K -THEORY OF CO-EXISTENTIALLY CLOSED CONTINUA

. We describe the possible values of K -theory for C ( X ) when X is a co-existentially closed continuum. As a consequence we also show that all pseudo-solenoids, except perhaps the universal one, are not co-existentially closed.


Introduction
A compactum is a compact Hausdorff space and a continuum is a connected compactum (not necessarily metrizable).Although model-theoretic methods cannot be directly applied to compacta or continua, there has nevertheless been an extensive study of model-theoretic properties of compacta in "dual" form; see Bankston [2] and subsequent papers.When X is 0-dimensional one can replace the study of X by the study of the Boolean algebra of clopen subsets of X , due to Stone duality.For more general compacta one can use continuous model theory to study commutative unital C*-algebras, which are dual to the category of compacta by Gel'fand duality.This is the approach taken in this paper.
Bankston [3] introduced co-existentially continua as the dual of the model-theoretic notion of existentially closed models of a theory (see Section 2 below for the definition).While co-existentially closed continua exist in abundance, only one concrete example is known.In Eagle, Goldbring and Vignati [16] it was shown that the pseudoarc is a co-existentially closed continuum.The pseudoarc was first constructed by Knaster [28] and later characterized by Bing [9] as the unique non-degenerate hereditarily indecomposable chainable continuum; Bing also showed that the pseudoarc is generic amongst subcontinua of R n for every n ≥ 2. In [16] it was shown that if the theory of continua, T conn , has a model companion then that model companion must be the theory of the and Usvyatsov [8]; the reader looking for background of continuous model theory, particularly in the context of C*-algebras, is referred to Farah, Hart, Lupini, Robert, Tikuisis, Vignati and Winter [17].We also assume that the reader is familiar with the elements of continuum theory (for which we suggest Nadler [35]) and C*-algebra K -theory (for which see Rørdam, Larsen and Laustsen [38]).
Fact 2.1 ([16, Fact 1.1 and Remark 1.2])There is a universally axiomatizable theory T conn in the language of unital C*-algebras such that M |= T conn if and only if M ∼ = C(X) for some continuum X .
The ultraproduct construction for commutative unital C*-algebras dualizes to the ultracoproduct of compact Hausdorff spaces, which we denote U X i .For our purposes it will suffice for us to know that U X i is the spectrum of the C*-algebra ultraproduct U C(X i ), so we omit the (somewhat lengthy) hands-on definition of U X i ; see [2].Banskton [3] introduces co-existentially closed continua in terms of mapping properties involving ultracopowers, but for our purposes it is more convenient to define a continuum X to be co-existentially closed if C(X) is an existentially closed model of T conn .We recall the precise definitions of existential closure in the setting of continuous logic: Definition 2.2 Let T be a theory in a signature L of continuous first-order logic.A model M |= T is called an existentially closed model of T if given any N |= T such that M ⊆ N , any quantifier-free L-formula φ(x, y), and any tuple b from M (of the appropriate length), we have: The equivalence of Bankston's topological definition of co-existentially closed continua to the one we give here can be found in the appendix to [16].
The next result was first proved by Bankston [4] using lattice bases, but can also be obtained by applying standard model-theoretic results about universally axiomatizable theories to T conn and then using Gel'fand duality.Fact 2.3 Every continuum is the continuous image of a co-existentially closed continuum of the same weight.

C J Eagle and J Lau
It will be useful to know that if X is a co-existentially closed continuum then Th ∀∃ (C(X)) is the maximal consistent ∀∃-theory extending T conn for non-degenerate continua.The proof relies on the following fact, originally due to K. P. Hart in [22]; see [16] for the details of how to translate the result from [22] into the version we state here.
Fact 2.4 Suppose that X and Y are continua and that Y is non-degenerate.Then there is a continuum Proof Apply the fact above to find a continuum Y ′ with C(Y ′ ) ≡ C(Y) and such that C(X) embeds in C(Y ′ ); by replacing C(X) by an isomorphic copy we may assume that in fact C(X) is a substructure of C(Y ′ ).Write σ as sup ∥x∥≤R φ(x), where φ is an existential formula.Fix any tuple a from C(X) so φ(a) C(X) = 0 as well.Since a was arbitrary, this shows that C(X) |= σ .
Throughout this paper when we refer to the dimension of a compactum we mean the covering dimension.
We denote by T the circle, which we often view as T = {z ∈ C : |z| = 1}.

