that is used in this paper follows [E]. We also use the powerful method of elementary submodels which is well-presented in [JuW] chapter Topolo

Transcription

1 A LINDELÖF SPACE WITH NO LINDELÖF SUBSPACE OF 1 PIOTR KOSZMIDER Λ AND FRANKLIN D. TALL Λ Abstract: A consistent example of an uncountable Lindelöf T 3 (and hence normal) space with no Lindelöf subspace of 1 is constructed. It remains unsolved whether extra set-theoretic assumptions are necessary for the existence of such a space. However, our space has 2 and is a P -space, i.e., G ffi 's are open, and for such spaces extra set-theoretic assumptions turn out to be necessary. 1. Introduction. The question of whether an uncountable T 2 space such as in the title exists is due to A. Hajnal and I. Juhász and has been open for almost thirty years. It first appeared in print in the list of problems in [R]. Hajnal and Juhász asked it after proving that under CH (the continuum hypothesis) there is no such compact T 2 example ([HJ]). The problem is interesting because all the other familiar properties equivalent to second countability in the class of metrizable spaces do reflect" down 1 (see [BT]). Until now, the only counterexample known for Lindelöfness failed even to be Hausdorff (see [BT]). Our example will be a subspace of the countable box topology on the product 2 copies of the two-point discrete space, and so will be zero-dimensional. It is still open whether a T 2 example can be constructed in ZFC. See [BT] for an assortment of results asserting that there are no such examples if various topological and set-theoretic conditions are assumed. Our example will be constructed by countably closed forcing. It in fact will be a P -space, i.e., G ffi 's are open. This simplifies the task of avoiding Lindelöf subspaces of 1 to that of just avoiding convergent! 1 -sequences. Notice that we speak of avoiding" Lindelóf subspaces rather than killing" them. That latter approach, while natural - do an iteration of length! 2 ; each subset of 1 appears at an initial stage; kill it by the next step of the forcing - will not work, since countably closed forcing preserves Lindelöfness for spaces of 1 (see [BT]). We also show that our example cannot be constructed in ZFC, since assuming for example the well-known combinatorial principle P 1 (see section 2 for the statement) which is a weak version of a generalized Martin's axiom (see [W]), we prove that every Lindelöf T 2 P -space of 2 has a Lindelöf subspace of 1. Before moving on to the forcing construction of our space in section 3, we first do some topology in section 2, to verify our assertion that we need only avoid convergent! 1 -sequences, and to prove the result assuming P 1. In the last section 4, we include some remarks and questions related to our result. The set-theoretic terminology used in this paper is standard; for example if A; B are sets of ordinals we write A < B to indicate that ff < fi for all ff 2 A and fi 2 B. Other unexplained symbols or terms can be found in [K]. The terminology concerning general topology Λ Both authors were partially supported by the second's author grant A-7354 from the Natural Sciences and Engineering Research Council of Canada. AMS Classification: Primary 54020, 54A35, 54A35, Secondary 03E35. Key words: Lindelöf, nonreflection. 1

