AN INTRODUCTION TO APPLICATIONS OFELEMENTARY SUBMODELS TO TOPOLOGYALAN DOWAbstract. The purpose in writing this paper is to demonstrate that the no-tion of an elementary submodel is a simple but powerful tool which should bebrought into the mainstream of set-theoretic topology. The hope is that thiscan be accomplished simply by supplying the reader with proofs of results,ranging from very trivial ZFC results to some quite di�cult independence re-sults, in which elementary submodelsplay a key role. The topological questionswhich are discussed concern, for the most part, metrizable and �rst countablespaces. However one new result in the paper is a proof from PFA that compactspaces with countable tightness have some points of �rst countability.IntroductionThis paper is an expanded version of the author's talk given at the Spring Topol-ogy Conference in Gainseville. The main purpose of both the talk and the paperis to give examples to demonstrate the usefulness of elementary submodels to set-theoretically oriented topologists. The author is not alone in believing that elemen-tary submodels should become as familiar a part of the language of set-theoretictopology as is the pressing-down lemma for example. I believe that, for set-theoretictopologists, elementary submodels provide:1. a convenient shorthand encompassing all standard closing-o� arguments;2. a powerful technical tool which can be avoided but often at great cost in bothelegence and clarity; and3. a powerful conceptual tool providing greater insight into thestructure of the set-theoretic universe.I hope to convince some readers of the validity of these points simply by (over-)using elementary submodels in proving some new and old familiar results. Thispaper is not a survey of their use nor an adequate (or even rigorous) introduction tothe concept { it is intended solely as a demonstration of how useful they can be evenin some rather unexpected applications. The author's primary reference is Kunen'stext [K] and the reader is directed there for both an introduction and to discoverwhat I probably should really have said in many of the proofs and discusssions.There are two new results worth mentioning in the paper. The �rst is thatit follows from the consistency of large cardinals that it is consistent that non-metrizability re ects in the class of locally-@1 spaces. This result is similar toFleissner's results in [F] about left-separated spaces with point-countable bases.The second is that it follows from PFA that each compact space of countabletightness necessarily contains points of countable character. The second result1991 Mathematics Subject Classi�cation. 54A35.Key words and phrases. elementary submodels, metrizable, countable tightness, re ection.1

2 ALAN DOWis related to a question of Arhangel'skii [A2] and is just something that Fremlin,Nyik�os and Balogh \missed" in the papers [Fr], [FrN] and [B].In the �rst section we will introduce elementary submodels and establish someof the non-standard assumptions we will make in the remainder of the paper. Inthe three sections following we apply elementary submodels in increasingly di�-cult arguments. Most of the results in these sections concern metric spaces andthe remainder are concerned with spaces of countable tightness. None of the re-sults in these sections involve forcing or large cardinals (although their existenceis acknowledged). Section �ve concerns applications of elementary submodels toforcing arguments. Not surprisingly this is an area in which elementary submodelsare particularly useful { especially when proper forcing is involved. The last twosections discuss large cardinals and iterated forcing respectively.i PreliminariesFor a set or class M and a formula, ' , in the language of set theory, the formula'M is de�ned recursively (see IV of [K]). 'M is just the formula you get whenyou \restrict all the quanti�ers to M". However note that (x � y)M is really�(8a 2 M )(a 2 x ) a 2 y)� , since (x � y) is not in the language of set theory.However it does not take long for one to become accustomed to the meaning of 'Mespecially when M is a \model" of most of ZF . We say that M is a model of '(denoted M j= ') if 'M holds .De�nition: If fa1; : : : ; ang � M � N then '(a1; : : : ; an) is absolute for M;Nif 'M (a1; : : : ; an) holds i� 'N (a1; : : : ; an) holds.De�nition: M is an elementary submodel of N , denoted M � N , if M � Nand for all n < ! and formulas ' with at most n free variables and all fa1; : : : ; ang �M the formula '(a1; : : : ; an) is absolute for M;N .For a cardinal �, the set H(�) is the set of all \hereditarily < � sized sets". Thatis, H(�) is the set of all sets whose transitive closure has size less than �. Thesesets are useful because if � is regular thenH(�) j= ZF - P (see IV in [K]).In practice, when one is investigating a property of some objects, say hX; T ; Ci,one usually knows the largest possible size of any set at all relevant to the validityof the property. Therefore there is a cardinal � large enough and a formula ofset theory ' so that '(X; T ; C) expresses the property and such that '(X; T ; C) isabsolute for V;H(�) . (For example, see the Levy Re ection Theorem, IV in [K]).Throughout this paper we shall often choose such \large enough" � or H(�) withlittle or no discussion as to how large it needs to be.Once we have shrunk our model to a set (namely H(�)), we then have thedownward Lowenheim-Skolem theorem. The proof of this theorem makes verytransparent the concept of elementary submodels.

ELEMENTARY SUBMODELS 3Theorem 1.1: For any set H and X � H, there is an elementary submodelM of H, such that X � M and jM j � jXj � !.Another very useful notion and resulting basic fact concerns elementary chains.M is called an elementary chain if it is a chain when ordered by �. It is worthnoting that � is a transitive order.Theorem 1.2: If M is an elementary chain then M � SM for all M 2M.Corollary 1.3: If M is a chain under inclusion of elementary submodels ofH, then M is an elementary chain and SM� H.Corollary 1.4: For uncountable regular cardinals � � � and X 2 H(�) withjXj < �, f� < � : 9M � H(�)�X �M ; jM j < � and M \ � = ��gis a closed and unbounded set (cub) in �.Proof. Inductively build an elementary chain of length �, fM� : � < �g, so thatfor � a limit ordinal, M� = SfM� : � < �g.Note that, for regular cardinals �, if M � H(�) has cardinality less than � thenM 2 H(�). Therefore we could have built the elementary chain so thatM� 2M�+1{ this will be called an elementary 2-chain. A continuous elementary chain orelementary 2-chain is one in which, for each limit �, we have that M� = S�<�M� .Another corollary to theorems 1.1 and 1.2 which we shall use frequently is thefollowing.Theorem 1.5: For any regular � � 2! = c and any X � H(�) with jXj � c,there is an M � H(�) so that X � M ; jM j = c AND M! � M:A remark often made for its shock value is to suppose that M is a countableelementary submodel of H( (2c)+ ) such that the reals R are in M . Then M j=(R is uncountable) and yet R \M is only countable. There's no paradox here:M thinks R is uncountable not R\M . Indeed the set R\M is not even in Mso M can't think anything about it. The lesson here is that if M � H, then'M (m) () 'H (m) holds for elements of M , and that, in general, neither of theimplications X 2M ) X � M , X � M ) X 2M hold.However in some case X 2M does imply X � M .Theorem 1.6: If M � H(�), � regular, and � 2 M is a cardinal such that� � M , then for all X 2M with jXj � �, X is a subset of M . In particular, eachcountable element of M is a subset of M .

4 ALAN DOWProof. If jXj � �, then H(�) j= �9f � onto! X�. Since � ; X are both in M ,M j= �9f � onto! X� . That is, (9f � onto! X)M holds, hence there is an f 2 Msuch that (f maps � onto X)M . Now we are down to what is know as a �0-sentence (see IV in [K]) - these formulas are absolute in many circumstances; thatis f \really" is a function from � onto X. Indeed, M j= (f � � � X) so we showthat f � ��X as follows. M j= (f � ��X) really means M j= (:9x 2 f n��X){ hence H(�) j= (:9x 2 f n � � X). Similarly H(�) j= (f is a function ) sinceM j= (8� 2 �)(8x; y 2 X) (f(�; x); (�; y)g � f ) x = y). Also, of course, M j=(8x 2 X)(9� 2 �) (�; x) 2 f). Finally, we show that x 2 M for each x 2 Xfollows from � � M . Indeed, let x 2 X. Since f is \really" onto, we may choose� 2 � such that (�; x) 2 f . Now M j= (9y)(�; y) 2 f so choose y 2 M such thatM j= (�; y) 2 f . Clearly H(�) \thinks" (realizes?) that x = y.When one says \let hX; T i be a topological space" it is usually meant that T isthe topology on X. However we shall mean that T is a base for a topology on X.As we shall see below this is much more convenient.Suppose hX; T i is a topological space in some H(�). Our general procedure isto take some kind of submodel, M � H(�) (frequently an elementary submodel),such that hX; T i 2 M . We then consider the (generally much smaller) subsetXM = X \M . At this point there are two natural topologies to consider on XM .On the one hand we have the subspace topology generated by fU \XM : U 2 T g.And on the other hand, if M j= T is a base for a topology on X plus some basicaxioms, then we'd get the base TM = fU \XM : U 2 T \Mg. In general, thesegive very di�erent topologies on XM . For example, if X = �! and M is countablethen, of course, hXM ; TMi is a countable metric space.However, it is by comparing these two topologies that we prove our re ection re-sults. The game we play is to jump back and forth between M and H(�), comparingwhat M \thinks" with what H(�) \thinks".Most of the results in this article are what are known as re ection results. Are ection question in topology usually has the form \if a space X has property P ,then what is the size of the smallest subspace Y which also has property P ?".However it is usually the case that P is the negation of a nice property. So onemight rephrase the question as \if � is a cardinal and X is a space such that everysubspace of X of cardinality at most � has P , then does this guarantee that X hasP ?".We will adopt the following notation :If P is a class of spaces or a property (which de�nes the class of spaces having thatproperty) then for a space X�(X;P ) = minfjY j : Y � X and Y does not have property Pg ;(where we assume the minimum of the empty set is 1).There are not too many re ection results that hold for the class of all topo-logical spaces but, for example, if we consider the separation property T1 thenfor any space X we have �(X;T1) 2 f2;1g. Another less trivial example is that�(X; �rst countable) � !1 for all X such that �(X) = !1 (but not for all X such

