Non-axiomatizability of real spectra in <span class="mathjax-formula">$\mathcal{L}_\infty \lambda $</span>

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ANNALES

DE LA FACULTÉ DES SCIENCES

Mathématiques

TIMOTHYMELLOR, MARCUSTRESSL

Non-axiomatizability of real spectra inL∞λ

Tome XXI, n^o2 (2012), p. 343-358.

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pp. 343–358

Non-axiomatizability of real spectra in

L_∞λ

Timothy Mellor⁽¹⁾, Marcus Tressl⁽²⁾

ABSTRACT. —We show that the property of a spectral space, to be a spectral subspace of the real spectrum of a commutative ring, is not expressible in the infinitary first order languageL_∞λof its defining lattice.

This generalises a result of Delzell and Madden which says that not every completely normal spectral space is a real spectrum.

R^ÉSUMÉ.—Nous montrons que la propriété d’un espace spectral d’être un sous-espace spectral du spectre réel d’un anneau commutatif n’est pas exprimable dans le langage infinitaire du premier ordreL_∞λde son treillis de définition. Ceci généralise un résultat de Delzell et Madden qui dit qu’en général, un espace spectral complètement normal n’est pas un spectre réel.

For a while it was an open question whether real spectra are precisely the completely normal spectral spaces. In [3], a counterexample is given, hence a completely normal spectral space is constructed, which is not homeomorphic to the real spectrum ofany (commutative, unital) ring. We extend this result here in two ways:

1. We show that the spectral space constructed in [3] is also not homeomorphic to any spectral subspace ofa real spectrum.

2. We show that there is no inﬁnitary ﬁrst order description which char- acterises real spectra.

The second point needs explanation. By Stone duality, the category ofspectral spaces and spectral maps is anti-equivalent to the category of bounded distributive lattices and bounded lattice homomorphisms. Hence the question what the topological type ofreal spectra is, can also be asked on the lattice side: Classify all lattices that correspond to real spectra via Stone duality.

Now the class ofall bounded distributive lattices is first order axioma- tisable in the language ofposets (consisting ofone binary relation symbol) and we can rephrase our question in terms ofmodel theory: Is the class of those lattices that correspond to real spectra via Stone duality, first order axiomatizable? It should be mentioned here that this indeed generalises the original question, since the class ofall lattices corresponding to completely normal spectral spaces is easily seen to be first-order axiomatizable (by ex- pressing that for each elementaofthe latticeL, the lattice{b∧ ¬a|b∈L} is normal, cf. [8]).

Now our theorem 5.1 also negates this more general question in a strong way: For every cardinal λ, the class ofall lattices that correspond to real spectra via Stone duality is not ﬁrst order axiomatizable in the inﬁnitary languageL∞λ ofposets.

In this context it must be mentioned that the (specialisation-)order type ofreal spectra is know by [5]. Whereas the (specialisation-)order type of arbitrary spectral spaces is still unknown (this is called Kaplansky’s problem and asked in [9, chap. 1]). The topological type ofZariski spectra ofrings has been determined by Hochster in the ﬁrst place: these are precisely the spectral spaces (cf. [6]).

For model theoretic terminology see [7]; the deﬁnition of L∞λ can be found in [4, p.65]. For basic properties of spectral spaces we refer to [6], [2]

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and [8]; a summary can be found in [15, section 2]). The deﬁnition of the real spectrum and the fact that it is indeed a completely normal spectral space can be found in [1].

Acknowledgement. — We would like to thank Niels Schwartz for valu- able discussions and comments on the topic during the “Passauer Obersem- inar Reelle Geometrie”.

2. Spectral spaces from Dedekind complete orders

Recall that a totally ordered setX = (X,) isDedekind completeif every subset ofX has a supremum inX. In this case, every subset ofXhas an inﬁmum,X has a largest element, denoted byand a smallest element, denoted by⊥.