K 0 for co-existentially closed continua
In this section we show that if X is a co-existentially closed continuum then K 0 (C(X)) = Z.In fact, we show that if X is any 1-dimensional continuum then K 0 (C(X)) = Z; we suspect that this result may already have been known, but as we were unable to locate a reference in the literature, we provide a proof here.The map D in the previous fact is also often called dim, but to avoid confusion we reserve dim for the covering dimension of a compactum.
Note that K 0 (C(X)) = Z if and only if D is an isomorphism, in which case K 0 (C(X)) is generated by the class [1 X ], where 1 X : X → C is the function that is constantly 1.
For each n ∈ N, let Z n denote the wedge product of n circles at a common basepoint, and let Z 0 be a single point.We will need the following two facts: Fact 3.2 ([23, Example 1.22 and Section 1.A]) Every 1-dimensional compact connected CW-complex is homotopy equivalent to Z n for some n.
If dim X = 1, then by Fact 3.2 X is homotopy equivalent to Z n for some n ≥ 0. Since K 0 is invariant under homotopy equivalence, when n = 0 we are back in the previous case, while if n ≥ 1 then by Fact 3.
Proof The inverse sequence of spaces induces a sequence of commutative unital C*-algebras and unital * -homomorphisms and by continuity of the K 0 functor we have that ) is an isomorphism for every n, and hence Proof It is shown in [33, Corollary 1] that we can express X as for some compact polyhedra P b,i each having dim P b,i ≤ dim X = 1.(If X is nonmetrizable this limit may be over an uncountable indexing family.)Since X maps continuously onto each term of the inverse system, and X is connected, each P b,i is also connected.Thus each P b,i is, in particular, a connected compact CW-complex of dimension at most 1.For each fixed b, let X b = lim ← − i P b,i .By Proposition 3.5 we obtain, for each b, that: Finally, by continuity of K 0 we obtain: Finally, if X is a co-existentially closed continuum then Fact 2.6 says that X has dimension 1, so we obtain:

K 1 for co-existentially closed continua
Unlike the case for K 0 , when X is a co-existentially closed continuum there are many possible values of K 1 (C(X)).In this section we show that K 1 (C(X)) must be a torsion-free divisible abelian group, but also show that the rank of K 1 (C(X)) can be arbitrarily large when X is a co-existentially closed continuum.
In fact, we will work primarily with the first integral Čech cohomology group of X , Ȟ1 (X), in order to take advantage of existing results in the literature.Recall that Ȟ1 (X) can be identified with the collection of homotopy classes of continuous maps from X to the circle T (see Hatcher [23,Exercise 4.3.2]).The connection to K -theory for C(X) is the following: Proof By Blackadar [12, V.3.1.3],since dim(X) ≤ 1 the stable rank of C(X) is given by: In particular, applying Fact 2.6 we have: We make some definitions, motivated by the analogy between existentially closed structures and algebraically closed fields.Later in this section we will use these notions to show that when X is co-existentially closed, K 1 (C(X)) is a divisible group.

Definition 4.3 A C*-algebra
A is approximately algebraically closed if for every n > 0, every a 0 , . . ., a n−1 ∈ A, and every ε > 0, there is We cannot axiomatize C(X) being algebraically closed in continuous logic (see Corollary 4.9 below).We can, however, axiomatize being approximately algebraically closed, which we now do.
For each n ∈ N >0 and each K ∈ R >0 , define: C J Eagle and J Lau Proposition 4.4 For any compactum X , the following are equivalent: (1) C(X) is approximately algebraically closed.
Proof Suppose that (1) holds.Fix n and K .Suppose that a 0 , . . ., a n−1 ∈ C(X) are such that ∥a j ∥ ≤ K for all j.Consider any ε with 0 Fix any x ∈ X , and write By Cauchy's bound for roots of complex polynomials, we have: As this holds for every x ∈ X , we have ∥f ∥ ≤ 2 + K , as desired to show that C(X) |= σ n,K .Now suppose that (2) holds.Given any a 0 , . . ., Proof Since X is metrizable it is first-countable.By Fact 4.5, it suffices to show that X is not locally connected.By Fact 2.6, X is hereditarily indecomposable; we will show that an indecomposable continuum cannot be locally connected at any point. 1uppose to the contrary that X is locally connected at a point x.
Then V is a proper subcontinuum of X with non-empty interior; this contradicts the indecomposability of X , by Macías [32, Corollary 1.7.21].
We do not know if "metrizable" can be eliminated from the hypotheses of Proposition 4.7.
Question 4.8 Is there a co-existentially closed continuum X such that C(X) is algebraically closed?Suppose that X is a metrizable co-existentially closed continuum.Proposition 4.4 and Theorem 4.6 then imply that all models of Th(C(X)) are approximately algebraically closed.In countably saturated models the sentences σ n,K express algebraic closure, not just approximate algebraic closure.Thus if Y is such that C(Y) ≡ C(X) and C(Y) is countably saturated then C(Y) is algebraically closed, while by Proposition 4.7 C(X) is not algebraically closed.Therefore: Corollary 4.9 The property of being algebraically closed is not axiomatizable in the language of unital C*-algebras.

