). If she cannot, player ∀ wins. That is the end of the round, and the next round commences from position z. Player ∃ wins if she survives every round. It can be checked that (X, h), x |= t Δ (respectively, (X, h), x |= dt Δ) iff ∃ has a winning strategy in this game (respectively, the game where she must additionally choose z = y).
As an aside, a transitive Kripke frame (W, R) can be made into a topological space

Topological semantics in open subspaces
Let X be a topological space and Y a subspace of X. Each assignment h : .
(This holds vacuously if Y is empty.)

Hilbert systems
These are familiar, and we will be informal. A Hilbert system H in a given language L ⊆ L is a set of axioms, which are L-formulas, and inference rules, which have the form Similarly, Γ is said to be satisfiable in X if there exist an assignment h into X and a point x ∈ X such that (X, h), x |= Γ.
Let ϕ be an L μ t dt 2[d]∀ -formula. We say that ϕ is satisfiable in F, or in X, if the set {ϕ} is so satisfiable. We say that ϕ is valid in F (respectively, in X) if ¬ϕ is not satisfiable in F (respectively, in X). We may also say in this case that F or X validates ϕ.
We also say that ϕ is equivalent to a formula ψ in F (respectively, X) if ϕ ↔ ψ is valid in F (respectively, X).

Logics
Let K be a class of Kripke frames or topological spaces. In the context of a given language L ⊆ L μ t dt 2[d]∀ , the (L)-logic of K is the set of all L-formulas that are valid in every member of K. A Hilbert system H for L whose set of theorems is T , say, is said to be • sound over K if T is a subset of the logic of K (all H-theorems are valid in K), • weakly complete, or simply complete, over K if T contains the logic of K (all K-valid formulas are H-theorems), • strongly complete over K if every countable H-consistent set Γ of L-formulas is satisfiable in some structure in K. Recall that in this paper, 'countable' means 'of cardinality at most ω'. The restriction to countable sets will be discussed at the beginning of section 10.2.
The logic of a single frame F is defined to be the logic of the class {F}; similar definitions are used for the other terms here. We say that a Kripke frame F is an H frame, or that F validates H, if H is sound over F. To establish this, it is enough to check that each axiom of H is valid in F, and that each rule of H preserves F-validity (in the notation in (2.2) above, this means that if ϕ 1 , . . . , ϕ n are valid in F then so is ψ).
It can be checked that H is weakly complete over K iff every singleton H-consistent set is satisfiable in some structure in K. Hence, every strongly complete Hilbert system is also weakly complete. The main aim of this paper is to provide Hilbert systems that are (where possible) sound and strongly complete over various topological spaces, with respect to various sublanguages of L μ t dt

Hilbert systems for mu-calculus
We now present a very brief diversion on a Hilbert system 'S4μ' for the mu-calculus that is (sound and) complete over the class of finite reflexive transitive Kripke frames. This will be used in two places: in Theorem 8.3, to show that in the language L μ 2 , the system S4μ is complete over every dense-in-itself metric space; and in Corollary 4.7, together with deep results of Dawar-Otto [7] and the evident soundness of S4μ over topological spaces, to show that L μ 2 is no more expressive than L t 2 over topological spaces, a fact used in Theorem 9.3 to establish strong completeness of S4μ over every dense-in-itself metric space in the language L μ 2 . In this section, all formulas are L μ 2 -formulas, and all Hilbert systems are for this language. The well known substitution rule ϕ ϕ(ψ/q) is not always sound in the mu-calculus and is not needed in other systems, so we omit it.

Kμ:
this is K augmented with the following for each formula ϕ positive in q: • fixed point axiom: ϕ(μqϕ/q) → μqϕ, provided that no free occurrence of an atom in μqϕ gets bound in ϕ(μqϕ/q) -consequently, ϕ(μqϕ/q) is well formed. The idea is roughly that μqϕ is a pre-fixed point of ϕ.
• fixed point rule: ϕ(ψ/q) → ψ μqϕ → ψ , provided that no free occurrence of an atom in ψ gets bound in ϕ(ψ/q) -hence, ϕ(ψ/q) is well formed. The idea this time is roughly that μqϕ is the least pre-fixed point of ϕ.
We write Kμ ϕ if ϕ is a theorem of this system. It is well known (see, e.g., [6, §6]) that the system is equivalent to the original equational system of Kozen [19]. K4μ: this is Kμ plus the '4' scheme 2ϕ → 22ϕ. We write K4μ ϕ if ϕ is a theorem of this system. K4μ is not needed in our spatial completeness results, but it is used in proving equivalence of L μ [d] and L dt [d] over T D spaces (Remark 4.8). S4μ: this is Kμ plus the S4 schemes 2ϕ → ϕ, 2ϕ → 22ϕ. We write S4μ ϕ if ϕ is a theorem of this system.
The following combines some famous and difficult work in the mu-calculus.
We are going to extend it to show that S4μ is sound and complete over the class of finite reflexive transitive frames (and, much later, over every dense-in-itself metric space). First, a form of the substitution rule can be established. Proof (sketch). Let ϕ, ψ, p be as stipulated. For a formula α, write α † = α(ψ/p). We show that S4μ ϕ ⇒ S4μ ϕ † (when the stipulation holds) by induction on the length of a derivation of ϕ in S4μ.
Definition 3. 4. For a formula ϕ, define a new formula ϕ * by induction: • p * = p for p ∈ Var; • − * commutes with the boolean connectives and μ. That is, The formula ϕ * is plainly well formed, for all ϕ ∈ L μ 2 .
Lemma 3. 5. Let ϕ be any formula. Then for every Kripke model (W, R, h) and w ∈ W , we have Proof. The proof is by induction on ϕ. The atomic and boolean cases are easy. Assuming the result for ϕ, it is a well-known exercise in the mu-calculus to check that (W, R, h), w |= (2ϕ) * iff (W, R, h), u |= ϕ * for every u ∈ R * (w). Inductively, this is iff (W, R * , h), u |= ϕ for every u ∈ R * (w), iff (W, R * , h), w |= 2ϕ as required. Remark 3. 6. For each formula ϕ, define a formula ϕ + in the same way as for ϕ * but using the clause (2ϕ) + = νq2(ϕ + ∧ q). It can then be shown that (W, R, h), w |= ϕ + iff (W, R + , h), w |= ϕ, for every formula ϕ, Kripke model (W, R, h), and w ∈ W , where R + is the transitive closure of R.
Proof. Soundness is easily checked. Conversely, assume that ϕ is consistent with S4μ. By Lemma 3.9, ϕ * is consistent with S4μ and hence with Kμ as well. By Fact 3.2, there is a finite Kripke model M = (W, R, h) and a world w ∈ W , with M, w |= ϕ * . We do not know that (W, R) is reflexive or transitive. However, by Lemma 3.5 we have (W, R * , h), w |= ϕ as well, and R * is reflexive and transitive. 2 Remark 3. 11. Continuing Remark 3.6, it can be shown in a similar way to Lemma 3.9 that K4μ ϕ ↔ ϕ + for each formula ϕ. We leave this as an exercise, since we will use it only in Remark 4. 8. It then follows as in Theorem 3.10 that K4μ is sound and complete over the class of finite transitive frames.

Translations
The language L μ t dt 2[d]∀ has some redundancy. We can express 2 with [d], and t with dt (but not vice versa). We can also express t , dt with μ -and often vice versa, using results of Dawar and Otto [7].
Later, we will need translations that work in both topological spaces and (possibly restricted) Kripke models. In this section, we will explore translations -but only to the extent needed later.

Translating d and dt to μ
This is the simplest case. We have already seen the idea, in the equivalence of t -and dt -formulas to ν-formulas given in (2.1).
∀ -formula ϕ μ as follows: These formulas can be checked to be well formed. The translation simply replaces t by an expression using μ and 2, and similarly for dt Proof. An easy induction on ϕ. We consider only the case t Δ (for finite Δ = ∅), in Kripke semantics (the case dt Δ is of course identical). Assume the lemma for each δ ∈ Δ. Take any transitive Kripke By the post-fixed point characterisation of greatest fixed points given in section 2.5, this holds iff ( * ) there is S ⊆ W with w ∈ S and such that for every s ∈ S and δ ∈ Δ, there is t ∈ S with sRt and M, t |= δ.
Then w ∈ S, and for each w n ∈ S and δ ∈ Δ, there is m > n with M, w m |= δ. Then w m ∈ S, and by transitivity of R we have w n Rw m . So ( * ) holds. [d]∀ -formula equivalent to ϕ in all Kripke frames. But the two are not equivalent in topological spaces, so we seek a better translation that works in both semantics.
[d]∀ -formula ϕ d as follows: Again, ϕ d is always well formed. It turns out that the translation − d is faithful in reflexive frames and T D spaces (recall from section 2.3 that a space is T D if the derivative of every singleton is closed).
Proof. An easy induction on ϕ. To show, e.g., that 2ϕ implies (2ϕ) d , we need reflexivity. We also note that Δ and d Δ both imply t Δ in reflexive Kripke models. 2 Proof. Let X be a T D topological space. We prove by induction on ϕ that each L μ t dt We consider only two cases: 2ϕ and t Δ. Inductively assume the result for ϕ and each formula in the finite set Δ of formulas, let h be an assignment into X, and let x ∈ X. In the proof, we write 'x |= ' as short for '(X, h), x |=', and for a formula ϕ, we write JϕK = {y ∈ X : y |= ϕ}.
We prove that It remains to prove that Proof of claim. Plainly, x ∈ S . For the other half, let y ∈ S and δ ∈ Δ be arbitrary; we show that Since O was arbitrary, this shows that y ∈ d (JδK ∩ S ). Since inductively, JδK = Jδ d K, this proves the claim. 2 By definition of the semantics, Claim 4.5.1 immediately yields x |= dt Δ d as required. This completes the induction and the proof that each ϕ is equivalent to ϕ d . (The reader may like to construct an alternative proof using the games described in Remark 2. 2.) Conversely, to show that the T D hypothesis is necessary, we first prove Proof of claim. For the first part, since d {x} ⊆ cl{x} and the latter is closed, This proves the claim. 2 Now suppose the space X is not T D . Then there is some point x of X with d {x} not closed. By Claim 4. 5

Translating μ to t
We use this translation only to prove strong completeness for L μ 2 in Theorem 9.3(2). Fortunately, most of the hard work involved has already been done by others. We will need only the fact below, but its proof was a major enterprise. To lift this to topological spaces, we will use the proof theory from section 3.
Now it is easy to check that S4μ is sound over every topological space. (The S4 axioms are sound by definition of the topological semantics of 2, and the fixed point axiom and rule are sound by the semantics of μ.) Hence, ϕ ↔ (ϕ t ) μ is valid in every topological space. But by Lemma 4.2, (ϕ t ) μ is equivalent to ϕ t in every topological space. We conclude that ϕ is equivalent to ϕ t in every topological space, as required.