ELEMENTARY SUBMODELS 5that �(X) > !) . We shall use such self-explanatory abbreviations for classes ofspaces as `� = !', `w = !' and `t � �' for `�rst countable', `countable weight'and `tightness at most �' respectively. The reader is, of course, referred to theHandbook of Set-theoretic Topology for all topological de�nitions and basic facts.ii Some Elementary Applications.In this section we prove a few simple theorems as an introduction to elementarysubmodel arguments.Example 2.1 The Delta System Lemma. Let � be a regular cardinal and let fF� j� < �g � [�]<!. Of course we want to show that there aren < ! ; F 2 [�]<! and I 2 [�]�so that jF�j = n and F� \ F� = F for all �; � 2 I with � 6= �. Let M be anelementary submodel of H(�+) such that fF� j � < �g is in M and jM j < �. Let� = sup(M \�) and choose any � 2 ���. We have found our n and F ; let n = jF�jand F = F� \M . Then one notes thatM j= 8 < �9�(jF�j = n ^ (F� \ (max(F [ f g) + 1) = F ):To see this, note that the set S = f j (9� 2 �)F� \ = Fg is an element of M .Furthermore � 2 S hence M j= S is co�nal in �.It follows that we may pick, by induction on � < �, F� so thatF� \ maxF� = F for all � < �. Alternatively, we may choose an elementary chainfM� : � < �g of elementary submodels of cardinality less than � so that M = M0and choose F� 2M�+1 so that F� \M� = F :In the next example we prove Arhangel'skii's famous result that the cardinalityof a Lindel�of �rst countable space is at most c .Example 2.2: A Lindel�of space with countable pseudocharacter and countabletightness has cardinality at most c .Let T be a base for a Lindel�of topology on X which has countable tightnessand pseudocharacter. Let hX; T i 2M � H(�) such that M! � M and � is \largeenough".Claim. M \X = XIndeed, suppose not, and choose z 2 X nM .Subclaim 1. For each y 2 X \M9Uy 2 T \M such that y 2 Uy and z =2 Uy .Proof of Subclaim 1. H(�) j= (9fUn : n < !g � T such that fyg = \fUn : n < !g).Therefore M is a model of this , so let fUn : n < !g 2M be such that M j= fyg =\nUn. Now since fUngn 2 M it follows that V j= fyg = \nUn , hence we maychoose Uy as required.Subclaim 2. X \M is closed (hence Lindel�of)

6 ALAN DOWProof of Subclaim 2. Assume x 2 X \M . By countable tightness, choose a count-able set Y � X \M so that x 2 Y . Fix a set fUngn2! � T exhibiting that Xhas countable pseudocharacter at x . Next choose, for each n 2 ! a collectionfUn;mgm2! � T such that x =2 [fUn;mgm2!andX n Un � [m2!fUn;mg. It followsthat Y n fxg = SfYn;m : n;m 2 !g where, for each n;m 2 ! Yn;m = Y \ Un;m.But since fYn;m : n;m 2 !g is a countable collection of countable subsets of M ,the collection and each member of it is an element of M . Now if x were not in Mwe would have M j= Y =[fYn;m : n;m 2 !gwhereas H(�) j= x 2 Y n Yn;m for each n;m 2 ! :Now by subclaim 1 U = fUy : y 2 X \Mg forms an open cover of X \M butnot of X. By subclaim 2 , U has a countable subcollection W which still coversX \M . Now W is a countable subset of M and therefore is an element of M . Butthis is a contradiction since M j= W covers X. The result now follows from thefact that we may assume that jM j = c.Proposition 2.3: If a space X with base T has a point-countable base andhX; T i 2M � H(�) then T \M is a base for each point of X \M .Proof. Let M � H(�) with hX; T i 2M . SinceH(�) j= hX; T i has a point-countable baseand M is an elementary submodel, there must be a set B 2M such thatM j= B is a point-countable base for hX; T i :It is straightforward to check that absoluteness guarantees that B is a base forhX; T i (in H(�)). Also H(�) j= B is point-countable since this follows from M j=8x 2 X fB 2 B j x 2 Bg is countable. Now let x be any point of X \M andsuppose B 2 B is a neighbourhood of x. Choose U 2 T and W 2 B so thatx 2 W � U � B. Now choose y 2 W \M which we may do since x 2 X \M .Since B is point-countable and fy;Bg 2 M it follows that fS 2 B : y 2 Sg � M ;hence , in particular, fB;Wg � M . Furthermore, since U 2 T and W � U � B ,it follows that M j= 9T 2 T such that W � T � B :Therefore there is a T 2 T \M such that x 2 T � B which was to be shown.As we shall see later, the hidden strength of the previous result is that the base Tis not assumed to be point-countable (recall our assumption that T denotes a base,not the whole topology). The next result uses some compactness in the topologicalsense to �nd when T \M is not a base at all points of X \M .Proposition 2.4: Let hX; T i be a countably compact space which is an elementof a countable elementary submodel, M , of some su�ciently large H(�).if T \M is not a base for hX; T ithen 9z 2 X \M such that T \M is not a base at z.

ELEMENTARY SUBMODELS 7Proof. Clearly we may as well assume that X \M is not dense in X , so chooseany z 2 X nX \M . Now if T \M does contain a base for all points of X \Mthen there is a cover U � T \M of X \M whose union does not contain z. Butnow X \M is countably compact and U is a countable cover of it (since T \M iscountable) . Therefore there is a �nite subcover , sayW � U , of X \M and henceof X \M . But now W 2M and M j= [W = X while H(�) j= z =2 [W.The following non-trivial result is an immediate consequence of the previous twopropositions.Example 2.5 Mis�cenko's Lemma: A countably compact space with a a point-countable base has a countable base.iii Elementary Chains and the !-covering propertyAs we saw in the proof of Arhangel'skii's theorem it is a very powerful assump-tion to have that your elementary submodel is \closed under ! -sequences". Alsowe cannot expect that countable elementary submodels can \trap" a great deal.Indeed a typical inductive construction usually carries through without much di�-culty through the countable limit ordinals (discounting the problems of \trapping"the uncountable sets). On the other hand most constructions have considerable dif-�culty passing !1, so we can expect some non-trivial re ection by taking elementarysubmodels of cardinality !1 even in the absence of CH .A useful property, which can to some extent replace \closed under !-sequences",is the !-covering property. We shall say that a set M has the !-covering propertyif for each countable A � M there is a countable B 2 M such that A � B . IffM� : � 2 !1g is an elementary 2-chain of countable elementary submodels ofsome H(�) such that for each � 2 !1 M� 2 M�+1 then clearly the union of theM�'s is an !-covering elementary submodel of H(�) of cardinality !1.In this section we shall present several proofs that use elementary submodels ofcardinality !1 which satisfy the !-covering property. It can be shown that suchelementary submodels are exactly those which are uncountable and are the unionof an elementary 2-chain of countable elementary submodels.Theorem 3.1: If every subspace of cardinality !1 of a countably compact spaceis metrizable, then the space itself is metrizable.It is convenient to make a few preliminary remarks before actually proving thetheorem. To give a slick proof using elementary submodels it seems to be necessaryto �rst prove that such a space is necessarily �rst countable, or at least that wemay assume that if there is a counterexample then there is a �rst countable one.This can be done directly with relative ease | because of countable compactnessa counterexample would have a subspace with density at most !1 which was alsoa counterexample. However it seems more appropriate to proceed by �rst provingthe following surprising result of Hajnal and Juhasz (the result for regular spaceswas proven by Tkacenko [Tk]). This was proven during their systematic study ofcardinal functions on unions of chains of spaces which is very similar to investigatingre ection properties of the cardinal functions. We state this result twice in orderto recall our notation introduced in xI.

