A widely connected topological space made from diamond
aa r X i v : . [ m a t h . GN ] J u l A BARELY CONNECTED TOPOLOGICAL SPACE MADE FROMDIAMOND
SAMUEL M. CORSON
Abstract.
Motivated by the Knaster-Kuratowski Fan, we give the construc-tion of an infinite topological space with unusual characteristics. The spaceis regular, separable, and connected, but removing any nonempty open setleaves the remainder of the space totally disconnected (in fact, totally sepa-rated). The space is also strongly Choquet and has a basis with nice properties.The construction utilizes Jensen’s diamond principle ♢ . Introduction
The goal of this paper is to produce an unusual space which, though connected,is “barely connected.” We assume some familiarity with topological terminologyin this introduction, but many definitions will be provided in Section 2. We adoptthe fairly standard convention that regular spaces are also Hausdorff.A point x in a connected space X is a dispersion point if the subspace X ∖ { x } is totally disconnected [6]. The famous Knaster-Kuratowski Fan (also known asCantor’s Leaky Tent) is a subspace of the plane which is infinite and connectedand contains a dispersion point [5]. One could imagine a connected space to bequite tenuously connected if it contained many dispersion points. Unfortunately aconnected space with at least three points can have at most one dispersion point.A reasonable alternative to having many dispersion points in a connected space X is to have the removal of any neighborhood O of any point ensure that thesubspace X ∖ O is totally disconnected. An example of such a space is an infiniteset with the cofinite topology. Unfortunately this example does not satisfy veryimpressive separation axioms; it is T but not Hausdorff.We produce a space which is dispersed by the removal of any neighborhood andwhich is regular. We use Jensen’s ◇ principle (see Section 3) in the constructionin addition to the standard Zermelo-Fraenkel axioms of set theory with the axiomof choice (ZFC). As ZFC + ◇ is consistent if and only if ZFC is consistent [3], theexistence of such a space is consistent with ZFC. Theorem 1.1. (ZFC + ♢ ) There exists a topological space ( X, τ ) , having a basis B , which is(1) regular;(2) separable;(3) connected; Mathematics Subject Classification.
Primary 54G15, 54A35, 54D05; Secondary 03E05,03E35, 03E50.
Key words and phrases. connected space, regular space, diamond principle.This work was supported by the Severo Ochoa Programme for Centres of Excellence in R&DSEV-20150554. (4) of cardinality ℵ ;and(5) for every nonempty O ∈ τ the subspace X ∖ O is totally separated (hencetotally disconnected);(6) every nonempty O ∈ τ is uncountable;(7) ∣B∣ = ℵ ;(8) every countable cover of X by elements of B has a finite subcover; and(9) every nesting decreasing countable sequence { B n } n ∈ ω of basis elements has ⋂ n ∈ ω B n ≠ ∅ (hence X is strongly Choquet).We note that this result is reasonably sharp. A space satisfying (1)-(9) abovecannot have any of the following additional properties: ● metrizable ● second countable ● Lindel¨of ● there exists a nonempty open set with compact closure(see Proposition 2.5). One also cannot strengthen (5) by replacing it by(5’) for each nonempty open O ⊆ X the subspace X ∖ O is zero-dimensional(see Proposition 2.6). Thus the space is topologically small (8) and narrow (2), butno space satisfying (1) - (9) can be so small as to be Lindel¨of, or so narrow as tobe second countable. The topology is reasonably fine (1), but no space satisfying(1) - (9) can be metrizable. Properties (1), (3) and (5) say that although the spaceis connected, it is barely connected. Moreover the disconnectedness occasioned byremoving any region is about as strong as allowable in light of (1). Despite thetenuous connectedness of the space, it is topologically thick (9).One also cannot replace (4) with “of cardinality ℵ ” since regular connectedspaces with at least two points must be of cardinality at least ℵ , but interestinglyone can produce (from ZFC) regular separable connected spaces of cardinality ex-actly ℵ [1]. Any (strongly) Choquet space having at least two points and satisfying(1) and (3) is necessarily of cardinality ≥ ℵ (see Proposition 2.9), and this togetherwith (4) highlights the well known fact that ♢ implies ℵ = ℵ (see Remark 3.22).It may be that one can strengthen (1) by replacing “regular” by “completely reg-ular” or “normal.” A completely regular connected space with at least two pointsmust have cardinality at least 2 ℵ , but since ♢ impies ℵ = ℵ this strengtheningmight be possible under our set theoretic assumptions. Another potential strength-ening would be to replace (8) with “ X is countably compact.” One could also tryproducing the space of Theorem 1.1, with all instances of ℵ replaced with 2 ℵ ,from ZFC alone.The space is constructed by induction over ℵ . At certain steps of the inductionwe begin the construction of a new basis element, making committments as tohow the basis elements interact with each other. New points are assigned to theincreasingly-defined basis elements, but we will not have finished the construction ofany basis element until the very end of the the induction. Connectedness is assuredby detecting potential disconnections (using ♢ ) and blocking them by determiningthat certain sequences of points will converge. The ability to block the detecteddisconnections will be assured by model-theoretic techniques. BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 3
We provide some topological preliminaries in Section 2, and some model-theoreticpreliminaries in Section 3. The space is constructed in Section 4 and the variousproperties claimed in Theorem 1.1 are verified in Section 5.2.
Topology preliminaries
We will review some very basic topology terminology and prove some easypreparatory results. We will let ω denote the set of natural numbers and adopt thestandard convention that an ordinal number is the set of ordinal numbers whichare strictly smaller than itself (e.g. 3 = { , , } and ω + = { , , . . . , ω } ). If α is asuccessor ordinal we let α − Definitions 2.1.
Recall that a space X is Hausdorff if for distinct x , x ∈ X thereexist disjoint open sets U , U ⊆ X such that x i ∈ U i . We say X is regular if it isHausdorff and for any closed subset C ⊆ X and x ∈ X ∖ C there exist disjoint opensets U x and U C such that x ∈ U x and C ⊆ U C . A space is separable if it includes acountable dense subset, and Lindel¨of if every open cover has a countable subcover. Definition 2.2. If X is a topological space and B is a subset of X , we say that B is a basis for the topology on X if all elements of B are open and each open set in X is a (possibly empty) union of elements of B .It is a standard exercise to show that if B is a collection of subsets of a set X such that ⋃ B = X , and for any B , B ∈ B and x ∈ B ∩ B there exists B ∈ B suchthat x ∈ B ⊆ B ∩ B , then the collection of unions of elements in B is a topologyon X with basis B .We now give some relevant definitions with regard to (dis)connectedness. Definitions 2.3.
Let X be a topological space. A writing X = V ⊔ V of X as adisjoint union of open sets V , V ⊆ X is a disconnection . We call a disconnection trivial is at least one of the V i is empty. The space X is connected if each discon-nection of X is trivial. A component of a space is a maximal connected subspace.Components are necessarily closed. We say X is totally disconnected if all com-ponents are of cardinality at most 1, and X is totally separated if for any distinct x, y ∈ X there exists a disconnection X = V ⊔ V such that x ∈ V and y ∈ V . Also, X is zero-dimensional if there exists a basis for X consisting of clopen sets. Clearlyzero-dimensional implies totally separated implies totally disconnected.Given a subset Y of a topological space X we’ll use Y to denote the closure of Y in X and ∂Y to denote the boundary Y ∩ ( X ∖ Y ) of Y . Lemma 2.4.
Suppose X is a connected topological space and X , X ⊆ X arenonempty disjoint open subsets such that X ∪ X is dense in X . Then there existsa point x ∈ ( X ∪ X ) ∖ ( X ∪ X ) . Proof.
Since X = X ∪ X = X ∪ X we must have X ∩ X ≠ ∅ , else X = X ⊔ X is a nontrivial disconnection. Since X is open and X ∩ X = ∅ we know that X ∩ X = ∅ , and similarly X ∩ X = ∅ . Selecting x ∈ X ∩ X we therefore have x ∈ ( X ∪ X ) ∖ ( X ∪ X ) . (cid:3) SAMUEL M. CORSON
Next we’ll provide some justification for why the properties of the space claimedin Theorem 1.1 are fairly sharp.
Proposition 2.5.
If a space ( X, τ ) with basis B satisfies the properties (1) - (9)of Theorem 1.1 then it cannot satisfy any of the following additional properties:(a) metrizable;(b) second countable;(c) Lindel¨of;(d) compact;(e) there exists a nonempty open set with compact closure. Proof.
The fact that (a) ∧ (2) ⇒ (b), (b) ⇒ (c), (c) ∧ (8) ⇒ (d), and (d) ⇒ (e)is fairly routine. We’ll derive a contradiction from (e) ∧ (1) ∧ (3) ∧ (4) ∧ (5)(in fact we will not use (1) but rather the weaker Hausdorff condition). Supposethat X is Hausdorff, connected, satisfies (5), has at least two points, and O ⊆ X is a nonempty open set such that O is compact. We can assume without loss ofgenerality that O is a proper subset of X , since X is Hausdorff and has at leasttwo points. As X is connected, we have ∂O ≠ ∅ . Fix x ∈ O . As O is totallyseparated, there exists for each y ∈ ∂O a pair of open sets ( U x,y , U y,x ) in X suchthat U x,y ∩ U y,x ⊆ X ∖ O and x ∈ U x,y and y ∈ U y,x . As ∂O is compact, there exists afinite subset { U y ,x , . . . , U y k ,x } such that ⋃ ki = U y i ,x ⊇ ∂O . Letting U = O ∩⋂ ki = U x,y i and U = ⋃ ki = U y i ,x we obtain a nontrivial disconnection X = U ⊔ (( X ∖ O ) ∪ U ) ,contradicting (3). (cid:3) Proposition 2.6. If X is a connected space and U and U are nonempty disjointopen sets then X ∖ U cannot be zero-dimensional. Proof.
Suppose otherwise for contradiction. Then we may write U as a union U = ⋃ i ∈ I O i where each O i is clopen in X ∖ U . Since U ≠ ∅ we select i ∈ I forwhich O i ≠ ∅ . As O i ⊆ U is open in X ∖ U , it is also open in U , and thereforeopen in X . As O i ⊆ U ⊆ X ∖ U and O i is closed in X ∖ U , we know O i is closed in X . Thus ∅ ≠ O i ⊆ X ∖ U ≠ X is clopen, contradicting the connectedness of X . (cid:3) Definitions 2.7. (see [4, Chapter 8]) Recall that a space is
Baire if any countableintersection of open dense sets is dense. Given a space X , the strong Choquet game is an infinitary game between a Player I and a Player II, a run of which we describe.Player I chooses a point and neighborhood ( x , U ) , then Player II chooses an openneighborhood x ∈ V ⊆ U , Player I chooses a point and neighborhood ( x , U ) such that U ⊆ V , Player II chooses a neighborhood x ∈ V ⊆ U , etc. Player I winsthe run if ⋂ n ∈ ω U n = ∅ and otherwise Player II wins. The Choquet game is similar,but the players are only selecting nonempty open sets so that U ⊇ V ⊇ U ⊇ ⋯ ,and again Player I wins if ⋂ n ∈ ω U n = ∅ , with Player II winning otherwise. A spaceis (strongly) Choquet if Player II has a winning strategy in the (strong) Choquetgame. Strongly Choquet implies Choquet implies Baire. Lemma 2.8.
If a space X has a basis such that any nesting decreasing countablesequence { B n } n ∈ ω of basis elements has nonempty intersection, then X is stronglyChoquet. Proof.
Assume the hypotheses. We define a winning strategy for Player II: givena move ( x n , U n ) we let V n be any basis element such that x n ∈ V n ⊆ U n . Byassumption we get ⋂ n ∈ ω V n ≠ ∅ , so this is a winning strategy. (cid:3) BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 5
Proposition 2.9. If X is a nonempty, Hausdorff, Choquet space with no isolatedpoints then ∣ X ∣ ≥ ℵ . In particular the conclusion applies if X is a regular, con-nected, strongly Choquet space with at least two points. Proof.