More topology
Not surprisingly, for our completeness theorems we will need some simple and standard topological definitions and results. They are collected here. We will see some more substantial ones in the next section. We begin with the following very simple fact.
Lemma 5. 1. Let X be a topological space, and suppose that N ⊆ X has empty interior.
The second part follows from the first, since X \cl

The d operator on sets
for every open neighbourhood O of x}, the set of strict limit points of S. The d operator has the following basic properties.
Proof. We prove only part 4, leaving the other parts to the reader. Recall from section 2.3 that X is T D if d {x} is closed for every x ∈ X. Aull and Thron showed in [2, theorem 5.1] that in fact, X is T D iff d S is closed for every S ⊆ X. (They say that this theorem is due to C. T This leads to the following.
. This contradicts the minimality of O. 2

Regular open sets
Let X be a topological space. A regular open subset of X is one equal to the interior of its closure. We will mainly be interested in regular open subsets of open subspaces of X, so we give definitions directly for such situations.
As 'int' is multiplicative and U is open, it is equivalent to say that S = U ∩ int cl S, and we sometimes prefer this formulation. In such a case, It is known (see, e.g., [13, chapter 10]) that for every open subset U of X, the set RO(U ) of regular open subsets of U is closed under the operations +, ·, −, 0, 1 defined by and (RO(U ), +, ·, −, 0, 1) is a (complete) boolean algebra. We will also use the notation RO(U ) to denote this boolean algebra. The standard boolean ordering ≤ on RO(U ) coincides with set inclusion, because for We will need the following general lemma.

Every regular open subset of S is a regular open subset of U .
Proof.
1. The first two points follow from boolean algebra considerations, and can easily be shown directly; of course, the first point is equivalent to

Normal spaces
Definition 5. 6. A topological space X is said to be Hausdorff (or T2) if for every two distinct points Hausdorff and for every two disjoint closed subsets It is standard that every T2 (and hence every normal) space is T1, and hence also T D .
The following is well known (see, e.g., [32, III, 6.1]), but is so important for us that we include a quick proof.
Lemma 5. 8. Every metric space is normal. Proof. Let X be a metric space. It is easy to check that X is Hausdorff, and we leave this to the reader. Let We proceed to prove ( * ).

Tarski's 'dissection theorem' and relatives
A 'dissection' of a space (or a non-empty open subset of it) is a partition of it into subsets that have topological relationships allowing them to represent the structure of certain Kripke frames. The original dissection results of Tarski, developed further by others, involved finitely many partition sets. Here we strengthen the analysis by allowing (countably) infinitely many partition sets; permitting each partition set to contain any given starting set, so long as any union of the starting sets is closed and nowhere dense; and making each partition set be within some prescribed distance of any point. We also develop a closely related result in which the subsets need not partition the space but each of them has the same predetermined set of limit points.
We will use these results in Proposition 7.10, to represent finite Kripke frames. We will state them and discuss them in section 6.1, and prove them in section 6.2. Section 6.3 contains a corollary also needed in Proposition 7. 10.

The dissection theorems
The first 'dissection theorem' is as follows. We will use it in Corollary 6.5, and in our main Proposition 7.10 to handle frames with irreflexive roots. It is also used in [16].
and a real number ε > 0. Then there are pairwise disjoint non-empty sets Example 6.2. We give an instance of Theorem 6.1 for X = R, ignoring the E i and ε. Suppose that T = (0, 1) and U = (0, 2). Choose pairwise disjoint infinite sets K i (i ∈ E) of positive integers, and let I i = {1/n : n ∈ K i } for each i ∈ E. Then the I i are non-empty and pairwise disjoint subsets of T with d Next is our second dissection result, which will be used in Proposition 7.10 to handle frames with reflexive roots. Recall that a subset N ⊆ X is nowhere dense if int cl N = ∅. Any such set plainly has empty interior.
This theorem is largely known, and has a long history. Paraphrasing slightly, Tarski [41, satz 3.10] proved the following, which was perhaps the original 'dissection theorem'. (He credited the proof to Samuel Eilenberg, noting that he had originally proven the result himself for R and its dense-in-themselves subspaces.) Let X be a dense-in-itself normal topological space with a countable basis of open sets (see below). Then for every r < ω, every non-empty open subset G of X can be partitioned into non-empty open sets G 1 , . . . , G r and a non- Here and below, the empty intersection (when r = 0) is taken to be X. This statement is equivalent to the statement of Theorem 6.3 (parts 2-4) above, when r = |G| < ω, |B| = 1, and with d B j replaced by cl B j . We observe that if r ≥ 2 then actually cl A topological space (X, τ ) has a countable basis of open sets iff there is countable τ 0 ⊆ τ such that τ is the smallest topology on X containing τ 0 . Given this and normality, Urysohn's theorem [43] yields that τ = τ d for some metric d on X. Any metric space is normal, and has a countable basis of open sets iff it is separable (see section 2.3). So Tarski's stipulation on X boils down to stipulating that X is a separable dense-in-itself metric space. Example 6. 4. We give an instance of Tarski's result for X = R and r = 1. Take a copy of R. Replace each rational in it by a copy of the real interval [0, 1]. Let G 1 be the union of the interiors of these intervals, and B 0 be the set of all other points. (Formally, let I x = [0, 1] for x ∈ Q, I x = {0} for x ∈ R \ Q, and consider I = x∈R ({x} × I x ), ordered lexicographically by x, p < y, q iff x < y or (x = y and p < q). Let G 1 = Q × (0, 1) and B 0 = I \ G 1 .) The resulting linear order (formally, (I, <)) is Dedekind complete, separable, and without endpoints, and hence order-isomorphic to the open interval G = (0, 1) of R (see, e.g., [33, theorem 2.30]). We identify it with this interval. It can be checked that cl Removing the restriction to |B| = 1 but with the same hypotheses on X, McKinsey and Tarski [27, theorem 3.5] proved that for every r, s < ω, every non-empty open set G can be partitioned into non-empty open sets G 1 , . . . , G r and non-empty sets B 0 , . .
This statement is equivalent to parts 2-4 of the statement of Theorem 6.3 above, with r = |G| < ω, s = |B| < ω, and with d B j replaced by cl B j . It was used in [27] to prove (in modal terminology) that the L 2 -logic of X is S4; a form of the result 'readily available to those whose main interest lies in sentential calculus rather than in topology or algebra' was given in [29, theorem 1.3].
Removing the assumption of separability, Rasiowa and Sikorski [32, III, 7.1] proved parts 2-4 of Theorem 6.3 essentially as formulated above, but for finite G, B and with d B j replaced by cl B j . Our use of d B j strengthens this, since part 4 implies that But it is only a formal strengthening, since the same effect can be achieved by first obtaining disjoint sets B k j with cl B k j = D for j ∈ B and k = 0, 1, and then defining So d B j = D as required.
In this paper, we will not need the ε-conditions (Theorem 6.3(5)), nor infinite index sets. However, they may be useful in future work. Indeed, analogous ε-conditions (for finite G, B) have already been proved by Kremer [20, lemma 6.1], and infinite index sets can also be helpful in arguments like Kremer's.
Other recent related results include [23, proposition 6.7], which (roughly speaking and among other things) replaces part 5 in Theorem 6.3 by the statement that each G i is the union of pairwise disjoint open balls O ik (k ∈ K i ) such that every open neighbourhood of every point in j∈B B j contains some O ik .