2 that is used in this paper follows [E]. We also use the powerful method of elementary submodels which is well-presented in [JuW] chapter Topological results. We say that a transfinite sequence fx ο : ο <! 1 g converges to a point x (and call it an! 1 -convergent sequence) whenever every open neighbourhood of x contains all but countably many points x ο. For example, if the character of a point x is 1 and the point is not a G ffi -set, then by picking x ff 6= x from the intersection of the first ff basic open sets, we can construct an! 1 -convergent sequence. We will work with subspaces of 2! 2 with the countable box topology which we denote by T, and which is generated by a basis of sets of the form [ff] = ff 2 2! 2 : ff fg; for ff a function with a countable domain included in! 2 and range 2. Note that the intersection of two sets of the form [ff] is empty or again of the same form. Thus the family above is a basis for a topology. Let us note some simple facts. FACT 1. (2! 2 ; T ) is a zero-dimensional, Hausdorff and hence regular space with a countably closed basis, i.e., the intersection of countably many basic open sets is open. PROOF: We leave it to the reader. Note that an! 1 -convergent sequence together with its limit forms a Lindelöf subspace of 1. Thus to obtain a space with no Lindelöf subspaces of 1 we need to at least eliminate all! 1 -convergent sequences. As the following fact shows, sometimes that is all we need to do. FACT 2. Let X be a Lindelöf, Hausdorff space with a countably closed basis. Then Lindelöf subspaces of X are closed. Furthermore, if a subspace of 1 is Lindelöf, then it includes a convergent! 1 -sequence. PROOF: Suppose F X is Lindelöf. Given x 2 X F, for each y 2 F let U y ;V y be disjoint basic open sets containing T x and y respectively. Then there exists fy n : n 2!g such that fv yn : n 2!g covers F. Then fu yn : n 2!g is an open set containing x and disjoint from F. Thus X F is open. Next we claim that if F is a Lindelöf subspace of X of 1 and p 2 F, then either p is isolated in F or χ(p; F ) 1. As intersections of countably many basic open sets are open, there is no nonisolated p 2 F with χ(p; F ) 0. So to see that either p is isolated in F or χ(p; F ) 1, it is enough to prove the following CLAIM: χ(p; F 1. PROOF: Let fv ff : ff <! 1 g be a family of basic open sets containing p such that for each q 2 F fpg there are ff 2! 1 and a basic U(ff(q);q) such that V ff U(ff(q);q) = ; and q 2 U(ff(q);q). We will show that fw fi : fi <! 1 g forms a basis for p in F, where W fi = T fv ff : ff < fig. First note that these sets are open as countable intersections of basic open sets. Then 2

3 choose any open V containing p. F V is closed, hence Lindelöf and covered by fu(ff(q);q) : q 2 F V g, so choose a countable subcover fu(ff(q n );q n ) : n 2!g. If fi = supfff(q n ) : n 2!g then W fi [ fu(ff(q n );q n ) : n 2!g = ; which gives that W fi V, completing the proof of the fact that fw fi : fi <! 1 g forms a basis for p in F and completing the proof of the claim. Now, since F is Lindelöf and uncountable, it cannot be discrete, so some point has 1 (and that point cannot be G ffi, because it would be isolated) and hence is a limit of a convergent! 1 -sequence. Since an! 1 -sequence which converges in a subspace converges in the whole space, we have completed the proof of Fact 2, and have: FACT 3. If a subspace X of 2! 2 with the topology T is Lindelöf and has no! 1 -convergent sequences then it has no Lindelöf subspace of 1. Remark: Fact 2 and 3 are straightforward generalizations to Lindelöf P-spaces of standard facts about compact Hausdorff spaces; however our main result shows that another standard fact does not generalize, namely, that compact Hausdorff spaces of cardinality < 1 are sequentially compact (see e.g. [vd]). Our Lindelöf P -space has 2 < 2 and yet has no convergent! 1- sequences. It also has no points of character less 2, showing that the»cech-posp»sil Theorem doesn't generalize. Although we cannot prove the consistency of all Lindelöf (P -) spaces having a Lindelöf subspace of 1, we can show this for Lindelöf P -spaces of 2, at the cost however of assuming 1 2. The following theorem can also be considered as a case when some generalization from compact to Lindelöf P -spaces does occur with respect to the existence of convergent sequences. For i = 0; 1, let P i be the statement (see [W] 5.12.) : Given a collection of fewer than i subsets of! i such that the intersection of any subcollection of cardinality less i has i, there is a subset B! i of i such that for any member A of the collection we have jb Aj i. By [vd], 6.2. P 0 is equivalent to the assertion that 0 is the minimal cardinal» such that there is a compact space X with χ(x) =» and a countable subset A X such thatthere is a cluster point x of A but no sequence of points of A which converges to x. Thus P 0 gives lots of convergent sequences in compact spaces of character less than 2!. The following is a generalization of this fact to Lindelöf P -spaces. THEOREM 4. It is consistent with CH plus 1 2 that every T 2 Lindelöf P -space of 2 has an! 1 -convergent sequence and so has a Lindelöf subspace of 1. PROOF: We use the fact that P 1 plus CH plus 1 2 is consistent (see [K], p. 286). Thus it remains to prove that P 1 implies the statement of the theorem. Let X be a T 2 Lindelöf P -space of 2. By a variation of the proof of Fact 2, it is routine to verify that < 1 by CH. Choose any Y X of 1. The Lindelöf property implies that Y has a complete accumulation point, call it x. Now the basic neighbourhoods of x obtained from the claim produce a countably closed uniform filter on Y of cardinality < 1. Y can be identified with! 1, thus P 1 can be applied. The B obtained from P 1 is an! 1 -convergent sequence; together with x it forms a Lindelöf subspace of X of 1. 3

