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TOPOLOGYPROCEEDINGSVolume 25, Spring 2000

Pages 225-249

UPWARDS PRESERVATION BY ELEMENTARYSUBMODELS

LUCIA R. JUNQUEIRA

Abstract. Given a topological space X and an elementarysubmodel M we can define a new topological space XM . Weinvestigate for which topological properties P it is true thatif XM has P , then X has P . We first look at this questionin general and then we impose conditions on M . In partic-ular, we show some preservation results assuming M to beω-covering and we also show that, under CH, the proper-ties of being ω-covering and countably closed are equivalentfor any elementary submodel M . After, we investigate howmuch we can weaken the hypothesis of M being ω-covering.

1. Introduction

Let 〈X, T 〉 be a topological space and let M be an elementarysubmodel of some “large enough” H(θ) (see e.g. [11], [3] or [20]).In [10] we considered the new topological space XM = X ∩M withthe topology generated by T ¹ M = {U ∩M : U ∈ T ∩M} and weinvestigated, in particular, which topological properties of X werepreserved when we take XM . This was a first step in the systematicstudy of topological spaces induced by elementary submodels. Themotivation of this study was the fact that elementary submodelshave been used in set-theoretic topology with increasing frequencyover the past 20 years; see for example [3], [4], [2], [20], [1].

2000 Mathematics Subject Classification. Primary 03C62, 54A65, 54D20,54D35, 54E35; Secondary 54D15, 54D30.

Key words and phrases. elementary submodel, Lindelof, first countable,second countable, separable, pointwise countable type.

This work was partially supported by Fapesp grant 97/09813-0.225

226 LUCIA R. JUNQUEIRA

In this paper we take a slightly different approach than the oneconsidered on [10]. There the approach was “downwards”, i.e. go-ing from X to XM , here it is “upwards”. More precisely, we inves-tigate for which topological properties P the statement “XM has Pimplies X has P” is true. Through this article, when this statementis true we will say that P is “preserved upwards”.

This approach was considered also in [19], [14], [16], where someupwards preservation results were proved and some questions aboutrecovering X from XM were answered. We refer the reader to anyof the papers mentioned above for a more detailed introduction onthe topic.

In this paper we investigate the upwards preservation of somebasic topological properties as separability, compactness, secondcountability, etc. In [19], F. D. Tall proved that T2 and T3 are pre-served upwards. It is easy to see that T0 and T1 are also preservedupwards (this is similar to the T2 case). On the other hand, it iseasy to see that T3 1

2and T4 are not preserved in general. For in-

stance, let X be any regular space that is not T3 12

(or not T4) andtake M a countable elementary submodel. Then XM will be a reg-ular second countable space (since M is countable) and therefore itis metrizable. This example also shows that we cannot expect manyproperties to be preserved in general. For a less trivial example ofnon-preservation of normality (one where XM is uncountable) wehave the following:

Example 1.1. There is a non-normal space X and an elementarysubmodel M such that XM is uncountable and normal.

Proof: Let A and B be two disjoint stationary subsets of somecardinal κ such that for each α < κ we have that A ∩ α and B ∩ αare non-stationary. Let M be any elementary submodel such thatM ∩ κ is an ordinal γ < κ.

We take X = A× B. By a result of [12], we have that X is notnormal (it is a product of two stationary disjoint subsets). However,the same result implies that XM = (A ∩ γ)× (B ∩ γ) is normal.

Remark 1.2. A consistent first countable example is to take X tobe the Niemytzky plane. Assume MA + ¬CH and take M of sizeℵ1. Then XM is normal because M ∩ R is a Q-set (see e.g. [18]).

UPWARDS PRESERVATION 227

However, there are some general positive preservation resultswhich are shown in the second section of this paper. In the thirdsection we show some preservation results assuming M to be ω-covering and in the fourth section we investigate how much wecan weaken this hypothesis on M . In the last section we connectpreservation of non-metrizability to Hamburger’s question.

I would like to thank G. Gruenhage and R.G.A. Prado for giv-ing me permission to include their results (in section 3 and 4) inthis paper. I would also like to thank Frank D. Tall for his usefulcomments and corrections on previous versions of this article.

2. Some general preservation results

In this section we show that there are some properties that areupwards preserved for any elementary submodel M .

Theorem 2.1. Let X be a topological space and let M be an ele-mentary submodel. If XM is compact, then X is compact.

Proof: Suppose X is not compact. This means that

∃C ∈ P(T ) such that⋃C = X and ∀C′ ⊆ C finite ∃x ∈ X \

⋃C′.

By elementarity, we can get C ∈ M . Fix such C. We then have that

H(θ) |=⋃C = X and ∀C′ ⊆ C finite ∃x ∈ X \

⋃C′.

Since C ∈ M , by elementarity, we have that⋃

(C ∩M) ⊇ X ∩M .But then C ∩M is an open cover of X ∩M without finite subcover(by elementarity, C ∩M has no finite subcover in M , but all finitesubsets of C ∩M are in M). This contradicts the compactness ofXM and we are done.

Note that a similar argument will show:

Theorem 2.2. Let X be a topological space and M be an elemen-tary submodel. If XM is countably compact, then X is countablycompact.


Proof: Just have to repeat the previous proof but assuming Ccountable.

The main theorem of section 3 in [10] implies, in particular, thatif X is compact Hausdorff and M is an elementary submodel withX ∈ M , then there is Y ⊆ X such that XM is a perfect preimageof Y . To show this result, it was defined, for every x ∈ XM , the set

Kx =⋂{V ∈ T ∩M : x ∈ V }.

Using this notation and the proof of this theorem we can showthe following result (M will always be an elementary submodelsatisfying X ∈ M):

Lemma 2.3. If XM is compact, then for every x ∈ X there isx′ ∈ X ∩M such that x ∈ Kx′.

Proof: Suppose not. Then there is x ∈ X such that for everyx′ ∈ X ∩M we have x /∈ Kx′ . Fix such x. By the definition of Kx′ ,this implies that

for all x′ ∈ X ∩M there is Vx′ ∈ T ∩M such that

x′ ∈ Vx′ and x /∈ Vx′ .