Proof of Theorems 6.1 and 6.3
We will sometimes use without mention Lemma 5.2(1-2) and the consequent additivity and monotonicity of d and cl. We will get to the theorems shortly, but first, fix a non-empty countable index set E and pairwise disjoint subsets E i ⊆ X (i ∈ E) such that i∈S E i is closed for every S ⊆ E. (Hence each E i is closed.) Two players, ∀ (male) and ∃ (female), play a game, G(E i : i < ω), to build pairwise disjoint subsets The game is co-operative, with no winners or losers. It has ω rounds, numbered 0, 1, 2, . . . . At the start of round n (for each n < ω), subsets I n i ⊆ X (i ∈ E), whose closures are pairwise disjoint, are in play. Initially -at the start of round 0 -we put are pairwise disjoint by assumption on the E i . Round n is played as follows. 1 So ∃ must include cl I n i n in I n+1 i n , but all other points she includes must lie in O n . These inclusions are needed in Claim 6.2. 3. As the game requires that cl I n+1 j (j ∈ E) are pairwise disjoint, she must also ensure that cl I n+1 i n is disjoint from cl I n j for each j ∈ E \ {i n }. Since she can satisfy these requirements by simply playing I n+1 i n = cl I n i n , she never gets stuck.
That completes the round, and the sets I n+1 i (i ∈ E) are passed to the start of round n + 1.
The following claim will be useful later.
But i∈S\N E i is closed by assumption on the E i , so i∈S cl I n i itself is a finite union of closed sets, and so closed. The equivalence to cl( i∈S I n i ) = i∈S cl I n i is easy. This proves the claim. 2 After ω rounds, the game ends. Its outcome is the sequence (I i : i ∈ E) of subsets of X, where I i = n<ω I n i for each i ∈ E. Since the I n i (i ∈ E) are pairwise disjoint for each n, the I i (i ∈ E) are also pairwise disjoint, and E i = I 0 i ⊆ I i for each i. We say that ∀ plays well if: A2. O n = ∅ whenever n < ω and i n = i m for every m < n, In short, ∀ chooses each index in E infinitely often, and his choices of O n whenever he picks a particular index form a decreasing chain whose largest member is non-empty and whose intersection is independent of the index. Now fix ε > 0. Let ε n = ε/(n + 2) for each n < ω. Then ε > ε 0 > ε 1 > · · · > 0 and lim n→∞ ε n = 0. We say that ∃ plays well (this notion is dependent on ε but we do not make ε explicit in the notation) if in each round n, she defines chooses (using Zorn's lemma) a maximal subset Z n ⊆ P n such that d(x, y) ≥ ε n for each distinct x, y ∈ Z n , and ensures that Z n ⊆ I n+1 i n . Let us make some observations about P n and Z n . Z1. d Z n = ∅ (because for each x ∈ X, the set N ε n /2 (x) ∩ Z n has at most one element). So by Lemma 5.2(3,5), Z n is closed and int Z n = ∅. If O n = ∅ then plainly P n = Z n = ∅. Suppose then that O n = ∅. By the claim, P n = ∅, and: Z2. Z n is non-empty (because P n is non-empty and any singleton subset of P n satisfies the ε n condition), Z3. d(x, Z n ) < ε n for every x ∈ P n (else x can be added to Z n , contradicting its maximality). Recall that Clearly, if ∃ plays well, the Z n (n < ω) are pairwise disjoint and Z m ⊆ I n i m for all m < n < ω.
Claim 6.2. 3. Suppose that both players play well. Then for each i ∈ E: Proof of claim. Fix i ∈ E. For part 1, using A1, let n < ω be least such that i n = i. Let δ > 0. By A1 and since the Z n are pairwise disjoint, we may choose n < ω such that ε n < δ, i n = i, and x / ∈ Z n . By assumption, This holds for all δ > 0, and it follows that x ∈ d I i . So n<ω cl O n ⊆ d I i as required.
To prove the converse inclusion d  This holds for all n with i n = i, so the last step using A4. This proves the claim. 2 With these results in hand, we can prove our two theorems. First, Theorem 6.1.
Proof of Theorem 6. 1. As in the theorem's statement, let there be given open sets and a real number ε > 0. By Lemma 5.2(3), every subset of i∈E E i is closed, so certainly i∈S E i is closed for every S ⊆ E. We are also given that E is non-empty and countable. So ∀ and ∃ can play G(E i : i ∈ E). ∃ will play so that d i∈E I n i = ∅ for each n < ω. By ( 6.4), this trivially satisfies (6.1). He also arranges to let i n = i for infinitely many n, for each i ∈ E, so condition A1 will hold. A2 holds because T = ∅. Given A1, it follows from (6.5) that A3 and A4 also hold, and thus, he plays well. ∃ responds to ∀'s move in round n by choosing Z n as above and letting I n+1 i n = I n i n ∪ Z n (so she plays well), and I n+1 i = I n i for every i ∈ E \ {i n }. We check (6.2). By (6.4) for n and Lemma 5.2(3), I n i n is closed; by Z1, Z n is closed too; and by (6.3), Z n ⊆ P n ⊆ O n . Hence, cl I n i n = I n i n ⊆ I n i n ∪ Z n = I n+1 i n ⊆ cl I n i n ∪ O n , as required for (6.2). Also, Z n ⊆ P n = O n \ i∈E cl I n i , and it follows that cl I n+1 i (i ∈ E) are pairwise disjoint. Moreover, since d Z n = ∅ by Z1, her move preserves (6.4).
Let the outcome of play be the sets I i (i ∈ E). We check that these sets meet the conditions of the theorem. Being the game's outcome, they are pairwise disjoint, and E i ⊆ I i for each i. Let i ∈ E. First, we check that ∅ = I i ⊆ T and d(x, I i ) < ε for each x ∈ cl T. By Claim 6.2.3, I i = ∅. Let n < ω be least such that i n = i. By the first clause of (6.5), O n = T, and Claim 6.2.3 yields d(x, I i ) < ε for every x ∈ cl T. Also, Second, we check that d I i = cl(T) \ U. By (6.4), d I n i = ∅ for each n. So Claim 6.2.3 yields d I i = n<ω cl O n . It is therefore sufficient to prove the next claim. Proof of claim. Certainly, each x ∈ cl(T) \ U lies in cl O n for each n, because for every δ > 0, Now we prove the converse, n<ω cl O n ⊆ cl(T) \ U. First note that if cl(T) \ U = ∅ then by A1 and (6.5), infinitely many O n are empty as well, so n<ω cl O n = ∅ and we are done.
Suppose then that cl( As U is open, we can choose δ > 0 with N δ (x) ⊆ U. As ∀ played well, we can pick m < n < ω such that i m = i n and ε n < δ.
this is a contradiction, and proves the claim, and the theorem. 2 We can also prove Theorem 6. 3.
Proof of Theorem 6. 3 Condition (6.1) is trivially met. He chooses i 0 = b, and also arranges to choose each index in E infinitely often. As we will see below, he plays well. ∃ will play so that the following properties hold for each n < ω: is open for each i ∈ G ∩ {i m : m < n} (that is, for each i ∈ G that ∀ already picked in some round earlier than n).
When n = 0, we have I 0 i = E i for each i ∈ E. P1 and P2 are then given, and P3 holds vacuously. Assume that P1-P3 hold for n. ∃ responds to ∀'s move (O n , i n ) in round n as follows. Of course she sets I n+1 i = I n i for i ∈ E \ {i n }, and defines Z n as described above.
This clearly satisfies (6.2). The set Z n is a closed (by Z1) subset of G (since Z n ⊆ O n ⊆ G by (6.3) and (6.6)), and it is disjoint from cl i∈E I n i (by (6.3) and Z n ⊆ P n ), so P1 for n + 1 follows. P2 is kept, since int cl i∈B I n+1 i = int cl i∈B\{i n } I n i ∪ cl I n i n ∪ Z n by ∃'s move = int cl( i∈B I n i ) ∪ Z n ) by additivity of cl, and Z1 = int cl i∈B I n i by Z1 and Lemma 5.1(1) = ∅ by P2 for n.
P3 is unchanged because i n / ∈ G. Case 2: i n ∈ G. Then ∃ chooses an open set I n+1 We need to check some things here. First, L is closed, since (by Z1) Z n is closed. Second, R is open, since G is open and (by Claim 6.2.1) j∈E\{i n } cl I n j is closed. Third, L ⊆ R, since cl I n i n ⊆ R by P1, and Z n ⊆ P n = O n \ cl i∈E I n i ⊆ G \ i∈E cl I n i ⊆ R by definition of Z n , (6.3), (6.6), and (6.7). So as X is normal (Lemma 5.8), an open set Q with L ⊆ Q ⊆ cl Q ⊆ R can be found. ∃ lets I n+1 i n be any such Q, so satisfying (6.7).
Next we check that ∃'s move satisfies (6.2). We have cl I n i n ⊆ I n+1 i n by (6.7). Also, This confirms (6.2). Finally we check that P1-P3 still hold. P1 follows from (6.7) and since it holds for n. P2 is unchanged, and P3 holds since ∃ chose I n+1 i n to be open.
Since Z n ⊆ I n+1 i n in both cases, ∃ plays well. We will soon see that ∀ plays well too, but first, a handy claim.
Proof of claim. Let n < ω be given. By (6.6) and Lemma 5. The claim now follows. 2 We can now see that ∀ plays well. He chooses each i ∈ E infinitely often, so A1 holds. A2 holds because every O n is non-empty, as we now show. Write E = i∈E E i . By ( 6.6) is non-empty because it contains I 1 b . Also, by P1 and Claim 6.2.5 we have I n b ⊆ G \ i∈G cl I n i ⊆ cl O n . It follows that cl O n , and hence O n , are non-empty as required. So indeed A2 holds. Since I n i ⊆ I n+1 i for each n, i, it follows from (6.6) that O 0 ⊇ O 1 ⊇ · · · . So, using A1, we see that A3-A4 hold, and therefore ∀ indeed plays well.
At the end of the game we define I i = n<ω I n i for i ∈ E, and let We check that the requirements of the theorem are met. Being the game's outcome, the I i are pairwise disjoint and E i ⊆ I i for each i ∈ E. As both players played well, by Claim 6.2.3 each I i is non-empty. By P1, I i = n<ω I n i ⊆ G for each i. It follows by ( 6.9) that I b ⊆ B b -so I j ⊆ B j for every j ∈ B. This is enough to show that the G i , B j are non-empty and pairwise disjoint subsets of G. So (G i , B j : i ∈ G, j ∈ B) is a partition of G. Also, E i ⊆ I i = G i for each i ∈ G, and E j ⊆ I j ⊆ B j for each j ∈ B. For each i ∈ G, by A1 there is m < ω with i m = i; then G i = {I n i : m < n < ω}; by P3, this is a union of open sets, and so is open.
For the remaining requirements (points 4-5) of the theorem, we need a claim.
Proof of claim. Let n < ω be arbitrary. By Claim 6.2.5, G ⊆ cl(O n ) ∪ i∈G cl I n i . By Claim 6.2.1, i∈G cl I n i is closed, so we actually have cl G ⊆ cl(O n ) ∪ i∈G cl I n i . So by (6.9), Similarly, take i ∈ G. Since the G l (l ∈ G) are pairwise disjoint open subsets of G, we have cl G i ⊆ cl(G) \ l∈G\{i} G l and hence cl(G i ) \G i ⊆ cl(G) \ l∈G G l = D. Conversely, by Claim 6.2.6 and Lemma 5.2 Finally, for point 5, let x ∈ cl G. By (6.8), x ∈ cl O 0 . Since i 0 = b, Claim 6.2.3 yields d(x, I b ) < ε. So there is y ∈ I b with d(x, y) < ε. Now let i ∈ G. We showed that I b ⊆ B b ⊆ D ⊆ cl G i . So y ∈ cl G i , and hence we can take z ∈ G i with d(y, z) < ε − d(x, y). Then d(x, G i ) ≤ d(x, z) ≤ d(x, y) + d(y, z) < ε as required. The proof that d(x, B j ) < ε for j ∈ B is similar, using that D ⊆ cl B j . 2