Obviously the second sentence of the proposition follows from the first, sincea connected space with at least two points cannot have isolated points. To provethe claim in the first sentence we let X satisfy the hypotheses. Let Σ be a winningstrategy for Player II. Let { , } < ω denote the tree of functions s ∶ n → { , } where n ∈ ω . For s ∈ { , } < ω we let ∣ s ∣ denote the cardinality of the domain of s , and for i ∈ { , } let s ⌢ i denote the function t ∶ ∣ s ∣ + → { , } such that t ↾ ∣ s ∣ = s and t (∣ s ∣) = i .We define a scheme of open sets labeled by elements of 2 < ω by induction. For ∣ s ∣ = U s = X . Assuming that we have already defined U s for all ∣ s ∣ = n we let U s ⊇ V s be the move dictated by Σ under the partial run of the game ( U s ↾ , V s ↾ , U s ↾ , . . . , U s ) . As V s ≠ ∅ we select disjoint nonempty open sets U s ⌢ , U s ⌢ ⊆ V s .For each element σ of the Cantor set { , } ω we see that ( U σ ↾ , V σ ↾ , . . . ) is a runof the game with Player II employing strategy Σ, and so ⋂ n ∈ ω U σ ↾ n ≠ ∅ , and fordistinct σ ∈ { , } ω these sets are disjoint. (cid:3) The relevance of the remaining material in this section will be made clear inSection 3. Given a collection S of subsets of a set X , the collection B = { X } ∪ {∩ J ∣ J ⊆ S , ∣ J ∣ < ∞} is a basis for a topology, which we denote τ ( S ) , on X . The topology τ ( S ) is the smallest (under inclusion) topology on X under which all elements of S are open. Definition 2.10. If X is a topological space whose topology equals τ ( S ) then wecall S a subbasis for the topology on X .It is easy to see that if x , x ∈ X and for all S ∈ S we have x ∈ S implies x ∈ S then for all open sets O ∈ τ ( S ) we have x ∈ O implies x ∈ O . Lemma 2.11.
Suppose X is a connected topological space. Letˆ X = {( x, i, j ) ∈ X × { , } ∣ i ≠ ∨ j ≠ } Let O = {( x, i, j ) ∈ ˆ X ∣ i = } and O = {( x, i, j ) ∈ ˆ X ∣ j = } . Endowing ˆ X withthe topology given by subbasis {( x, i, j ) ∈ ˆ X ∣ x ∈ O } O ⊆ X open ∪ { O , O } , the spaceˆ X is connected. Proof.
Let ˆ X = ˆ X ∖( O ∪ O ) . It is fairly easy to see that each of ˆ X , O , and O is homeomorphic to X (using projection to the X coordinate). Particularly eachof ˆ X , O , and O is a connected subspace of ˆ X . Notice as well that any open setcontaining an element ( x, , ) ∈ ˆ X must also contain the elements ( x, , ) ∈ O and ( x, , ) ∈ O .If ˆ X = V ⊔ V is a disconnection, we suppose without loss of generality that V ∩ ˆ X ≠ ∅ . As ˆ X is connected we know ˆ X ⊆ V , and therefore O , O ⊆ V .Thus ˆ X = ˆ X ∪ O ∪ O ⊆ V and we are done. (cid:3) Lemma 2.12.
Suppose X is a connected topological space and W , W ⊆ X areopen with W ∩ W ≠ ∅ (we allow W = W ). Letˆ X = {( x, i ) ∈ X × { , } ∣ i = ⇒ x ∈ W ∩ W } SAMUEL M. CORSON
Let O = {( x, i ) ∈ ˆ X ∣ i = } . Endowing ˆ X with the topology given by subbasis {( x, i ) ∈ ˆ X ∣ x ∈ O } O ⊆ X open ∪ { O } , the space ˆ X is connected. Proof.
Notice that the set ˆ X = ˆ X ∖ O is homeomorphic to X (using projectionto the X coordinate). Thus ˆ X is connected. Any open set containing an element ( x, ) ∈ ( W ∩ W ) × { } will also contain ( x, ) . If ˆ X = V ⊔ V is a disconnection,without loss of generality V ∩ ˆ X ≠ ∅ , we have V ⊇ ˆ X , and therefore V ⊇ O aswell. Therefore ˆ X = ˆ X ∪ O = V . (cid:3) Lemma 2.13.
Suppose X is a connected topological space and W ⊆ X is anonempty open subset. Letˆ X = {( x, i, j ) ∈ X × { , } ∣ [ x ∈ W ∨ i = ∨ j = ] ∧ [ i = j = ⇒ x ∈ W ]} Let O = {( x, i, j ) ∈ ˆ X ∣ i = } and O = {( x, i, j ) ∈ ˆ X ∣ j = } . Endowing ˆ X with thetopology given by subbasis {( x, i, j ) ∈ ˆ X ∣ x ∈ O } O ⊆ X open ∪ { O , O } , the space ˆ X is connected. Proof.
Notice that the subspaces ˆ X = {( x, i, j ) ∣ i = , j = } and ˆ X = {( x, i, j ) ∣ i = , j = } are each homeomorphic to X (using projection to the X coordinate). Thusˆ X and ˆ X are connected. Also, if x ∈ W then any open set containing ( x, , ) must also contain ( x, i, j ) for all choices i, j ∈ { , } .Suppose ˆ X = V ⊔ V is a disconnection. Let x ∈ W and suppose without loss ofgenerality that ( x, , ) ∈ V . Then ( x, i, j ) ∈ V for all choices i, j ∈ { , } . As ˆ X isconnected we have ˆ X ⊆ V , and similarly ˆ X ⊆ V . Then for any x ′ ∈ W we have ( x ′ , i, j ) ∈ V for i, j ∈ V . Then ˆ X = ˆ X ∪ ˆ X ∪ {( x, i, j ) ∈ ˆ X ∣ x ∈ W } ⊆ V and weare done. (cid:3) Lemma 2.14.
Suppose X is a connected topological space and W ⊆ X is anonempty open set. Letˆ X = {( x, i, j ) ∈ X × { , } ∣ [ x ∈ W ∨ j = ] ∧ [ i = ⇒ [ x ∈ W ∧ j = ]]} Let O = {( x, i, j ) ∈ ˆ X ∣ i = } and O = {( x, i, j ) ∈ ˆ X ∣ j = } . Endowing ˆ X with thetopology given by subbasis {( x, i, j ) ∈ ˆ X ∣ x ∈ O } O ⊆ X open ∪ { O , O } , the space ˆ X is connected. Proof.
Notice that O is homeomorphic to X (using projection to the first coordi-nate) and therefore O is connected. Given x ∈ X , any open set containing ( x, , ) will also contain ( x, , ) . Given x ∈ W , any open set containing ( x, , ) will alsocontain ( x, , ) . Suppose ˆ X = V ⊔ V is a disconnection. As W is nonempty weselect x ∈ W . If, without loss of generality, ( x, , ) ∈ V then ( x, , ) ∈ V , and since O is connected we have O ⊆ V . Then for arbitrary x ′ ∈ W we have ( x ′ , , ) ∈ V as well. Then W × { } ∪ O ⊆ V , and as W × { } ⊆ V we have W × { } × { } ⊆ V .Then ˆ X ⊆ V . (cid:3) Model Theory Preliminaries
In the course of our construction we will consider first-order logical statementsregarding unary relations. More specifically we’ll construct an uncountable I Rel ⊆ ℵ which will index a collection { B α } α ∈ I Rel of unary relations. A collection of first-order sentences { Θ α } α ∈ I Log , indexed by a subset I Log ⊆ I Rel , will be constructedsimultaneously. Each of the B α will be first introduced in some sentence Θ α ′ , and BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 7 each Θ α ′ will introduce one or more new relations B α which has not yet beenmentioned. These occur under one of the following circumstances.(a) Θ α ≡ [ B α ∩ B α = ∅ ] ∧ [ B α ≠ ∅ ] ∧ [ B α ≠ ∅ ] , where α = α + B α nor B α has yet occurred in a sentence Θ α ′ with α ′ ∈ I Log ∩ α .(b) Θ α ≡ [ B α ∩ B α ⊇ B α ] ∧ [ B α ≠ ∅ ] , with α , α < α , and B α has alreadyoccurred in some sentence Θ α ′ with α ′ ∈ I Log ∩ α , and B α has already occurredin some sentence Θ α ′′ with α ′′ ∈ I Log ∩ α , and B α has not yet occurred in anysentence Θ α ′′′ with α ′′′ ∈ I Log ∩ α .(c) Θ α ≡ [ B α ∩ B α ⊆ B α ] ∧ ( ∀ x )[ x ∈ B α ∪ B α ∪ B α ] ∧ [ B α ≠ ∅ ] ∧ [ B α ≠ ∅ ] ,with α + = α , and α < α , and B α has already occurred in some Θ α ′ with α ′ ∈ I Log ∩ α , and neither B α nor B α has occurred in any sentence Θ α ′′ with α ′′ ∈ I Log ∩ α .(d) Θ α ≡ ( ∀ x )[ x ∈ B α ∪ B α ] ∧ [ B α ⊆ B α ] ∧ [ B α ∩ B α = ∅ ] ∧ [ B α ≠ ∅ ] ∧ [ B α ≠ ∅ ] ,with α = α + α < α , and B α has already occurred in a sentence Θ α ′ with α ′ ∈ I Log ∩ α , and neither B α nor B α has occurred in a sentence Θ α ′′ with α ′′ ∈ I Log ∩ α .Consider the map h ∶ I Rel → I Log with h ( α ) being the smallest ordinal such that B α occurs in Θ h ( α ) . It is clear that α ≤ h ( α ) ≤ α +
1. We will let I Log ,β = I Log ∩ β ,Γ β = { Θ α } α ∈ I Log ,β and I Rel ,β = { α ∈ I Rel ∣ h ( α ) < β } . We will say that Γ satisfies † , or that the subset Γ β satisfies † , provided that each sentence is of form (a), (b),(c), or (d) and that the order on the set of relations and the order on the set ofsentences interact according to the conditions specified in each of (a), (b), (c) or(d). Remark 3.1.
Suppose that a structure ( χ, { ℶ α } α ∈ I Rel ,β ) models B α ≠ ∅ for each α ∈ I Rel ,β and that Γ β satisfies † . If it is not a model of Γ β then there exists x ∈ χ which witnesses this. More particularly suppose Θ α is violated. In case Θ α is of type(a) then ( ∃ x ∈ χ )[ x ∈ ℶ α ∩ℶ α ] . In case of type (b), ( ∃ x ∈ χ )[ x ∈ ℶ α ∖ ( ℶ α ∩ℶ α )] .In case of type (c), ( ∃ x ∈ χ )[[ x ∈ ( ℶ α ∩ ℶ α ) ∖ ℶ α ] ∨ [ x ∉ ℶ α ∪ ℶ α ∪ ℶ α ]] . In caseof type (d), ( ∃ x ∈ χ )[[ x ∉ ℶ α ∪ ℶ α ] ∨ [ x ∈ ℶ α ∖ ℶ α ] ∨ [ x ∈ ℶ α ∩ ℶ α ]] . Lemma 3.2.
Suppose that a structure ( χ, { ℶ α } α ∈ I Rel ,β ) models Γ β , which itselfsatisfies † . Suppose that ( χ ∗ , { ℶ ∗ α } α ∈ I Rel ,β ) is such that χ ∗ ⊇ χ and ℶ ∗ α ⊇ ℶ α . Alsosuppose that for each x ∗ ∈ χ ∗ there exists a model ( χ ∗∗ , { ℶ ∗∗ α } α ∈ I Rel ,β ) of Γ β andpoint x ∗∗ ∈ χ ∗∗ such that for all α ∈ I Rel ,β we have x ∗∗ ∈ ℶ ∗∗ α if and only if x ∗ ∈ ℶ ∗ α .Then ( χ ∗ , { ℶ ∗ α } α ∈ I Rel ,β ) models Γ β . Proof.
Assume the hypotheses. Notice that ( χ ∗ , { ℶ ∗ α } α ∈ I Rel ,β ) models B α ≠ ∅ forall α ∈ I Rel ,β , since ℶ ∗ α ⊇ ℶ α and ( χ, { ℶ α } α ∈ I Rel ,β ) models B α ≠ ∅ . By Remark 3.1, if ( χ ∗ , { ℶ ∗ α } α ∈ I Rel ,β ) fails to model Γ β then there are x ∗ ∈ χ ∗ and α ′ ∈ I Log ,β such that x witnesses the failure of Θ α ′ . Select a model ( χ ∗∗ , { ℶ ∗∗ α } α ∈ I Rel ,β ) of Γ β and point x ∗∗ ∈ χ ∗∗ such that for all α ∈ I Rel ,β we have x ∗∗ ∈ ℶ ∗∗ α if and only if x ∗ ∈ ℶ ∗ α . Then x ∗∗ witnesses the failure of Θ α ′ in ( χ ∗∗ , { ℶ ∗∗ α } α ∈ I Rel ,β ) , a contradiction. (cid:3) Construction 3.3.