C J Eagle and J Lau
Continuing with the setup from the paragraph above, Y is also hereditarily indecomposable (because hereditary indecomposability is axiomatizable and X is hereditarily indecomposable by Fact 2.6), and hence Y is not locally connected.Thus Fact 4.5 cannot be extended to non-metrizable continua in general, and the method of proof of Proposition 4.7 cannot be used to answer Question 4.8.There is a notion of a continuum being almost locally connected that does follow from algebraic closure even in the non-metrizable setting (see Countryman [15,Theorem 2.4]).The same argument as above shows that this almost local connectedness is compatible with hereditary indecomposability, but we do not know if it is compatible with being co-existentially closed.
Remark 4.10 Concerning the question of whether or not T conn has a model companion, we observe that a negative answer to Question 4.8 would strongly refute the existence of such a model companion, since the existence of a model companion is equivalent to co-existential closure being preserved by co-elementary equivalence.
We now return to gathering information about K 1 (C(X)) when X is co-existentially closed.Recall that an abelian group (G, +) is called n-divisible (for some fixed n ∈ N) if for every g ∈ G there is x ∈ G such that nx = g, where nx is the sum of n copies of x.A group is divisible if it is n-divisible for all n ≥ 1.For each n ∈ N, the following are equivalent: (1) Ȟ1 (X) is n-divisible.
It follows immediately that if X is a continuum such that C(X) is approximately algebraically closed and dim(X) ≤ 1 then Ȟ1 (X) is a divisible group.In particular, by Theorem 4.6 and Fact 2.6 those hypotheses are satisfied when X is co-existentially closed.The group Ȟ1 (X) is always torsion-free abelian, so we conclude: Theorem 4.12 If X is a co-existentially closed continuum then Ȟ1 (X) is a torsion-free divisible abelian group.Remark 4.13 In general, extracting K -theoretic information about a C*-algebra A from Th(A) is a non-trivial matter (see [17,Sections 3.11 and 3.12]).By Proposition 4.1 and standard translations between topological properties and C*-algebraic ones, we can view the proof of Theorem 4.12 as showing that if T is the theory of commutative unital projectionless C*-algebras of real rank at most 1, then if A |= T , for each n ≥ 1 the property of having K 1 (A) be n-divisible is detected by Th(A).Because our argument depends heavily on working with continua of dimension at most 1, we do not know if this result can be extended to more general classes of C*-algebras.
For the pseudoarc P we have K 1 (C(P)) = Ȟ1 (P) = 0; this follows from the continuity of Čech cohomology, since P can be represented as an inverse limit of arcs, and arcs have trivial first cohomology.In light of the fact that all co-existentially closed continua have the same K 0 as P (Corollary 3.7), and the fact that the proof of Theorem 4.12 does not appear to use the full power of having C(X) be approximately algebraically closed, one might therefore conjecture that Ȟ1 (X) = 0 for every co-existentially closed continuum X .Our next goal is to show that this is not the case even if X is metrizable, and moreover (allowing X to be non-metrizable) the rank of Ȟ1 (X) can be arbitrarily large.
Proof Using Fact 2.3, let X be a co-existentially closed continuum with w(X) = w(Y) and with a continuous surjection f : X → Y .Every continuous surjection onto a hereditarily indecomposable continuum is confluent; this was originally proved in the metric case by Cook [14], but see Bankston [6,Theorem 2] for a proof in the non-metric setting.In particular, f is confluent.It then follows by Lelek [31,Corollary 2] that f induces an injective homomorphism f * : Ȟ1 (Y) → Ȟ1 (X).As injective homomorphisms preserve independence, this gives rank( Ȟ1 (X)) ≥ rank( Ȟ1 (Y)).
Proof Let Y be a hereditarily indecomposable continuum with Ȟ1 (Y) = Q (one example of such is the universal pseudo-solenoid described in Section 5 below).Then Theorem 4.14 produces a metrizable co-existentially closed continuum X with rank( Ȟ1 (X)) ≥ rank( Ȟ1 (Y)) = 1, and in particular Ȟ1 (X) ̸ = 0.By Theorem 4.14, in order to show that Ȟ1 (X) can have rank at least κ, for some infinite κ, it suffices to find a hereditarily indecomposable continuum Y where the rank of Ȟ1 (Y) is at least κ.The Y we produce will be an ultracoproduct of a metrizable continuum whose first Čech cohomology group is Q.We begin by proving the following, which may be of independent interest.C J Eagle and J Lau Proposition 4.16 Let (X i ) i∈I be a family of compacta and let U be an ultrafilter on an index set I .Then U Ȟ1 (X i ) is a quotient of the group Ȟ1 U X i .
Proof It is convenient to phrase this proof categorically.
The ultraproduct construction can be represented categorically as Recall that in the category of compact Hausdorff spaces the coproduct operation is taking the Stone-Čech compactification of the disjoint union of spaces; that is, for any A ⊆ I : By Gel'fand duality applied to the categorical description of U C(X i ), the ultracoproduct U X i is where for A, B ∈ U with A ⊆ B, the map π A,B is the natural embedding of i∈A X i into i∈B X i .By continuity and contravariance of Čech cohomology we therefore have that where π * A,B denotes the map induced on Čech cohomology by π A,B .For each A ∈ U , we define a map η ie f and g are homotopic maps, then the restriction of a homotopy from f to g to any X i is a homotopy from f | Xi to g| Xi , and hence ([f | Xi ]) i∈A = ([g| Xi ]) i∈A .That is, η A is well-defined.It is straightforward to verify that η A is a group homomorphism.We show that η A is surjective.Suppose we are given ([ Since each f i is continuous so is f , and therefore f extends to the Stone-Čech compactification as a continuous map f β : i∈A X i → T; we then have