8 ALAN DOWProposition 3.2 [J]: If every subspace of cardinality at most !1 has countableweight then the space itself has countable weight.Proposition 3.2 [J]: For any space X, �(X;w = !) > !1 implies w(X) = !.Proof. Let hX; T i 2M where M is an !-covering elementary submodel of H(�) ofcardinality !1 . We must �rst show that T \M is a base for the subspace topologyon X\M . Indeed suppose x 2 X\M and U is an open neighbourhood of x . Since�(X;w = !) > !1 , X \M n U has a countable dense subset D. Since M has the!-covering property we may choose a countable D0 2M so that D � D0 � X. NowM j= w(D0 [ fxg) = ! hence there is T 2 T \M such that x 2 T and T \D0 � U .So we now have X \M nU = D � D0 n T � X n T , hence M \T � U as was to beshown.It now follows that there is a countable subset B of T \M which is a base forX \M since w(X \M ) = ! and T \M forms a base. We may suppose B 2M bythe !-covering property . But now M j= w(X) = ! , hence the result follows byelementarity.Proof of 3.1. Let T be a base for the topology on X and assume that hX; T i isnot metrizable. Let hX; T i 2 M where M is an !1-sized, !-covering elementarysubmodel of some H(�). We shall show that X \M with the subspace topology isnot metrizable; hence �(X;metriz) = !1. By 3.2, we know that X has a subspaceZ with jZj = !1 and w(Z) > !. By elementarity, there is such a set Z in M , soassume Z 2 M . Since X is countably compact and w(Z) is uncountable we knowthat Z is not metrizable | hence we may as well assume that X = Z .We may also assume that, for each x 2 X, Z [ fxg is metrizable, hence �rstcountable. Therefore M j= Z [ fxg is metrizable. If X is not regular at x then Mwill re ect this since Z 2M is dense. Indeed, suppose U 2 T is a neighborhood ofx such that V nU 6= ; for each neighborhood of x. By elementarity, we may assumethat U 2 M . Assume though that x has a neighborhood such that V \M � U(i.e. X \M is regular). Since Z [ fxg is �rst-countable and in M we may chooseW 2 M such that W \ Z � V . Therefore W � V � U . Since W and U are bothmembers of M , this is a contradiction since M j= V � U while H(�) j= V n U 6= ;.So we may assume that X is regular at x and therefore it follows that X is �rstcountable at x and T \M contains a local base at x. Therefore it su�ces to showthat hX \M; T \M i is not metrizable.Let fM� : � < !1g be a continuous �-chain of countable elementary submodelsof M with hX; T i 2 M0 and whose union is all of M . For each � 2 !1, we havethat 9x 2 X \M� such that T \M� does not contain a base at x. But sincefX; T ;M�g 2M�+1 there is in fact an x 2M�+1\X \M� such that T \M� doesnot contain a base at x.Finally, let us suppose that hX\M; T \M i has a point-countable base and obtaina contradiction to Proposition 2.3 . Let N be a countable elementary submodel ofH(�) such that each of X;M; T and fM� : � 2 !1g are in N . Let � = N \ !1and consider a point x 2 M \ X \M� such that T \ M� does not contain aneighbourhood base at x as discussed in the previous paragraph. But now N j=

ELEMENTARY SUBMODELS 9M = SfM� : � 2 !1g hence(T \M ) \N =[fT \M� : � 2 �g:Therefore (T \M )\N does not contain a local base for x 2 X \M \N , which isthe contradiction we seek.A noteworthy aspect of the above proof is the double usage of elementary sub-models. That is we developed some of the properties of the model M and then putM itself into a countable submodel.Clearly one of the awkward things about the above proof is that we had to �rstshow that the space would have to be �rst countable in order to deduce that T \Myielded the subspace topology on X \M . We shall now discuss the situation forre ecting countable character.It is easy to see that �(X;� = !) > !1 6) �(X) = !:Indeed remove the limit ordinals having co�nality !1 from !2+ 1 and observe thatthis example shows that even�(X;� = !) > !1&X is countably compact 6) �(X) = !:Therefore we could not have proceeded directly in 3.1 . But for which spaces does�(X;� = !) > !1 imply �rst countability?Proposition [J]: For compact spaces X,�(X;� = !) > !1 ) �(X) = !:It makes sense to ask howmuch compactness you need to obtain the above result.A space is called initially !1 � compact if every cover by !1 open sets has a �nitesubcover. This condition is, of course, equivalent to each of the conditions \there isno free closed �lter base of size !1" and \each set of size at most !1 has a completeaccumulation point". Let us �rst observe that this is how much compactness oneneeds to prove Arhangel'skii's result relating free sequences and countable tightness.Recall that a sequence fx� : � < �g is called a free sequence of length � if for each� < � it is the case that fx� : � < �g is disjoint from fx� : � � �g. When we sayfree sequence we shall assume the length is !1.Proposition 3.3: If a countably compact space does not have countable tight-ness then it contains free sequences. In addition, for an initially !1-compact spaceX , t(X) = ! i� X has no free sequences .Note that 3.3 is actually a re ection type result as well since it has as an imme-diate Corollary the fact that�(X; t = !) > !1 ) t(X) = !for all initially !1-compact spaces.

10 ALAN DOWProposition 3.4: For initially !1-compact regular spaces X ,�(X;� = !) > !1 ) �(X) = !:Proof. Let hX; T i be a regular initially !1-compact space such that �(X;� = !) >!1. By the remark following 3.3 we have that t(X) = !. Let M be an !-coveringelementary submodel of some H(�) so that hX; T i 2MandjM j = !1. It su�ces toshow that M j= �(X) = !.As in 3.2 it su�ces to show that T \ M induces the subspace topology onX \M . Let x 2 X \M and Tx = fT 2 T j x 2 Tg. Let U 2 Tx and suppose thatT \M n U 6= ; for all T 2 Tx \M . Using initial !1-compactness we may choosez 2\fT \M n U : T 2 Tx \Mg:Using t(X) = !, choose a countable set D � X \M n U so that z 2 D. Again, by!-covering of M we can �nd T 2 Tx \M so that x 2 TandT \D = ;. Now, sincewe are assuming that X is regular and T 2M we may choose T 0 2 Tx \M so thatT 0 � T , hence T 0 \ D = ;. This is a contradiction since z is supposed to be inT 0 \D.I do not know if one needs to assume that X is regular in the previous result.If there is a non-compact �rst-countable initially !1-compact space then there isan example to show that the assumption of regularity in 3.4 is necessary. Onthe other hand, it easy to see that one does not need to assume regularity if CHholds. Indeed, this is because under CH ( and it is consistent with :CH) that everyinitially !1-compact Hausdor� space of countable tightness is compact!Proposition 3.5: Let hX; T i be an initially !1-compact Hausdor� space ofcountable tightness. Then every maximal free �lter of closed sets has a base ofseparable sets. Furthermore if CH holds then the space is compact.Proof. Suppose that F is a maximal free �lter of closed subsets of hX; T i. LetM be an !-covering elementary submodel of some appropriate H(�) such thatfX; T ;Fg 2 MandjM j = !1. If CH holds we assume in addition that M! � M .Choose any z 2 FM = TfF \M : F 2 F \Mg, which we may do since jF \M j =!1. Let A 2 F \M be arbitrary and, by countable tightness, choose a countableset D � A \M so that z 2 D. Since M has the !-covering property and A 2 Mwe may assume that D 2 M . Since z 2 F \D = F \D for each F 2 F \M itfollows that M j= D \ F 6= ; for each F 2 F :Therefore, by elementarity and the maximality of F , D 2 F , F has a base ofseparable sets. It also shows that fF \M j F 2 F \ Mg � F , FM 2 F andfurthermore that jFM j > !1 since F is a free �lter and X is initially !1-compact.Now suppose that M is closed under !-sequences and that z0 is any other pointof FM . Let UzandUz0 be disjoint neighbourhoods of z and z0. Let Dz = D \ Uzand Dz0 = D\Uz0 . Now just as we showed that D was in F , the same proof showsthat both Dz and Dz0 are in F since they are both in M . However this contradictsthat z 2 F \M for all F 2 F \M since z =2 Dz0 .

ELEMENTARY SUBMODELS 11One can prove even a stronger result than the above one but the proof does notbene�t by the use of elementary submodels and can be proven by a simple inductionof length !1 .Proposition 3.5A [Fremlin]: If hX; T i contains no free sequences then foreach countably complete maximal �lter F of closed sets and each set H 2 F+ =fZ � X : Z\F 6= ; for each F 2 Fg there is a countable H0 � H so that H0 2 FHowever an interesting feature of the proof of 3.5 is that it gives us a prettygood idea of how the consistency results in both directions must go. For exampleto show that it is consistent with :CH we can imagine that M is an inner modelof CH and there are more reals to be added. It must be the case that new subsetsof X \M are added which can serve as the pair Uz; Uz0 mentioned above. Thereare a lot of properties that we can show the pair must have | for example theyboth meet every countable set in M whose closure is a member of F \M and thatM j= F is a countably complete �lter. We then investigate which kinds of forcingswhich add reals could not possibly add such a pair. It turns out that Cohen forcingis such a forcing but we shall not give the details here. In sections 5{7 we shallprove the result, due to , Fremlin and Nyik�os that assuming the Proper ForcingAxiom, each initially !1-compact space of countable tightness is compact. As forthe consistency of there being such spaces the above analysis indicates that we haveto plan for those inner models of CH and be building a space in such a way that itis possible to add the necessary sets. This is still open.Another question which suggests itself is whether or not we could replace `com-pact' in the character re ection result with `countably compact & countable tight-ness'. It turns out that if there are large cardinals then it is consistent that simply`countable tightness' will su�ce and no compactness is necessary at all. This willbe proven in section 6. However it is consistent that these two properties do notsu�ce.Example 3.6: In the constructible universe, L, there is a countably compactspace of countable tightness and uncountable character such that each subspace ofcardinality !1 has countable character.It is shown in [DJW], that there is a family of functions ff� : � < !2g in L sothat1. f� : �! ! for each � < !22. � < � < !2 implies f < � : f�( ) 6= f�( )g is �nite3. 8f : !2 ! ! 9� < !2 such that f < � : f( ) 6= f�( )g is in�nite.For each � < !2, let A�;0 = f(�;m) 2 � � ! : m � f�(�)g and for n > 0 letA�;n = f(�; n+f�(�)) : � < �g. By a straightforward `Ostaszewski-type' inductionone can de�ne a locally countable, locally compact topology on !2 � ! so that foreach � < !2 with uncountable co�nality the subspace � � ! is countably compactand furthermore ensure that for each n < ! the set A�;n is clopen.Next one de�nes, just as in [DJW] , a topology on X = fpg[!2�! by declaringthat !2 � ! is endowed with the above topology and U is a neighbourhood of fpgproviding p 2 U and 8� 2 !29n 2 ! so that U � SfA�;m : n < m 2 !g.