IfX is a totally ordered set, then theDedekind completion ofX is deﬁned to be “the” Dedekind complete set X, containing X as an ordered subset, such that X is dense in X, i.e. for all y1, y2 ∈ X with y1 < y2, if there is no point x ∈ X with y1 < x < y2, then y1, y2 ∈ X. An explicit description ofX is

X :=X∪ {+∞,−∞} ∪ {the non-principal cuts ofX}

together with its natural order. Ofcourse we won’t add±∞ifXhas already a largest or a smallest element. Recall that a cut (L, R) of X is called principal, ifthe setLhas a supremum inX∪ {±∞}; these are the cuts of the form±∞or ofthe formx⁺ orx⁻ for somex∈X.

We recall a few easy facts related to Dedekind complete (total) orders.

Firstly, recall that the interval topology (or order topology) ofa totally ordered set (X,) is deﬁned to have the subbasis ofopen sets ofthe form (−∞, x) and (x,+∞), wherex∈X.

Fact 2.1. — For every totally ordered setX, the following are equivalent:

(i) X is compact in the interval topology.

(ii) X is Dedekind complete.

Fact 2.2. — IfX andY are Dedekind complete (total) orders, then the lexicographical productX×Y is again Dedekind complete.

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Fact 2.3. — The following are equivalent for every totally ordered setX: (i) X is compact and connected in the interval topology.

(ii) X is Dedekind complete and there is nojumpinX, i.e. there are no x < yin X such that the open interval(x, y)is empty.

Fact 2.4. — The following are equivalent for every totally ordered setX: (i) X is boolean (i.e. X is compact and every connected subset of X is

a singleton) in the interval topology.

(ii) X is Dedekind complete andjump dense, i.e. for allx < yfrom X there arexa < by such that (a, b) =∅.

If this is the case, then the clopen subsets ofX are the ﬁnite unions of closed intervals[a, b]such that ahas an immediate predecessor inX∪ {±∞}and b has an immediate successor inX∪ {±∞}.

Lemma 2.5. — Let X be a Dedekind complete totally ordered set and letY be a Dedekind complete, totally ordered set that is boolean in the order topology and that has at least two elements. ThenX×Y is again Dedekind complete and boolean in the lexicographical order topology.

Proof. — By 2.2 and 2.4 we only need to show that X ×Y is jump dense. This is obvious.

Let S be a ﬁnite totally order set with at least 2 elements. We will now consider X ×S lexicographically ordered together with the induced order topology.

By 2.5 the lexicographic product X ×S is a boolean and Dedekind complete (total) order. It is worth noting that the order topology onX×S is in general incompatible with the product topology onX×S, i.e. neither reﬁnes nor coarsens this topology.

Definition 2.6. — LetS be an arbitrary boolean space. A partial order onS is called aspectral order onS if for allx, y∈X withxy there is a clopen subset C of X such that x ∈ C, y ∈ C and such that C is a ﬁnal segment of , i.e. for all c, z∈X,c∈C andcz impliesz∈C.

The pair(S,)is called aPriestley space(cf. [10]).

A morphism between Priestley spaces(X,^X)and(Y,^Y)is a contin- uous map f :X −→Y which preserves the spectral orders, i.e. aX b⇒ f(a)Y f(b).

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We shall use basic properties and terminology ofspectral spaces. IfXis a spectral space, thenK(X) denotes the set ofclosed and constructible subsets of X. Note that K(X) is the bounded distributive lattice corresponding to X via Stone duality (in our setup, such a lattice L is mapped to the spectral space SpecL ofprime ﬁlters ofL which has the sets VL(a) :={∈

SpecL|a∈L}as a basis of closed sets). We refer to [15, section 2] for more details.

Theorem 2.7. — The functor from the category of spectral spaces together with spectral maps to the category of Priestley spaces, which sendsX to(Xcon,X)is an isomorphism. Here Xcon denotes the patch space of X andX denotes specialisation inX.

Proof. — This is the content of[11].

Example 2.8. — If X = (X,) is a totally ordered set and boolean in the order topology, then there is a (unique) spectral topologyτ onX such that

(a) The constructible topology ofτ is the order topology ofX.

(b) For allx, y∈X,xy Xifand only ifyis in the closure ofxw.r.t.τ.

Hence the category ofDedekind complete, totally ordered and jump dense sets is isomorphic to the category ofspectral spaces which have a total specialization order.

Proposition 2.9. — LetX be a Dedekind complete total order and let S be a ﬁnite spectral space. Let be any total order of S such that ⊥^S and^S are minimal points in the spectral topology of S (of course, such a total order only exists ifS is a singleton or not irreducible). Let X×S be equipped with the order topology of the lexicographic order ofX and(S,).