But then {Vx′ : x′ ∈ X ∩M} is an open cover of XM . Therefore,by compactness of XM , it has a finite subcover V. Since V ⊆ Mand V is finite, we have that V ∈ M . Also, M |= V covers X. Thus,by elementarity, V covers X, a contradiction because x /∈ ⋃V.

We can now show:

Theorem 2.4. If XM is compact and T2, then there is a perfectmap from X to XM .

Proof: Take Y as in the proof of Theorem 3.2 in [10], namelyY =

⋃{Kx : x ∈ XM}. By the previous lemma we have thatY = X. Also, by theorem 2.1, X is compact and therefore regular(it is T2 because XM is T2 and being T2 is upwards preserved [19]).We can now apply Theorem 3.2 in [10] to get the perfect map fromX to XM .

We then have the following:


Corollary 2.5. XM is compact T2 if and only if X is compact T2

and XM is a perfect image of X.

F. D. Tall showed in [19] that the property of being locally com-pact is also preserved upwards for any elementary submodel M .It is not hard to show that connectedness is also always preserved(this was independently shown by R.G.A. Prado and F. D. Tall[17]):

Theorem 2.6. For any elementary submodel M , if XM is con-nected then X is connected.

Proof: Suppose X is not connected. Then there are two disjointclopen subsets of X, A and B, such that X = A ∪ B. By elemen-tarity, there are such A and B in M . But then XM = (A∪B)∩M ,and A ∩M and B ∩M are clopen in XM , a contradiction.

3. M ω-covering

Some basic topological properties involving countable sets, likeseparability, second countable, etc, are not preserved in general(for instance we can take X regular and not having one of theseproperties and M countable, as done in the introduction). But theywill be preserved if we make extra assumptions on M . The morenatural assumption would be to assume M countably closed, whichseems a bit strong. We will show that it is enough to assume Mω-covering, which is in general weaker (we can have M ω-coveringof size ℵ1 [3]). However, we will also show that, under CH, theproperties of being ω-covering and countably closed are equivalent.

Definition 3.1. An elementary submodel M is countably closed ifit is closed for countable subsets, i.e. ωM ⊆ M .

Definition 3.2. An elementary submodel M is ω-covering if forevery countable subset A of M there is a countable set B ∈ M suchthat A ⊆ B.

In [10] we proved some results in which we assumed M to becountably closed and raised the question if this hypothesis could beweakened to M ω-covering. We had examples that showed that in


some cases it cannot, if one assumes ¬CH, but there was still thecase of what happens under CH. We show here that, under CH,these two properties are equivalent, which solves the problem.

We separate the following lemma from the proof because it willbe used later:

Lemma 3.3. If M is an ω-covering elementary submodel, thenω1 ⊆ M .

Proof: Suppose that there is α < ω1 such that M ∩ω1 = α (notethat M ∩ω1 is always an ordinal). But then, by ω-covering, there isa countable A ∈ M such that α ⊆ A. Since A is countable, A ∈ Mimplies that A ⊆ M . Also, we can suppose that A ⊆ ω1 (just takeA′ = A ∩ ω1). Now, α /∈ M and A ∈ M imply that α ( A. Thus,there is β ∈ (M ∩ ω1) \ α, contradicting the choice of α.

Theorem 3.4. CH is equivalent to every ω-covering elementarysubmodel of Hθ is countably closed.

Proof: It is clear that if ¬CH, then there is an elementarysubmodel M which is ω-covering but not countably closed (just takeany ω-covering elementary submodel of size ℵ1). Assume now thatwe have CH and that M is an ω-covering elementary submodel.

By CH, Hθ |= there is a bijection f : ω1 −→ P(ω). Then,by elementarity, there is f ∈ M , such that f : ω1 −→ P(ω) is abijection. Since f ∈ M and ω1 ⊆ M , we then have that P(ω) ⊆ M .

Fix A ⊆ M countable. We have to show that A ∈ M . By ω-covering, there is B ∈ M countable such that A ⊆ B. Since B iscountable, there is a bijection g : B −→ ω. By elementarity we cantake g ∈ M (since B ∈ M).

Let C = {n ∈ ω : there is a ∈ A such that g(a) = n} = f(A).Then C ⊆ ω and therefore C ∈ M (since P(ω) ⊆ M). But g ∈ M ,so g−1(C) = A ∈ M and we have what we wanted.

We now give some preservation results. The result for secondcountable spaces is essentially proved in [3] and the result for count-able tightness is due to R.G.A. Prado and is in [16].

Theorem 3.5. The following topological properties are preservedupwards if we assume M to be ω-covering:

(a) second countability;(b) first countability;


(c) Lindelofness;(d) separability;(c) countable tightness.

Proof: (a) Since XM is second countable, there is B ⊆ T ∩ Mcountable such that it is a base of XM . But then, by ω-covering,there is B′ ∈ M countable such that B ⊆ B′. We can take B′ ⊆ T(since T ∈ M).

Since B ⊆ B′, we have that B′ is also a base of XM . Thus, usingB′ ∈ M we have

M |= B′ is a base of X,

which, by elementarity, implies that B′ is a base of X. Thus, X issecond countable.

(b) This is similar to (a) but we work with a base Bx for x ∈X ∩M instead of B.

(c) We carry out the proof as in the compact case (theorem 2.1),but with a countable subcover instead of a finite subcover. The onlydifference will be that we do not necessarily have that all countablesubsets of C ∩M are in M . But we can use ω-covering to get thatfor every C′ ⊆ C countable, there is C′′ ∈ M countable satisfyingC′ ⊆ C′′. Since C ∈ M we can take C′′ also satisfying C′′ ⊆ C.So, if C ∈ M is the open cover witnessing that X is not Lindelof,then C ∩ M has no countable subcover in M . But XM Lindelofimplies that there is a countable subcover C′ of C ∩M , so there isa countable subcover C′′ ∈ M as above, a contradiction.