12 ALAN DOWiv More on metric spaces { Hamburger's question.Peter Hamburger has asked a natural question about metric spaces which canbe asked in our terminology as \Does there exist a �rst countable non-metrizablespace, X, such that �(X;metriz) > !1?". If the existence of large cardinals isinconsistent, then the answer is known to be \yes". In fact the example would justbe a special kind of subspace of the ordinal space !2 { called an E-set . An E- setis what is known as a non-re ecting stationary set. A set E of ordinals is calledan E-set if E is stationary in its supremum, (8� 2 E) cf(�) < !1 and for each� < sup(E) with cf(�) > ! E \ � is not stationary in �.As mentioned above if there are no large cardinals then in fact there is an E-setcontained in !2 (see [De2]). In section 6 we shall discuss the consistency, from alarge cardinal, of there being no E-sets. Therefore Hamburger's question for ordinalspaces is resolved. We shall shall show that the situation is the same for locally-@1spaces. Recall that X is locally-� if every point has a neighbourhood of cardinalityat most �.We proceed by analyzing the inductive step: \if X is a locally small space, does�(X;metriz) � �) �(X;metriz) > �?"The singular case holds in ZFC and the consistency of the regular case follows from(and implies) the consistency of large cardinals. The main tools will be Proposition2.3 and elementary chains.Theorem 4.1: Suppose ! = cf(�) � � < � and that X is a locally-� space.Then �(X;metriz) 6= �.Proof. We may as well assume that X has cardinality �. Fix a base B for Xconsisting of open sets of cardinality at most �.Choose a regular cardinal � much larger than � and an elementary 2-chainfMn : n < !g so that � [ f�; hX;Big � M0 � H(�), jMnj < � for each n 2 !,and X � SMn. By assumption, Xn = Mn \ � is metrizable for each n 2 !.Furthermore, for each B 2 B \Mn, Theorem 1.6 implies that B � Xn - hence Xnis open inX. Therefore X has a point-countable base. Furthermore, B 2 B\Mn )B � Xn, hence B \Mn does not contain a base for any point of X nXn. Also by2.3, B \M contains a base for all points of Xn. Therefore Xn is a clopen subset ofX and fXn+1 nXn : n 2 !g is a partition of X into clopen metrizable pieces.Theorem 4.2: Suppose cf(�) � � < � and that X is a locally-� space.Then �(X;metriz) 6= �.Proof.The former (published) proof of Theorem 4.2 is very wrong. I would like to thankPaul Szepticky for bringing this to my attention. In addition I am very grateful toStavros Christodoulou for supplying a correct proof.The idea is to generalize the proof of Theorem 4.1. One must �rst provethat Proposition 2.3 can be generalized by replacing \point-countable" by \point-lambda" and demanding that � be a subset of M .

ELEMENTARY SUBMODELS 13Proposition 2.3(a). If a space X with a base T has a point-� base and hX;T i 2M � H(�) and � � M , then T \M is a base for each point of X \M .Next one must show that an X as in 4.2 has a point-� base. To see this, letfM� : � 2 cf(�g be an as in the \proof" of 4.2. For each �, let T� be a point-countable base for the open metrizable space M�\X. Then one can easily see thatT = ST� is a point-cf(�)g base for X.Now one �nishes the proof by showing that each M� \X is closed just as in 4.1.`Locally-�' can be replaced by `locally-< �' in 4.1, but I don't know if it can bein 4.2. Also we leave as an open question, the regular cardinal version of 4.2.Question 4.3: If � is regular and there is a locally-< � topology on the set �in which every subspace of cardinality less than � is metrizable , does it follow thath�; T i is not metrizable i� f� < � j cf(�) = !and � 6= �g is stationary ?The proof of the next result must be delayed until 6.1.Theorem 4.4: If it is consistent that there is a supercompact cardinal then itis consistent that, for the class of locally-@1 spaces ,�(X;metriz) > !1 ) X is metrizableRecall that a space is said to be (@1-)CWH (for Collection-Wise Hausdor�) ifevery (@1-sized) discrete set can be separated by disjoint open sets. Shelah has alsoproven that it is consistent (subject to a large cardinal) that a locally-@1 �rst count-able space which is @1-CWH is CWH. However when the local smallness conditionin this and the above results on metric spaces are dropped no such re ection resultsare known to hold. It is known that the situation is di�erent since an example in[F] shows that 4.4 does not hold if local smallness is dropped. To �nish this sectionwe will �rst formulate a combinatorial principal on !2 and then construct a spacefrom it. I do not know whether or not this combinatorial principle is a consequenceof GCH or even 2!1 = !2 | it is consistent with these assumptions.Let S02 be the co�nality ! limits in !2 and let (y) denote the statement:(y) 9fA� j � 2 S02g and fg� j � 2 S02g so that:1. A� is a co�nal increasing sequence in �;2. g� is a function from A� into !;3. 8� < !2 9h� : �! ! such that 8� 2 � \ S02f� 2 A� j h�(�) � g�(�)g is �nite; and4. 8g : !2 ! ! 9� 2 S02 so thatf� 2 A� j g(�) � g�(�)g is in�nite.Example 4.5: (y) implies there is a �rst countable space which is @1-CWHand for which subspaces of size @1 are metrizable but which is not CWH and notmetrizable.

14 ALAN DOWLet fA� : � 2 S02g and fg� : � 2 S02g be as in (y) . We shall de�ne a topology onthe set !2 [ !2�!2�! so that !2 is closed discrete and unseparated and the restof the space is open and discrete.For each � 2 S02 let f��n : n 2 !g list A� in increasing order. For each point� 2 !2 we de�ne a countable neighbourhood base U (�; n) as follows:for � =2 S02 U (�; n) = f�g [ f�g � !2 � (! n n) ;for � 2 S02U (�; n) = f�g [ f�g � !2 � (! n n) [ f(��m; �; g�(��m)) : n < m 2 !g.The simplicity of the space ensures that a subspace will be metrizable if and onlyif it is CWH. To see that the space is @1-CWH , let � < !2 and choose h� as in (y) .For each � 2 �\S02 , de�ne h0(�) = h�(�)+ j where j is such that h�(��n) > g�(��n)for all n > j. Otherwise de�ne h0 equal to h�. It is easy to check that this h0 as afunction from � into the neighbourhood bases yields a separation of �.Let us now show that the space is not CWH. Indeed suppose that g : !2 ! !is such that U (�; g(�)) is disjoint from U (�; g(�)) for each � < � < !2. Choose� 2 S02 so that A0 = fm 2 ! : g(��n) � g�(��n)g is in�nite. Let m = g(�) and choosem < n 2 A0. But now the point (��n ; �; g�(��n)is in both the sets U (��n ; g(��n)) and U (� ; g(�)).v Elementary Submodels in Forcing ProofsForcing, of course, is the technique developed by Cohen which takes a (ground)model of set theory, together with a `new' desired set, and canonically constructs amodel of set theory (the extension) containing the new set and the ground model.The di�cult part of most forcing arguments is to show what sets are not added.That is, one must prove some kind of preservation argument. For example, it isfrequently important that the ordinal which is !1 in the ground model remains soin the extension | we would say that \!1 is preserved". Some other examples ofproperties which we may want preserved include: \being an ultra�lter over !"; \atree having no co�nal branches"; \a Souslin tree remaining Souslin".If V is the ground model, and P 2 V is a poset then we assume the existenceof G � P { a generic �lter (i.e. for each dense open D � P with D 2 VG\D 6= ;). V [G] is just the model obtained by adding G to V and using the axiomof comprehension to interpret all the P -names from V . (The fact that this worksis remarkable and di�cult to prove but to apply it is not as di�cult as I suspectis commonly assumed ). Therefore we now have two models of set theory, V andV [G]. If T 2 V is a base for a topology on X 2 V , then we can still discuss hX; T iin V [G] | it will be the same topological space but it may have di�erent secondorder properties. That is, we would want to discuss the preservation of topologicalproperties such as: the countable compactness of hX; T i, the non- normality ofhX; T i , etc. .In this section we give some examples of how elementary submodels can beutilized in proving such preservation results. We begin with Cohen-real forcing.Recall that the poset Fn(I; 2) = fs : s is a function into 2 ,dom(s) 2 [I]<! g andis ordered by s < t () s � t.