Then the partial order

(x, s)(y, t) :⇐⇒ x=y andst inS (i.e.t∈ {s}) is a spectral order on the boolean space X×S.

Proof. — X×S is boolean by 2.5.

Takeα= (x, s), β= (y, t)∈(X, S) withαβ. We have to ﬁnd a clopen subset ofX×Sthat is a ﬁnal segment w.r.t.such thatα∈Candβ∈C.

Case 1. x < y.

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Ifx < yis a jump ofX, then takeC:= (−∞,(x,^S)] = (−∞,(y,⊥^S)).

If x < y is not a jump of X, then there is some z ∈ X with x < z < y.

Letv∈S be the predecessor of^S in the total order ofS. Then, as^S is minimal in the spectral order ofS, the clopen subset C:= (−∞,(z, v)] = (−∞,(z,^S)) ofX×S is a ﬁnal segment w.r.t..

Case 2. y < x.

Ify < xis a jump ofX, then takeC:= [(x,⊥^S),+∞) = ((y,^S),+∞).

If y < x is not a jump of X, then there is some z ∈ X with y < z < x.

Let u∈S be the successor of ⊥^S in the total order of S. Then, as ⊥^S is minimal in the spectral order of S, the clopen subset C := [(z, u),+∞) = ((z,⊥^S),+∞) ofX×S is a ﬁnal segment w.r.t..

Case 3. x=y. Hence, since αβ we knowst inS.

We ﬁrst claim that for everyr∈S, there is a clopen setCr⊆X×Ssuch thatCr\({x}×S) is closed underand such thatCr∩({x}×S) ={(x, r)}:

• Ifr =^S and r=⊥^S, then the point (x, r) is isolated in the order topology ofX×S, henceCr:={(x, r)} has the required properties.

• If r = ^S, then let v ∈ S be the predecessor of ^S in the total order ofS. ThenCr= [(x, r),+∞) = ((x, v),+∞) has the required properties.

• If r =⊥^S, then then let u ∈ S be the successor of ⊥^S in the total order ofS. ThenCr= (−∞,(x, r)] = (−∞,(x, u)) has the required properties.

So we can findCras claimed and we define C as the finite union

C:=

r∈S,sr

Cr.

Then C is clopen in X×S andC\({x} ×S) is closed under (as each Crhas this property). ButC∩({x} ×S) is equal to{x} × {s}^S, which is a ﬁnal segment w. r. t. , too. ThusC is a ﬁnal segment w.r.t. such that α∈C andβ∈C.

Definition 2.10. — Let X be Dedekind complete and letS be a ﬁnite spectral space which is not irreducible. Let be any total order of S such that ⊥^S and^S are minimal points in the spectral topology ofS.

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We deﬁne

X#(S,)

as the spectral space whose patch space is the Dedekind complete total order X ×S and whose specialisation order is the partial order from 2.9. If S is the spectral space consisting of three elements−1,0,1 with specialization order −1,10 and total order −1<0<1, then we write

X^# instead ofX#(S,).

For x ∈ X we write x⁻ := (x,−1), x⁺ := (x,1) and we identify X with X× {0} inX^#.

Proposition 2.11. — LetX = (X,)be Dedekind complete and densely totally ordered. A subsetY of X^#is closed and constructible if and only if Y is a ﬁnite union of sets of the form [(x,0),(x,0)], where x, x ∈X and xx.

In particularK(X^#) is order isomorphic to the lattice generated by the closed intervals ofX in the powerset of X

Proof. — This is clear with the characterization ofthe constructible subsets ofX^# in 2.4: Observe that the constructible subsets ofX^#are by deﬁnition the clopen subsets ofthe lexicographic orderX×{−1,0,1}.

3. Completely normal spectral spaces not occurring in real spectra

We use standard notation from commutative algebra: Let A be a ring (this means, commutative and unital always). Forf ∈A, we denote byV(f) the set ofall prime ideals ofAcontainingf andD(f) = SpecA\V(f).

Following [3] we shall work with Zariski spectra ofreal closed rings (in the sense ofN. Schwartz) instead ofreal spectra ofcommutative rings.