Suppose that 0 ≤ β ≤ ℵ and Γ β satisfies † . Consider the subset X β ⊆ { , } I Rel ,β given by σ ∈ X β if and only if there exists a model ( χ, { ℶ α } α ∈ I Rel ,β ) of Γ β and x ∈ χ such that σ ( α ) = ⇔ x ∈ ℶ α . Given Y, Z ⊆ I Rel ,β we let X β,Y,Z = { σ ∈ X β ∣ σ − ( ) ⊇ Y and σ − ( ) ⊇ Z } . For each α ∈ I Rel ,β we let b βα = X β, ∅ , { α } . Lemma 3.4.
Let Γ β satisfy † . Then Γ β is consistent if and only if ( X β , { b βα } α ∈ I Rel ,β ) models Γ β . SAMUEL M. CORSON
Proof.
Certainly the reverse direction is clear. For the forward direction we supposethat Γ β is consistent. Let ( χ, { ℶ α } α ∈ I Rel ,β ) be a model. For each α ∈ I Rel ,β we select x α ∈ ℶ α and let σ α ∈ X β be such that σ α ( α ) = ⇔ x α ∈ ℶ α . For each α ∈ I Rel ,β we see that σ α ∈ b α β , and so ( X β , { b βα } α ∈ I Rel ,β ) models B α ≠ ∅ for each α ∈ I Rel ,β .Thus by Remark 3.1 we know that if ( X β , { b βα } α ∈ I Rel ,β ) fails to model Γ β thenthere is some point witnessing this. For example, if Θ α is false and of type (a)then we select σ ∈ X β for which σ ∈ b βα ∩ b βα , in the notation of type (a). Let ( χ ∗ , { ℶ ∗ α } α ∈ I Rel ,β ) be a model with x ∗ ∈ χ ∗ such that σ ( α ) = ⇔ x ∗ ∈ ℶ ∗ α . Then x ∗ witnesses that Θ α fails in ( χ ∗ , { ℶ ∗ α } α ∈ I Rel ,β ) , a contradiction. The comparablearguments are clear in types (b), (c), and (d). Thus ( X β , { b βα } α ∈ I Rel ,β ) is indeed amodel of Γ β . (cid:3) Remark 3.5.
Let Γ β satisfy † . Notice that if I Log ,β = ∅ then Γ β = ∅ = I Rel ,β . Inparticular we have X β = { , } ∅ consists of a single element, the empty functionfrom ∅ to { , } , and X β is a nonempty model of Γ β .In case Γ β is inconsistent we have I Log ,β ≠ ∅ and X β = ∅ . If Γ β is consistentthen either Γ β = ∅ , in which case X β ≠ ∅ as we have seen, or Γ β ≠ ∅ in which casewe are satisfying some sentence of type (a), (b), (c), or (d) and therefore X β ≠ ∅ .Thus X β is empty if and only if Γ β is inconsistent. Lemma 3.6.
Let Γ β satisfy † and suppose that Γ β is consistent. For each σ ∈ X β and β ≤ β we have σ ↾ I Rel ,β ∈ X β . Proof.
Given σ ∈ X β we select a model ( χ, { ℶ α } α ∈ I Rel ,β ) and x ∈ χ such that x ∈ ℶ α ⇔ σ ( α ) =
1. Notice that the reduct ( χ, { ℶ α } α ∈ I Rel ,β ) models Γ β , and so σ ↾ I Rel ,β ∈ X β follows. (cid:3) Lemma 3.7.
Let Γ β satisfy † and suppose that β ∈ I Log ,β and that Θ β is of type(a). Then Γ β + is consistent if and only if Γ β is, and in either case we have X β + = {( σ, i, j ) ∈ X β × { , } I Rel ,β + ∖ I Rel ,β ∣ i = ∨ j = } Proof.
Let J a denote the set on the right-hand-side. If Γ β is inconsistent then X β + = ∅ = X β by Remark 3.5 and clearly J a = ∅ as well. If Γ β is consistent then X β ≠ ∅ . For each α ∈ I Rel ,β + we let ℶ α ⊆ J a be the set { ρ ∈ J a ∣ ρ ( α ) = } .We claim that ( J a , { ℶ α } α ∈ I Rel ,β + ) is a model of Γ β + . To see this, first noticethat B α ≠ ∅ holds for all α ∈ I Rel ,β + , for if α ∈ I β we have some σ ∈ b β α and ( σ, , ) ∈ ℶ α , and for α ∈ I Rel ,β + ∖ I Rel ,β we take σ ∈ X β (since this is nonempty)and notice that ( σ, , ) ∈ ℶ α and ( σ, , ) ∈ ℶ α . Next, it is clear that J a modelsΘ β by how it is defined. If J a fails to model some Θ α for α ∈ I Rel ,β then we havesome ρ ∈ J a as in Remark 3.1, and it is easy to see that the restriction ρ ↾ I Rel ,β witnesses that Θ α is false in X β , a contradiction.Since ( J a , { ℶ α } α ∈ I Rel ,β + ) is a model of Γ β + we know in particular that Γ β + isconsistent. Thus by Lemma 3.6 it is clear that X β + ⊆ J a . As ( J a , { ℶ α } α ∈ I Rel ,β + ) is a model of Γ β + we have in fact J a ⊆ X β + (by how X β + is defined) and so J a = X β + . (cid:3) Lemma 3.8.
Let Γ β satisfy † and suppose that β ∈ I Log ,β and that Θ β is of type(b). Then Γ β + is consistent if and only if both Γ β is consistent and b β α ∩ b β α ≠ ∅ .If Γ β + is consistent then X β + = {( σ, i ) ∈ X β × { , } I Rel ,β + ∖ I Rel ,β ∣ i = ⇒ [ σ ( α ) = σ ( α ) = ]} BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 9
Proof.
Suppose Γ β + is consistent. Then certainly Γ β must be consistent as well.Suppose for contradiction that b β α ∩ b β α = ∅ . We have ( X β + , { b β + α } α ∈ I Rel ,β + ) as a model of Γ β + by Lemma 3.4. As Θ β asserts that B α ≠ ∅ , we have some σ ∈ b β + α ⊆ b β + α ∩ b β + α , but now σ ↾ I Rel ,β ∈ X β by Lemma 3.6 and this is anelement of b β α ∩ b β α , a contradiction. Thus the forward directin of the first sentenceis established.Now suppose that Γ β is consistent and that b β α ∩ b β α ≠ ∅ . We let J b denotethe set {( σ, i ) ∈ X β × { , } I Rel ,β + ∖ I Rel ,β ∣ i = ⇒ [ σ ( α ) = σ ( α ) = ]} and ℶ α = { ρ ∈ J a ∣ ρ ( α ) = } for each α ∈ I Rel ,β + . We claim that ( J b , { ℶ α } α ∈ I Rel ,β + ) is a model of Γ β + . That B α ≠ ∅ is satisfied for α ∈ I Rel ,β is quite clear. Since b β α ∩ b β α ≠ ∅ we select σ in this intersection and notice that ( σ, , ) ∈ ℶ α ∩ ℶ α , so B β ≠ ∅ holds as well. If Γ β + fails in ( J b , { ℶ α } α ∈ I Rel ,β + ) then by Remark 3.1 thereis a ρ ∈ J b witnessing this. Arguing as in Lemma 3.7 we obtain a contradiction.Thus ( J b , { ℶ α } α ∈ I Rel ,β + ) ⊧ Γ β + .Now we have established that Γ β + is consistent (the reverse direction of the firstsentence). We also notice by Lemma 3.6 that X β + ⊆ J b . As ( J b , { ℶ α } α ∈ I Rel ,β + ) is a model of Γ β + we have in fact J b ⊆ X β + , by how X β is defined, and so J b = X β + . (cid:3) Lemma 3.9.
Let Γ β satisfy † and suppose that β ∈ I Log ,β and that Θ β is of type(c). Then Γ β + is consistent if and only if Γ β is, and in either case we have X β + = {( σ, i, j ) ∈ X β × { , } I Rel ,β + ∖ I Rel ,β ∣ [ σ ( α ) = ∨ i = ∨ j = ] ∧ [ i = j = ⇒ σ ( α ) = ]} Proof.
Let J c denote the set on the right-hand-side. If Γ β is inconsistent then X β + = ∅ = X β and so J c = ∅ . If Γ β is consistent then X β ≠ ∅ . Define ℶ α = { ρ ∈ J c ∣ ρ ( α ) = } . We claim that ( J c , { ℶ α } α ∈ I Rel ,β + ) ⊧ Γ β + . For each α ∈ I Rel ,β wehave ( J c , { ℶ α } α ∈ I Rel ,β + ) ⊧ B α ≠ ∅ , since we select σ ∈ b β α and notice that ( σ, , ) ∈ℶ α . For α , α ∈ I β + ∖ I β we select σ ∈ b α and see that ( σ, , ) ∈ ℶ α ∩ ℶ α . Thus ( J c , { ℶ α } α ∈ I Rel ,β + ) ⊧ B α ≠ ∅ for all α ∈ I Rel ,β + .Certainly ( J c , { ℶ α } α ∈ I Rel ,β + ) ⊧ Θ β , simply by how J c is defined, and in casean earlier Θ α fails we obtain a witness ρ and argue as in Lemma 3.7. Thus ( J c , { ℶ α } α ∈ I Rel ,β + ) ⊧ Γ β + and J c = X β + follows. (cid:3) Lemma 3.10.
Let Γ β satisfy † and suppose that β ∈ I Log ,β and that Θ β is oftype (d). Then Γ β + is consistent if and only if Γ β is, and in either case we have X β + = {( σ, i, j ) ∈ X β × { , } I Rel ,β + ∖ I Rel ,β ∣ [ σ ( α ) = ∨ j = ] ∧ [ i = ⇒ [ σ ( α ) = ∧ j = ]]} Proof.
The argument, which we omit, goes more-or-less as in Lemmas 3.7 and3.9. (cid:3)
Lemma 3.11.
Let Γ β satisfy † . If I Log ,β has no largest element, then X β is equalto { σ ∈ { , } I Rel ,β ∣ ( ∀ β ∈ I Log ,β )[ σ ↾ I Rel ,β ∈ X β ]} Proof.
Call the set in question J L and let ℶ α = { σ ∈ J L ∣ σ ( α ) = } . If Γ β isinconsistent then Γ β is inconsistent for some β ∈ I Log ,β , so that X β = X β = ∅ = J L . Suppose that Γ β is consistent. Then ( X β , { b β α } α ∈ I Rel ,β ) ⊧ Γ β by Lemma 3.4.Moreover, by Lemma 3.6 we have J L ⊇ X β . This implies that ( J L , { ℶ α } α ∈ I Rel ,β ) ⊧ B α ≠ ∅ for each α ∈ I Rel ,β . If ( J L , { ℶ α } α ∈ I Rel ,β ) fails to model Γ β then we have a witness σ as in Remark3.1. If σ witnesses that Θ α is violated in J L then we select β ∈ I Log ,β largeenough that α ∈ I Log ,β and notice that σ ↾ I Rel ,β witnesses that Θ α fails in X β , acontradiction. Thus ( J L , { ℶ α } α ∈ I Rel ,β ) ⊧ Γ β , and by how X β is defined we obtain J L ⊆ X β and so J L = X β . (cid:3) Lemma 3.12.
Let Γ β satisfy † and suppose that Γ β is consistent and 0 ≤ β ≤ β .For each σ ∈ X β there exists ρ ∈ X β such that ρ ↾ I Rel ,β = σ . Proof.