(d) The idea is the same as in the previous proofs. If XM isseparable, then there is a countable set D ⊆ X ∩M dense in XM .But M ω-covering gives us a countable set E ∈ M such that D ⊆ Eand we can take E ⊆ M ∩ X. Since D ⊆ E, E is dense in XM ,and therefore M |= X has a countable dense subset. Thus, byelementarity, X is separable.

(e) Suppose not, then there is A ⊂ X and there is x ∈ cl(A)such that for every B ⊂ A, |B| ≤ ℵ0, we have that x /∈ cl(B). Byelementarity, we can take A ∈ M and x ∈ M . Note that x ∈ cl(A)implies x ∈ clτM (A ∩M) (also by elementarity).

As t(XM ) = ℵ0 , there is B ⊂ A ∩ M such that |B| = ℵ0 andx ∈ clτM (B). Using that M is ω−covering, there is C ∈ M, |C| =


ℵ0 and B ⊂ C. Call D = C ∩A. Notice that D ∈ M,D ⊂ A, |D| =ℵ0 and x ∈ clτM (D ∩M).

To finish the proof note that x ∈ cl(D). For otherwise

∃V ∈ T such that x ∈ V and V ∩D = ∅.This implies,

M |= ∃V ∈ T such that x ∈ V and V ∩D = ∅.Pick this V in T ∩ M . The previous sentence tells us that V ∩D ∩M = ∅. This contradicts the fact that x ∈ clτM (D ∩M). Sot(X) = ℵ0.

Corollary 3.6. If XM is separable metrizable and M is an ω-covering elementary submodel with X ∈ M , then X is separablemetrizable.

Proof: If XM is separable metrizable, then XM is second count-able and regular and therefore X is second countable regular. ThusX is separable metrizable.

It is natural to ask if we can get Lemma 2.2 for X Lindelof andM ω-covering. The same proof shows that the result is true if weassume M to be countably closed:

Theorem 3.7. For a countably closed elementary submodel M anda space X of pointwise countable type (with X ∈ M), if XM isLindelof, then XM is a perfect image of X.

We next show that, for regular spaces, pointwise countable typeis also preserved upwards if M is ω-covering:

Theorem 3.8. Suppose X is regular, M is an ω-covering elemen-tary submodel with X ∈ M and that XM has pointwise countabletype. Then X has pointwise countable type.

Proof: Fix x ∈ XM . Then there is a compact K (in XM ) suchthat x ∈ K and χ(K,XM ) = ℵ0. Let B = {Un : n ∈ ω} be acountable outer base of K. Note that each Un is an open set in XM

but we may have Un /∈ M .We claim that without loss of generality we can suppose B ⊆

T ∩M . To see that, fix n ∈ ω. For each y ∈ K, there is Vy ∈ T ∩Msuch that y ∈ Vy ⊆ Un. Then V = {Vy : y ∈ K} is an open cover of


K and therefore it has a finite subcover V ′. Since V ′ ⊆ M and it isfinite, V ′ ∈ M , so

⋃V ′ ∈ M . Also⋃V ′ ⊇ K and

⋃V ′ ⊆ Un. Takethen Vn =

⋃V ′. Clearly {Vn : n ∈ ω} is still an outer base of K.Now, M ω-covering, B ⊆ M and B countable imply that there is

B′ ∈ M countable such that B ⊆ B′ ⊆ T ∩M . Define Bx = {B ∈B′ : x ∈ B}. Note that Bx ∈ M , since x and B′ are in M , and thatBx is countable. We can then enumerate (in M) Bx = {Bn : n ∈ ω}.Also, by taking finite intersections we can make Bx decreasing andwe still have a family in M . Therefore, without loss of generality,we can suppose Bx is decreasing.

Still working in M , using the regularity of the space, we caninductively pick Vn ∈ T ∩M such that x ∈ Vn ⊆ Vn ⊆ Bn ∩ Vn−1,for each n ∈ ω. Let B′x = {Vn : n ∈ ω}. Thus B′x ∈ M and forevery n ∈ ω, we have Vn+1 ⊆ Vn+1 ⊆ Vn.

We then have that K ′ =⋂B′x is closed and it is in M , since

B′x ∈ M . Also K ′ ⊆ K (since Vn ⊆ Bn and Bx ⊇ B). Thus, K ′ is aclosed subset of K, so it is compact (in XM ), and it has countablecharacter.

We have shown that for every x ∈ X ∩ M there is a compact(in XM ) K ′ ∈ M such that x ∈ K ′ and χ(X ′, XM ) = ℵ0. Byelementarity, this is true in Hθ which implies that X has pointwisecountable type.

The next example shows that some assumption on M is neededin the previous theorem:

Example 3.9. Let X = Nω1 and suppose M is such that M ∩ω1 iscountable. Then X does not have pointwise countable type but XM

does.

Proof: X does not have pointwise countable type because itis an uncountable power of a non-compact space (see e.g. [6]).Since M ∩ ω1 is countable, we have that XM is homeomorphicto a subspace of Nω, which is first countable and therefore it haspointwise countable type.

Remark 3.10. We can have M ∩ ω1 countable without M beingcountable. This is the case if we have Chang’s Conjecture. But ifone assumes that 0# does not exist, M uncountable implies ω1 ⊆ M[14].


We finish this section with the following preservation results (forM countably closed) and examples from R.G.A. Prado’s thesis [16].Since these results do not appear elsewhere we include the proofshere (with R.G.A. Prado’s kind permission):

Theorem 3.11. Suppose that M is a countably closed elementarysubmodel and that X ∈ M is a topological space. If XM is sequentialthen X is sequential.

Proof: Suppose not, i.e. suppose there are A sequentially closedand x ∈ cl(A) \ A. By elementarity, we can pick such A and x inM .

Note that A ∩M is sequentially closed in XM . To see that, letf : ω −→ A∩M be a convergent sequence to y ∈ X ∩M (we meanconvergence in the TM sense) and call B = {f(n) : n ∈ ω}. Wethen have that f ⊂ ω ×B ⊂ M and f is countable. So, using thatM is countably closed we can conclude that f ∈ M and B ∈ M .Then,

M |= f : ω −→ B converges to y ( in the T sense) ,

so by elementarity, f : ω −→ B ⊂ A converges to y (in the Tsense). As A is sequentially closed, y ∈ A, so y ∈ A ∩M .