Note that every real spectrum is (naturally) homeomorphic to the Zariski spectrum of a real closed ring. We refer to [12] and [13] for real closed rings.

Proposition 3.1. — LetA be a real closed ring and let X ⊆SpecA.

Suppose we are given

(I) an open quasi-compact subset U of SpecA such that U∩X is connected and such that there are pointsx, y∈U∩X which do not have a common specialization inU;

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(II) an ordinalλand for eachα < λ,fα∈A such that for allα < β < λ we have

fα(u)> fβ(u)>0 (u∈U∩X).

Then there is a collection(Uα)α<λ of open, nonempty and disjoint subsets ofU∩X.

Proof. — SinceAis real closed, there is somes∈Asuch thatU =D(s).

Sincex, y∈X do not have a common specialization inU, there areϕ, ψ∈A withx∈V(ϕ)∩D(s),y∈V(ψ)∩D(s) andV(ϕ)∩V(ψ)∩D(s) =∅. Then ϕ²+ψ² is a unit inAs, hence there is someh∈Asuch that

sⁿ·(h·(ϕ²+ψ²)−s^2k) = 0.

Deﬁne

g:=f0·ϕ²·h.

Theng(x) = 0 andg(y) =f0(y)·s^2k(y). We take

Uα:={u∈U∩X |s^2k·fα(u)< g(u)< s^2k·fα⁺(u)} (α < λ).

Clearly theUαare open and disjoint subsets ofX. It remains to show that eachUαis nonempty.

OtherwiseU∩X ⊆ {gs^2k·fα}∪{gs^2k·f_α⁺}is covered by two closed subsets ofSpecA, which have empty intersection inU∩X by assumption.

Since g(x) = 0 s^2k(x)fα(x) andg(y) = s^2k(y)·f0(y) s^2k(y)·fα⁺(y) both sets are nonempty. This contradicts the assumption that U ∩X is

connected.

Corollary 3.2. — Let A be a real closed ring and let X ⊆ SpecA contain at least two points. Suppose

(I) X is quasi-compact, connected and there is no specialization inside X.

(II) There are an ordinalλ and for eachα < λ,fα∈Asuch that for all α < β < λwe have

fα(u)> fβ(u)>0 (u∈X).

Then there is a collection (Uα)α<λ of open, nonempty and disjoint subsets of X.

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Proof. — Take x, y∈X, x=y and suppose x, yhave a common specialization in SpecA. Let z be the ﬁrst common specialization of x, y in SpecA.

Then for eachw∈Xwe havezw, hence there is an open quasi-compact subset Uw ofSpecA with w∈ Uw and z ∈ Uw. Then X ⊆

w∈XUw and sinceX is quasi-compact, there is an open quasi-compact subsetUofSpecA containingXand not containingz. Sincezis the ﬁrst common specialization of x, y in SpecA,x, y do not have a common specialization in U. Now we may apply 3.1.

Recall that for an inﬁnite cardinalκ, aκ-set X is a totally ordered set with the property that for allA, B⊆Xofsize strictly less thanκ, ifA < B, then there is some x∈X withA < x < B. For example everyκ-saturated totally ordered set is aκ-set.

Proposition 3.3. — LetD=E×I, whereE is the Dedekind completion of a κ-set Y,κan inﬁnite cardinal, andI is Dedekind complete.

SupposeD^#is a proconstructible subset ofSperAfor some ringA. Then there are fα ∈A and xα, yα ∈Y (α < κ) such that for all α < β < κ we have

xα< xβ< yβ< yα and

for all u∈D with (xβ,⊥^I)< u <(yβ,^I) :fα(u)> fβ(u)>0.

Proof. — Letf0 = 1 and supposexµ < xν < yν < yµ,fµ > fν >0 on [(xν,⊥^I),(yν,^I)], have already been constructed for µ < ν < λ, λ < κ.

We define xν < xλ < yλ< yν andfλ ∈A with fν > fλ >0 on [(xλ,⊥Î), (yλ,Î)] as follows:

Pick u, v ∈ Y with xν < u < v < yν (ν < λ). Then the set [(u,^I)⁺, (v,⊥^I)⁻] ⊆D^# is open and quasi-compact, there is some fλ ∈ A that is

>0 on that set and 0 on the complement ofthis set inD^#.