Let σ ∈ X β be given and let σ β = σ . Suppose that we have defined σ β forall β ≤ β < β ≤ β such that if β ≤ β ≤ β < β we have σ β ↾ I Rel ,β = σ β .Firstly, if I Rel ,β = I Rel ,β for some β < β then we let σ β = σ β . Suppose that I Rel ,β = I Rel ,β for all β < β . If β = β + β ∈ I Log ,β and X β is computedfrom X β by one of Lemma 3.7, 3.8, 3.9, or 3.10 and in each case we see that wecan produce such a σ β . If β is a limit then either I Rel ,β = ∅ , in which case we let σ β be the unique element in X β , or β = sup I Rel ,β and we let σ β ( α ) ⎧⎪⎪⎨⎪⎪⎩ σ β ( α ) = β ∈ I Rel ,β large enough that α ∈ I Rel ,β σ β ( α ) = β ∈ I Rel ,β large enough that α ∈ I Rel ,β and σ β ∈ X β by Lemma 3.12. (cid:3) Lemma 3.13.
Let Γ β satisfy † . The set X β is compact as a subspace of { , } I Rel ,β under the product topology. Proof.
If Γ β is inconsistent then X β = ∅ , which is compact. Otherwise, supposingthat σ ∈ { , } I Rel ,β ∖ X β we let J = X β ∪ { σ } and define ℶ α = { σ ∈ J ∣ σ ( α ) = } for each α ∈ I Rel ,β . As J ⊇ X β we see that ( J, { ℶ α } α ∈ I Rel ,β ) ⊧ B α ≠ ∅ for each α ∈ I Rel ,β . Since J ≠ X β we know that ( J, { ℶ α } α ∈ I Rel ,β ) is not a model of Γ β (byhow X β is defined). By Remark 3.1 we have a witness for this, and it is easy tosee that this witness is σ . Thus if Θ α , say of type (b), is violated we have that { σ ∈ { , } I Rel ,β ∣ σ ↾ { α , α , α } = σ ↾ { α , α , α }} ⊆ { , } I Rel ,β ∖ X β . The othercases in types (a), (c), (d) are comparable. Thus X β is closed in { , } I Rel ,β , and asthe latter space is compact we are finished. (cid:3) Lemma 3.14.
Let Γ β satisfy † . The topological space ( X β , τ ({ b βα } α ∈ I Rel ,β )) iscompact and connected. Proof.
If Γ β is inconsistent then X β is empty, and therefore compact and connected.Suppose that Γ β is consistent. The space ( X β , τ ({ b βα } α ∈ I Rel ,β )) is compact because X β is compact as a subspace of { , } I Rel ,β under the product topology, and each b βα is the intersection of an open subset in { , } I Rel ,β with X β .We prove connectedness by induction on 0 ≤ β ≤ β . For β =
0, or more generallyif I Log ,β = ∅ , then X β is a single point, and so the space is necessarily connected.Suppose that X β is connected for all β < β . Suppose further that I Log ,β hasa maximal element, β . If β = β +
1, then Θ β is of type (a), (b), (c), or (d),and so X β is defined from X β by Lemma 3.7, 3.8, 3.9, or 3.10 respectively. Thenwe conclude that X β is connected by Lemma 2.11, 2.12, 2.13, or 2.14 respectively.Otherwise we have X β = X β + .Suppose that I Log ,β is nonempty and does not have a maximal element. Sup-pose for contradiction that X β = V ⊔ V is a nontrivial disconnection. Let V = BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 11 ⋃ j ∈ J ( b β α ,j ∩ ⋯ ∩ b β α nj,j ) and similarly V = ⋃ j ∈ J ( b β α ,j ∩ ⋯ ∩ b β α nj,j ) . Since X β is compact we may assume that both J and J are finite. Select β < β whichis large enough that α i,j ∈ I Rel ,β for all j ∈ J ∪ J and 0 ≤ i ≤ n j . Let V β =⋃ j ∈ J ( b β α ,j ∩ ⋯ ∩ b β α nj,j ) and V β = ⋃ j ∈ J ( b β α ,j ∩ ⋯ ∩ b β α nj,j ) .Notice that neither V β nor V β is empty, for if, say, σ ∈ b β α ,j ∩⋯∩ b β α nj,j for some j ∈ J , then σ ↾ I Rel ,β ∈ b β α ,j ∩ ⋯ ∩ b β α nj,j ⊆ V β , and a similar argument applies for V β .The sets V β and V β are obviously open in X β . If σ ∈ X β ∖ ( V β ∪ V β ) then byLemma 3.12 select ρ ∈ X β such that ρ ↾ I Rel ,β = σ , but clearly ρ ∈ X β ∖ ( V ∪ V ) ,which cannot be. If σ ∈ V β ∩ V β then again we select ρ ∈ X β with ρ ↾ I Rel ,β = σ and it is easy to see that ρ ∈ V ∩ V , which is also impossible. Thus X β = V β ⊔ V β is a nontrivial disconnection, contradicting the induction assumption. (cid:3) Definitions 3.15.
Let Γ β satisfy † . Given a model ( χ, { ℶ α } α ∈ I Rel ,β ) of Γ β thereis a function Coor β ∶ χ → X β (which we’ll call the coordinates function ) given byCoor β ( x ) = σ , where σ ( α ) = x ∈ ℶ α . We’ll say that a model ( χ, { ℶ α } α ∈ I Rel ,β ) of Γ β is Boolean saturated if for every finite
Y, Z ⊆ I Rel ,β such that X β,Y,Z ≠ ∅ the set Coor − β ( X β,Y,Z ) is infinite. Lemma 3.16.
Let Γ β satisfy † . The coordinates function Coor β ∶ χ → X β iscontinuous, where χ is given the topology τ ({ ℶ α } α ∈ I Rel ,β ) and X β is given thetopology τ ({ b βα } α ∈ I Rel ,β ) . The function Coor β is an embedding if injective. Theimage Coor β ( χ ) is dense in X β if χ is Boolean saturated. Proof.
For the first two sentences we have x ∈ ℶ α ∩⋯∩ℶ α n if and only if Coor β ( x ) ∈ b βα ∩ ⋯ ∩ b βα n , and this is sufficient. For the third sentence, if χ is Boolean saturatedthen Coor β ( χ ) is dense in the set X β under the subspace topology inherited by { , } I Rel ,β , and since this topology is finer than τ ({ b βα } α ∈ I Rel ,β ) we are done. (cid:3) Lemma 3.17.
Let β < ℵ and Γ β satisfy † . Suppose Γ β is consistent and that β is the largest element in I Log ,β . Let ( χ, { ℶ α } α ∈ I Rel ,β ) be a Boolean saturatedmodel of Γ β .If Θ β is of type (a) and x , x ∈ χ are distinct, we can define ℶ α , ℶ α ⊆ χ so that ( χ, { ℶ α } α ∈ I Rel ,β ) is a Boolean saturated model of Γ β and x ∈ ℶ α and x ∈ ℶ α . Proof.
Let {( Y n , Z n )} n ∈ ω be an enumeration of all ordered pairs ( Y, Z ) with Y, Z ⊆ I Rel ,β finite and X γ ,Y,Z ≠ ∅ , and such that for each n ∈ ω there are infinitely many n ′ in each of 3 ω , 3 ω +
1, and 3 ω + ( Y n , Z n ) = ( Y n ′ , Z n ′ ) . We’ll inductivelyconstruct a sequence of finite sets F − ⊆ F ⊆ F ⊆ ⋯ as well as assign points to ℶ α , ℶ α , χ ∖ ( ℶ α ∪ ℶ α ) .Let x ∈ ℶ α and x ∈ ℶ α and let F − = { x , x } . If n ≥ F n − we select x ∈ Coor − β ( X Y n ,Z n ,β ) ∖ F n − and assign x in the followingway: ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ x ∈ ℶ α if n ∈ ωx ∈ ℶ α if n ∈ ω + x ∈ χ ∖ ( ℶ α ∪ ℶ α ) if n ∈ ω + and let F n = F n − ∪ { x } . For x ∈ χ ∖ ⋃ n ∈ ω F n we let x ∈ ℶ α . The check that ( χ, { ℶ α } α ∈ I Rel ,β ) is a Boolean saturated model of Γ β is straightforward. (cid:3) Lemma 3.18.
Let β < ℵ and Γ β satisfy † . Suppose Γ β is consistent and that β is the largest element in I Log ,β . Let ( χ, { ℶ α } α ∈ I Rel ,β ) be a Boolean saturatedmodel of Γ β .If Θ β is of type (b) and x ∈ ℶ α ∩ ℶ α , we can define ℶ α ⊆ χ so that ( χ, { ℶ α } α ∈ I Rel ,β ) is a Boolean saturated model of Γ β and x ∈ ℶ α . Proof.
Let {( Y n , Z n )} n ∈ ω be an enumeration of all ordered pairs ( Y, Z ) with Y, Z ⊆ I Rel ,β finite and X γ ,Y,Z ≠ ∅ , and such that for each n ∈ ω there are infinitelymany n ′ in each of 2 ω and 2 ω + ( Y n , Z n ) = ( Y n ′ , Z n ′ ) . We’ll inductivelyconstruct a sequence of finite sets F − ⊆ F ⊆ F ⊆ ⋯ as well as assign points to ℶ α and to χ ∖ ℶ α .Let x ∈ ℶ α and let F − = { x } . If n ≥ F n − weselect x ∈ Coor − β ( X β ,Y n ,Z n ) ∖ F n − and assign x in the following way: ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ x ∈ ℶ α if x ∈ ℶ α ∩ ℶ α and n ∈ ωx ∈ χ ∖ ℶ α if x ∈ ℶ α ∩ ℶ α and n ∈ ω + x ∈ χ ∖ ℶ α if x ∉ ℶ α ∩ ℶ α and let F n = F n − ∪ { x } . Assign x ∈ χ ∖ ⋃ n ∈ ω F n by the rule ⎧⎪⎪⎨⎪⎪⎩ x ∈ ℶ α if x ∈ ℶ α ∩ ℶ α x ∈ χ ∖ ℶ α otherwiseAgain the check that the conclusion is fulfilled is left to the reader. (cid:3) Lemma 3.19.
Let β < ℵ and Γ β satisfy † . Suppose Γ β is consistent and that β is the largest element in I Log ,β . Let ( χ, { ℶ α } α ∈ I Rel ,β ) be a Boolean saturatedmodel of Γ β .If Θ β is of type (c) and x , x are distinct points in χ ∖ ℶ α , we can define ℶ α , ℶ α ⊆ χ so that ( χ, { ℶ α } α ∈ I Rel ,β ) is a Boolean saturated model of Γ β and x ∈ ℶ α and x ∈ ℶ α . Proof.
Let {( Y n , Z n )} n ∈ ω be an enumeration of all ordered pairs ( Y, Z ) with Y, Z ⊆ I Rel ,β finite and X β ,Y,Z ≠ ∅ , and such that for each n ∈ ω there are infinitely many n ′ in each of 4 ω , 4 ω +
1, 4 ω +
2, and 4 ω + ( Y n , Z n ) = ( Y n ′ , Z n ′ ) . We’llinductively construct a sequence of finite sets F − ⊆ F ⊆ F ⊆ ⋯ as well as assignpoints to ℶ α ∖ ℶ α , ℶ α ∖ ℶ α , ℶ α ∩ ℶ α and χ ∖ ( ℶ α ∪ ℶ α ) .Let x ∈ ℶ α ∖ ℶ α , x ∈ ℶ α ∖ ℶ α and set F − = { x , x } . If n ≥ x ∈ Coor − β ( X β ,Y n ,Z n ) ∖ F n − and assign ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ x ∈ ℶ α ∖ ℶ α if x ∉ ℶ α and n ∈ ωx ∈ ℶ α ∖ ℶ α if x ∉ ℶ α and n ∈ ω + x ∈ ℶ α ∩ ℶ α if x ∈ ℶ α and n ∈ ωx ∈ ℶ α ∖ ℶ α if x ∈ ℶ α and n ∈ ω + x ∈ ℶ α ∖ ℶ α if x ∈ ℶ α and n ∈ ω + x ∈ χ ∖ ( ℶ α ∪ ℶ α ) if x ∈ ℶ α and n ∈ ω + F n = F n − ∪ { x } . Assign x ∈ χ ∖ ⋃ n ∈ ω F n by x ∈ ℶ α ∖ ℶ α . (cid:3) BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 13
Lemma 3.20.
Let β < ℵ and Γ β satisfy † . Suppose Γ β is consistent and that β is the largest element in I Log ,β . Let ( χ, { ℶ α } α ∈ I Rel ,β ) be a Boolean saturatedmodel of Γ β .If Θ β is of type (d) and x ∈ ℶ α , we can define ℶ α , ℶ α ⊆ χ so that ( χ, { ℶ α } α ∈ I Rel ,β ) is a Boolean saturated model of Γ β and x ∈ ℶ α . Proof.