C J Eagle and J Lau
This completes the proof since Ȟ1 (X) is a torsion-free divisible abelian group (Theorem 4.12) and when such groups are uncountable their rank is equal to their cardinality.
Having just shown that there are co-existentially closed continua with Ȟ1 (X) ̸ = 0, it is tempting to hope that models of Th(C(P)) must have Ȟ1 (X) = 0, as this would give us a way to show that Th(C(P)) is not the model companion of T conn .Unfortunately, this strategy does not work.Proposition 4.18 If U is any countably incomplete ultrafilter and X is any continuum with dim(X) = 1 then Ȟ1 U X ̸ = 0. Proof As a consequence of the above, if the connected component of the unitary group is definable then K 1 (M) is in M eq .It then follows from [17, Proposition 3.12.1(iii)]that if B is any C*-algebra with C(P) ⪯ B, then K 1 (C(P)) ⪯ K 1 (B).However, K 1 (C(P)) = 0, while Proposition 4.18 (along with Proposition 4.1) shows that ultrapowers of C(P) often have non-zero K 1 , giving the desired contradiction.
For planar continua we can improve Theorem 4.12.
Theorem 4.21 If X is a planar co-existentially closed continuum then Ȟ1 (X) = 0, and hence X does not separate the plane.
Proof By Krasinkiewicz [29, Theorem 3.3], Ȟ1 (X) is a finitely divisible group, meaning that if g ∈ Ȟ1 (X) and g ̸ = 0 then {n ∈ N : g is divisible by n} is finite.(Note, in particular, that on the definition given in [29] the trivial group is finitely divisible.)On the other hand, Theorem 4.12 shows that Ȟ1 (X) is a divisible group, meaning every element is divisible by every n ∈ N.This combination of properties is possible only when Ȟ1 (X) = 0.The claim about not separating the plane then follows directly from Lau [30, Theorem 1].
There are 2 2 ℵ 0 distinct hereditarily indecomposable one-dimensional plane continua that do not separate the plane (see Ingram [26,Theorem 2]), and moreover many of these are not continuous images of the pseudoarc.We do not know if any of these are co-existentially closed.
Problem 4.22 Is there a planar co-existentially closed continuum other than the pseudoarc?
5 Continua that are not co-existentially closed As an application of our results on K 1 (C(X)) for co-existentially closed continua X , we show that various continua cannot be co-existentially closed.The main result of this section is that the only pseudo-solenoid that could be co-existentially closed is the universal one.The continua considered in this section will all be metrizable, but since that was not the case in earlier sections we will include the metrizability hypothesis in our theorem statements.
Recall that if C is a continuum then a continuum X is C-like if it can be written as an inverse limit of copies of C. In particular, arc-like continua (also known as the metrizable chainable continua) are inverse limits of copies of [0, 1], while circle-like continua (also known as the metrizable circularly chainable continua) are inverse limits of copies of the circle T.