Pickµ < λ. Then{fµ> fλ} ∩D^#is open and quasi-compact containing (u,Î). Since the open quasi-compact subsets ofD^#are finite unions ofsets ofthe form [d⁺₀, d⁻] withd0, d∈D there is somedµ∈Dwith (u,Î)< dµ

such that [(u,^I), d⁻_µ)⊆ {fµ > fλ}. From (u,^I)< dµ we get somee∈E with u < eµ dµ. Since u∈Y and E is the Dedekind completion ofthe dense setY, there is someuµ∈Y withu < uµ< dµ.

Hence fµ > fλ >0 on [(u,^I)⁺,(uµ,^I)]. SinceY is a κ-set, there is some yλ ∈ Y with u < yλ < xµ for allµ < λ. Now choose xλ ∈ Y with u < xλ< yλ.

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Theorem 3.4. — LetD=E×I, whereE is the Dedekind completion of a κ⁺-set Y, κ an inﬁnite cardinal, and I is a compact and connected Dedekind complete total order with at least two points, such that there is no collection of nonempty, open and disjoint subsets of I of size κ. Then the completely normal spectral space D^# is not homeomorphic to any spectral subspace of the real spectrum of any ring.

The main example here isI= [0,1]⊆andκ=ℵ¹.

Proof. — By 3.3, there are fα ∈ A and xα, yα ∈ Y such that for all α < β < κwe have xα< xβ < yβ< yαand

fα(u)> fβ(u)>0

whenuranges in the interval ((xβ,⊥^I),(yβ,^I)) ofD. SinceY is aκ⁺-set, there arex, y∈Y withxα< x < y < yα for allα < β < κ. Hence

fα(u)> fβ(u)>0 (u∈D,(x,⊥^I)< u <(y,^I)).

Pick z∈Y withx < z < y and letX :={z} ×I. Then fα(u)> fβ(u)>0 (u∈X).

SinceX is homeomorphic toI,X is compact and connected by assumption.

Hence we may apply 3.2, which gives a collection (Uα)α<κofopen, nonempty and disjoint subsets ofX. SinceX is homeomorphic toI, this contradicts our assumption onI.

4. Back and forth equivalence of lattices generated by closed intervals

LetX = (X,) be a totally ordered set. Aclosed intervalofX is any subset ofthe form [a, b], [a,+∞), (−∞, b] or (−∞,∞) with a, b∈X. The boundary pointsof[a, b] are defined to bea,bifab, the only boundary point of[a,+∞) is defined to bea(providedXhas no largest element) and the only boundary point of(−∞, b] is defined to be b (provided X has no smallest element). The empty set and X (providedX has no smallest and no largest element) do not have boundary points. Observe that a boundary point according to this definition is in general not a boundary point w.r.t.

the order topology ofX (X might be discrete).

LetL(X) be the set ofﬁnite unions ofclosed intervals ofX. Obviously, L(X) is the sublattice ofthe powerset ofX generated by the set ofclosed intervals.

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Forα∈L(X), a closed interval ofX contained inαand maximal with this property is called a component of α; observe again that this is in general not a connected component ofαin the sense ofthe order topology ofX. However, it is clear thatαis the disjoint union ofits components and there are only ﬁnitely many ofthem.

Forα∈L(X), aboundary point of αis a boundary point ofone of its components. We write bd(α) for the (ﬁnite) set of boundary points ofα.

It is useful to notice the following:

Observation 4.1. — Letα∈L(X).

(i) Ifnα=I1∪...∪In with closed intervalsI1, ..., In ofX, then bd(α)⊆

k=1bd(Ik).

(ii) There aren∈N0 and nonempty closed intervalsI1, ..., In ofX with α=I1∪...∪In such that for each k∈ {1, ..., n−1} there is some x∈X withIk< x < Ik+1.

Moreover, wheneverαis represented in this form, thenI1, ..., In are precisely the components ofα.

GivenS⊆X, we deﬁne

L(X, S) :={α∈X | bd(α)⊆S}.

Remark4.2. — L(X, S) is a bounded sublattice of L(X) and for every other set S⊆X we have S⊆S ⇐⇒ L(X, S)⊆L(X, S).