A corollary
We will use the following corollary in Proposition 7.10 to handle non-rooted frames. In the simple case where S 0 = S 1 = ∅ and (so) T = U, it says that any non-empty open set U has regular open subsets U 0 , U 1 whose closures (1) are disjoint within U and (2) contain all 'boundary points' of U (points in cl U \ U). It is proved using Lemma 5.7 (essentially normality) to 'fatten' two sets (obtained from Theorem 6.1) whose derivatives are exactly the set of boundary points.
Proof. Since T is a non-empty open subset of U, we can use Theorem 6.1 to choose disjoint non-empty subsets I 0 , We now work in the subspace U. Recall that cl U denotes the closure operator in the subspace topology on U, so cl U K = U ∩ cl K for subsets K ⊆ U. The sets cl U S 0 , cl U S 1 , I 0 , I 1 are pairwise disjoint (by assumptions) and closed in U. (Each I i is closed in U because by Lemma 5.2(1), Hence, I 0 ∪ cl U S 0 and I 1 ∪ cl U S 1 are disjoint closed subsets of U. The subspace U is a metric space in its own right, and so, by Lemma 5.8, normal. Using Lemma 5.7 in U (taking Q = U), we can find regular open subsets U 0 , U 1 of U with and cl U U 0 ∩ cl U U 1 = ∅. Working back in X again, this says that For each i = 0, 1, write T i = U i \ cl S i . By (6.11), I i ⊆ U i , and by definition of With (6.12), (6.13), and T i = ∅, this proves the corollary. 2

Representations of frames over topological spaces
Our next aim is to use the results of the preceding section to construct a 'representation' from an arbitrary dense-in-itself metric space to any given finite connected locally connected transitive serial Kripke frame. The notion of representation is chosen so as to preserve L μ [d]∀ -formulas, and this will allow us to prove completeness theorems in the next two sections.
Until the end of section 7.5, we fix a topological space X and a finite Kripke frame F = (W, R). We will frequently regard the elements of W as propositional atoms.
One more equivalent formulation, which we will use frequently, is Here, ρ −1 assigns an atom w ∈ W to the possibly empty subset {x ∈ X : ρ(x) = w} of X. The condition says that for every x ∈ X, the set of points of W with preimages under ρ in every open neighbourhood of x but distinct from x itself is precisely R(ρ(x)).
Note that ρ need not be surjective. Indeed, the empty map is vacuously a representation of F over the empty space -and we definitely do allow empty representations.
It can be checked that if ρ : X → W is a representation then rng ρ is the domain of a transitive generated subframe of F. Endow W with the topology generated by {R(w) : w ∈ W } (so the open sets are those A ⊆ W such that a ∈ A implies R(a) ⊆ A). Then every representation of F over X is an interior map from X to W : that is, a map that is both continuous and open. The converse fails in general. For example, let F be the two-world reflexive frame ({0, 1}, ≤), and let X = R with its usual topology. Let ρ : R → {0, 1} be given by ρ(x) = 0 if x ∈ Z, and ρ(x) = 1 otherwise. Then ρ is an interior map, but not a representation, We remark that if F is reflexive, then ρ : X → W is an interior map iff ρ −1 (R −1 (w)) = cl ρ −1 (w) for each w ∈ W . Indeed, interior maps are suitable notions of representation for L 2 , and many topological completeness proofs use them. See [3,25] for more information.
Although Shehtman uses the term 'd-p-morphism' (when ρ is surjective), here we call ρ a 'representation' because it is closely related to the representations of algebras of relations seen in algebraic logic. Indeed, if ρ is a surjective representation of (W, R) over X then ρ −1 induces an embedding from ℘(W ) into ℘(X) that preserves the algebraic structure with which these power sets can be naturally endowed.

Representations over subspaces
Our main interest is in representations over X itself, but representations over subspaces are also useful in proofs. Given a subspace U of X, a map ρ : U → W induces a well defined assignment ρ −1 : W → ℘(X) by ρ −1 (w) = {x ∈ X : x ∈ U and ρ(x) = w}, for w ∈ W . Put simply, preimages under ρ of elements of W are obviously subsets of U , but they are also subsets of X, and so ρ −1 can be regarded equally as an assignment into U or X, as appropriate. The following easy lemma gives some connections between the two views. It is a specialisation of a more general result in which ρ −1 is replaced by any assignment and w by any atom (see also Lemma 2.3).
Lemma 7.2. Let U be a subspace of X and let ρ : U → W be a map. Let x ∈ U and w ∈ W be arbitrary.  R(ρ(x), w) for every x ∈ U and w ∈ W . So we can work in (X, ρ −1 ) instead of (U, ρ −1 ). To avoid too much jumping around between subspaces, we will do this below, often without mention. Part 3 of the next lemma makes it a little more explicit. The lemma gives some general information on the relationships between representations of different generated subframes of F over different subspaces of X. Lemma 7. 3. Let G = (W , R ) be a generated subframe of F. Let T , U , and U i (i ∈ I) be open subspaces of X, with T ⊆ U = i∈I U i . Finally, let ρ : U → W be a map. Then: Proof. Simple. 2

Representations preserve formulas
Here, we will show that surjective representations preserve all formulas of L μ [d]∀ . Since representations are like p-morphisms, albeit between different kinds of structure, this is entirely expected and the proof is essentially quite standard -see [34, lemma 20] and [3, corollary 2.9], for example. We do need, however, that F is finite. We will be able to handle larger sublanguages of L μ t dt 2[d]∀ by using the translations of section 4. Let us explain the setting. Suppose we are given a representation ρ : X → W of F over X. Recall that Var is our fixed base set of propositional variables, or atoms. For each assignment h : Var → ℘(W ) of atoms in Var into W , the map ρ −1 • h : Var → ℘(X) is an assignment of atoms into X, given of course by So ρ, or rather ρ −1 , gives us a way to transform an assignment into F to one into X, and then to evaluate a formula in the resulting model on X. Clearly, we would like to get the same result as in the original model on F, and this leads to the following definition.
We are now ready for our main preservation result.

Basic representations
Certain very primitive representations called basic representations will play an important role later, because they can easily be extended to more interesting representations. Definition 7. 6. Let S, U be open subspaces of X, with S ⊆ U , and let σ : Note that we use 3 and not d here.
Remark 7. 7. In the setting of this definition: 1. Vacuously, if σ is empty then it is U -basic. 2. More generally, but equally trivially, if rng σ is contained in a nondegenerate cluster C in F, then σ is U -basic. For, (X, σ −1 ), x |= 3w ∧ 3v implies that w, v ∈ rng σ ⊆ C, and so Rwv as C is a nondegenerate cluster.
We remark (but will not formally use) that σ is U -basic iff rng σ is a (possibly empty) union of R-maximal clusters in F whose preimages under σ have pairwise disjoint closures within U . Moreover, each such preimage is a clopen subset of S.

Full representations and full representability
In induction proofs, we often need a stronger inductive hypothesis than formally required for the final result. This will be the case in Proposition 7.10 below, where we build a representation by combining several 'smaller' representations obtained inductively. For this to work, we will need these smaller representations to be well behaved on the boundaries of their domains. The following definition will help to do this.
Every surjective representation is vacuously ∅-full. Definition 7. 9. We say that F is fully representable (over X) if whenever then σ extends to a T -full representation ρ : U → W of F over U . Moreover, if F is rooted then for any given root w 0 of F and x 0 ∈ T , we can choose ρ so that ρ(x 0 ) = w 0 .
Notice that in the boolean algebra RO(U ) of regular open subsets of U , we have T = −S, so {S, T } is a partition of 1. That is, S, T ∈ RO(U ), S · T = 0, and S + T = 1.
In Proposition 7.10 below, we will fulfil our main aim, to prove (surjective) representability of every finite connected locally connected serial transitive frame. We are going to do it by induction on the size of the frame. We appear to need a stronger inductive hypothesis, namely full representability, than is needed for the conclusion. T -fullness and extending σ are mainly to do with this, but the extending of σ is also helpful in the proof of strong completeness in Theorem 9.1 later. Note that if F is fully representable over X, and X = ∅, then by taking U = X and S = σ = ∅, we see that there exists a surjective representation of F over X. So we do obtain our desired conclusion from the stronger hypothesis of full representability.
Proposition 7. 10. Suppose that X is a dense-in-itself metric space. Then every finite connected locally connected serial transitive frame F = (W, R) is fully representable over X.
Proof. The proof is by induction on the number of worlds in F. Let F = (W, R) be a finite connected locally connected serial transitive frame, and assume the result inductively for all smaller frames. Recall from sections 2.1-2.2 that we write and, for w ∈ W , Let U ⊆ X be open, let S be a regular open subset of U , and let σ : S → W be a U -basic representation of F over S. Write T = U \ cl S, and suppose that T = ∅. We need to extend σ to a T -full representation ρ : U → W of F over U . Further, if F is rooted and we are given a root w 0 of F and x 0 ∈ T , we wish to choose ρ so that ρ(x 0 ) = w 0 . There are three cases.
Choose any such w 0 (it may not be unique). Then w 0 is a root of F, and since w 0 is reflexive, R(w 0 ) = W and w 0 ∈ R • (w 0 ). So R • (w 0 ) = ∅. Since T is clearly a non-empty open set, we can use Theorem 6.3