As XM is sequential, we have that A∩M is closed in XM . But,by our choice of A and x and by elementarity, we have M |= x ∈cl(A) \ A. This implies that x /∈ A ∩M and, by the definition ofTM , also implies that x ∈ clτM (A ∩M) = A ∩M , a contradiction.

Theorem 3.12. Suppose that M is a countably closed elementarysubmodel and that X ∈ M is a topological space. If XM is Frechetthen X is Frechet.

Proof: Suppose that X is not a Frechet space. Then M |= X isnot a Frechet space. We get then, M |= ∃A ⊂ X, ∃x ∈ X such thatx ∈ cl(A) but no sequence contained in A converges to x. Again,pick x and A in M .

First note that x ∈ clτM (A ∩M). Otherwise ∃V ∈ T ∩M suchthat x ∈ V and V ∩ A ∩M = ∅. This is equivalent to M |= x ∈ Vand V ∩A = ∅, contradicting our assumption.

As XM is Frechet we can pick f : ω −→ A ∩ M a sequenceconverging to x. Now, M is countably closed, f ⊂ M and f iscountable so f ∈ M . As in the previous proof M |= f converges tox (in the T sense), a contradiction.


It is important to notice that some restriction on the elementarysubmodel is necessary for the above results to hold. For a trivialexample, we can again take any countable elementary submodel Mand any X ∈ M with uncountable tightness.

The condition that M is not only ω−covering but countablyclosed required for Theorem 3.11 and Theorem 3.12 is also (consis-tently) necessary (by theorem 3.4, under CH these two conditionsare equivalent). In [16], there is an example, due to A. Dow, assum-ing p > ω1, of a topological space X and an ω−covering elementarysubmodel M of size ℵ1 containing X such that X is not sequentialbut XM is Frechet.

4. Weakening the hypothesis of M being ω-covering

As we mentioned before none of the properties of theorem 3.5 arepreserved in general, since we can take M countable. One couldask what happens if we assume M uncountable. Since the mainpoint there was not the size of M but the size of M ∩ T , we canhave an example similar to 3.9:

Example 4.1. Suppose there is an uncountable elementary sub-model M such that M ∩ ω1 is countable. Then there is X not firstcountable such that XM is a separable metrizable space.

Proof: Let X = 2ω1 . Since M ∩ ω1 is countable, XM is homeo-morphic to a subspace of 2α, where α is a countable ordinal [10],and therefore it is a separable metrizable space.

A natural question then is to ask if we can weaken “M ω-covering” to “ω1 ⊆ M” in theorem 3.5. Recall that in lemma 3.3we showed that M ω-covering implies ω1 ⊆ M . Also, 0# does notexist and |M | ≥ ℵ1 imply ω1 ⊆ M [14]. We will show that we can,if we make extra assumptions on the space. We will start by givingsome examples, which shows not only that these assumptions arenecessary, but also that the main point actually seems to be thecofinality of M ∩ κ, for a certain cardinal κ.

Note that, if κ is a cardinal, we can take M =⋃

n∈ω Mn, whereeach Mn is an elementary submodel including ω1 such that


Mn ⊂ Mn+1 and Mn ∩κ = αn < κ. Then γ = M ∩κ has countablecofinality.

Example 4.2. There is a regular space X and an elementary sub-model M such that ω1 ⊆ M and X is not first countable, but XM

is.

Proof: Let X = κ+1, where κ is a regular cardinal, κ ≥ ω2, withthe following topology: κ is discrete and the basic open sets at κare of the form (α, κ], for α < κ. Clearly X is not first countable.However, if M is an elementary submodel such that M ∩ κ hascountable cofinality, then XM is first countable.

Remark 4.3. Note that in the previous example, X does not havecountable tightness, but XM does.

Example 4.4. There is a non-Lindelof space X and an elementarysubmodel M with ω1 ⊆ M such that XM is Lindelof.

Proof: Take X = ω2 with the ordinal topology and M an elemen-tary submodel such that ω1 ⊆ M and M∩ω2 = α, α with countablecofinality. Then X is not Lindelof, but XM = α is Lindelof, sinceit is σ-compact (if {αn : n ∈ ω} is a sequence cofinal in α, then wecan write α =

⋃n∈ω αn + 1).

Remark 4.5. The above example is also an example of a space Xand an elementary submodel M such that X is not σ-compact butXM is an uncountable σ-compact space.

Example 4.6. There is a space X and an elementary submodel Msuch that XM is separable uncountable but X is not separable.

Proof: Let X ⊆ 2ω2 be such that f ∈ X if and only if there isα < ω2 such that f(β) = 0 for every β > α. Note that X is notseparable: if D = {fn : n ∈ ω} is a countable subset of X, then forevery n ∈ ω there is αn < ω2 such that fn(β) = 0, for every β > αn;but now for α > sup{αn : n ∈ ω}, V = {f ∈ X : f(α) = 1} is openand V ∩D = ∅, which implies that D is not dense.

Fix an elementary submodel M such that M ∩ ω2 = δ, wherecf(δ) = ω. We will show that XM is separable.

Let {δn : n ∈ ω} ⊆ M be an increasing sequence of ordinalscofinal in δ. Note that by the definition of X and by elementarity,f ∈ X ∩ M if and only if there is γ < δ such that f(β) = 0, forevery β > γ.


For each n ∈ ω, we have δn ∈ M and therefore 2δn ∈ M . Also,δn < ω2, so 2δn is separable. By elementarity, for each n ∈ ω, wecan then take An ∈ M such that An is a countable dense subset of2δn . Define

Dn = {f ∈ X ∩M : f ¹ δn ∈ An and f(β) = 0 for every β ≥ δn}.Let D =

⋃n∈ω Dn. Note that each Dn is countable (since An is

countable) and therefore D is countable.It just remains to show that D is dense in XM . Let Vp = {f ∈

X ∩M : p ⊆ f}, p finite, be a basic open set of XM . If n ∈ ω issuch that δn > sup{β ∈ dom(p) : p(β) 6= 0}, then Dn ∩Vp 6= ∅, andtherefore D ∩ Vp 6= ∅.