Let {( Y n , Z n )} n ∈ ω be an enumeration of all ordered pairs ( Y, Z ) with Y, Z ⊆ I Rel ,β finite and X β ,Y,Z ≠ ∅ , and such that for each n ∈ ω there are infinitely many n ′ in each of 3 ω , 3 ω +
1, and 3 ω + ( Y n , Z n ) = ( Y n ′ , Z n ′ ) . We’ll inductivelyconstruct a sequence of finite sets F − ⊆ F ⊆ F ⊆ ⋯ as well as assign points to ℶ α ∖ ℶ α , ℶ α ∖ ℶ α , ℶ α ∩ ℶ α and χ ∖ ( ℶ α ∪ ℶ α ) .Let x ∈ ℶ α and F − = { x } . For n ≥ x ∈ Coor − β ( X β ,Y n ,Z n ) ∖ F n − and assign ⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩ x ∈ χ ∖ ( ℶ α ∪ ℶ α ) if x ∈ ℶ α and n ∈ ωx ∈ ℶ α ∖ ℶ α if x ∈ ℶ α and n ∈ ω + x ∈ ℶ α ∖ ℶ α if x ∈ ℶ α and n ∈ ω + x ∈ ℶ α ∖ ℶ α if x ∉ ℶ α and let F n = F n − ∪ { x } . For x ∈ χ ∖ ⋃ n ∈ ω F n we assign x ∈ ℶ α ∖ ℶ α . (cid:3) Definitions 3.21. (see [2]) A subset E of an ordinal α is bounded in α providedthere exists β < α which is an upper bound on E . A subset C of ℵ is club if itis unbounded in ℵ and closed under the order topology in ℵ . The intersectionof two club sets in ℵ is again a club set. A subset E of ℵ is stationary if it hasnonempty intersection with every club subset of ℵ . The intersection of a club setand a stationary set is again stationary.We quote Jensen’s ♢ principle (see [3, Lemma 6.5] or [2, Theorem 13.21]): ♢ : There exists a sequence { S α } α <ℵ such that S α ⊆ α and for any J ⊆ ℵ the set { α < ℵ ∣ J ∩ α = S α } is stationary in ℵ .The sequence which is claimed in ♢ is often called a ◇ -sequence . Jensen provedthat ♢ holds in G¨odel’s constructible universe L . As a result, if ZFC is consistentthen so is ZFC + ♢ . Remark 3.22.
An obvious consequence of ◇ is that ℵ = ℵ , since the function2 ω → ℵ given by J ↦ min { ω ≤ α < ℵ ∣ S α = J } is injective. Another well knownconsequence is the following: Lemma 3.23. (ZFC + ♢ ) Let T ⊆ ℵ be a club. There exists a sequence {( S α, , S α, )} α ∈ T such that for any pair ( J , J ) of sets such that J , J ⊆ ℵ the set { α ∈ T ∣ α ∩ J = S α, and α ∩ J = S α, } is stationary. Proof.
Give the product ℵ × { , } the lexicographic order (initially comparing leftcoordinates). The unique order isomorphism f ∶ ℵ × { , } → ℵ has f ( α, ) = α foreach limit ordinal α < ℵ . For i ∈ { , } define f i ∶ ℵ → ℵ by f i ( α ) = f ( α, i ) , thus f ( ℵ ) ⊔ f ( ℵ ) = ℵ . Moreover for each limit ordinal α we have f i ( J ∩ α ) = f i ( J ) ∩ α .Let { S α } α <ℵ be a ♢ -sequence on ℵ . Let T ′ ⊆ T be the set of limit ordinals in T ; this is also a club. Let sequence {( S α, , S α, )} α ∈ T ′ be defined by S α,i = f − i ( S α ) .Given J , J ⊆ ℵ we let J = f ( J ) ⊔ f ( J ) . We have { α ∈ T ′ ∣ J ∩ α = S α, and J ∩ α = S α, } = { α ∈ T ′ ∣ f ( J ∩ α ) ⊔ f ( J ∩ α ) = S α } = { α ∈ T ′ ∣ ( f ( J ) ∩ α ) ⊔ ( f ( J ) ∩ α ) = S α } = { α ∈ T ′ ∣ J ∩ α = S α } = T ′ ∩ { α < ℵ ∣ J ∩ α = S α } and this set is stationary as the intersection of a club and a stationary. Letting ( S α, , S α, ) = ( ∅ , ∅ ) for α ∈ T ∖ T ′ we obtain the desired sequence. (cid:3) The Construction
We will be defining collections and sets by induction. These will include sets I Log , I
Rel , I
Conn ⊆ ℵ , a collection Sequ of ordered pairs of strictly increasing ω -sequences, countable collections { B β } β <ℵ of subsets of ℵ , and a collection of firstorder sentences for unary predicates Γ. We’ll explain the interactions among thesesets.The set Γ will be indexed by I Log : Γ = { Θ α } α ∈ I Log and Γ will satisfy † . Theset I Rel will satisfy I Rel ⊇ I Log and will index the set { B α } α ∈ I Rel of unary predicatesymbols which appear in Γ. Each of the statements Θ α which appears in Γ willsatisfy one of (a) - (d) of † , and the interactions with the symbols { B α } α ∈ I Rel willbe as determined in † . Thus, as in Section 3 we let h ∶ I Rel → I Log map α to theminimal α ′ such that B α appears in Θ α ′ . As noted, this will satisfy α ≤ h ( α ) ≤ α + ≤ β < ℵ we let I Log ,β = I Log ∩ β and I Rel ,β = { α ∈ I Rel ∣ h ( α ) < β } ,noting that this will sometimes be a proper subset of I Rel ∩ β .For each 0 ≤ β < ℵ the collection B β will be indexed by I Rel ,β : B β = { B βα } α ∈ I Rel ,β .Importantly, the superscript β in this case is only an index; it is not a set of functionsfrom β to a set B or B α . Moreover B βα ⊆ β will hold for each α ∈ I Rel ,β . Also wewill have B β α = B βα ∩ β whenever α ∈ I Rel ,β and β ≤ β < ℵ .We will let I Conn ,β = I Conn ∩ β . The elements of Sequ will be indexed by I Conn × { , } : Sequ = {({ s α, ,q } q ∈ ω , { s α, ,q } q ∈ ω )} α ∈ I Conn . For 0 ≤ β < ℵ and α ∈ I Conn ,β , both of { s α, ,q } q ∈ ω and { s α, ,q } q ∈ ω will be strictly increasing such that α = sup q ∈ ω s α, ,q = sup q ∈ ω s α, ,q .Now we enumerate the inductive hypotheses explicitly. Let T ⊆ ℵ be the secondderived subset (i.e the set of limit points of limit points in ℵ under the ordertopology). As the set T is club in ℵ we let {( S α, , S α, )} α ∈ T be a sequence as inLemma 3.23. Let T ∗ = T ∪ { } . Each ordinal γ < ℵ may be uniquely written as γ = t + ωl + n where t ∈ T ∗ and l, n ∈ ω . Let R denote the set of ordinals γ < ℵ which are of form γ = t + ω k + n with γ ∉ ω and n ≥
2. As we are assuming ♢ wehave ℵ = ℵ . Thus, let f ∶ R → ⋃ δ <ℵ { , } δ be a function such that each elementof the codomain has uncountable preimage.The following will hold for all 0 ≤ β < ℵ .(i) The set I Log ,β is such that I Log ,β = I Log ,β ∩ β for each β ≤ β . Also, Γ β = { Θ α } α ∈ I Log ,β and satisfies † . Also, Γ β = { Θ α } α < β ⊆ Γ β for β ≤ β .(ii) The set I Rel ,β is such that I Rel ,β = I Rel ,β ∩ β for each β ∈ I Log ,β . Also, thecollection B β = { B βα } α ∈ I Rel ,β is such that B βα ⊆ β , and for β ≤ β and α ∈ I Rel ,β we have B β α = B βα ∩ β .(iii) B β ⊧ Γ β .(iv) If ω ≤ β then ( ω, { ω ∩ B βα } α ∈ I Rel ,β ) is a Boolean saturated model of Γ β . BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 15 (v) The set I Conn ,β is such that I Conn ,β = I Conn ,β ∩ β for each β ≤ β , and I Conn ,β ⊆ T . The set Sequ β consists of ordered pairs of strictly increasingsequences Sequ β = {({ s α, ,q } q ∈ ω , { s α, ,q } q ∈ ω )} α ∈ I Conn ,β such that sup q ∈ ω s α, ,q = α = sup q ∈ ω s α, ,q and { s α, ,q } q ∈ ω , { s α, ,q } q ∈ ω ⊆ α ∖ ( T ∪ ω ) . Also, Sequ β = {({ s α, ,q } q ∈ ω , { s α, ,q } q ∈ ω )} α ∈ I Conn ,β ⊆ Sequ β for each β ≤ β .(vi) For each α ∈ I Conn ,β and α ∈ I Rel ,β we have that α ∈ B βα implies that thereexists N ∈ ω for which { s α , ,q } q ≥ N , { s α , ,q } q ≥ N ⊆ B βα .We’ll consider cases in the construction: N (for “n”atural number), L (for “l”imitordinal), A, B, C, D (corresponding to introduction of logical sentences and basiselements according to (a), (b), (c), (d) in † ), and also Cases E and F. Case N: γ ∈ ω . For γ ∈ ω we let Γ γ = B γ = Sequ γ = I Log ,γ = I Rel ,γ = I Conn ,γ = ∅ . Itis easy to see that (i) - (vi) hold at each such γ .For the remaining cases we suppose that (i) - (vi) holds for all β < γ , where ω ≤ γ < ℵ . Case L: γ is a limit ordinal. Suppose γ is a limit ordinal. We let I Log ,γ =⋃ β < γ I Log ,β and Γ γ = ⋃ β < γ Γ β . Let I Rel ,γ = ⋃ β < γ I Rel ,β . For each α ∈ I Rel ,γ we let B γα = ⋃ β < γ,α ∈ I Rel ,β B βα , and write B γ = { B γα } α ∈ I Rel ,γ . Let I Conn ,γ = ⋃ β < γ I Conn ,β andSequ γ = ⋃ β < γ Sequ β . We’ll check that conditions (i) - (vi) are satisfied.Certainly (i) and (ii) are clear. To see (iii), we let α ∈ I Rel ,γ be given. We have α ∈ I Rel ,β for some β < γ and since B β ⊧ B α ≠ ∅ , we have ∅ ≠ B βα = B γα ∩ β , so thatin particular B γ ⊧ B α ≠ ∅ . Now if B γ ⊧ Γ γ fails, say B γ ⊧ Θ α fails, we know byRemark 3.1 that there is an existential witness for this. For example if Θ α is oftype (a) then we have some x ∈ γ such that x ∈ B γα ∩ B γα , but now we can select β < γ with x ∈ B βα ∩ B βα and α ∈ I Log ,β which contradicts B β ⊧ Θ α . If Θ α is of type(b), (c), or (d) then the check is similar. Thus (iii) holds.For (iv) we consider one of two possibilities. If γ = ω then ( ω, ω ∩ { B γα } α ∈ I Rel ,γ ) = ( ω, ∅ ) and this is certainly a Boolean saturated model of Γ ω = ∅ . Suppose now that γ > ω . By (iii) we know that Γ γ is consistent. The check that ( ω, ω ∩ { B γα } α ∈ I Rel ,γ ) isa model of Γ γ follows along precisely the same lines as the check in (iii). We checkthat this model is Boolean saturated. Let Y, Z ⊆ I Rel ,γ be finite such that X γ,Y,Z ≠∅ . Select ω < β < γ large enough that Y, Z ⊆ I Rel ,β . Since ( ω, { ω ∩ B βα } α ∈ I Rel ,β ) isa Boolean saturated model of Γ β , we use the coordinate map Coor β ∶ ω → X β andhave by assumption that Coor − β ( X β,Y,Z ) is infinite. But letting Coor γ ∶ ω → X γ bethe γ -coordinate map, it is clear that Coor − γ ( X γ,Y,Z ) ⊇ Coor − β ( X β,Y,Z ) . We haveshown (iv).The claim (v) is clear by induction.For (vi) we let α ∈ I Conn ,γ and α ∈ I Rel ,γ and suppose that α ∈ B γα . As γ isa limit ordinal we select α , α < β < γ such that α ∈ I Conn ,β and α ∈ I Rel ,β . Wehave α ∈ B γα ∩ β = B βα , so by inductive hypothesis there exists N ∈ ω such that { s α , ,q } q ≥ N ∪ { s α , ,q } q ≥ N ⊆ B βα ⊆ B γα . This completes Case L. Case A: γ = t + ω ( k + ) + n, n ≠ . Let β = t + ω ( k + ) . By assumption theconditions (i) - (vi) hold at and below β . We will take care of these γ = β + n byinduction on n ≥