ELEMENTARY SUBMODELS 15Lemma 5.1: If G is Fn(I; 2)-generic over V and A 2 V [G] is a subset of !,then both FA = fB � ! : B 2 V andA � Bg and IA = fB � A : B 2 V gare countably generated.Proof. Let _A be a Fn(I; 2)-name for A and let _A 2M � H(�).Claim: If B 2 V and p j̀ _A � B , then 9B0 2M with p j̀ _A � B0 � B:To prove the claim, let p0 = p \M and B0 = fn : 9q < p0 so that q j̀ n 2_Ag. Clearly B0 � B and B0 2 M . Furthermore p0 j̀ _A � B0. This proves theclaim and that F is countably generated. That I is countably generated is provenanalogously.Lemma 5.2: Suppose hX; T i is a space and x 2 X is such that t(x;X) = !then 1 j̀ Fn(I;2) t(x;X) = !.Proof. Suppose 1 j̀ (x 2 _A) and _A 2 M � H(�). We shall complete the proof byshowing that 1 j̀ x 2 _A \M:Assume that p j̀ U \ ( _A \M ) = ; where x 2 U 2 T . Let p0 = p \M and de�neAp0 = fy 2 X : 9q < p0 q j̀ y 2 _Ag and note that x 2 Ap0 and Ap0 2 M . SinceM j= t(x;X) = ! , x is in the closure of some countable subset B of Ap0 which isan element of M . Now choose y 2 U \ B and, by elementarity, p0 > q 2 M sothat q j̀ y 2 _A. Finally, we have our desired contradiction since dom(q)\dom(p) =dom(p0), hence p [ q 2 Fn(I; 2), and (p [ q) j̀ y 2 U \ _AAlthough countably closed forcing does not preserve countable tightness in gen-eral, it is often the case that additional hypotheses on the space are required toprove the desired preservation result.Lemma 5.3: If X is a space of countable tightness which is �rst countable oncountable subsets then the countable tightness of X is preserved by countably closedforcing.Proof. Let P be a countably closed forcing , _A a P -name and assume p j̀ x 2 _A forsome x 2 X. Let M be a countable elementary submodel such that fp; x; _A;Xg 2M . With Aq de�ned as above, we have that x 2 Aq \M for each q 2M \ P . Byassumption X \M is �rst countable hence choose fUn : n 2 !g a neighbourhoodbase for x in the subspace X \M . Within M choose a descending sequence fpn :n 2 !g � P with p0 = p and for each n 2 ! there is an xn 2 X \M so thatpn j̀ xn 2 Un \ _A. Finally since P is countably closed there is a q 2 P withq < pn for each n, and q j̀ fxn : n 2 !g � _A. This completes the proof sincex 2 fxn : n 2 !g.

16 ALAN DOWAnother preservation result for Cohen forcing we'll need is 5.4. This result isproven in [DTW] and we shall not give a proof here. The proof uses a combinatorialstructure on the Cohen poset called an endowment and elementary submodels donot play a role.Proposition 5.4: If hX; T i is a space such that for some set I,1 j̀ Fn(I;2)hX; T i has a �-discrete base then X must already have one.A poset P is de�ned to be proper [S] if for each � > ! the stationarity of eachstationary S � [�]! is preserved by forcing with P . Recall that C � [�]! is closedif the union of each countable chain contained in C is again in C. The elementarysubmodel approach makes the concept of properness much easier to understandand to use. In fact properness can be viewed as a condition which guaranteesthat many elementary submodels in V will extend to elementary submodels inV [G]. If � is a large enough cardinal and if � = jH(�)j we can identify [�]! and[H(�)]!. Furthermore the set of countable elementary submodels of H(�) is closedand unbounded. Since P is proper , it can be shown that if G is P-generic overV , then in V [G] the set fM 2 H(�) : M � H(�)andM \ V 2 V g is stationaryin [H(�)]!. Therefore there are \stationarily many" such M such that P;G 2 M .Now we haveH(�) j= G \D 6= ; for each dense open D � P such that D 2 Vhence by elementarity M j= G \D 6= ; for each dense open subset D of P suchthat D 2 V . It also follows that M \ V is an elementary submodel of the H(�) ofV . Any condition q 2 P which forces that G \M meets each dense open subsetfrom M \ V is called a (P;M \ V )-generic condition. Combinatorially, in V , thistranslates to q 2 P is (P;M )-generic if for each r < q and each dense open D 2Mthere is a condition p 2 D \M such that r is compatible with p.As a result there is an equivalent de�nition of proper which is the one we shallwork with. P is proper i� for each regular � > 2jP j and each countable elementarysubmodel, M , of H(�) which includes P , there is a (P;M )-generic condition beloweach p 2 P \M (see [S]).Lemma 5.5: If P is an !1-closed poset, then P is proper and furthermore, ifX 2 V and G is P -generic over V then [X]! � V .Proof. Let fDn : n 2 !g list the dense open subsets of P which are in M - acountable elementary submodel. Let p0 = p be any element of P \M and choose adescending sequence pn , n 2 ! so that pn 2 Dn \M . Since P is countably closed,there is a q 2 P so that q < pn for each n 2 !. This q is clearly an (M;P )-genericcondition. Furthermore, this q has the property that for each element of M whichis a P -name of a function from ! into V , q forces it to equal a function in V Thisis how one proves [X]! � V .A useful generalization of countably closed forcing is the iteration of Cohenforcing followed by countably closed forcing. There are many preservation resultsfor the iteration which do not hold for countably closed forcing itself.

ELEMENTARY SUBMODELS 17Lemma 5.6: Suppose _Q is a Fn(I; 2)-name of a countably closed poset andthat I is uncountable. If hX; T i has countable tightness at x 2 X, then1 j̀ Fn(I;2)�QhX; T i has countable tightness at x :Proof. By 5.2 we may begin by assuming that G is Fn(I; 2)-generic over V andhX; T i 2 V . Let _A be a Q-name of a subset of X , q 2 Q , and assume that q j̀ x 2_A. Let M be a countable elementary submodel containing fX; T ; I; Q; q; A; xg.Now since I is uncountable and M is not, there are, in V [G], �lters on Fn(!; 2)which are generic over V [G \M ]. That is, if P is any countable atomless posetwhich is an element of V [G \M ], then there is a �lter H � P so that H 2 V [G]and H \D 6= ; for all dense open D � P with D 2 V [G \M ]. Well, such a P isQ\M and so we choose such an H � Q \M . Since Q is countably closed, chooseq0 2 Q so that H � fp 2 Q : q0 < pg.Claim: x 2 fy 2 X \M : q0 j̀ y 2 _Ag :Proof of Claim. Let U 2 T be an open neighbourhood of x and let p 2 Q \M .Recall the de�nition of Ap = fy 2 X : 9p0 < p p0 j̀ y 2 _Ag. Since Ap 2 M andM j= t(x;X) = !, there is a countable B � Ap such that x 2 _A and B 2 M .Therefore U \ B 6= ; and furthermore, by elementarity for each y 2 B there is ap0 2M so that p0 j̀ y 2 _A. This shows thatDU = fp 2 Q \M : 9y 2 U \M such that p j̀ y 2 _Agis a dense open subset of Q \M . Furthermore U 2 V , hence U 2 V [G \M ] , andso H \DU 6= ;. Since q0 is below every member of H, this completes the proof ofthe claim and the Lemma.The condition [X]! � V in 5.5 gives us a kind of !- absoluteness for V relative toV [G] which is similar to what we had when we were taking elementary submodelsclosed under !-sequences. For example we have the following result.Lemma 5.7: If G is generic over a countably closed poset P and hX; T i 2 Vis a countably compact space having no free !1-sequences, thenV [G] j= hX; T i is countably compact, with no free !1-sequences .Proof. It is a trivial consequence of 5.5 that X is still countably compact since, forexample, there are no new countable subsets of X and each countable subset fromV still has all its limit points. Now suppose that f _x� : � < !1g is a P-name so that1 j̀ f _x� : � < !1g is a free sequence in X:Since P is countably closed we can choose, in V , a descending sequence fp� : � <!1g � P and fy� : � < !1g so that , for each �, p� j̀ y� = _x�. It follows thatfy� : � < g \ fy� : � � < �g = ; , for each � < !1 , since p� j̀ y� = _x� for each� < �. But now the sequence is fy� : � 2 !1g is a free sequence since, by 3.3, Xhas countable tightness in V .

18 ALAN DOWTodor�cevi�c pioneered the use of elementary submodels as \side conditions" inbuilding proper posets. The following result is due to Fremlin for some special casesand the general result is due to Balogh.Proposition 5.8: If hX; T i is a non-compact, countably compact space , thenthere is a proper poset Q so thatj̀ QhX; T i contains a copy of the ordinal space !1:Proof. It turns out that the proof splits into two essentially di�erent cases, depend-ing on whether or not X contains free sequences. As we only plan to use the casewhen X does not, we shall only prove the result for this case and refer the reader to[B], or [D] for a proof of the other case. Since the iteration of proper posets is againproper, we may assume, by 5.7, that we have already forced with the countablyclosed collapse of the cardinal jT j + jXj. Therefore we may assume that X hascardinality and character !1. Choose a free maximal closed �lter F and de�ne Qas follows. q 2 Q if q = hgq; Hq ; Mqi where:1. Hq 2 [T ]<! ;2. Mq is a �nite elementary 2-chain of countable elementarysubmodels of H(�) such that fX; T ;Fg 2M for each M 2Mq;3. gq is a function whose domain,Eq, is a subset of f� 2 !1 : 9M 2Mq M\!1 =�g;4. for each � 2 Eq and eachM;M 0 2Mq, � 2M 0nM ) gq(�) 2M 0\TfF\M :F 2 F \Mg :The actual de�nition of the conditions is designed to make the �nding of (M;Q)-generic conditions a triviality. Indeed, if � > 2jP j and M � H(�) with P 2 Mand p 2M \ P , then, we will show below that, q = hgp;Hp;Mp [ fM \H(�)gi is(M;P )-generic.We take care to ensure that the range of the union of the �rst coordinates overthe generic �lter will yield a copy of !1 by de�ning q < p providing: gq � gp ; Hq �Hp ; Mq �Mp and, gq(�) 2\fU 2 Hp : gp(��) 2 Ugfor each � 2 Eq\ maxEp , where�� = min(Ep n �) :It is not too di�cult to show that, if P does not collapse !1 and G is P -genericthen !1 �[frange(gp) : p 2 Gg � hX; T i:The main di�culty to this claim is in showing that8p 2 G; 8M 2Mp; 9q 2 G;M \ !1 2 Eq :However anyone who reads the rest of the proof can easily do this. One may �nd iteasier to slightly change the de�nition of the conditions by allowingHq 2 [T [!1]<!and for q < p add the condition that� > max(Hq\��) for each � 2 Eq\max(Ep) and �� = min(Epn�):The result of thisis that, if G is Q-generic, E = SfEp j p 2 Gg is a cub in !1 and g = Sfgp j p 2 Ggis a homeomorphism. That is, if �� 2 Ep, then p j̀ �� 2 E and if p j̀ E \ �� hasno maximum then p j̀ (E \��) co�nal in �� (keep adding things to Hp \��) and