In what follows, unless mentioned otherwise, X will always bea topological space and M an elementary submodel with X ∈ M .The next result will show that, in the previous examples, it wasessential that we always took elementary submodels M such thatM ∩ κ had countable cofinality, for a certain κ:

Theorem 4.7. Any of the following implies cf(κ ∩M) = ω:(a) XM is second countable and w(X) = κ;(b) XM is first countable and χ(X) = κ;(c) XM is separable, d(X) = κ and |X| = κ.

Proof: (a) Since w(X) = κ, there is B ∈ P(T ) such that B is abase of X of size κ (and therefore for every B′ ⊆ B of size < κ, B′is not a base of X). By elementarity we can take B ∈ M . Also,B ∩ M is a base for XM . So XM second countable implies thatthere is B′ ⊆ B ∩M countable such that B′ is a base of XM .

Now, |B| = κ, so there is a bijection f from κ onto B. Again, byelementarity, we can take f ∈ M and we have that

M |= f : κ −→ B is a bijection.

Therefore,f : κ ∩M −→ B ∩M is a bijection.

Note that B ∈ M , for every B ∈ B′ ⊆ B ∩M . Since f ∈ M , thisimplies that f−1(B) ∈ M . Thus the set I = {f−1(B) : B ∈ B′} ⊆M ∩ κ and it is countable.

Suppose cf(κ ∩M) > ω. Then I is bounded in κ ∩M and wecan fix γ ∈ κ ∩M such that γ > α, for every α ∈ I. Thus I ⊆ γ.


Now, define in M , B′′ as the image of γ by the function f . Sinceγ and f are in M , we have that B′′ ∈ M . Also note that B′ ⊆ B′′,because I ⊆ γ. Thus, B′′ is a base of XM .

We then have that

M |= B′′ is a base of X.

Thus, by elementarity, B′′ is a base of X. But B′′ ⊆ B′ and |B′′| ≤|γ| < κ, a contradiction.

(b) Similar to (a), but work with a point x ∈ X ∩M and a baseBx at x instead of B.

(c) Since |X| = κ, there is a bijection f : κ −→ X and, usingelementarity, we can take f in M . Separability of XM gives us acountable set D ⊆ X ∩M dense in XM .

Suppose cf(κ ∩M) > ω. Then, since D is countable, f−1(D) isbounded in κ ∩ M . Therefore we can pick α ∈ κ ∩ M such thatα > sup(f−1(D)). Since α ∈ M , we have that E = f”α ∈ M . Butalso D ⊆ E and thus E is dense in XM . We then have that thereis a set E in M which is dense in XM and f−1(D) is bounded inκ ∩M . This implies that M |= X has a dense subset of size < κ.Thus, by elementarity, d(X) < κ, a contradiction.

The following corollaries of 4.7 (b) and (c) will be used later:

Corollary 4.8. If XM is first countable, χ(X) ≤ ℵ1 and ω1 ⊆ M ,then X is first countable.

Corollary 4.9. If |X| = ℵ1, XM is separable and ω1 ⊆ M , thenX is separable.

In [19], F. D. Tall showed that hereditary separability and hered-itary Lindelofness are upwards preserved whenever ω1 ⊆ M .

The assumption |X| = κ in 4.7 (c) seems, at a first glance, tobe extra. But the following modification of example 4.6 shows thatthe hypotheses |X| = ℵ1 cannot be weakened to d(X) = ℵ1 in theprevious corollary:

Example 4.10. There is a space X and an elementary submodelM such that ω1 ⊆ M and d(X) = ℵ1, but XM is separable.

Proof: Let D = {fδ : δ < ω1} be any dense subset of 2ω2 ofsize ℵ1. Fix γ < ω2 of cofinality ω1 and {γα : α < ω1} a cofinal


increasing sequence converging to γ. For each δ < ω1 and α < ω1

define:

gδα(β) =

{0 if β ∈ [γα, γ)fδ(β) otherwise

Then E = {gβα : α < ω1, δ < ω1} is a dense subset of 2ω2 of size ℵ1

without countable dense subsets (in 2ω2).We now take X as in example 4.6 and Y = X ∪E. Then d(Y ) =

ℵ1, but we can take again M an elementary submodel such thatM ∩ ω2 is an ordinal of countable cofinality and proceed as in 4.6to show that YM is separable.

We have the following corollaries of 4.7 (b):

Corollary 4.11. Suppose M and X are such that ω1 ⊆ M , XM isseparable metrizable and w(X) ≤ ω1. Then X is separable metriz-able.

Corollary 4.12. (CH) If XM is separable metrizable and ω1 ⊆ M ,then X is separable metrizable.

It is natural to ask if CH is necessary in the previous corollary.We make this discussion below, but first we investigate how a coun-terexample should look.

Lemma 4.13. Let M be an elementary submodel with ω1 ⊆ M .Suppose there is a space X ∈ M such that X is not separable metriz-able but XM is. Then there is a space Y ∈ M with Y = ω∪{x} suchthat χ(x, Y ) > ℵ1 but YM is first countable (and therefore separablemetrizable).

Proof: Suppose there is X ∈ M not separable metrizable. ThenX is not second countable (note that since XM is metrizable, it isregular, so X is also regular), so there is Z ⊂ X of cardinality ℵ1

such that Z is not second countable (see e.g. [9]). By elementaritywe can take Z ∈ M . Since we are assuming ω1 ⊆ M , this impliesthat Z ⊆ M . Also, ZM is separable and |Z| = ω1, so corollary 4.9give us that Z is separable.

Now, ZM is separable metrizable but Z is not and Z∩M = Z. Sothe topologies of ZM and Z are obviously different. Thus, Z cannotbe first countable (otherwise we would have that ZM is a subspaceof Z [10], and therefore they would have the same topology). Note


that since ω1 ⊆ M , by corollary 4.8, we actually have that χ(Z) >ℵ1 and therefore we can fix x ∈ Z such that χ(x,Z) > ℵ1.