1, dealing with n = p + n = p + n ∈ ω + n will be consideredat the same time. For all γ ∈ [ β , β + ω ) we will let I Conn ,γ = I Conn ,β and Sequ γ = Sequ β .Let {( x m , y m )} m ∈ ω be an enumeration of all ordered pairs of distinct points x, y ∈ β . We have n ∈ ω + m = n − β + n ≡ [ B β + n − ∩ B β + n = ∅ ] ∧ [ B β + n − ≠ ∅ ] ∧ [ B β + n ≠ ∅ ] . We note byLemma 3.7 that Γ β + m ∪ { Θ β + n } is consistent. We are assuming by induction that(i) - (vi) hold at and below β + m .We select distinct points x, y ∈ ω , where x = x m if x m ∈ ω and similarly for y and y m . By Lemma 3.17 together with induction condition (iv) there exist J β + m , J β + m + ⊆ ω with x ∈ J β + m and y ∈ J β + m + such that ( ω, { ω ∩ B β + mα } α ∈ I Rel ,β + m ∪ { J β + m , J β + m + }) is a Boolean saturated model of Γ β + m ∪ { Θ β + n } . Subcase A.1: x m , y m ∈ β ∖ I Conn ,β . Notice that ( β + m, { B β + mα } α ∈ I Rel ,β + m ∪ { J β + m ∪ { x m } , J β + m + ∪ { y m }}) is a model of Γ β + m ∪ { Θ β + n } . We let B β + mβ + m = J β + m ∪ { x m } and for α ∈ I Rel ,β + m ∪ { β + m, β + m + } we let B β + m + α = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ B β + mα if α < β + m and x m ∉ B β + mα B β + mα ∪ { β + m } if α < β + m and x m ∈ B β + mα J β + m + ∪ { y m } if α = β + m + B β + m + α = ⎧⎪⎪⎨⎪⎪⎩ B β + m + α if y m ∉ B β + m + α B β + m + α ∪ { β + m + } if y m ∈ B β + m + α We let I Log ,β + m + = I Log ,β + m ∪ { β + m + } and Γ β + m + = Γ β + m ∪ { Θ β + m + } ,and I Rel ,β + m + = I Rel ,β + m ∪ { β + m, β + m + } . By Lemma 3.2 we have ( β + m + , { B β + m + α } α ∈ I Rel ,β + m + ) ⊧ Γ β + m + . The check that conditions (i), (ii), (iv), (v)and (vi) also hold at γ = β + m + = β + n and γ = β + m + = β + n + Subcase A.2: x m , y m ∈ I Conn ,β . We have I Conn ,β ⊆ T and { s x m , ,q } q ∈ ω ∪ { s x m , ,q } q ∈ ω ⊆ β ∖ ( T ∪ ω ) , and similarly for y m . If without loss of general-ity x m > y m (considered as elements of β ) we select N ∈ ω large enough that q ≥ N implies x x m ,i,q > y m for i ∈ { , } . Notice that ( β + m, { B β + mα } α ∈ I Rel ,β + m ∪ { J β + m ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N , J β + m + ∪ { y m } ∪ { s y m ,i,q } i ∈{ , } ,q ∈ ω }) is a modelof Γ β + m ∪ { Θ β + n } . Now we let B β + mβ + m = J β + m ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N andfor α ∈ I Rel ,β + n ∪ { β + m, β + m + } we let B β + m + α = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ B β + mα if α < β + m + x m ∉ B β + mα B β + mα ∪ { β + m } if α < β + m + x m ∈ B β + mα J β + m + ∪ { y m } ∪ { s y m ,i,q } i ∈{ , } ,q ∈ ω }) if α = β + m + B β + m + α = ⎧⎪⎪⎨⎪⎪⎩ B β + m + α if y m ∉ B β + m + α B β + m + α ∪ { β + m + } if y m ∈ B β + m + α BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 17
We define I Log ,β + m + , Γ β + m + , and I Rel ,β + m + as in Subcase A.1. That (i) -(vi) hold at γ = β + m + γ = β + m + Subcase A.3: Exactly one of x m , y m is in I Conn ,β . Without loss of generality x m ∈ I Conn ,β . Select N ∈ ω large enough that y m ∉ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N (if x m < y m then of course we can let N = x m > y m then we are using the factthat x m = lim q → ∞ s x m ,i,q ).Notice that ( β + m, { B β + mα } α ∈ I Rel ,β + m ∪ { J β + m ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N , J β + m + ∪ { y m }}) is a model of Γ β + m ∪ { Θ β + n } . Let B β + mβ + m = J β + m ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N andfor α ∈ I Rel ,β + m ∪ { β + m, β + m + } we let B β + m + α = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ B β + mα if α < β + m + x m ∉ B β + mα B β + mα ∪ { β + m } if α < β + n + x m ∈ B β + mα J β + m + ∪ { y m } if α = β + m + B β + m + α = ⎧⎪⎪⎨⎪⎪⎩ B β + m + α if y m ∉ B β + m + α B β + m + α ∪ { β + m + } if y m ∈ B β + m + α Update the parameters as in Subcase A.1 and (i) - (vi) follow.
Case B: γ = t + ω ( k + ) + n, n ≠ . Let β = t + ω ( k + ) . We know that conditions(i) - (vi) hold at and below β . We will induct on n ≥
1. For all γ in this case wewill have I Conn ,γ = I Conn ,β and Sequ γ = Sequ β Let {( B β α m, , B β α m, , x m )} m ∈ ω be an enumeration of all triples such that x m , α m, , α m, < β and x m ∈ B β α m, ∩ B β α m, . Let m = n − β + m ≡ [ B α m, ∩ B α m, ⊇ B β + m ] ∧ [ B β + m ≠ ∅ ] . We note by Lemma 3.8 that Γ β + m ∪ { Θ β + m } is consis-tent, since ∅ ≠ Coor β + m ( B β + mα m, ) ∩ Coor β + m ( B β + mα m, ) ⊆ b β + mα m, ∩ b β + mα m, .We have by induction that ( ω, { ω ∩ B β + mα } α ∈ I Rel ,β + m ) is a Boolean saturatedmodel of Γ β + m and we select x ∈ ω ∩ B β α m, ∩ B β α n, = ω ∩ B β + mα m, ∩ B β + mα m, , with x = x m in case x m ∈ ω . By Lemma 3.18 we let J β + m ⊆ ω , with x ∈ J β + m , be suchthat ( ω, { ω ∩ B β + mα } α ∈ I Rel ,β + m ∪ { J β + m }) is a Boolean saturated model of Γ β + m ∪ { Θ β + m } . Subcase B.1: x m ∈ β ∖ I Conn ,β . It is clear that ( β + m, { B β + mα } α ∈ I Rel ,β + m ∪ { J β + m ∪ { x m }}) is a model of Γ β + m ∪ { Θ β + m } . We let B β + mβ + m = J β + m ∪ { x m } and for α ∈ I Rel ,β + m ∪ { β + m } we let B β + m + α = ⎧⎪⎪⎨⎪⎪⎩ B β + mα if x m ∉ B β + mα B β + mα ∪ { β + m } if x m ∈ B β + mα Let I Log ,β + m + = I Log ,β + m ∪ { β + m } , Γ β + m + = Γ β + m ∪ { Θ β + m } , and I Rel ,β + m + = I Rel ,β + m ∪ { β + m } . By Lemma 3.2 we have that B β + m + ⊧ Γ β + m + , and the checksfor the remaining (i), (ii), (iv), (v), and (vi) are straightforward. Subcase B.2: x m ∈ I Conn ,β . As property (vi) holds at β + m we select N ∈ ω large enough that both { s x m ,i,q } i ∈{ , } ,q ≥ N ⊆ B α m, and { s x m ,i,q } i ∈{ , } ,q ≥ N ⊆ B α m, . It is clear that ( β + m, { B β + mα } α ∈ I Rel ,β + m ∪ { J β + m ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N }) is a model of Γ β + m ∪ { Θ β + m } . We let B β + mβ + m = J β + m ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N and for α ∈ I Rel ,β + m ∪ { β + m } we let B β + m + α = ⎧⎪⎪⎨⎪⎪⎩ B β + mα if x m ∉ B β + mα B β + mα ∪ { β + m } if x m ∈ B β + mα One updates the parameters as in the end of Subcase B.1. That (i) - (vi) holdat γ = β + m + = β + n is easy to see. Case C: γ = t + ω ( k + ) + n, n ≠ . Let β = t + ω ( k + ) . As in Case A we willdeal with these γ by induction on n ≥
1, and consider n = p + n = p + n ∈ ω +
1. For all ofthe γ being considered in this case we let I Conn ,γ = I Conn ,β and Sequ γ = Sequ β Let {( B β α m , x m , y m )} m ∈ ω be an enumeration of all triples such that α m ∈ I Rel ,β , x m , y m ∈ β , x m ≠ y m , and x m , y m ∉ B β α m . We have n ∈ ω + m = n − β + n ≡ [ B β + m ∩ B β + m + ⊆ B α m ] ∧ ( ∀ x )[ x ∈ B β + m ∪ B β + m + ∪ B α m ] ∧ [ B β + m ≠ ∅ ] ∧ [ B β + m + ≠ ∅ ] . We note by Lemma 3.9 that Γ β + m ∪ { Θ β + n } isconsistent.We know by induction that ( ω, { ω ∩ B β + mα } α ∈ I Rel ,β + m ) is a Boolean saturatedmodel of Γ β + m . Select distinct points x, y ∈ ω ∖ B β + mα m such that x = x k or y = y k if x k or y k is in ω . That such points exist is implied by the fact that ( ω, { ω ∩ B β + mα } α ∈ I Rel ,β + m ) is Boolean saturated, for Coor β + m ({ x m , y m }) ⊆ X β + m, { α m } , ∅ .By Lemma 3.19 there exist J β + m , J β + m + ⊆ ω such that x ∈ J β + m , y ∈ J β + m + and ( ω, { ω ∩ B β + mα } α ∈ I Rel ,β + m ∪ { J β + m , J β + m + }) is a Boolean saturated model of Γ β + m ∪ { Θ β + n } . Subcase C.1: x m , y m ∈ β ∖ I Conn ,β . Notice that ( β + m, { B β + mα } α ∈ I Rel ,β + m ∪ { J β + m ∪ { x m } , J β + m + ∪ (( β + m ) ∖ ( ω ∪ { x m }))}) is a model of Γ β + m ∪ { Θ β + n } . We let B β + mβ + m = J β + n ∪ { x m } and for α ∈ I Rel ,β + m ∪ { β + m, β + m + } we let B β + m + α = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ B β + mα if α < β + m + x m ∉ B β + mα B β + mα ∪ { β + m } if α < β + m + x m ∈ B β + mα J β + m + ∪ (( β + m ) ∖ ( ω ∪ { x m })) if α = β + m + B β + m + α = ⎧⎪⎪⎨⎪⎪⎩ B β + m + α if y m ∉ B β + m + α B β + m + α ∪ { β + m + } if y m ∈ B β + m + α Let I Log ,β + m + = I Log ,β + m ∪ { β + n } and Γ β + m + = Γ β + m ∪ { Θ β + n } , and I Rel ,β + m + = I Rel ,β + m ∪ { β + m, β + m + } . By Lemma 3.2 we have that B β + m + ⊧ Γ β + m + , and checking that the other induction hypotheses at β + m + β + m + Subcase C.2: At least one of x m , y m is in I Conn ,β . Let, without loss ofgenerality, x m ∈ I Conn ,β . We have I Conn ,β ⊆ T and { s x m ,i,q } i ∈{ , } ,q ∈ ω ⊆ β ∖ ( T ∪ ω ) . BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 19 As x m , y m are distinct we select N ∈ ω large enough that y m ∉ [ x x m ,N, , x m ) ∪ [ x x m ,N, , x m ) .Notice that ( β + m, { B β + mα } α ∈ I Rel ,β + m ∪ { J β + m ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N , J β + m + ∪ (( β + m ) ∖ ( ω ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N ))}) is a model of Γ β + m ∪ { Θ β + n } . We let B β + mβ + m = J β + n ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N and for α ∈ I Rel ,β + m ∪ { β + m, β + m + } we let B β + m + α equal ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ B β + mα if α < β + m + x m ∉ B β + mα B β + mα ∪ { β + m } if α < β + m + x m ∈ B β + mα J β + m + ∪ (( β + m ) ∖ ( ω ∪ { x k } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N )) if α = β + m + B β + m + α = ⎧⎪⎪⎨⎪⎪⎩ B β + m + α if y m ∉ B β + m + α B β + m + α ∪ { β + m + } if y m ∈ B β + m + α Update the parameters as in Subcase C.1 and the check that the inductive con-ditions hold is straightforward.