to partition T into non-empty open sets
is connected (as it is rooted) and locally connected, serial, and transitive (as it is a generated subframe of F). Since w 0 is a world of F but not of F * (v • ), the frame F * (v • ) is smaller than F. By the inductive hypothesis, F * (v • ) is fully representable over X. So, taking the regular open subset 'S' of G v • to be ∅ and 'T ' to be for each x ∈ U . The map ρ is well defined because the G v • , the B v • , and S are pairwise disjoint, and plainly it is total, extends σ, and satisfies ρ(x 0 ) = w 0 . We aim to show that ρ is a T -full representation of F over U . The following claim will help.
Proof of claim. Let x ∈ D and w ∈ W be given. There are two cases. The first is when w ∈ R • (w 0 ). Now (7.5) gives x ∈ cl G w \ G w . As ρ w is a G w -full representation of F * (w), a frame of which w is a world, Since ρ B w has constant value w, we obtain again that (X, ρ −1 ), x |= d w. This proves the claim. 2 We now check that ρ is a representation of F over U . Let x ∈ U and w ∈ W . We require (X, ρ −1 ), x |= d w iff R(ρ(x), w). There are four cases.
We aim to use the inductive hypothesis on these sets and σ : S → F(w 0 ), so we check the necessary conditions.  Proof of claim. First we show that σ : S → R(w 0 ). We know that σ : S → W = {w 0 } ∪ R(w 0 ). Assume for contradiction that there is some x ∈ S with σ(x) = w 0 . Then plainly, x ∈ U and (X, σ −1 ), x |= 3w 0 . As σ is a U -basic representation of F over S, we obtain Rw 0 w 0 , contradicting the choice of w 0 as irreflexive. So indeed, rng σ ⊆ W \ {w 0 } = R(w 0 ). Since σ is a representation of F over S, by Lemma 7.3 it is also a representation (over S) of the generated subframe F(w 0 ) of F. It is trivially U -basic, since if x ∈ U , w, v ∈ R(w 0 ), and (X, σ −1 ), x |= 3w ∧ 3v, then x ∈ U and w, v ∈ W as well, so Rwv since σ is U -basic. This proves the claim . 2 In summary, U is open, S is a regular open subset of U , σ is a U -basic representation of F(w 0 ) over S, and T = U \ cl S = ∅. Now F(w 0 ) is smaller than F (since w 0 / ∈ R(w 0 )), connected (since F is locally connected), and locally connected, serial, and transitive (since it is a generated subframe of F). By the inductive hypothesis, F(w 0 ) is fully representable over X. So σ extends to a T -full representation ρ : U → R(w 0 ) of F(w 0 ) over U .
By T -fullness, We extend ρ to a map ρ : U → W by defining for x ∈ U . This is plainly well defined and total, with ρ(x 0 ) = w 0 . Since ρ extends ρ , it also extends σ. We will show that ρ is a T -full representation of F over U . To do it, we need another claim.
Proof of claim. By (7.6) and Lemma 5.2(1), we have cl T \ U = d I ⊆ cl I.
Using openness of T = T ∪ I, the assumption that X is dense in itself, and Lemma 5.2(5, 2), we have Hence, cl I ⊆ cl T . Since I ∩ U = ∅ and U is open (Claim 7. 10.2), we have cl I ∩ U = ∅. So cl I ⊆ cl T \ U , proving the claim. 2 Claim 7. 10. 5. ρ is a representation of F over U .
We also require that (X, The second case is when x / ∈ I. In this case, x ∈ U , an open set, and ρ U = ρ , a representation over U of the generated subframe F(w 0 ) of F. By Lemma 7.3, (X, ρ −1 ), x |= d w iff R(ρ(x), w) for every w ∈ W , as required. The claim is proved. 2 Claim 7. 10. 6. ρ is T -full.
We must also show that ρ(U ) = W . Well, I = ∅. Take x ∈ I. Then ρ(x) = w 0 , and by the proof of Claim 7. 10.5, (X, ρ −1 ), x |= d w for every w ∈ R(w 0 ). This can only be if ρ is surjective.
This proves the claim and completes case 2 of Proposition 7. 10. Only case 3 remains, but this is the hardest case. 2

Case 3: otherwise -that is, F is not rooted.
By the case assumption, F has proper connected generated subframes -for example, F * (w) for any w ∈ W . So let F 0 = (W 0 , R W 0 ) be a maximal proper connected generated subframe of F. Then Claim 7. 10.7. F 0 and F 1 are proper connected generated subframes of F. Also, Proof of claim. By definition, F 0 is a proper connected generated subframe of F. Since F 1 is rooted, it is connected, and a proper subframe of F (which by the case assumption is not rooted). It is a generated subframe of F by definition of F * (a). Clearly b ∈ W 0 ∩ W 1 , so W 0 ∩ W 1 = ∅. The subframe (W 0 ∪ W 1 , R W 0 ∪ W 1 ) of F is plainly a generated subframe of F. It is connected, since F 0 , F 1 are connected and W 0 ∩ W 1 = ∅. It properly extends F 0 since a ∈ W 1 \ W 0 . By maximality of F 0 , we have W 0 ∪ W 1 = W . This proves the claim. 2 Being generated subframes, F 0 and F 1 are locally connected serial transitive frames. Since they are proper connected subframes of F, by the inductive hypothesis they are fully representable over X. Our plan is to combine suitable representations of them to give a representation of F over U .
Recall that S is a regular open subset of U and σ : S → W is a U -basic representation of F. We use W 0 , W 1 to split S (and, later, σ) in two. Let Proof of claim. We prove the last point first. Suppose for contradiction that there is some x ∈ U ∩ cl(S 0 ) ∩ cl(S 1 ). As x ∈ cl S 0 , we have (X, σ −1 ), x |= 3 w∈W 0 w. As 3 is additive and W 0 finite, it follows that there is some w 0 ∈ W 0 such that (X, σ −1 ), x |= 3w 0 . Similarly, as x ∈ cl S 1 and σ(S 1 ) ⊆ W \ W 0 , there is some As σ is a U -basic representation, we obtain Rw 0 w 1 . Since F 0 is a generated subframe of F, this implies that w 1 ∈ W 0 , a contradiction. So U ∩ cl(S 0 ) ∩ cl(S 1 ) = ∅ as required.  The claim and the assumption at the outset that T = ∅ are more than enough to apply Corollary 6.5, to obtain open subsets U i , T i of U , for i = 0, 1, satisfying the following conditions: We now work in the boolean algebra RO(U ) of regular open subsets of U . By C5, we have U 0 , U 1 ∈ RO(U ). We define further elements of RO(U ): The main property of these sets is as follows.
Proof of claim. Let i < 2. By Claim 7. 10.8 and condition C5 above, S i , U i ∈ RO(U ). By this and condition C3, We aim to apply the inductive hypothesis to V i , M + S i , T i , F i , for each i = 0, 1. We will construct a V i -basic representation of F i over M + S i , and extend it inductively to a representation over V i . We will arrange that these two representations over V 0 and V 1 agree on M , so their union will be our desired representation over U .
Our first step, then, is to find a V i -basic representation of F i over M + S i , and the next claim helps us get one.
By the claim, in order to find a V i -basic representation of F i over M + S i , all we need is to find suitable representations over M and S i and take their union. By Claim 7. 10.7, is a proper subframe of F. It is obviously connected (since rooted), and a generated subframe of F, so a locally connected serial transitive frame. By the inductive hypothesis, it is fully representable over X. So we can find an (M -full) representation of F * (b 0 ) over M , by using the definition of 'fully representable' if M is non-empty, and trivially by taking β = ∅ if M is empty. Also, for each i < 2 let To prove that it is V i -basic, let x ∈ V i and v, w ∈ W i be given, and suppose that (X, (β ∪ σ i ) −1 ), x |= 3w ∧ 3v. We require Rwv.
Plainly, x ∈ cl(M ∪ S i ) = cl M ∪ cl S i , and x ∈ V i ⊆ U . But U ∩ cl M ∩ cl S i = ∅ by Claim 7. 10. 10. So there are two possibilities.
The first one is that x / ∈ cl M . In this case, we must have (X, The other possibility is that , so Rwb 0 , and since Rb 0 v, we deduce Rwv by transitivity. (Essentially we are using that F * (b 0 ) is a nondegenerate cluster.) This proves the claim . 2 In summary, for each i < 2: • V i is open (by Claim 7. 10.9). 10. 9. In a boolean algebra, 10.11).
Proof of claim. Let i < 2. Then ρ V i = ρ i , a representation of F i over V i . By Lemma 7.3(1), this is also a representation of F over V i , which is an open set by Claim 7. 10. 9. By (7.12), U = V 0 ∪ V 1 , so by Lemma 7.3(2), ρ is a representation of F over U , proving the claim. 2 Claim 7. 10. 13. ρ is T -full.
Proof of claim. Let x ∈ cl T \ U . We require (X, ρ −1 ), x |= d w for every w ∈ W . For each i < 2, as cl T \ U ⊆ cl T i by condition C4 above, and x / ∈ U ⊇ V i , we have x ∈ cl T i \ V i . Since ρ i ⊆ ρ, it follows from (7.10) that (X, ρ −1 ), x |= d w for every w ∈ W i . This holds for each i = 0, 1. Since Finally, we show that ρ(U ) = W . Since each ρ i is a T i -full representation of F i over V i , it is surjective, and by (7.12) we obtain ρ( This proves the claim, and with it, Proposition 7.10 (as F is not rooted, the value of ρ(x 0 ) is immaterial in this case). 2 Remark 7. 11. We end with some technical remarks on the definition of 'fully representable' (Definition 7.9) and its relation to the proof just completed. They are not needed later, and the reader can of course skip them if desired.
It is very helpful throughout the proof that U is open -see, e.g., Lemma 7. 3. However, we cannot assume in Definition 7.9 that U is regular open in X. For if we did, then in case 2 of the proof, we have cl I ⊆ cl T ⊆ cl U by Claim 7. 10.4 and T ⊆ U , so U = U = int cl U = int(cl U ∪ cl I) = int cl U . Therefore, U is not regular open in X, and we can not apply the inductive hypothesis to it. We use that X is dense in itself to show that I ⊆ cl T , as well as to use the results of section 6. At least according to the construction we gave, S should be open. In case 1, if S is not open then there is x ∈ S \ int S ⊆ cl(U \ S), and a little thought shows that (X, ρ −1 ), x |= d w 0 for any such x. For ρ to be a representation, we would need R(ρ(x), w 0 ). Since ρ ⊇ σ and x ∈ S, this says that R(σ(x), w 0 ), which we have no reason to suppose is true.
The problem if S is not regular open in U is that, again in case 1, we used that U \ S = cl T . If this were to fail, there may be points x ∈ U \ (S ∪ cl T ) (so x ∈ U ∩ int cl S). We have to define ρ on these x, and defining ρ(x) = w 0 as in the proof may not give a representation. However, as σ is U -basic, it is possible to define ρ(x) using σ instead. This effectively extends σ to U ∩ int cl S. So we can assume without loss of generality that S is regular open in U . It is therefore easier to do so and avoid the problem completely.
We could just suppose in Definition 7.9 that S is regular open in X, but we cannot suppose this of U , and we have to work in RO(U ), so there is little gain in doing so.
We need that σ is U -basic in order that in case 3, the subsets S 0 , S 1 have disjoint closures in U (Claim 7. 10.8). This in turn is needed to apply normality in the proof of Corollary 6. 5.
We cannot assume instead in Definition 7.9 that σ is X-basic, because in case 3, we cannot guarantee that β ∪ σ i is X-basic. This is because we do not know that M ∩ cl S i = ∅, but only that U ∩ M ∩ cl S i = ∅. We could solve this problem by assuming further that cl S ⊆ U (which implies that S is regular open in X), but this weakens the proposition sufficiently to cause trouble in Theorem 9.1 later, where we would need to ensure that cl S n ∪ cl S n+1 ⊆ U n for each n.
We require that T = ∅ in Definition 7.9 because Proposition 7.10 trivially fails without this condition, unless σ is already surjective. We include surjectivity in the definition of 'full representation' (Definition 7.8) because surjective representations preserve ∀ (see Proposition 7.5). We might try to drop surjectivity from Definition 7.8 and simply prove it from the second part of the definition, as in cases 1 and 2 of the proof, but it is not clear how to do this in case 3. Finally, we mention that actually ρ(T ) = W -not only ρ but also ρ T is surjective.