Proof. — By 4.1(i) it is clear thatL(X, S) is a bounded sublattice of L(X). The equivalence follows with the observation{s} ∈L(X, S).

Now letX, Y be totally ordered sets and letS ⊆X,T ⊆Y be arbitrary sets. Suppose we are given an order isomorphismf :S −→T. We deﬁne a map Ff :L(X, S)−→L(Y, T) as follows: For a closed interval αof X we deﬁne the closed intervalFf(α) ofY by

Ff(α) =











∅ ifα=∅

Y ifα=X

[f(s),+∞) ifα= [s,+∞) andsis not the smallest element ofX

(−∞, f(s)] ifα= (−∞, s] andsis not the largest element ofX

[f(s1), f(s2)] ifs1s2, α= [s1, s2] ands1 is not the smallest element ofX ands2is not the largest element ofX

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Now we deﬁneFf on all ofL(X, S) by Ff(α) =

{Ff(γ)|γis a component ofα}.

In the situation above, we say that the order isomorphism f :S −→T is faithfulifthe following two conditions are satisﬁed:

(a) For everys∈S,sis the smallest or largest element ofX ifand only iff(s) is the smallest or largest element ofY, respectively.

(b) For alls1, s2∈S we have

s1< s2 is a jump inX ⇐⇒ f(s1)< f(s2) is a jump inY.

Proposition 4.3. — If f is faithful, then Ff is a lattice isomorphism L(X, S) −→ L(Y, T) and the inverse is Ff⁻¹. If S ⊆ X, T ⊆ Y and f : S −→T is another faithful order isomorphism, then f extends f if and only ifFf extendsFf.

Proof. — On the level ofclosed intervals, the mapFf⁻¹ is inverse toFf, since f satisfies condition (a) ofthe definition of”faithful”. Using 4.1(ii) and becausef satisfies condition (b) of the definition of “faithful”,Ff maps a component γ of α∈L(X, S) to the componentF(γ) of F(α). It is then clear that Ff⁻¹ is inverse to Ff. Since both maps obviously are monotone, the first assertion follows.

For the equivalence, only⇐ needs a proof. However ifFf extends Ff

and s∈S then {f(s)}=Ff({s}) = Ff({s}) ={f(s)}and so f extends

f.

Recall that aback-and-forth systemofﬁrst order structuresM and N (in an arbitrary ﬁrst order language) is a non-empty family (fi:Mi−→

Ni |i ∈ I) ofisomorphisms fi between substructures Mi of M and Ni of N, respectively, satisfying the following conditions:

Forth: For all i ∈I and a∈M there is j ∈I witha ∈Mj such thatfj

extends fi; hence alsoMi⊆Mj, Ni⊆Nj.

Back: For all i ∈ I and b ∈ N there is j ∈ I with b ∈ Nj such that fj

extends fi.

Ifthere is a back and forth system betweenM and N, thenM andN are calledback and forth equivalent.

Ifλis a cardinal, then a back and forth system (fi:Mi−→Ni |i∈I) is said to have the λ-extension property iffor allJ ⊆I ofcardinality

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< λ such that the family (fj : Mj −→ Nj |j ∈ J) is totally ordered by extension, there is some i ∈ I such that fi extends fj for all j ∈ J. So trivially, every back and forth system satisﬁes theℵ⁰-extension property.

Ifthere is a back and forth system ofM andN that has theλ-extension property, then M and N are calledstrongly λ-back and forth equivalent.

LetXandY again be totally ordered sets and let (fi :Si−→Ti |i∈I) be a back and forth system ofX andY.

LetFi :=Ffi in the notation introduced before 4.3. Since (fi : Si −→

Ti |i∈I) is a back and forth system ofX andY, it is clear that every fi

is faithful. By 4.3,

Fi:L(X, Si)−→L(Y, Ti)

is a lattice isomorphism and we claim that (Fi:L(X, Si)−→L(Y, Ti)|i∈ I) is a back and forth system ofL(X) andL(Y).