Case D: γ = t + ω ( k + ) + n, n ≠ . Let β = t + ω ( k + ) . As in Cases A and Cwe will induct on n ≥ n = p + n = p + n ∈ ω +
1. For all of the γ being considered inthis case we let I Conn ,γ = I Conn ,β and Sequ γ = Sequ β Let {( B β α m , x m )} m ∈ ω be an enumeration of all pairs where x m ∈ B β α m . We have n ∈ ω + m = n − β + n ≡ ( ∀ x )[ x ∈ B α m ∪ B β + m + ] ∧ [ B β + m ⊆ B α m ] ∧ [ B β + m ≠ ∅ ] ∧ [ B β + m + ≠∅ ] . We note by Lemma 3.10 that Γ β + n ∪ { Θ β + n + } is consistent.We know ( ω, { ω ∩ B β + mα } α ∈ I Rel ,β + m ) is a Boolean saturated model of Γ β + m .Select a point x ∈ ω ∩ B β + mα m , with x = x m if x m ∈ ω . By Lemma 3.20 we can select J β + m , J β + m + ⊆ ω with x ∈ J β + m and ( ω, { ω ∩ B β + mα } α ∈ I Rel ,β + m ∪ { J β + m , J β + m + }) is a Boolean saturated model of Γ β + m ∪ { Θ β + n } . Subcase D.1: x m ∈ β ∖ I Conn ,β . Notice that ( β + m, { B β + mα } α ∈ I Rel ,β + m ∪ {{ x m } ∪ J β + m , J β + m + ∪ (( β + m ) ∖ ( ω ∪ { x m }))}) is a model of Γ β + m ∪ { Θ β + n } . We let B β + mβ + m = J β + m ∪ { x m } and for α ∈ I Rel ,β + m ∪ { β + m, β + m + } we let B β + m + α = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ B β + mα if α < β + m + x m ∉ B β + mα B β + mα ∪ { β + m } if α < β + m + x m ∈ B β + mα J β + m + ∪ (( β + m ) ∖ ( ω ∪ { x k })) if α = β + m + B β + m + α = ⎧⎪⎪⎨⎪⎪⎩ B β + m + α if x m ∉ B β + m + α B β ++ α ∪ { β + m + } if x m ∈ B β + m + α Let I Log ,β + m + = I Log ,β + m ∪ { β + n } and Γ β + m + = Γ β + m ∪ { Θ β + n } , and I Rel ,β + m + = I Rel ,β + m ∪ { β + m, β + m + } . By Lemma 3.2 we have that B β + m + ⊧ Γ β + m + , and checking that the other induction hypotheses are met is straightfor-ward. Subcase D.2: x m ∈ I Conn ,β . By induction we know that we can select N ∈ ω suchthat { s x m ,i,q } i ∈{ , } ,q ≥ N ⊆ B α m . Notice that ( β + m, { B β + mα } α ∈ I Rel ,β + m ∪ {{ x m } ∪ J β + m { s x m ,i,q } i ∈{ , } ,q ≥ N , J β + m + ∪ (( β + m ) ∖ ( ω ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N ))}) is a model of Γ β + m ∪ { Θ β + n } . We let B β + mβ + m = J β + m ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N and for α ∈ I Rel ,β + m ∪ { β + m, β + m + } we let B β + m + α equal ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ B β + mα if α < β + m + x m ∉ B β + mα B β + mα ∪ { β + m } if α < β + m + x m ∈ B β + mα J β + m + ∪ (( β + m ) ∖ ( ω ∪ { x m } ∪ { s x m ,i,q } i ∈{ , } ,q ≥ N )) if α = β + m + B β + m + α = ⎧⎪⎪⎨⎪⎪⎩ B β + m + α if x m ∉ B β + m + α B β + m + α ∪ { β + m + } if x m ∈ B β + m + α Update the other parameters as in Subcase D.1. Again, one can check that theinductive conditions are met for γ = β + m + γ = β + m + Case E: γ = t + ω ( k + ) + n, n ≠ . Let β = t + ω ( k + ) . For γ = β + n we will let I Log ,γ = I Log ,β I Rel ,γ = I Rel ,β I Conn ,γ = I Conn ,β Γ γ = Γ β Sequ γ = Sequ β We know by assumption that (iii) holds at β , so that in particular Γ β is consistent.Let {( Y m , Z m )} m ∈ ω be an enumeration of all pairs ( Y, Z ) with Y, Z ⊆ I Rel ,β finiteand X β ,Y,Z ≠ ∅ and such that ( Y m , Z m ) = ( Y m ′ , Z m ′ ) for infinitely many m ′ ∈ ω .Select σ m ∈ X β ,Y m ,Z m for each m ∈ ω .We proceed by induction on n ≥ m = n −
1. For each α ∈ I Rel ,β welet B β + m + α = ⎧⎪⎪⎨⎪⎪⎩ B β + mα if σ m ( α ) = B β + mα ∪ { β + m } if σ m ( α ) = B β + m + = { B β + m + α } α ∈ I Rel ,β + m . By Lemma 3.2, we have B β + m ⊧ Γ β = Γ β + m for all m ∈ ω by induction on m ∈ ω . Thus (iii) holds at each β + n . All otherinduction requirements (i), (ii), (iv), (v), (vi) are quite obvious.We observe that the above construction has guaranteed that ([ β , β + ω ) , {[ β , β + ω ) ∩ B β + ωα } α ∈ I Rel ,β + ω ) is a Boolean saturated model of Γ β + ω = Γ β . Case F: γ = t + ω k + n, n ≥ . Let β = t + ω k . For all such γ we let I Log ,γ = I Log ,β Γ γ = Γ β I Rel ,γ = I Rel ,β Subcase F.1: γ = t + with t ∈ T and there exist nonempty sets L , L ⊆ I Rel ,t such that t = ( ⋃ α ∈ L B tα ) ⊔ ( ⋃ α ∈ L B tα ) , ⋃ α ∈ L B tα = S t, and ⋃ α ∈ L B tα = S t, . Fix BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 21 such L , L ⊆ I Rel ,t . We first notice that ( t, B t ) is a Boolean saturated model of Γ t ,since ( ω, { ω ∩ B tα } α ∈ I Rel ,t ) is a Boolean saturated model.Next we notice that ⋃ α ∈ L b tα is disjoint from ⋃ α ∈ L b tα . Were disjointness to fail,we would have α ∈ L and α ∈ L with b tα ∩ b tα ≠ ∅ , but since ( t, B t ) is Booleansaturated we would have B tα ∩ B tα ≠ ∅ .Also, ( ⋃ α ∈ L b tα ) ∪ ( ⋃ α ∈ L b tα ) is dense in X t . This follows from the fact thatCoor t ( t ) is dense in X t (by Lemma 3.16) and Coor t ( t ) ⊆ ( ⋃ α ∈ L b tα ) ∪ ( ⋃ α ∈ L b tα ) .As X t is connected (Lemma 3.14), by Lemma 2.4 we select σ ∈ X t which is a limitpoint of ⋃ α ∈ L b tα and also a limit point of ⋃ α ∈ L b tα . Let L = { α ∈ I Rel ,t ∣ σ ∈ b tα } .Notice that L is nonempty, since X t models some statements of type (c) (by CaseC of our induction). Let L = { α q } q ∈ ω be an enumeration (we allow repetitionof indices for the same element of L ). For each q ∈ ω we select α ,q ∈ L and α ,q ∈ L such that b tα ,q ∩ ⋂ qi = b tα i ≠ ∅ and b tα ,q ∩ ⋂ qi = b tα i ≠ ∅ . Let { ǫ q } q ∈ ω be a strictly increasing sequence such that sup q ∈ ω ǫ q = t . Pick ω < β < t suchthat β is of form β = t + ω ( k + ) and β > α , α , , α , , ǫ . Assuming β q has been selected we select β q + which is of form β q + = t q + + ω ( k q + + ) and β q + > β q , α , . . . , α q , α ,q , α ,q , ǫ q .We have by construction that { β q } q ∈ ω is a strictly increasing sequence withsup q ∈ ω β q = t . Since ([ β q , β q + ω ) , {[ β q , β q + ω ) ∩ B β q + ωα } α ∈ I βq + ω ) is Boolean saturatedfor each q ∈ ω we select s t, ,q , s t, ,q ∈ [ β q , β q + ω ) such thatCoor β q + ω ( s t, ,q ) ∈ b β q + ωα ,q ∩ ⋂ qi = b β q + ωα i Coor β q + ω ( s t, ,n ) ∈ b β q + ωα ,q ∩ ⋂ qi = b β q + ωα i For all α ∈ I Rel ,γ = I Rel ,β we let B γα = ⎧⎪⎪⎨⎪⎪⎩ B tα if σ ( α ) = B tα ∪ { t } if σ ( α ) = ( γ, { B γα } α ∈ I Rel ,γ ) is a model of Γ γ by Lemma 3.2. Let I Conn ,γ = I Conn ,t ∪ { t } and Sequ γ = Sequ t ∪ {({ s t, ,q } q ∈ ω , { s t, ,q } q ∈ ω )} . That (i) - (vi) hold at γ is astraightforward to check. For example, for any α ∈ I Rel ,t + , if t ∈ B t + α then thesequence { s t, ,q } q ∈ ω and the sequence { s t, ,q } q ∈ ω are eventually in B t + α . Subcase F.2: γ = t + ω k + , γ ∉ ω and Subcase F.1 fails. Pick σ ∈ X t and forall α ∈ I Rel ,γ we let B γα = ⎧⎪⎪⎨⎪⎪⎩ B β α if σ ( α ) = B β α ∪ { β } if σ ( α ) = ( γ, { B γα } α ∈ I Rel ,γ ) is a model of Γ γ , and the other inductiveproperties are easily seen to still hold. Subcase F.3: γ = t + ω k + n, γ ∉ ω, n ≥ . We will let I Conn ,γ = I Conn ,γ − . Recallthe function f ∶ R → ⋃ δ <ℵ { , } δ chosen at the onset of our construction. If J isa well-ordered set (for example, a subset of an ordinal) we let ord ( J ) denote theunique ordinal which is order isomorphic to J . Given any function g ∈ ⋃ δ <ℵ { , } δ we let dom ( g ) denote the domain. We will look at two possibilities.Suppose dom ( f ( γ )) ≤ ord ( I Rel ,γ ) . Let ι ∶ dom ( f ( γ )) → ord ( I Rel ,γ ) be the uniqueorder embedding of dom ( f ( γ )) as an initial interval of ord ( I Rel ,γ ) . If there exists σ ∈ X γ − = X β such that σ ( ι ( ǫ )) = ( f ( γ ))( ǫ ) for all ǫ ∈ dom ( f ( γ )) , then we selectsuch a σ ∈ X β and for all α ∈ I Rel ,β = I Rel ,γ − = I Rel ,γ we let B γα = ⎧⎪⎪⎨⎪⎪⎩ B γ − α if σ ( α ) = B γ − α ∪ { γ − } if σ ( α ) = ( f ( γ )) ≤ ord ( I Rel ,γ ) or there does not exist such a σ ∈ X γ , we selectan arbitrary σ ∈ X γ and again for all α ∈ I Rel ,γ we let B γα = ⎧⎪⎪⎨⎪⎪⎩ B γ − α if σ ( α ) = B γ − α ∪ { γ − } if σ ( α ) = γ < ℵ . Let I Rel = ⋃ β <ℵ I Rel ,β . For each α ∈ I Rel we let B α = ⋃ α ≤ β B βα . Let B = { B α } α ∈ I Rel . Let I Log = ⋃ β <ℵ I Log ,β andΓ = ⋃ β <ℵ Γ β . Let Sequ = ⋃ β <ℵ Sequ β . Let τ denote the topology τ ( B ) on ℵ . Wecheck the properties that ought to hold for ( ℵ , τ ) (including that B is a basis for τ ) in the following section.5. Verification of Theorem 1.1
In this section we finish the proof of Theorem 1.1 by verifying the various claimsabout the space ( ℵ , τ ) . Remark 5.1.