Weak completeness
We are now ready to prove our first tranche of main results, showing that Hilbert systems for various sublanguages of L are sometimes sound and always complete over any non-empty dense-in-itself metric space. Several of the proofs use the translations − d and − μ of section 4. We establish only weak completeness here. We will discuss strong completeness later, in section 9.4.

The Hilbert systems
We will use the Hilbert systems for the mu-calculus in Definition 3.1, and also the following ones. The two basic systems are
As usual, we denote particular Hilbert systems extending K or S4 by sequences of letters and numbers indicating the axioms mustered. For example, S4.UC denotes the extension of S4 by the axioms generated by the two schemes given in U and C below. The relevant schemes are as follows. Recall that 2 * ϕ abbreviates ϕ ∧ 2ϕ.

4:
all instances of the '4' scheme 2ϕ → 22ϕ D: 3 t: all instances of the following schemes, sometimes referred to as the tangle axioms.
We have formulated these Hilbert systems using 2 and t , and we will refer to them below as the 2-form of the systems.

Soundness
Soundness of (the 2-form of) S4μ over the class of all topological spaces was already observed in Corollary 4.7. We proceed to examine soundness of the Hilbert systems just introduced. First, we consider validity of the axioms. (We postpone discussion of G 1 to Remark 8. 6.) For short, we say that a scheme is valid in a topological space X (or class K of spaces) if all instances of the scheme are valid in X (resp., K). Proof. It is easy to check that in both 2and [d]-forms, the axioms of K.U are valid in every topological space. Every instance of 2ϕ → ϕ is trivially valid in every topological space. (This is not true for [d]ϕ → ϕ, of course.) The 4-scheme in its 2-form is easily seen to be valid in every topological space, while its [d]-form was shown to be valid in precisely the T D spaces by Esakia (see, e.g., [8, proposition 2]). Turning to the D axiom, plainly 3 is valid in every space, and d is valid in precisely the dense-in-themselves spaces.
Next, working in any model (X, h) on an arbitrary topological space X, we show that the tangle axioms in both 2and [d]-forms are true at all points. We write JϕK for the set {x ∈ X : (X, h), x |= ϕ}, as usual. The result for Fix is immediate from the fixed-point semantics, which tells us that  This lemma is sufficient to prove soundness theorems, for the following reason. Let K be a class of topological spaces, and H a Hilbert system whose rules are at most those listed above (modus ponens and the two generalisation rules). These rules plainly preserve validity over K. So as mentioned in section 2.12, if the axioms of H are valid in K then H is sound over K. For example, if G 1 in [d]-form is valid in K then we can conclude from the lemma that Kt.UG 1 in [d]-form is sound over K.

Finite model property
Given a class K of frames, a Hilbert system is said to have the finite model property over K if it is sound and complete over the class of finite frames in K.
Our completeness theorems rely critically on several results on the finite model property. Two of them come from Fact 3.2 and Theorem 3.10, but the majority were proved in [14,15], using special kinds of filtration: we recall the relevant ones in Fact 8.2 below. Related earlier results on the finite model property include [10,12], [34, theorem 15], [36, theorem 10], and [45].  Table 1 hold.
So armed, we can proceed to prove soundness and completeness theorems. The pioneering result in this field was the theorem of [27] that the L 2 -logic of every separable dense-initself metric space is S4. The assumption of separability was removed in [32]. We begin by generalising this theorem, establishing (weak) completeness results for L μ 2 and L t 2 over any dense-in-itself metric space. We will go on to prove strong completeness in Theorem 9.3. Theorem 8. 3. Let X be a non-empty dense-in-itself metric space.

The Hilbert system
S4μ is sound and complete over X for L μ 2 -formulas. 2. The Hilbert system S4t is sound and complete over X for L t 2 -formulas. Proof. For part 1, soundness is easy to check and indeed we have already mentioned it in Corollary 4. 7. For completeness, let ϕ be an L μ 2 -formula that is not a theorem of S4μ. By Theorem 3.10, we can find a finite reflexive transitive frame F = (W, R), an assignment h into F, and a world w ∈ W with (W, R, h), w |= ¬ϕ. By replacing F by F(w), we can suppose that w is a root of F -this can be justified in a standard way using Lemma 2.1. Since F is rooted, it is clearly connected. Since it is reflexive and transitive, it is locally connected and serial. So by Proposition 7.10, it is fully representable over X. So, taking U = X and S = σ = ∅ in the definition of 'fully representable' (Definition 7.9), for any x ∈ X we may choose an X-full, hence surjective, representation ρ of F over X with ρ(x) = w. Then We obtain (X, ρ −1 • h), x |= ¬ϕ. Thus, ϕ is not valid in X, proving completeness.
The proof of part 2 is similar. The differences are: for soundness, use Lemma 8.1; for completeness, ϕ is assumed to be an L t 2 -formula that is not a theorem of S4t; we use Fact 8.2 in place of Theorem 3.10 to obtain a finite reflexive transitive Kripke model satisfying ¬ϕ at a root; and having obtained, for any x ∈ X, a surjective representation ρ of F over X with ρ(x) = w, we use the additional translation − μ from section 4, as follows. Note that ϕ ∈ L Note that we have shown that in each case, any consistent formula is satisfiable in X at any chosen point. 2

Weak completeness for L 2∀ and L t 2∀
Completeness for languages with ∀ follows the same lines, although soundness requires that the space be connected.
Theorem 8. 4. Let X be a non-empty dense-in-itself metric space. 1. The Hilbert system S4.UC is complete over X for L 2∀ -formulas, and sound if X is connected. 5 2. The Hilbert system S4t.UC is complete over X for L t 2∀ -formulas, and sound if X is connected.
Proof. For part 1, soundness was shown in Lemma 8. 1. For completeness, even when X is not connected, suppose that ϕ ∈ L 2∀ is not a theorem of S4.UC. By Fact 8.2, or by [36, theorem 10], S4.UC has the finite model property, so we can find a finite reflexive (hence serial and locally connected) transitive connected frame F = (W, R), an assignment h into F, and a world w ∈ W such that (W, R, h), w |= ¬ϕ. (F may not be rooted.) As in Theorem 8.3, we may take a surjective representation ρ of F over X. Using surjectivity, take x ∈ X with ρ(x) = w. Then as before, (X, Part 2 is proved similarly. 2 We have no results for L μ 2∀ because we are not aware of any completeness theorem for this language with respect to finite reflexive transitive connected frames. If one is proved in future, we could take advantage of it. 8. 6. Weak completeness for L [d] and L dt [d] In one way this is even easier, as we do not need the translation ϕ d . But again, soundness requires a condition on the space. Theorem 8. 5. Let X be a non-empty dense-in-itself metric space. 1 Proof. For part 1, soundness follows from Lemma 8.1 and the assumed validity of G 1 . For completeness, even when X does not validate G 1 , suppose that ϕ ∈ L [d] is not a theorem of KD4G 1 . Now KD4G 1 has the finite model property (see [34, theorem 15] or Fact 8.2), so we can find a finite serial transitive locally connected frame F = (W, R), an assignment h into F, and a world w ∈ W such that (W, R, h), w |= ¬ϕ. As usual, by replacing F by F * (w), we can suppose that F is connected and w is a root of it. Let x ∈ X be arbitrary. By Proposition 7.10, F is fully representable over X, so there is a surjective representation ρ of F over X with ρ(x) = w. Then (X, ρ −1 • h), x |= ¬ϕ by Proposition 7.5. So ϕ is not valid in X.
The proof of part 2 is similar, but in order to apply Proposition 7.5, we first use the translation − μ to turn ϕ ∈ L dt [d] into an L μ [d] -formula ϕ μ equivalent to ϕ in transitive frames and in X. Again, we have shown that any consistent formula is satisfiable at any given point of X. 2 Remark 8. 6. Theorem 8.5(1) is related to earlier work of Shehtman [34]. In [34, theorem 23, p. 39], the following is proved for the language L Shehtman's results (i), (ii) above follow from Theorem 8.5(1). We remark that the converse of his lemma 2 fails in general -the reader may check that the subspace R 2 \ {(1/n, y) : n a positive integer, y ∈ R} of R 2 validates G 1 , but for no open neighbourhood V of (0, 0) is V \ {(0, 0)} connected. [25, theorems 3.12, 3.14] give a characterisation of when a topological space validates G n , for n ≥ 1.
Shehtman [34, p. 43] also states two open problems:  8.5(1) appears to resolve problem 2 and the second part of problem 1, both positively.
Shehtman also proved in [34, theorem 29] that the L [d] -logic of any separable zero-dimensional densein-itself metric space is KD4. This does not follow from Theorem 8. 5. The separability assumption was removed, and the result extended to tangled closure operators, in [16].

Weak completeness for L
The following is now purely routine.
Theorem 8. 7. Let X be a non-empty dense-in-itself metric space. 1 Proof. There are no new elements in the proof, so we leave it to the reader. 2

Strong completeness
Here, we will prove that KD4G 1 t is strongly complete over any non-empty dense-in-itself metric space X: any countable KD4G 1 t-consistent set of L dt [d] -formulas is satisfiable over X. The analogous results for L μ 2 and the weaker languages L [d] and L t 2 will follow. The analogous result for L 2 also follows, but this is a known result, proved recently by Kremer [20]. 6 We will then show that strong completeness frequently fails for languages with ∀. 6 Kremer's argument does not appear to work in our situation. One difficulty is that strong completeness even for L t 2 fails in Kripke semantics (an example in [14, §5] can be used to show this). Even without the tangled closure operators, satisfying an infinite set of formulas over a connected locally connected frame presents further difficulties.