Using the fact that Z is separable, we can now take D a countabledense subset of Z. By elementarity (since Z ∈ M), we can takeD ∈ M . Then D ∪ {x} is a subspace of Z satisfying the conditionswe want for Y (note that χ(x,Z) = χ(x,D∪{x}) because D is densein Z). So we can take Y = ω ∪ {x} homeomorphic to D ∪ {x}.

The previous lemma gives one of the implications of the followingequivalence, which was formulated by G. Gruenhage (he proved thisimplication independently; the proof of the other direction is alsodue to him).

Lemma 4.14. The following are equivalent:(a) there is a non metrizable space X and an elementary sub-

model M satisfying ω1 ⊆ M and X ∈ M such that XM is anuncountable separable metrizable space;

(b) there is a filter F on ω which is not countably generated (ofsize ≥ ℵ2) and an uncountable elementary submodel M such thatω1 ⊆ M , F ∈ M and FM = F ∩M is countably generated.

Proof: Suppose we have (a) and take Y = ω ∪ {x} as in theprevious lemma. Look at the filter F on ω determined by theneighborhoods of x. Then χ(x, Y ) > ℵ1 implies that F is notcountably generated, but YM first countable implies that FM is.

For the other implication, let F and M be as in (b) and defineX = ω ∪ {F}, with the points of ω being isolated and the neigh-borhoods of F being given by {F} ∪ F , for F ∈ F . Clearly X isnot first countable since F does not have a countable base, so Xis not metrizable. But FM is countably generated, which impliesthat XM is first countable and therefore it is separable metrizable.To get an uncountable space we can replace each isolated point bya copy of the real line, i.e., we can take Y as the topological sumof countably many copies of the real line plus the point F , wherea basic neighborhood of F is F with F many copies of the realline for F ∈ F . Then YM is second countable because it is firstcountable at F and therefore it is metrizable, but, as before, Y isnot metrizable.

The following example is also due to G. Gruenhage:

Example 4.15. (t = ℵ2) There is a space X such that XM isseparable metric but X is not metrizable.


Proof: Suppose T is a tower of size ℵ2 and let F be the filtergenerated by T . Since we are assuming t = ℵ2, F is not countablygenerated. Take M any elementary submodel of size ℵ1 such thatδ = M ∩ ω2 has countable cofinality. We then have that FM iscountably generated, so we can now use the previous lemma to getthe example.

In view of corollary 4.12 and the previous example, it is natu-ral to ask if ¬CH is consistent with “for any M with ω1 ⊆ M ,XM separable metrizable implies X separable metrizable”. This isequivalent to ask if the negation of (b) in lemma 4.14 is consistentwith ¬CH. To answer this question we will use the following resultfrom [5]. I would like to thank A. Dow and I. Juhasz for pointingit out to me.

Lemma 4.16. Let V be a model of CH and G be a Fn(κ, 2)-genericfilter. If, in V [G], F is a filter on ω which is not countably gener-ated, then there are filters Fα (α < ω1) such that every member ofF is a member of all but countably many of the Fα’s and for everyα < ω1 there is some a ∈ Fα such that ω \ a ∈ F .

Using this lemma we can prove the following:

Theorem 4.17. If V , G and F are as in the previous lemma and(in V [G]) M is an elementary submodel with ω1 ⊆ M and F ∈ M ,then F ∩M is not countably generated.

Proof: Suppose F∩M is countably generated and take {An : n ∈ω} generating F ∩M . By the lemma above and by elementarity,there is {Fα : α < ω1} ∈ M as in the conclusion of the lemma.Since ω1 ⊆ M we have that Fα ∈ M for every α < ω1.

Now, for each n ∈ ω, An ∈ F ∩M and therefore An ∈ Fα, forevery α ∈ ω1 \ In, for some countable set In. Thus there is α ∈ ω1

such that An ∈ Fα for every n ∈ ω. Then F ∩ M ⊆ Fα, whichimplies that F ∩M ⊆ Fα ∩M . Now, Fα ∈ M , so M models thatF ⊆ Fα and, by elementarity, we have that F ⊆ Fα. But there isa ∈ Fα such that ω \ a ∈ F , a contradiction.

This and lemma 4.14 then give us the following:

Corollary 4.18. Let V be a model of CH and G be a Fn(κ, 2)-generic filter. In V [G], if XM is separable metrizable and M is anelementary submodel with X ∈ M and ω1 ⊆ M , then X is separablemetrizable.


We finish this section proving that for the Lindelof property andpointwise countable type we have results similar to theorem 4.7:

Theorem 4.19. If XM is Lindelof and L(X) = κ, for some suc-cessor cardinal κ, then cf(κ ∩M) = ω.

Proof: Suppose cf(κ ∩ M) > ω. Since L(X) = κ and κ is asuccessor cardinal, there is an open cover C of X with no subcoverof size < κ (otherwise we would have L(X) ≤ λ, for λ such thatκ = λ+). Since L(X) = κ, we can suppose that C has size κ. Byelementarity, we can take C ∈ M and we have that C ∩ M is anopen cover of XM . Thus it has a countable subcover D.

Fix a bijection f : κ −→ C. We can take f ∈ M , by elementarity,and we have that f : κ ∩M −→ C ∩M is a bijection.

Since f ∈ M , f−1(B) ∈ M for every B ∈ D. By assumption,cf(κ∩M) > ω and D is countable. So there is α ∈ κ∩M such thatα > sup{β ∈ κ ∩M : β = f−1(B), B ∈ D}. Let (in M) D′ = f”α.Then D′ is a subcover of C ∩M and it is in M . But this impliesthat

M |= C has a subcover of size < κ

which, by elementarity implies that C has a subcover of size < κ, acontradiction.

The following corollary follows immediately from the previousresult:

Corollary 4.20. XM Lindelof, L(X) ≤ ℵ1 and ω1 ⊆ M imply thatX is Lindelof.

For pointwise countable type, we have the following:

Theorem 4.21. If X is a regular space with w(X) = ℵ1 and Mis such that ω1 ⊆ M and XM has pointwise countable type, then Xhas pointwise countable type.