ELEMENTARY SUBMODELS 19p j̀ fg(�) j � 2 E \ ��g converges to g(��) - again keep adding neighbourhoods ofg(��) to Hp \ T .However the hard part of the proof is to show that P is proper (hence preserves!1). Let � > 2jP j and let P 2M � H(�) with jM j = ! and let p 2 P \M . De�nep0 = hgp;Hp;Mq [ fM \H(�)gi. We must �rst show that p0 2 P and that p0 < p.First of all P 2 M and � = supf� : 9q 2 P; 9M 0 2 Mq � 2 M 0g 2 M , henceM\H(�) � H(�). Furthermore, ifM 0 2Mp, then M 0 2M andM j= M 0 � H(�) ,hence M 0 � M \H(�). It follows that p0 < p.Now consider r < p0 andD 2M such that D is a dense open subset of P ; withoutloss of generality we may assume r 2 D. Let r0 = hgr \M ; Hr \M ; Mr \M iand note that r0 2 P \M and that r � r0. Let ~D = fhgq;Hqi : r0 � q 2 Dg 2H(�) \M . Let us �rst note that it su�ces to �nd a pair hg;Hi 2 ~D \M suchthat range(g n gr) � U� where U� = TfU 2 Hr : gr(��0) 2 Ug and ��0 = minEr nM . Indeed, if hg;Hi 2 ~D \M , then by elementarity there is q 2 D \M sothat hgq;Hqi = hg;Hi and r0 < q. One easily checks that q and r are compatible.Let ErnM = f�0 ; : : : ; �n�1g listed in increasing order. For expository purposes,�rst suppose that n = 1. Then, by de�nition of r 2 P we know thatgr(�0) 2\fF : F 2 F \Mg :Now, by 3.5A , we may assume that F is just a base for the �lter which consistsof separable sets. Therefore F = F \M for each F 2 F \M and, since F iscountably complete TfF : F 2 F \Mg 6= ;. Since ~D 2 M \ H(�) , it followsthat Z = fx 2 X : 9hg;Hi 2 ~D fxg = range(g � gr0 )g 2 M \H(�). Therefore ifZ =2 F+, then there is some F 2 F \M such that F \Z = ;. But this contradictsthat gr(�0) 2 F \ Z. Therefore it follows that gr(�0) 2 Z Hence we may choosesuch an x in U� and a ~q 2 ~D \M such that fxg = range(g~q � gr0).The idea of the elementary chains is that we can then handle the case n > 1.For i = 0; 1; : : : ; n� 1, let Mi 2 Mr be so that Mi \ !1 = �i and let gr(�i) = xi.Also, for i = 0; : : : ; n� 1 ,let gi = gr \Mi and Hi = Hr \Mi and let gn = gr andHn = Hr.Just exactly as in the case n = 1, but usingMn�1 in place ofM = M0, we obtainthat cl[fx : 9hg;Hi 2 ~D s:t: g = g_hmax(dom(g0)); xi&Hn�1 � Hg] 2 F :De�ne, for i = 0; : : : ; n� 1 :~Dn�(i+1) = fhg;Hi : cl�Zn�(i+1)[hg;Hi]� 2 Fg;where for any hg;Hi we let(1) Zn�(i+1)[hg;Hi] =fx : 9hg0;H0i 2 ~Dn�i s:t: g0 = g_hmax(dom(g0)); xi and H � H0g :Note that ~Dn�i 2 M0 for all i = 0; : : : ; n, where ~Dn = ~D. Furthermorewe have noted above that hgn�1;Hn�1i 2 ~Dn�1. Assume that i < n and thathgn�i;Hn�ii 2 ~Dn�i. Now since xn�(i+1) 2 Zn�(i+1)[hgn�(i+1);Hn�(i+1)i] andZn�(i+1) 2Mn�(i+1), we again obtain thatclZn�(i+1) 2 F :

20 ALAN DOWTherefore hg0;H0i 2 ~D0 2M0.We may now pick, h�0; y0i; : : : ; h�n�1; yn�1i and H01; : : : ;H0n�1 all in M0, so that fy0; : : : ; yn�1g � U� . These are picked recursively so that for eachi = n � 1; : : : ; 0 , if g0n�i = g0n�(i+1)_h�n�(i+1); yn�(i+1)i ;then H 0n�i � H0n�(i+1) and hg0n�i;H0n�ii 2 ~Dn�i. To carry out the inductive stepwe note that since hg0n�i;H0n�ii 2 ~Dn�i , we have that clZn�i+1[hg0n�i;H0n�ii] 2 F .Therefore U� \Zn�i+1[hg0n�i;H0n�ii] 6= ; and we may choose yn�(i+1) 2M0 \U� \Zn�i+1[hg0n�i;H0n�ii]. Then by elementarity we can choose �n�i+1 and Hn�i+1 inM as required above.vi. Large Cardinals and Reflection AxiomsWe have seen lots of examples where we had a space hX; T i 2 H(�) and aproperty or formula '(v1; v2; v3) so that when we took M � H(�) and a parameterA 2M , we had H(�) j= '(X; T ; A) (hence M j= '(X; T ; A) )but H(�) j= :'(X \M; T \M;A \M ) :In fact the whole point of re ection is to �nd conditions on M which are su�cientto guarantee that ' does re ect, as opposed to the situation described above.If a cardinal � is supercompact and hX; T i; A 2 H(�) and'(v1; v2; v3) is any formula such that H(�) j= '(X; T ; A) then there is an M �H(�) such thatjM j < � & fX; T ; Ag 2M & H(�) j= '(X \M; T \M;A \M )(see [KaMa]). When we combine this with forcing we get re ection results at \small"cardinals which need large cardinals. To get the most out of this, one would wantto master the techniques described in such articles as [De1], [KaMa] and [DTW].We shall just take the results after the fact as Axioms.PFA is, of course, the Proper Forcing Axiom: Given a proper poset P and afamily fD� : � < !1g of dense open subsets of P , there is a �lter G � P such thatG \D� 6= ; for all � < !1.Fleissner has an axiom called AxiomR: If S � [X]! is stationary andC � [X]<!2is t.u.b. then 9Y 2 C such that S \ [Y ]! is stationary in [Y ]!.The set C � [X]<!2 is said to be t.u.b. if it is unbounded and if the union of everychain of length !1 from C is again a member of C.Axiom R is a speci�c case of a scheme (see [DTW] for more details). Roughlyspeaking, if P is a nice class of forcing notions, then we could have Axiom P� : If'(v1; v2; v3) is a (local + structural) property which is preserved by forcings fromP and if (a; b; c) 2 H(�) is such that '(a; b; c) holds | then 9Y 2 [H(�)]<� suchthat '(a \ Y; b \ Y; c \ Y ) holds and Y \ � 2 �.For example, for axiom R, take X;S;C 2 H(�) and P is the class of properposets of cardinality < � = !2. Then'(X;S;C; �) � \S is stationary in [X]! , C is unbounded in [X]<� and closed

ELEMENTARY SUBMODELS 21under unions of chains of length !1 ". Now note that proper forcing preservesstationarity { hence '.PFA+ is what you get when you combine PFA and axiom R. For what it's worth,the author �nds it easiest to apply PFA by recalling how the consistency of PFA+is proven. That is, the model in which PFA holds is obtained by forcing with aniteration of length � of proper posets. When you are considering a space hX; T i inthe extension you know, from the fact that � is a large cardinal, that this space andany of its properties will re ect to an inner model { but there will be more forcingto be done. But now the di�erence between PFA and the above axiom scheme isthat you get to choose the next forcing in the iteration. The idea then is to choosethe next forcing so that the iteration of it with any other proper poset will preservethe properties of interest.Other axioms which are frequently used (but not as axioms) are:\the Cohen forcing Axiom" = Axiom Cohen2! ;\Mitchell forcing Axiom " = Axiom (Cohen�!1-closed)!2 ; where Cohen�!1-closed denotes the class of posets which are of the form Fn(!2; 2)�P and P is forcedto be a countably closed poset by the Cohen posets.\Levy forcing Axiom" = Axiom !1-closed!2 .Using these axioms together with a judicious choice of ' we can obtain !1-sized ,!-covering elementary submodels of some H(�) together with some instances of�11-re ection. For example as we promised in xIV .Proposition 6.1: Axiom R ) �(X;metriz) < !2 for X in the class of locally-@1 spaces. Hence, in particular, Axiom R implies there are no E-sets.Proof. Let X be a locally-@1 space and assume that X is not metrizable. By 4.1and 4.2, we may assume that � = jXj = X is a regular cardinal. Let B be a base forX consisting of open sets of cardinality at most @1. For our application of AxiomR we de�ne S = fs 2 [X]! j s n sup(s) 6= ;g :Next, �x a su�ciently large cardinal � and let M denote the set of !1-sized, !-covering elementary submodels of the structure (H(�); X;B;2). LetC = fY 2 [X]!1 : (9 M 2M)(M \X = Y )g :Since X is �rst-countable and locally-@1, it follows from an easy generalizationof 1.4 that C is a t.u.b. subset of [X]!1 . Before we show that S is stationaryin [X]!, let us suppose that it is and show how to deduce the result from AxiomR. By Axiom R, we may choose Y 2 C such that S \ [Y ]! is stationary in [Y ]!.Let M 2 M be such that M \X = Y . Since M is !1-covering we may choose acontinuous elementary 2-chain fM� : � 2 !1g whose union is M . Choose a large �0and a continuous 2-chain fN� : � 2 !1g of elementary submodels of H(�0) so thatfM� : � 2 !1g 2 N0. Since S\ [Y ]! is stationary, we may choose a � 2 !1 such thatM�\Y = N�\Y 2 S. By the de�nition of S, there is an x 2 Y \N��sup(Y \N�).We �rst observe that, by elementarity, there is such an x in Y . Indeed, �rst noticethat, since M� and X are in M�+1 and M� \X = M� \Y = N� \Y , it follows thatN�+1 j= (9x) (x 2 Y \M� � sup(Y \M�)).Finally we use 2.3 again to contradict that Y is supposed to be metrizable. Forthis we will show that x =2 S(B \ N�), where x is chosen as above. For eachB 2 B \ N� , the element sup(B) is a member of N� , hence B � sup(N� \ X).