It is straightforward to check that ( ℵ , B ) ⊧ Γ and also that ( ω, { ω ∩ B α } α ∈ I Rel ) is a (Boolean saturated) model of Γ, arguing precisely as in Case L ofour construction. That Γ satisfies † is also clear by construction. Lemma 5.2. B is a basis for the topology τ = τ ( B ) on ℵ . Proof.
Since ( ℵ , B ) ⊧ Γ and by Case C of our construction, it is clear that ⋃ B = ℵ .Suppose that x ∈ B α ∩ B α . Select β < ℵ which is of form β = t + ω ( k + ) largeenough that x ∈ β and α , α ∈ I Rel ,β . We have B β α = β ∩ B α and B β α = β ∩ B α ,so in particular x ∈ B β α ∩ B β α . By the treatment of Case B there exists β ≤ α < β + ω such that x ∈ B α α ⊆ B α α ∩ B α α and Θ α ∈ Γ α + with Θ α ≡ [ B α ∩ B α ⊇ B α ] ∧ [ B α ≠ ∅ ] . Since ( ℵ , B ) models Γ, we have x ∈ B α ⊆ B α ∩ B α . (cid:3) Lemma 5.3. (Property (1)) The space ( ℵ , τ ) is regular. Proof. If x, y ∈ ℵ are distinct then we can select β < ℵ which is of form β = t + ω ( k + ) with x, y < β . By the treatment of Case A there exist β ≤ α < α < β + ω such that B α α ∩ B α α = ∅ , x ∈ B α α and y ∈ B α α , and accompanying Θ α of type (a).That x ∈ B α , y ∈ B α and B α ∩ B α = ∅ is immediate. Thus the space is Hausdorff,and regularity is argued along similar lines, using Case D and the fact that B is abasis for τ . (cid:3) Lemma 5.4. (Property (2)) The space ( ℵ , τ ) is separable. Proof.
Since ( ω, { ω ∩ B α } α ∈ I Rel ) ⊧ Γ, we know more particularly that B α ∩ ω ≠ ∅ for all α ∈ I Rel , which demonstrates that ω is dense in ( ℵ , τ ) . (cid:3) Towards property (3) we prove the following.
Lemma 5.5.
Suppose that ∅ ≠ L ⊆ I Rel . The set J = { β ∈ ℵ ∣ ∅ ≠ ⋃ α ∈ I Rel ,β ∩ L B βα = β ∩ ⋃ α ∈ L B α } is club in ℵ . BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 23
Proof.
To see that J is unbounded we let β < ℵ be given, without loss of generalitysuch that I Rel ,β ∩ L ≠ ∅ . Supposing that β n has already been defined we select ℵ > β n + > β n large enough that ( β n ∩ ⋃ α ∈ L B α ) ∖ ⋃ α ∈ I Rel ,βn ∩ L B β n α ⊆ ⋃ α ∈ I Rel ,βn + ∩ L B β n + α (this is possible since the set on the left is countable). Letting β = sup n ∈ ω β n it iseasy to see that β ∩ ⋃ α ∈ L B α ⊆ ⋃ α ∈ I Rel ,βn + ∩ L B βα , and the reverse inclusion holdsnecessarily, and clearly the set β ∩⋃ α ∈ L B α is nonempty as I Rel ,β ∩ L ⊇ I Rel ,β ∩ L ≠ ∅ .Thus β ∈ J .To see that J is closed we let { β n } n ∈ ω be an increasing sequence in J and β = sup n ∈ ω β n . Then ⋃ α ∈ I Rel ,β ∩ L B βα ⊇ ⋃ n ∈ ω ( ⋃ α ∈ I Rel ,βn ∩ L B β n α ) = ⋃ n ∈ ω ( β n ∩ ( ⋃ α ∈ L B α )) = β ∩ ( ⋃ α ∈ L B α ) ⊇ ⋃ α ∈ I Rel ,β ∩ L B βα and as ⋃ α ∈ I Rel ,β ∩ L ≠ ∅ , the set ⋃ α ∈ I Rel ,β ∩ L B βα = β ∩ ( ⋃ α ∈ L B α ) is nonempty. Thus β ∈ J . (cid:3) Lemma 5.6. (Property (3)) The space ( ℵ , τ ) is connected. Proof.
Suppose for contradiction that ℵ = V ⊔ V is a nontrivial disconnection.Then we can select nonempty L , L ⊆ I Rel such that V = ⋃ α ∈ L B α and V =⋃ α ∈ L B α . For i ∈ { , } we let J i = { β ∈ ℵ ∣ ∅ ≠ ⋃ α ∈ I Rel ,β ∩ L i B βα = β ∩ ⋃ α ∈ L i B α } and by Lemma 5.5 each of J , J is club in ℵ . By our selection of sequence {( S α, , S α, )} α ∈ T we know that the set J = { t ∈ T ∣ [ t ∩ V = S t, ] ∧ [ t ∩ V = S t, ]} is stationary in ℵ . Then we can select t ∈ J ∩ J ∩ J , and t + t ∈ I Conn ,t + and sequences { s t, ,q } q ∈ ω and { s t, ,q } q ∈ ω such that { s t,i,q } q ∈ ω ⊆ S t,i for i ∈ { , } . Either t ∈ V or t ∈ V , so that either ∅ ≠ S t, ∩ V ⊆ V ∩ V or ∅ ≠ V ∩ S t, ⊆ V ∩ V , and in either case we obtain acontradiction. Thus ( ℵ , τ ) is connected. (cid:3) Property (4), that the space is of cardinality ℵ , is quite obvious. Lemma 5.7. (Property (5)) For every nonempty open O in the space ( ℵ , τ ) thesubspace ℵ ∖ O is totally separated. Proof.
Clearly it suffices to check the claim in the case where O is a basis element.Let α ∈ I Rel be given, as well as distinct x, y ∈ ℵ ∖ B α . Select β < ℵ of form β = t + ω ( k + ) which is large enough that x, y ∈ β and α ∈ I Rel ,β . By thetreatment of Case C there exist α , α ∈ [ β , β + ω ) such that x ∈ B α α , y ∈ B α α ,and B α α ∩ B α α ⊆ B α α , and accompanying Θ α of type (c). Thus x ∈ B α , y ∈ B α , B α ∩ B α ⊆ B α , and B α ∪ B α ∪ B α = ℵ , as ( ℵ , B ) ⊧ Γ. (cid:3) Lemma 5.8. (Property(6)) Each nonempty open subset in ( ℵ , τ ) is uncountable. Proof.
It suffices to prove that each basis element is uncountable. Given α ∈ I Rel wenotice that for each β < ℵ of form β = t + ω ( k + ) large enough that α ∈ I Rel ,β we have B β + ωα ∩ [ β , β + ω ) ≠ ∅ since ([ β , β + ω ) , {[ β , β + ω ) ∩ B β + ωα } α ∈ I Rel ,β + ω ) is a (Boolean saturated) model of the set of sentences Γ β which includes a sentencerequiring B α ≠ ∅ . That B α is uncountable follows immediately. (cid:3) Lemma 5.9. (Property (7)) ∣ B ∣ = ℵ . Proof.
That ∣ B ∣ ≤ ℵ is obvious since the elements are indexed by I Rel ⊆ ℵ . Insteadof directly verifying that B is uncountable, we will use the forthcoming property(8). Were B countable, ( ℵ , τ ) would be Lindel¨of, and by (8) we would have ( ℵ , τ ) compact, which together with (1), (3), (4) and (5) gives a contradiction (see theproof of Proposition 2.5). (cid:3) Towards properties (8) and (9) we give the following.
Lemma 5.10.
For each β < ℵ and σ ∈ X β there exists β > β and x ∈ β such thatCoor β ( x ) ↾ I Rel ,β = σ . Proof.
Let δ = ord ( I Rel ,β ) and ι ∶ δ → I Rel ,β be the unique order isomorphism.Let g ∈ { , } δ be given by g ( ǫ ) = σ ( ι − ( ǫ )) . Select β ∈ R such that β > β and f ( β ) = g . By Subcase F.3 we have Coor β ( β − ) ↾ I Rel ,β = σ . (cid:3) Lemma 5.11. (Property (8)) Every countable cover of ℵ by elements of B has afinite subcover. Proof.
Let { α n } n ∈ ω ⊆ I Rel be such that ⋃ n ∈ ω B α n = ℵ . Select β < ℵ large enoughthat α n ∈ I Rel ,β for all n ∈ ω . We claim that { b βα n } n ∈ ω is a cover of X β . To see this,we suppose otherwise and let σ ∈ X β ∖ ⋃ n ∈ ω b βα n . By Lemma 5.10 we select β > β and x ∈ β such that Coor β ( x ) ↾ I Rel ,β = σ . Clearly x ∈ β ∖ ⋃ n ∈ ω B β α n , and so x ∈ ℵ ∖ ⋃ n ∈ ω B α n , a contradiction.Thus { b βα n } n ∈ ω is a cover of X β , and as ( X β , τ ({ b βα } α ∈ I Rel ,β ) is compact (byLemma 3.14) we may select a finite subcover { b βα k , . . . , b βα kn } . Given arbitrary x ∈ ℵ we select β ≥ β, x and notice that Coor β ( x ) ↾ I Rel ,β ∈ ⋃ ni = b βα ki . Thus { B α k , . . . , B α kn } is a subcover. (cid:3) Lemma 5.12. (Property (9)) If { B n } n ∈ ω is a nesting decreasing sequence of ele-ments of B then ⋂ n ∈ ω B n ≠ ∅ , and so ( ℵ , τ ) is strongly Choquet. Proof.
That ( ℵ , τ ) is strongly Choquet will follow immediately from the first claim(see Lemma 2.8). Supposing that { α n } n ∈ ω is a sequence in I Rel such that B α ⊇ B α ⊇ ⋯ we let β < ℵ be large enough that α n ∈ I Rel ,β for all n ∈ ω and of form β = t + ω ( k + ) . By Case E we have ([ β , β + ω ) , {[ β , β + ω ) ∩ B β + ωα } α ∈ I Rel ,β + ω ) is a Boolean saturated model of Γ β + ω .Notice that b β + ωα ⊇ b β + ωα ⊇ ⋯ , for if there existed σ ∈ b β + ωα n + ∖ b β + ωα n we wouldhave, by Boolean saturation, some x ∈ [ β , β + ω ) with x ∈ B β + ωα n + ∖ B β + ωα n , so x ∈ B α n + ∖ B α n , contrary to assumption.By Lemma 3.13 we know that X β + ω is compact as a subset of { , } I Rel ,β + ω .Each b β + ωα n is the intersection of X β + ω with a closed subset of { , } β + ω . As ( X β + ω , { b β + ωα } α ∈ I Rel ,β + ω ) ⊧ Γ β + ω (Lemma 3.4), we know more particularly that BARELY CONNECTED TOPOLOGICAL SPACE MADE FROM DIAMOND 25 b β + ωα n ≠ ∅ . Thus ⋂ n ∈ ω b β + ωα n ≠ ∅ by compactness. Selecting σ ∈ ⋂ n ∈ ω b β + ωα n , we pickby Lemma 5.10 an element β > β + ω and x ∈ β such that Coor β ( x ) ↾ I Rel ,β + ω = σ .Clearly x ∈ ⋂ n ∈ ω B β α n ⊆ ⋂ n ∈ ω B α n , so we are done. (cid:3) Acknowledgement
The author is enormously grateful to Jan van Mill for feedback on a draft of thisarticle.
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