Solenoids
Consider the circle as T = {z ∈ C : |z| = 1}.For each n ∈ N, let µ n : T → T be the map µ n (z) = z n .
C J Eagle and J Lau Definition 5.1 A solenoid is a metrizable continuum that is not arc-like and that is circle-like, where each map in the inverse system is a map of the form µ n (with possibly different values of n for different maps).
Definition 5.2 Let N = (n 1 , n 2 , . ..) be a sequence of positive natural numbers.The N -adic solenoid S N is the inverse limit of the system: It is immediate from the definition that every solenoid is the N -adic solenoid for some N .
Definition 5.3 Let (n 1 , n 2 , . ..) be a sequence of positive integers.The corresponding supernatural number is the formal product ∞ i=1 n i .We say that two supernatural numbers N and M are equivalent, and write N ∼ M , if there are finite values 1 ≤ n 0 , m 0 < ∞ such that m 0 • N = n 0 • M .Note that any supernatural number can be expressed as a product of the form i∈N p ki i , where p i is the ith prime and k i ∈ {0, 1, 2, . . ., ∞}.The equivalence of two supernatural numbers means that these formal products differ on at most finitely many primes, each of which appears to a finite power (see the discussion before and after Hurder and Lukina [25, Definition 1.1] for more details).We denote by [∞] the supernatural number [∞] = i∈N p ∞ i .Bing [10] showed that if N = (n 1 , n 2 , . ..) and M = (m 1 , m 2 , . ..) are two sequences that produce equivalent supernatural numbers, then S N ∼ = S M .McCord [34] showed the converse.Thus solenoids are classified up to homeomorphism by equivalence of supernatural numbers: Every solenoid contains an arc (in fact, every proper subcontinuum of a solenoid is an arc) and therefore solenoids are not hereditarily indecomposable.Since every co-existentially closed continuum is hereditarily indecomposable (Fact 2.6) it follows that no solenoid can be co-existentially closed.Nevertheless, solenoids provide useful information about the class of pseudo-solenoids (described below), which are hereditarily indecomposable

Fact 3 . 1 ([ 38 ,
Example 3.3.5])Let X be any continuum.There is a surjective group homomorphism D : K 0 (C(X)) → Z that satisfies D([p]) = Tr(p(x)) (independently of the choice of x ∈ X ).Here we identify M n (C(X)) with C(X, M n (C)) and Tr is the standard trace on M n (C).

Fact 4 . 5 (
Kawamura and Miura [27, Theorem 1.1]) A first-countable continuum is algebraically closed if and only if it is locally connected, has dimension at most 1, and has Ȟ1 (X) = 0. Theorem 4.6 If X is a co-existentially closed continuum then C(X) is approximately algebraically closed.Proof The space interval [0, 1] is first-countable, locally connected, dimension 1, and has Ȟ1 ([0, 1]) = 0. Therefore by Fact 4.5 C([0, 1]) is algebraically closed.It is therefore also approximately algebraically closed, so C([0, 1]) |= σ n,K for all n and K by Proposition 4.4.Each σ n,K is a ∀∃-sentence, so by Lemma 2.5 we have C(X) |= σ n,K for all n and K as well, and therefore (again by Proposition 4.4) C(X) is approximately algebraically closed.Proposition 4.7 If X is a metrizable co-existentially closed continuum then C(X) is not algebraically closed.

Fact 5 . 4 (
[10,34]) Let N and M be supernatural numbers.Then S N ∼ = S M if and only if N ∼ M .Consequently, we often index solenoids by supernatural numbers instead of by sequences of natural numbers.
[1,Banakh, Bankston, Raines and Ruitenburg [1, Theorem 2.1] a compactum is 3-chainable if and only if it is a one-dimensional continuum with trivial first Čech cohomology group.Every ultracoproduct of compacta by a countably incomplete ultrafilter fails to be 3-chainable[1, Lemma 5.3].Since being a continuum and being one-dimensional are elementary properties they are preserved by ultracoproducts.