22 ALAN DOWTherefore it su�ces to show that sup(N� \X) � sup(M� \X). To see this observethat for each � 2 X \N�, sup(f� < !1 : fsup(M� \X) � �)g) 2 N� .Now to prove that S is stationary. Suppose that A is a closed and unbound-ed subset of [X]! such that A \ S = ;. Let fM� j � 2 �g be a continuouselementary 2-chain of elementary submodels of H(�) such that, for each � 2 �,!1 [ fX;B; A; Sg � M�, jM�j < �, and M� \ � 2 �. As in the proof of 4.2, eachX \M� = � \M� is an open metrizable subspace of X. Since X is not metriz-able, there is an � such that X \M� is not closed. Since X is �rst-countable,we may choose a countable set s � X \M� such that s nM� 6= ;. Therefores n sup(M� \�) 6= ;. Now let N be a countable elementary submodel of M� whichcontains s [ fAg; we claim that a = N \ X 2 A \ S. First a 2 S since a � sand sup(a) � sup(M�). Since N j= (A is an unbounded subset of [X]!), it followsthat there is a countable chain C � N \ A such that a = [C - hence a 2 A. Thiscontradicts that A was chosen to miss SNow we show that it is consistent that 3.4 can be improved.Proposition 6.2: It follows from the Mitchell Axiom (and the L�evy Axiom)that a space with countable tightness and uncountable character has a � !1-sizedsubspace with uncountable character. That is, �(X;� � !1) � !1 for any X withcountable tightness.Proof. Let hX; T i have countable tightness and assume that x 2 X has uncountablecharacter. Suppose further that the character of every countable subspace of X iscountable (for the L�evy Axiom we must assume that every @1-sized subspace hascountable character). Choose a regular cardinal � large enough to contain the powerset of the power set of X. De�ne the formula ' so that'(x;X; T ;H(�)) i� 8>>>><>>>>: t(x; hX; T i) = ! and �(x; hX; T i) > !�8a 2 [H(�)]!9~a 2 H(�) so thatj~aj = ! ; a � ~a ; andH(�) j= �(x; hX \ ~a; T i = !�:Now we must check that forcing by Cohen � !1-closed preserves that ' holds.Lemma 5.6 (and 5.3 for L�evy) proves that countable tightness is preserved. Thesecond line in the de�nition of ' is also preserved by any proper forcing but itdeserves more discussion. At �rst glance it seems a total triviality | but theimportant point is that we are talking about the set H = H(�) as opposed to thede�ned notion. All the second line is really saying is that \H has the !-coveringproperty " and we are simply asserting that this is preserved by proper forcing. Ofcourse if we had put a = ~a in line two { this would not have been preserved byany forcing which adds a real. Since H has the !-covering property, uncountablecharacter is preserved.Now either of the above Axioms gives us a set M with jM j = !1 (which we mayas well assume is a subset of H(�) ) so that'(x;X \M; T \M;M ) holds:Therefore, in the subset X 0 = X \M and with respect to the topology inducedby T 0 = T \M , the point x has countable tightness and uncountable character.

ELEMENTARY SUBMODELS 23Therefore to �nish the proof we wish to show, just as we were doing in x III , thatT 0 induces the subspace topology on X 0 { at least at x. So let x 2 U 2 T andassume that x 2 clT 0 [X0 n U ]. Since we have t(x; hX 0; T 0i) = !, we may choose acountable a � X 0 n U so that x 2 clT 0a. But now M has the !-covering propertyhence we may choose a countable ~a 2M so that a � ~a 2M . This contradicts thatM j= ~a has countable character with respect to the topology induced by T sinceM would then contain a base for the subspace topology at x .We �nish this section with the PFA results on initially !1-compact spaces ofcountable tightness. Fremlin and Nyik�os proved (i) and (ii) is due to Balogh. Infact Balogh proved that, under PFA, compact spaces of countable tightness aresequential but we do not include this result since it depends on the case in 5.7which we did not prove.Theorem 6.3: PFA implies that if X is an initially !1-compact space of count-able tightness then :(i) X is compact;(ii) X is sequentially compact; and(iii) X is �rst countable at some of its points.Proof. Let hX; T i be an initially !1-compact space with countable tightness. Let Pbe the usual countably closed collapse of jXj . In the extension obtained by forcingwith P , the space hX; T i will not be compact if any of the conditions (i) - (iii)failed to hold. Indeed, if (ii) fails to hold then clearly X contains a closed subspacein which there are no points of �rst-countability { so we may as well assume that(iii) fails. For each x 2 X, �x a closed G�, Fx such that x =2 Fx. Let~P = fg 2 <!1X j \�2dom(g)Fg(�) 6= ;g:If ~G is a ~P -generic branch, then T�2!1 F ~G(�) = ; since each non-empty G� subsetof X must contain many points. One can now observe that forcing with P will adda generic branch through ~P ; ( or force with ~P in the �rst place, or even that wemay assume without loss of generality that each non-empty G� subset of X has thesame cardinality as X hence ~P �= P ).In the extension hX; T i is still countably compact and contains no free sequencesby 5.7 . Now use 5.8 to �nd a proper poset Q in the extension so that there is aP �Q-name _g so that1 j̀ _g is a homeomorphism from !1 into hX; T i :Therefore there are also P �Q-names ff _W�; _U�g : � 2 !1g such that(2) 1 j̀ P�Q8� 2 !1 f _W�; U�g � T and_g([0; �]) � _W� � _W� � _V� and _V� \ _g((�; !1)) = ; :Finally, we de�ne D� for � 2 !1 to be D� = fp 2 P � Q : 9x;W;U such thatp j̀ _g(�) = x; _W� = W; and _U� = Ug. Since the above statements are forced by 1,it follows that D� is dense for each � 2 !1. Use PFA to �nd a �lter G � P which

24 ALAN DOWmeets each D�. Pick, for each � 2 !1, x�;W�; U�; and p� so that p� 2 G \D�and p� j̀ _g(�) = x�; _W� =W�; and _U� = U�:Since G is a �lter, it follows that for � < �; x� 2W� and that x� =2 V� . Therefore(back in V ) fx� : � 2 !1g is a free sequence { since we have the same base for thetopology in both models, W� \X n V� must be empty. This contradicts the factthat X contains no free sequences (in V ) .Remark. The role of the pair fW�; V�g in the above proof is critical. It is nottrue, in general, that if you introduce a free sequence with proper forcing then youmust have had one to begin with. Perhaps the easiest way to see what is going on isto think of the above mentioned \author's-view" of PFA. When you meet !1-manydense sets from the poset P �Q, you are really forcing over some inner model. Wecan think of this forcing as introducing a sequence which is \free with respect tothe inner model space". However there are still points to be added to the spacewhich can destroy that freedom. Also there are still neighbourhoods to be addedof the points you do have and this is why we do not, and can not, assert that weget a copy of !1 in X.vii Submodels closed under !-sequences and ForcingIn this last section we will prove a few results that show that the techniquesinvolved when using large cardinals can be used even without the large cardinals.All that is going on in the results of the previous section is that a forcing statementis �rst re ected, then the forcing is factored and �nally a preservation result isproven. When countable objects seem to determine all the re ection that you needthen it is possible that a large cardinal is not needed. It may su�ce to re ectthe forcing statement (as in the above outline) by simply taking an elementarysubmodel closed under !-sequences. The more di�cult arguments (e.g. those usingPFA) may require the assumption of }(!2) because it sometimes depends on theorder in which you iterate your posets. If the forcing is simply an iteration of thesame poset then you probably just need to assume CH in the ground model as weshall demonstrate below with Cohen forcing. Frequently these results are provenusing the �-system lemma and other combinatorics .The general procedure is to let, say, f _A� : � < !2g , f _B� : � < !2g andf _C� : � < !2g be Fn(!2; 2)-names of subsets of !2 . Let A = f _A� : � < !2g andsimilarly de�ne B and C. Let M � H(!3) be such thatfA;B; Cg 2M ; M! �M ; and M \ !2 = � < !2 :Recall that a Fn(!2; 2)-name, say _A, of a subset of !2 can be assumed to be asubset of !2 � Fn(!2; 2) where p j̀ � 2 _A i� (�; p) 2 _A .Now we let _A�� = _A� \ �� � Fn(�; 2)� for each � < �. Similarly de�ne _B��and _C��. Using the facts that M � H(!3) and M! � M one can easily prove that1 j̀ Fn(�;2) _A�� = _A� \ � and many other re ection results of the form1 j̀ Fn(�;2)'� _A��; : : : ; C ��, 1 j̀ Fn(!2;2)'� _A�; : : : ; C � :