Proof: Let B0 ∈ M (by elementarity) be a base of size ℵ1 ofX, and we can suppose that B0 is closed under finite unions. Forx ∈ X ∩M , there is a compact set K (in XM ) such that x ∈ K andχ(K, XM ) = ℵ0. As in the proof of theorem 3.8, we now have towork to get a compact set K ′ ∈ M satisfying the same properties.

Using that K is compact, we can argue as in the proof of theorem3.8, and show that, without loss of generality, we can suppose that


there is B = {Un : n ∈ ω} ⊆ M an outer base for K such thatUn ∈ B0 for every n ∈ ω (this last condition is possible because weare assuming that B0 is closed under finite unions).

Now, there is a bijection f : ω1 −→ B0 and we can take it in M .Take α < ω1 ⊆ M , α > sup{f−1(Un) : n ∈ ω and B′ = f ′′α}. ThenB′ is countable, B′ ∈ M and B′ ⊇ B. We can now finish the proofas in theorem 3.8.

5. Preservation of non-metrizability and Hamburger’sproblem

We saw that if M is ω-covering, then the property of being “sepa-rable metrizable” is upwards preserved (corollary 3.6). What aboutonly metrizability? If the question is for any M and any X, theexamples given in the previous section for non-upwards preserva-tion of first countability, can also be used here (e.g. example 4.2).So the more interesting cases are when first countability is upwardspreserved, for instance if M is ω-covering (theorem 3.5), or when wesimply assume that X is first countable. In [19], F. D. Tall gives aconsistent example of a non-metrizable first countable space of sizeℵ2 such that XM is metrizable for every M of size ℵ1. This exampleis a first countable non-metrizable space with all subspaces of sizeℵ1 metrizable, i.e., it is an example for Hamburger’s problem (seee.g. [8]). We next show that any such example has to be of thisform. We first show:

Theorem 5.1. If X is a first countable space such that there is anelementary submodel M with X ∈ M , XM metrizable and ω1 ⊆ M ,then all subspaces of size ℵ1 of X are metrizable.

Proof: Since X is first countable, XM is a subspace of X [10].Suppose there is Y ⊆ X such that |Y | = ℵ1 and Y is not metrizable.We then have that

Hθ |= ∃Y ∈ P(X) such that |Y | = ℵ1 and (Y, T ) is not metrizable.

By elementarity, we can take Y ∈ P(X) ∩M such that

Hθ |= |Y | = ℵ1 and (Y, T ) is not metrizable.

Now, Y ∈ M and |Y | = ℵ1. Since ω1 ⊆ M , this implies thatY ⊆ X ∩M and therefore Y is a subspace of XM (because XM is


a subspace of X). But XM is metrizable and Y is not, a contradic-tion.

We have the following corollaries:

Corollary 5.2. Suppose X and M are such that M is an ω-covering elementary submodel with X ∈ M and XM is metrizable.Then X must have all subspaces of size ℵ1 metrizable.

Proof: Just note that, by theorem 3.5, M ω-covering impliesthat X is first countable (since XM metrizable implies XM firstcountable) and also implies that ω1 ⊆ M (by lemma 3.3).

Corollary 5.3. Assume M is an elementary submodel with ω1 ⊆M and |M | = ℵ1. There is a first countable non-metrizable spaceX such that X ∈ M and XM is metrizable if and only if there is afirst countable non-metrizable space X with all subspaces of size ℵ1

metrizable.

Proof: One direction is just the previous theorem. For the otherdirection, suppose there is a first countable non-metrizable spaceX with all subspaces of size ℵ1 metrizable. Then, if M is an el-ementary submodel of cardinality ℵ1 with X ∈ M , then XM is asubspace of X (since X is first countable [10]) of size ≤ ℵ1, andtherefore it is metrizable.

The previous result connected the question of preservation ofnon-metrizability by elementary submodels to Hamburger’s prob-lem. It is interesting to notice that the question of preservationof non-metrizability by countably closed forcing is also related toHamburger’s question:

Theorem 5.4. Suppose that X is a topological space with |X| = ℵ1

and that P is a countably closed partial order. If X is metrizable inV P, then X is metrizable in V .

Proof: First note that X must be first countable: since X ismetrizable in V P, X is first countable in V P; but countably closedforcing does not add new countable subsets (see e.g. [13]), so Xwas already first countable in V .

Suppose X = {xα : α < ω1} and let Vα = {Vn(α) : n ∈ ω} bea base at xα for each α < ω1. Let r ∈ P be such that r ° X is


metrizable. Then r ° X has a σ-discrete base. Fix names B, Bn

and s ≤ r such that

s ° B is a base of X such that B =⋃n∈ω

Bn and each Bn is discrete.

Also, fix ϕ a name for a function in the extension that choosesthe open sets of B, more precisely, fix a name ϕ and p ≤ s suchthat

p ° ϕ : ω1 × ω −→ T with ϕ′′ω1 × ω = B and ϕ′′ω1 × {n} = Bn.

Let h : ω1 −→ ω1 × ω be any bijection. For each α < ω1 andeach n ∈ ω define:

Dα,n = {q ∈ P : there is U ∈ T and γ < ω1 such that

(q ° xα ∈ U ⊆ Vn(α) and U = ϕ(h(γ)) )}.A standard density argument shows that Dα,n is dense below p forevery α ∈ ω1 and n ∈ ω.

Define a decreasing sequence {pβ : β < ω1} such that p0 ≤ p andpβ ∈ Dh(β). This is possible because P is ω1-closed and the sets aredense below p. For each β < ω1 we can fix an open set Uh(β) andγβ < ω1 witnessing pβ ∈ Dh(β).

Define

Uk = {Uh(β) : β is such that h(γβ) ∈ ω1 × {k} }and let U =

⋃k∈ω Uk.

We first show that U is a base for X. Fix α < ω1 and n ∈ ω. Letγ = h−1(α, n). Then pγ ° xα ∈ Uh(γ) ⊆ Vn(α) which implies thatxα ∈ Uh(γ) ⊆ Vn(α).