The problem
Let us outline a naïve approach to the problem. It does not work, but it will illustrate the difficulty we face and motivate the formal proof later.
Let Γ be a countable KD4G 1 t-consistent set of L dt [d] -formulas. For simplicity, assume that Γ is maximal consistent. Write Γ as the union of an increasing chain Γ 0 ⊆ Γ 1 ⊆ · · · of finite sets. Fix x ∈ X. As we saw at the end of the proof of Theorem 8.5, each Γ n (n < ω) is satisfiable at the point x. So we can find an assignment g n on X with (X, g n ), x |= Γ n . Suppose we could build a new assignment g that behaves like g n for larger and larger n, as we approach x. Then we might hope that (X, g), x |= Γ n for all n, and so (X, g), x |= Γ.
To define such a g, we choose a countable sequence X = S 0 ⊇ S 1 ⊇ · · · of open neighbourhoods of x, such that S1. every open neighbourhood of x contains some S n (that is, the S n form a 'base of open neighbourhoods' of x).
X is a metric space, so we can do this. Since we can make the S n as small as we like, and the Γ n are finite sets, we can suppose that for each n < ω: S2. for each [d]ϕ ∈ Γ n , we have (X, g n ), y |= ϕ for every y ∈ S n \ {x}, S3. for each d ϕ ∈ Γ n , there is y ∈ S n \ cl S n+1 with (X, g n ), y |= ϕ.
We can now define a new assignment g by 'using g n within S n ', for each n < ω. More precisely, we let for each atom p and each n < ω. We also need to define g at x itself, but we can use Γ to determine truth values of atoms there. Now we try to prove that ϕ ∈ Γ iff (X, g), x |= ϕ for all formulas ϕ, by induction on ϕ. The atomic and boolean cases are easy. Consider the case d ϕ.
If d ϕ ∈ Γ, then d ϕ ∈ Γ n for all large enough n, so by S3, there is y ∈ S n \ cl S n+1 with (X, g n ), y |= ϕ. As S n \ cl S n+1 is open and g n agrees with g on it, it follows that (X, g), y |= ϕ. This holds for cofinitely many n, so (X, g), x |= d ϕ.
Conversely, if (X, g), x |= d ϕ, then for infinitely many n, there is y ∈ S n \ S n+1 with (X, g), y |= ϕ. If we could find such a y ∈ S n \ cl S n+1 , then as above, (X, g n ), y |= ϕ, and it would follow by S2 and maximality of Γ that d ϕ ∈ Γ.
But it may be that we can only find such y ∈ cl S n+1 . The truth of ϕ at such y may not be preserved when we change from g to g n , because it may depend on points in S n+1 , and at such points, g agrees with g n+1 , not g n . (We cannot just make S n+1 smaller to take the witnesses y out of cl S n+1 , because g will then change, and we may no longer have (X, g), y |= ϕ.) So we would like to arrange a smooth transition between g n and g n+1 , avoiding unpleasant discontinuities. It would be sufficient if there is some closed T n+1 ⊆ S n+1 such that g n and g n+1 agree on the 'buffer zone' S n+1 \ T n+1 . Much of the formal proof below is aimed at achieving something like this for atoms occurring in Γ n -see Claim 9. 1 [d] -formulas. We show that Γ is satisfiable in X. We can suppose without loss of generality that Γ is maximal consistent. Since Γ is countable, we can write it as Γ = n<ω Γ n , where Γ 0 ⊆ Γ 1 ⊆ · · · is a chain of finite sets. Let Var n be the finite set of atoms occurring in formulas in Γ n , for each n < ω. So Var 0 ⊆ Var 1 ⊆ · · · . For each n < ω, as Γ n is KD4G 1 t-consistent, by Fact 8.2 there is a finite serial transitive locally connected Kripke model M n = (W n , R n , h n ) and a world w n ∈ W n with M n , w n |= Γ n .
We can assume without loss of generality that the W n (n < ω) are pairwise disjoint. For each n, fix an arbitrary e n ∈ W n with R n w n e n and such that e n is R n -maximal -that is, R • n (e n ) = ∅.
So tp i (w) is the 'atomic type' of w in M j with respect to the finite set Var i of atoms. We do not need to write tp j i (w) since the W n are pairwise disjoint so j is determined by w. And τ j i is the set of such types that occur as types of points in the cluster R j (e j ).
The following claim shows that we can actually assume without loss of generality that τ j i = τ i i whenever i ≤ j < ω, so that τ j i is independent of j.
Proof of claim. Essentially König's tree lemma. We will define by induction infinite sets ω = S −1 ⊇ S 0 ⊇ S 1 ⊇ · · · . We let s n = min S n , and we will arrange that 0 = s −1 < s 0 < s 1 < · · · and s n ≥ n for all n. Let n < ω and suppose that we are given S n−1 and s n−1 = min S n−1 ≥ n − 1 inductively. Using that ℘℘Var n is finite and S n−1 infinite, choose infinite S n ⊆ S n−1 \ {s n−1 } such that τ s n ∈ ℘℘Var n is constant for all s ∈ S n . The term τ s n is defined for all s ∈ S n , because s ≥ min S n > s n−1 ≥ n − 1 and so s ≥ n. Of course define s n = min S n . Then s n > s n−1 and s n ≥ n as required. This completes the definition. Then for any n ≤ m < ω we have s n ∈ S n and s m ∈ S m ⊆ S n , so τ s n n = τ s m n , as required. This proves the claim. 2 Now replace M n , w n , e n by M s n , w s n , e s n for each n < ω. Do not change Γ n or Var n . Since n ≤ s n , we have Γ n ⊆ Γ s n , and consequently we still have M n , w n |= Γ n for each n. Moreover, if i ≤ j < ω we have τ s i i = τ s j i , and consequently after replacement, τ i i = τ j i . For each n < ω, define the frames F n = (R n (w n ), R n R n (w n )), C n = (R n (e n ), R n R n (e n )).
F n is a generated subframe of (W n , R n ), so inherits its serial, transitive, and locally connected properties. Also, F n is connected since (W n , R n ) validates G 1 . The reason for considering F n instead of just (W n , R n ) or (R * n (w n ), R n R * n (w n )) will be seen in Claim 9. 1.5. As e n is R n -maximal, C n is a nondegenerate cluster, so trivially a connected serial transitive locally connected frame, and (as R n is transitive) a generated subframe of F n . We conclude from Proposition 7.10 that F n and C n are fully representable over X, for all n < ω.
See Fig. 1. It is easily seen that The following claim lists some other basic facts about our situation. Applying this for n + 1 and n gives cl S n+1 ∩ cl S n ⊆ cl O n+1 \ P n ⊆ P n \ P n = ∅. 4. O n and P n are regular open subsets of X, so by Lemma 5.5, S n = O n \ cl P n is a regular open subset of X too. Since cl S n ∩ cl S n+1 = ∅ by part 2, Lemma 5.5(2) yields that S n ∪ S n+1 is also a regular open subset of X. Since each of these three sets is a subset of U n by part 2, by Lemma 5.5(3) it is also regular open in U n . 5. By (9.3) (for n and n + 1), cl S n and cl S n+1 are disjoint from P n \ cl O n+1 , so by additivity of closure, Claim 9.1. 3. There are surjective representations ρ n of F n over U n (n < ω) such that tp n (ρ n (x)) = tp n (ρ n+1 (x)) for every x ∈ S n+1 . (9.4) Proof of claim. We define the ρ n by induction on n to satisfy (9.4) and additionally ( * ) ρ n S n+1 is a representation of C n over S n+1 .
First let n = 0. Since C 0 is fully representable over X, we can choose a representation σ : S 1 → C 0 . Because C 0 is a nondegenerate cluster, σ is actually a U 0 -basic representation (see Remark 7.7). By Claim 9.1.2, S 1 is a regular open subset of U 0 , and U 0 \ cl S 1 = ∅. Now F 0 is also fully representable over X, so σ extends to a surjective representation ρ 0 of F 0 over U 0 . Clearly, condition ( * ) above is met. Let n < ω and assume inductively that for each m ≤ n, a surjective representation ρ m of F m over U m has been constructed, such that ρ m S m+1 is a representation of C m over S m+1 , and tp m (ρ m (x)) = tp m (ρ m+1 (x)) for all x ∈ S m+1 when m < n. We will define ρ n+1 to continue the sequence.
Note first that since C n is a nondegenerate cluster, ρ n S n+1 is U n -basic -see Remark 7. 7. It is also surjective. For, let w ∈ R n (e n ) be given. Take x ∈ S n+1 (note that S n+1 is non-empty by Claim 9.1.2). As C n is a nondegenerate cluster, R n (ρ n (x), w), so as ρ n S n+1 is a representation, (S n+1 , (ρ n S n+1 ) −1 ), x |= d w. This certainly implies that ρ n (y) = w for some y ∈ S n+1 .
For each w ∈ R n (e n ), define See Fig. 2. Because ρ n S n+1 is surjective onto C n , each set D w is non-empty, and plainly, S n+1 is partitioned by the D w (w ∈ R n (e n )). Because τ n+1 n = τ n n , each H w is non-empty and w∈R n (e n ) H w = R n+1 (e n+1 ). (The sets H w may not be pairwise disjoint, but any two of them are equal or disjoint.) Observe that S n+1 ⊆ d D w for each w ∈ R n (e n ). (9.5) To see this, let x ∈ S n+1 and w ∈ R n (e n ). Because C n is a nondegenerate cluster, R n (ρ n (x), w). As ρ n S n+1 is a representation of C n over S n+1 , we have (S n+1 , (ρ n S n+1 ) −1 ), x |= d w. Since (ρ n S n+1 ) −1 (w) = D w , this says exactly that x ∈ d D w . Let w ∈ R n (e n ) and consider D w as a subspace of X. We show that it is dense in itself. Let  is open, we can suppose that O ⊆ S n+1 . But by (9.5 So D w is a dense-in-itself metric space in its own right. Since C n+1 is a nondegenerate cluster, so is its subframe H w . Hence, H w is trivially a finite connected locally connected serial transitive frame. So by Proposition 7.10, there is a surjective representation The sets D w partition S n+1 , so σ is a well defined and total map. It has the following property. Let x ∈ S n+1 . Writing ρ n (x) = w, say, we have x ∈ D w and σ(x) = σ w (x) ∈ H w , so tp n (σ(x)) = tp n (w) by definition of H w . That is, tp n (σ(x)) = tp n (ρ n (x)) for each x ∈ S n+1 . (9.7) We show that σ is a representation of C n+1 over S n+1 . Since C n+1 is a nondegenerate cluster, we need show only that (X, σ −1 ), x |= d v for every x ∈ S n+1 and v ∈ R n+1 (e n+1 ).
So take such x, v. Suppose that ρ n (x) = w, say, so x ∈ D w . Choose w ∈ R n (e n ) such that v ∈ H w (it may not be unique). By (9.5), x ∈ d D w . But by (9.6), (X, σ −1 ), y |= d v for every y ∈ D w . It follows that (X, σ −1 ), x |= d d v, and hence (X, σ −1 ), x |= d v as required.
Claim 9.1.4 applies: (X, g), x |= ϕ. We conclude that for every i ≥ n there is x ∈ U i with (X, g), x |= ϕ. As U i ⊆ O i \ {x 0 } and the O i form a base of neighbourhoods of x 0 , it follows that (X, g), x 0 |= d ϕ.
Conversely, suppose that (X, g), x 0 |= d ϕ. Since Γ is maximal consistent, either d ϕ ∈ Γ or ¬ d ϕ ∈ Γ. Choose n < ω such that either d ϕ ∈ Γ n or ¬ d ϕ ∈ Γ n . As O n is an open neighbourhood of x 0 , there is x ∈ O n \ {x 0 } with (X, g), x |= ϕ. Since O n \ {x 0 } = n≤i<ω U i by (9.2), we have x ∈ U i for some i ≥ n. The atoms of ϕ lie in Var i , and ρ i : U i → R i (w i ), so M i , v |= ϕ for some v ∈ R i (w i ) (by Claim 9. 1.4). So by Kripke semantics, M i , w i |= d ϕ (we defined F i as based on R i (w i ) rather than on W i or R * i (w i ) so that we can take this step). Since M i , w i |= Γ i , we have ¬ d ϕ / ∈ Γ i ⊇ Γ n . So d ϕ ∈ Γ n ⊆ Γ, proving the claim. 2 The general case now follows: dt Δ ↔ d dt Δ. It follows by soundness (Lemma 8.1) that dt Δ ↔ d dt Δ is valid in X, so (X, g), x 0 |= dt Δ iff (X, g), x 0 |= d dt Δ. By Claim 9. 1.5, this is iff d dt Δ ∈ Γ. Since Γ is maximal KD4G 1 t-consistent, this is iff dt Δ ∈ Γ, as required. The claim is proved. Hence, (X, g), x 0 |= Γ, so the theorem is proved as well. 2