ELEMENTARY SUBMODELS 25The �nal and crucial step after having obtained the validity of the appropriateforcing re ection is to prove that further Cohen forcing preserves the property.Let us begin with a well-know result of Kunen's.Proposition 7.1: In the model obtained by adding !2-Cohen reals to a modelof CH, there are no !2-chains in P(!) mod fin.Proof. Suppose A; f _A� j � < !2g , are Fn(!2; 2)-names such that1 j̀ Fn(!2;2)A = f _A� j � < !2g � P(!) and _A� �� _A� for � < � < !2:Also �x a name B for f _B� : � 2 !2g so that1 j̀ B = fB 2 P(!) j j _A� nBj < ! for all � < !2g:Let A ;B 2M � H(!3) with M! � M; jM j = !1 and M \!2 = �. For � < � ,we may assume that, in fact, _A� and _B� are Fn(�; 2)-names. NowM j= 1 j̀ f _B� : � < !2g = fB j (8� < !2) j _A� nBj < !g:Suppose _A is such that 1 j̀ Fn(�;2)(8� 2 �) j _A� n _Aj < !: Since Fn(�; 2) is ccc and! is countable, there is a _B 2M so that 1 j̀ Fn(�;2) _A = _B. Therefore,1 j̀ 9� < � such that _B = _B� :This application of \!-absoluteness" has shown that if G is Fn(!2; 2)-genericand if G� = G \ Fn(�; 2), thenV [G�] j= fval( _B�; G�) j � < �g = fB 2 P(!) j (8� < �)val( _A�; G�) �� Bg:Now in V [G], let A� = val( _A�; G) for � < !2 . By assumption, jA� n A�j < !for all � < � and jA� n B�j < ! for all � < �. Now refer to 5.1 and letI = fB � ! : B 2 V [G�] and B � A�g . Since I is countably generatedand cf(�) = !1, there is an I 2 I so that A� �� I for co�nally many � 2 �.Therefore, A� �� I for all � 2 �. But then, by the above, there is a � 2 �such that I = B� . But this implies that A� =� A�+1 (a contradiction) sinceB� = I � A� ��A�+1 �� B�This technique is also useful in proving Malykin's interesting new result. VanDouwen and vanMill have shown that it is consistent that (e.g. under PFA) !��fxgis C�-embedded in !� for any point x 2 !�. Malykin has shown that this is alsotrue in the Cohen model. I feel that this result demonstrates that there are stillinteresting consistency results to be obtained in the Cohen model.Proposition 7.2: If G is Fn(!2; 2)-generic over V , a model of CH, then, inV [G], !�-fxg is C�-embedded for each x 2 !�.

26 ALAN DOWSketch of Proof. Assume that f !�-fxg ! [0; 1] is continuous and that r0 < r1are such that x 2 f [0; r0] \ f [r1; 1]. It is well-known that R0 = f [0; r0] andsimilarly R1 are regular closed subsets of !�. Let fA�g�<!2 and fB�g�<!2 bethe subsets of ! whose remainders are contained in R0 and R1 respectively. FixFn(!2; 2)-names for the A�'s and the B� 's and �nd � < !2 just as in 7.1 (youwould also want to ensure that x was in M ). Using 5.1 and the fact that !�is an F-space one can show that R0 \ R1 � TfC� : C 2 x \ V [G�]g. Indeed,suppose that D � ! is such that D \ X 6=� ; for all X 2 x \ V [G�] and thatD� \R0 = ;. By 5.1, there are fXn : n 2 !g � x\ V [G�] which generate the �lterfY 2 V [G�] : D �� Y g. This set fXn : n 2 !g need not be in V [G�] in generalbut since x is a �lter we can enlarge the set fXng so that we may assume that itis in V [G�]. Let Z = TfX�n : n 2 !g and note that x 2 Z \R0 and that Z \R0 isagain regular closed. But now Z 2 V [G�], hence we may choose an � < � so thatA� 2 V [G�] and A�� � Z \R0. This contradicts that there should be an n so thatXn �� ! n A�. To �nish the proof then we just have to note that x \ V [G�] doesnot generate x.We �nish with a new proof of a result from [DTW]. The original proof of this(and the PMEA analogue) involved rather more di�cult �lter combinatorics.Proposition 7.3: If G is Fn(!2; 2)-generic over V , a model of CH, then,in V [G], a �rst countable space of weight !1 is metrizable if each of its @1-sizedsubspaces are metrizable.Proof. Let f _B� : � 2 !1g be Fn(!2; 2)-names of subsets of !2 so that 1 j̀ h!2; f _B� j� < !1gi is a �rst countable space in which each subspace of size !1 is metrizable.Let M � H(!3) be so that M! � M , jM j = !1 and f _B�g�<!1 2 M . Let� = M \ !2 ; G� = G \ Fn(�; 2) and let f _B��g�<!1 be as above. Then V [G�] j=h�; f _B��g�<!1i is a �rst countable space. By 5.4, 1 j̀ Fn(�;2)h�; f _B��g�<�i has a�-discrete base. Fix a Fn(�; 2)-name _U so that 1 j̀ _U � !1 � !1� ! and so that _U\codes" a �-discrete base for �. That is, the (�; n)th member of the base will bethe union of f _B�� : (�; �; n) 2 _Ug. We will show that the collection whose (�; n)thmember is the union of f _B� : (�; �; n) 2 _Ug will form a �-discrete base for thewhole space.We would be done if the name _U were a member of M but there is no reason tosuppose that this would be so. However, the trick is to isolate, for each remainingx 2 !2 a countable piece of the name _U which will do the job. This countable piecewill be in M which will allow us to play the !-absoluteness game.Let x 2 !2 and let �0 2 !1 be such that 1 j̀ f _B� : � < �0g contains a baseat x. Let N be a countable elementary submodel of H(!3) which contains the setfx; f _B� : � < �0g; _U;Mg. Now let � = N \ !1 and let _UN = _U \ N . Since M isclosed under !-sequences, _UN 2M .Let '(�; _U) denote the formula (with parameter f _B� j � 2 !1g):\1 j̀ � If y 2 !2 is such that f _B� j � 2 �g contains a base for y, then,(i) for each n 2 !, y has a neighbouhood meeting at most one member of the familyf[f _B� j (�; ; n) 2 _Ug j 2 !1g, and

ELEMENTARY SUBMODELS 27(ii) for each � 2 �, such that y 2 _B� , there are 2 ! and n 2 ! such thaty 2 [f _B� j (�; ; n) 2 _Ug � B� �."Now we observe that : M j= '(�0; _UN ):Therefore, H(!3) j= '(�0; _UN ):But now, since _UN = _U \N , we have:N j= '(�0; _U):And, �nally, since N � H(!3), we obtain that:H(!3) j= '(�0; _U):This completes the proof since it shows that, at least with respect to x, _U codesa �-discrete base. References[1] A. V. Arhangel'skii,On the cardinality of bicompacta satisfying the �rst axiom of countability, Soviet Math. Dokl. ,10 ,1969 ,951-955[2] ibid., Structure and classi�cation of Topological Spaces and Cardinal Invariants , Uspehi Mat.Nauk. ,33 ,23-84 ,1978[3] Zoltan Balogh, On the structure of compact spaces of countable tightness , to appear Proc.AMS[4] Keith J. Devlin, The Yorkshireman's guide to the Proper Forcing Axiom , Proc. 1978 Cam-bridge Summer School in Set Theory ,1978[5] ibid., The Axiom of Constructibility ,Springer Verlag ,1984[6] A. Dow, Removing large cardinals from the Moore-Mrowka solution , preprint[7] A. Dow, I. Juhasz and W. A. R. Weiss, On increasing chains of �rst countable spaces , toappear in Israel J. Math.[8] A. Dow, F. Tall and W. A. R. Weiss, New proofs of the Normal Moore space Conjecture , toappear in two parts in Top. Appl.[9] W.G. Fleissner, Left separated spaces with point-countable bases , Trans. AMS ,294 ,1986,665-677[10] David Fremlin, Perfect pre-images of !1 , to appear[11] David Fremlin and PeterNyikos, Initially !1-compact spaces are compact under PFA , PrivateCommunication[12] Istvan Juhasz, Cardinal Functions in Topology: Ten Years Later ,Math. Center Tracts ,Am-sterdam[13] A. Kanamori and M. Magidor, The evolution of large cardinal axioms in set theory , Proc.Conf. on Higher Set Theory ,Lect. Notes Math. ,669 ,1978 ,99-275[14] Kenneth Kunen, Set Theory: An Introduction to independence proofs ,North Holland ,1978[15] M. Magidor,On the Role of Supercompact and Extendible Cardinals in Logic , Israel J. Math.,10 ,1971 ,147-171Department of Mathematics, York University, North York, Ontario, Canada M3J1P3