It just remains to show that Uk is discrete, for each k ∈ ω. SinceX is first countable, it is enough to show that V is discrete forevery countable V ⊆ Uk. Suppose V = {Uh(αn) : n ∈ ω}. Since P isω1-closed, we can take q ≤ pαn , for each n ∈ ω. Then, we have

q ° Uh(αn) = ϕ(h(γαn)) and h(γαn) ∈ ω1 × {k},which implies q ° V = {Uh(αn) : n ∈ ω} ⊆ Bk. But q ≤ p andp ° Bk is discrete, therefore, q ° V is discrete. Since to be discretejust depends on basic open sets, it is easy to see that this impliesV discrete and we are done.

As a consequence we have:


Corollary 5.5. Suppose X is a non-metrizable space such that forsome countably closed forcing P, X is metrizable in V P. Then Xhas all subspaces of size ℵ1 metrizable.

Here, however, we do not have an equivalent result as we had inthe elementary submodel case. This can be shown, for instance, bythe next result, which is from [15]. In [15], the result and the proofare given using some notations defined there. The proof below isthe same proof, but here we present it in a more topological formand a slightly stronger version. This result shows, for instance, thatthe ladder system on an E-set example (due to Fleissner [7]) forHamburger’s problem cannot become metrizable after any count-ably closed forcing. We need the following definition:

Definition 5.6. We say that a topological space X is stationar-ily κ-collectionwise Hausdorff if for every closed discrete subspace{xα : α < S} indexed by a stationary S ⊆ κ, there is a stationaryT ⊆ S such that {xα : α ∈ T} can be separated by disjoint opensets. X is stationarily collectionwise Hausdorff if it is stationarilyκ-collectionwise Hausdorff for every κ.

Note the same proof that shows that the ladder system on an E-set is not collectionwise Hausdorff, also shows it is not stationarilycollectionwise Hausdorff. So by the next result, it cannot be collec-tionwise Hausdorff after a countably closed forcing, and thereforecannot be made metrizable.

Theorem 5.7. Suppose that P is countably closed and that θ is aregular cardinal ≥ ℵ2. If X is a first countable regular not stationar-ily θ-collectionwise Hausdorff space, then X is not θ-collectionwiseHausdorff in VP.

Proof: Suppose X is not stationarily θ-collectionwise Hausdorffand take a discrete set Y ⊆ X indexed by a stationary S ⊆ θwitnessing it. We identify Y = S. Suppose S can be separated inVP and we will work for a contradiction.

For every α ∈ S, fix Uα = {Un(α) : n ∈ ω} a decreasing base atα and let U =

⋃α∈S Uα. Because X is regular, we can suppose that

U is such that the set Hα,β = {(n,m) : Un(α)∩Um(β) 6= ∅} is finitefor each α, β ∈ S. To see this, just note that, since S is discrete,we can take, for each α ∈ S, U0(α) such that U0(α) ∩ S = {α}.


Now, we are assuming that S is separated in VP, so there isp0 ∈ P and a name for a function g such that

p0 ° g : S −→ ω and {Ug(α)(α) : α ∈ S} is a separation of S.

Let M be an elementary submodel of Hκ (for κ large enough)such that we have θ, S,U ,P, °P, p0, g ∈ M and such that M ∩ θ =β ∈ S (possible since X is stationary in θ).

If possible, choose p1 ≤ p0, α0 < β, p1 ∈ M such that

p1 ° Ug(α0)(α0) ∩ U0(β) 6= ∅.Note that this would imply p1 ° g(β) > 0.

Similarly, by induction, choose, if possible, p0 ≥ p1 ≥ . . . ≥ pn+1,αn < β, pn+1 ∈ M such that

pn+1 ° Ug(αn)(αn) ∩ Un(β) 6= ∅(and therefore pn+1 ° g(β) > n).

Note that the process must stop: otherwise, since P is ω1-closed,this would imply that there is s ∈ P such that s ° g(β) > n,for every n ∈ ω, a contradiction. Thus, there is n ∈ ω such thatpn ∈ M and for every q ≤ pn, for every α < β, q ∈ M implies thatq 1 Ug(α)(α) ∩ Un(β) 6= ∅.Claim 5.8. For every α < β, pn ° Ug(α)(α) ∩ Un(β) = ∅.Proof: Suppose not. Then there is α < β and there is q ≤ pn

such that q ° Ug(α)(α)∩Un(β) 6= ∅. Then there is α < β and thereis q ≤ pn such that q ° (g(α), n) ∈ Hα,β.

By our assumption on U , Hα,β is a finite subset of ω × ω andtherefore it is in M . Fix α < β. So

Hλ |= there is q ≤ pn such that q ° (g(α), n) ∈ Hα,β,

and pn, g, α, Hα,β are all in M . Therefore, by elementarity,

M |= there is q ≤ pn such that q ° (g(α), n) ∈ Hα,β.

Thus, there is q ∈ M , q ≤ pn, such that q ° Ug(α)(α) ∩ Un(β) 6= ∅,a contradiction.

Let

B = {γ ∈ S : for every α < γ pn ° Ug(α)(α) ∩ Un(γ) = ∅}.


Since β = θ ∩ M ∈ B (by the claim), a standard elementarysubmodel argument shows that B is stationary. Indeed, first notethat B has all parameters in M , so

M |= for every club C ⊆ θ there is γ ∈ C such that γ ∈ B

(since, for every club C in M , β = M ∩θ ∈ C and by elementarity).We then have that M |= B is stationary in θ, and therefore, byelementarity again, B is stationary in θ.

For each α ∈ S, let qα ≤ pn and mα ∈ ω such that qα ° g(α) =mα. Define (in V), f : S −→ ω by f(α) = mα + n.

It just remains to show that f gives a separation of B, and we willget a contradiction, by our choice of Y . Fix α, β in B. Withoutloss of generality, we can suppose α < β. Since β ∈ B, pn °Ug(α)(α) ∩ Un(β) = ∅. Therefore, by the definition of f we have

qα ° Uf(α)(α) ∩ Uf(β)(β) = ∅,which implies that the same is true in V and we are done.

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Universidade de Sao Paulo, Caixa Postal: 66281, Sao Paulo, SP,CEP: 05315-970, Brasil

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