By 4.2 we already know that L(X, Si) and L(Y, Ti) are sublattices of L(X), L(Y) respectively. By symmetry we then only need to check the

“Forth” condition. Let i ∈ I and α ∈ L(X). Then bd(α) is ﬁnite and by applying the “Forth” condition ofthe system (fi : Si −→ Ti |i ∈ I) ﬁnitely many times, there is some j ∈I such that bd(α)⊆Sj. By 4.1(ii), α∈L(X, Sj) and asfj extendsfi, also Fj extends Fi.

Thus we know that (Fi : L(X, Si)−→ L(Y, Ti)|i ∈ I) is a back and forth system ofL(X) andL(Y).

Now suppose (fi : Si −→ Ti |i ∈ I) is a back and forth system of X andY that has theλ-extension property for some cardinalλ.

Then also (Fi :L(X, Si)−→L(Y, Ti)|i∈I) has theλ-extension property. To see this, it is enough to recall that Fi extends Fj ifand only iffi

extends fj (cf. 4.3).

Theorem 4.4 [4, Thm 5.3.7, p. 316 and the notation before Thm.

5.3.2]. — If M, N are elementary equivalent,λ-saturated, ﬁrst order structures, then they are stronglyλ-backand forth equivalent.

Moreover, stronglyλ-backand forth equivalent structures are elementary equivalent in the inﬁnitary languageL∞λ.

Scholium 4.5. — Letλbe a cardinal and letX, Y beλ-saturated totally ordered sets. Suppose X and Y are elementary equivalent (for example if

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X andY are dense and totally ordered with endpoints). Then the lattices L(X) andL(Y) are stronglyλ- back and forth equivalent.

Moreover L(X) and L(Y) are elementary equivalent in the inﬁnitary languageL_∞λ.

Proof. — By 4.4,X and Y are stronglyλ - back and forth equivalent.

We have shown in this section that in this case, also the latticesL(X) and L(Y) are strongly λ - back and forth equivalent. By 4.4 again, L(X) and L(Y) are elementary equivalent in the inﬁnitary languageL∞λ.

5. Main Theorem

Theorem 5.1. — For every cardinal λ, there are bounded distributive lattices LandL, such that

(i) The lattices L and L are strongly λ-backand forth equivalent; in particular, they are elementary equivalent in the inﬁnitary language L_∞λ.

(ii) L=K(C)for some spectral spaceCthat is not homeomorphic to any proconstructible subset of any real spectrum of a ring.

(iii) L=K(SperA)for some ringA.

Proof. — Definition of L and A:

We choose aλ-saturated real closed ﬁeldR. LetAbe the ring ofcontin- uous semi-algebraic functions [0,1] −→R, where [0,1] is the unit interval in Rand deﬁneL =K(SperA).

Definition of L andC:

LetI be a densely totally ordered, λ-saturated set. Letκλbe such that there is no collection ofnonempty, open and disjoint subsets ofI of size κ. Let Y be aκ⁺-saturated total order. Now deﬁne L=K((Y ×I)^#) (recall thatY,I denote the Dedekind completions ofY,I respectively).

(i) We show thatL∼=L(Y×I),L∼=L([0,1]) and verify the assumptions of4.5 forY ×I and [0,1]:

• SinceIandY areλ-saturated, also the Dedekind completionsY and Iareλ-saturated. Thus,Y ×I isλ-saturated, too.

• By 2.11 we haveL∼=L(Y ×I).

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• Since R is λ-saturated, the totally ordered set [0,1] is λ-saturated, too.

• It is well known that L is naturally homeomorphic to the lattice L([0,1]) ofﬁnite unions ofclosed intervals from [0,1], thus L ∼= L([0,1]).

• Both [0,1] andY×Iare dense and totally ordered sets with endpoints, so they are elementary equivalent.

Hence all assumptions of4.5 are satisﬁed and we obtain (i) from 4.5.

(ii) holds by 3.4 applied toY,I and the choice ofC.

(iii) holds by deﬁnition ofL.

A possible answer to the question on the determination ofthe topological type ofreal spectra can thus not be formulated in terms ofinﬁnitary ﬁrst order languages; at least not in an obvious way.

An interesting alteration ofthe question is the following: Is every spectral subspace ofthe real spectrum ofa ringAitselfthe real spectrum ofa ring B? More importantly, can we even construct B in a natural way out ofA and the given set?

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