Strong completeness for L [d]
We can now easily derive the analogous result for 'modal' L [d] -formulas, essentially by showing that KD4G 1 t is a conservative extension of KD4G 1 .
Theorem 9. 2. Let X be a non-empty dense-in-itself metric space. Then the Hilbert system KD4G 1 is strongly complete over X for L [d] -formulas, and sound if G 1 is valid in X. Proof. For soundness, see Theorem 8. 5. For strong completeness, let Γ be a countable KD4G 1 -consistent set of L [d] -formulas. Let Γ 0 ⊆ Γ be finite and put γ = Γ 0 . Then γ is KD4G 1 -consistent, so by Fact 8.2 it is satisfiable in some finite serial transitive locally connected frame F. It is easily seen that F is a KD4G 1 t frame, and it follows that γ is KD4G 1 t-consistent. Since Γ 0 was arbitrary, Γ is KD4G 1 t-consistent. By Theorem 9.1, Γ is satisfiable over X. This also follows, using the translations − d and − t of section 4. Theorem 9. 3. Let X be any dense-in-itself metric space. 1. The Hilbert system S4t is sound and strongly complete over X for L t 2 -formulas. 2. The Hilbert system S4μ is sound and strongly complete over X for L μ 2 -formulas. 3. (Kremer,[20]) The Hilbert system S4 is sound and strongly complete over X for L 2 -formulas. Proof. Soundness is clear in all cases: cf. Theorem 8. 3. We prove strong completeness. For part 1, let ϕ be an S4t-consistent L [d] -formulas. Since F is reflexive, it follows from Lemma 4.4 that ϕ d is equivalent to ϕ in F. So ϕ d is satisfiable in F. Plainly, F is a KD4G 1 t frame, so ϕ d is KD4G 1 t-consistent.
Since − d commutes with ∧, it is now easily seen that if Γ ⊆ L t 2 is a countable S4t-consistent set then is a countable KD4G 1 t-consistent set. By Theorem 9.1, Γ d is satisfiable over X. Since X is T D , by Lemma 4.5 each γ ∈ Γ is equivalent to γ d in X, so Γ is also satisfiable over X.
For part 2, for a set Γ ⊆ L μ 2 we write Γ t = {γ t : γ ∈ Γ} ⊆ L t 2 , where the translation − t : L μ 2 → L t 2 is as in Fact 4. 6. Let Γ ⊆ L μ 2 be a countable S4μ-consistent set. Let Γ 0 ⊆ Γ be any finite subset. By assumption, the formula Γ 0 is S4μ-consistent. So by Theorem 3.10, there is a finite reflexive transitive frame F in which Γ 0 is satisfiable. By Fact 4.6, ϕ t is equivalent to ϕ in F, for each ϕ ∈ L μ 2 . So (Γ t 0 ) is also satisfiable in F. Since F is clearly an S4t frame, it follows that (Γ t 0 ) is S4t-consistent. As Γ 0 was arbitrary, Γ t is S4t-consistent.
By part 1, Γ t is satisfiable in X. But by Corollary 4.7, each γ ∈ Γ is equivalent to γ t in X. So Γ is also satisfiable in X.
Part 3 can be proved similarly, by showing in the same way that for L 2 -formulas, S4-consistency implies S4t-consistency, and then appealing to part 1. 2

Universal modality
We do not include the universal modality in our strong completeness results, for good reason.
Theorem 9. 4. There is a countable set Σ of L 2∀ -formulas such that for every non-empty compact locally connected dense-in-itself metric space X, each finite subset of Σ is satisfiable in X, but Σ as a whole is not.
Compact means that if S is a set of open sets with S = X, then X = S 0 for some finite S 0 ⊆ S. Every compact space X is sequentially compact -for every sequence x i (i < ω) of points of X, there is z ∈ X such that for every open neighbourhood O of z, the set {i < ω : x i ∈ O} is infinite. Locally connected means that every open neighbourhood of a point x contains a connected (in the subspace topology) open neighbourhood of x. An example of a compact locally connected dense-in-itself metric space is the subspace [0, 1] of R.
The model is shown in Fig. 3 -it goes off to the right forever, roughly repeating after every three steps. Of course R is reflexive. Note that the underlying frame (W, R) is connected.
We let Σ be the set comprising the following formulas: They are plainly true at every world in M. So for every finite subset Σ 0 ⊆ Σ, we have M, a 0 |= Σ 0 . As can be checked, the frame of M validates S4.UC, and it follows that Σ 0 is S4.UC-consistent. Hence, by Theorem 8.4, Σ 0 is satisfiable in X.
Assume for contradiction that Σ is true at some point of some model (X, h) on X. Below, we will write x |= ϕ instead of (X, h), x |= ϕ. By Σ1, for each i < ω there is x i ∈ X with x i |= 3p i ∧ 3r ∧ 3g. As X is compact, it is sequentially compact and contains a point z such that for every open neighbourhood O of z, the set {i < ω : x i ∈ O} is infinite. Then z |= 3r ∧ 3g as well. By Σ3, z |= 2¬b. As X is locally connected, there is a connected open neighbourhood N of z with y |= ¬b for all y ∈ N .
Take i < j < ω with x i , x j ∈ N . Let U = {x ∈ N : x |= 3p i }. Then U is an open subset of N , because for every u ∈ U we have u |= 3p i ∧ 2¬b, and Σ4 gives u |= 23p i . And N \ U is also open, because U = {x ∈ X : x |= 3p i } is closed and N \ U = N \ U . We have x i ∈ U , but by Σ2, x j ∈ N \ U . So N is the union of two disjoint non-empty open sets (U and N \ U ), contradicting its connectedness. 2 Corollary 9. 5. Let X be a non-empty compact locally connected dense-in-itself metric space, and L ⊆ L μ t dt 2[d]∀ a language containing L 2∀ or L [d]∀ . Then no Hilbert system for L is sound and strongly complete over X. Proof. Assume for contradiction that the Hilbert system H is sound and strongly complete over X. Let Σ be as in Theorem 9.4 (use the translation − d if necessary to ensure it is a set of L-formulas). Since every finite subset of Σ is satisfiable in X, and H is sound over X, it follows that Σ is H-consistent. But H is strongly complete over X, so Σ is satisfiable over X, contradicting the theorem. 2 This does not rule out the possibility of strong completeness of a system having inference rules (2.2) with infinitely many premises.

Conclusion
This paper has presented some completeness theorems for various spatial logics over dense-in-themselves metric spaces. Table 2 summarises them. The numbers in parentheses refer to our earlier results. The first line of the table is of course known, included here to give a more complete picture. For handy reference, Table 3 summarises the ingredients of each logic.
There are of course many problems left open by our work, and we present some of them here. For simplicity, in this section we take metric spaces to be non-empty.

Extensions
Problem 10. 1. Can the results be extended to more general topological spaces?
For example, consider the topological space T defined as follows. For ordinals α, β write α β for the set of all maps f : α → β. The set of points of T is n≤ω n 2, and the open sets are unions of sets of the form {f ∈ T : f ⊇ g} for some g ∈ n<ω n 2. This space is dense in itself, and T0 -that is, no two distinct points have the same open neighbourhoods. It is not even T D , but still it may be that the methods in this paper can be applied to it. So we ask: Table 2 Soundness and completeness for a non-empty dense-in-itself metric space X.

Language
Logic Sound Complete Strongly complete L 2 S4 yes yes [27] yes [20] L  Results of Kudinov [21,22] are relevant. Recently, Kudinov and Shehtman [23] proved numerous results about logics of topology with 2, [d], ∀, and the 'difference modality' [ =]. In particular, they determine the logic of R n for n ≥ 2 in the language with [d] and [ =]. However, results for general dense-in-themselves metric spaces appear to be lacking.

Strong completeness
Our strong completeness results for languages with [d] are limited to logics with G 1 . We could ask for more: Problem 10. 4. Let X be a dense-in-itself metric space and let L be L [d] or L dt [d] . Is the L-logic of X strongly complete over X?
By Theorems 9.1 and 9.2, the answer is 'yes' if X validates G 1 . We saw in Corollary 9.5 that in the language L 2∀ , there are many dense-in-themselves metric spaces over which S4.UC is not strongly complete. So we ask: Problem 10. 5. Can strong completeness for languages with ∀ be proved for each dense-in-itself metric space in some reasonably large class, and for R n for n ≥ 1?