4 3. The construction We think of the collections of functions 2 A for A! 2 as sets of branches through binary branching trees. We want to construct a set X 2! 2 and hence a subspace of (2! 2 ; T ). We force with countably many countable pieces of functions so that X will consist of! 2 of them. The requirements 1), 2), 5) on the forcing conditions and the forcing order are what one might expect. For the sake of the Lindelöf property of X, the projections ß A (X) on the coordinates from a countable set A! 2 need to be countable (thus the requirement 6) below). In fact, if A 2 [! 2 ]! 1 V, any ordering of A in the type of! 1 will make ß A (X) an! 1 -tree with no Aronszajn subtree (compare [JW]). Therefore we could call the combinatorial structure of X, a forest with no Aronszajn tree. Note that some kind of the requirement as in 4) below is necessary if the forcing is to satisfy 2 -c.c. The remaining technical requirements 3) and 7) are needed to avoid the! 1 -convergent sequences by splitting" the points arbitrarily high. We force over a model of CH, which is needed in two of the lemmas. DEFINITION 5. The forcing P consists of conditions of the form p = (a p ;b p ;F p ) which satisfy the following: 1) a p 2 [! 2 ]!, b p 2 [! 2 ]!. 2) F p = ff ο p : ο 2 b p g; f ο p : a p! 2. 3) 8ο 2 b p 8ff 2 a p 9 2 b p ff f ο p j(ff a p ) = f p j(ff a p ). 4) 8ο; 2 b p ; ο 6= 9ff 2 a p (max(ο; ) + 1) f ο p (ff) 6= f p (ff). The order is defined by p» q if and only if the following are satisfied: 5) a p a q ; b p b q ; 8ο 2 b q f ο p ja q = f ο q. 6)8ο 2 b p 9 2 b q f ο p ja q = f q : 7) 8ο; 2 b q 8ff 2 a q f ο q j(a q ff) = f q j(a q ff) ) f ο p j(a p ff) = f p j(a p ff). LEMMA 6. P is a ff-closed notion of forcing. PROOF: Suppose p n is a decreasing sequence of conditions of P. Define a p = S fa pn : n 2!g and b p = S fb pn : n 2!g. Finally for ο 2 b p put f ο p = S ff ο p n : n 2!g. It is not difficult to check that p works as a lower bound of the sequence. However we may focus the attention of the reader on the fact that even though we can be building a complete binary tree while taking the sequence (p n : n 2!), at the limit we include (in the limiting condition) only countably many branches of this tree (to satisfy 1). This is similar to forcing a Kurepa tree with no Aronszajn subtree with countable conditions as in [T] and such a tree is related to the Lindelöf property as shown in [JW]. We will exploit this relationship below. LEMMA 7. For any ff 2! 2 the set D ff = fp 2 P : ff 2 a p g is dense in P. PROOF: Take any q 2 P. We show that there is an extension p of q such that ff 2 a p. We may without loss of generality assume that ff 62 a q. If there is fi 2 a q such that fi > ff, then we define a p = a q [ fffg, b p = b q and fp ο = fq ο [ f< ff; 0 >g for all ο 2 b p. It is easy to see that all the conditions 1) - 7) are satisfied. Now suppose that a p < ff. In this case, fulfilling 3) will require more effort. Let Q 2 [ff;ff+!) denote the set of all functions with domain [ff; ff +!), range 2 and such that the preimage of 1 is finite. Let 0 Λ denote the one of these functions which is constantly equal to 0. Let ο 0 > b q ;ff+!. We define a p = a q [ [ff; ff +!), b p = b q [ [ο 0 ;ο 0 +!). To define F p we need to define fp ο for ο 2 b p. If ο 2 b q then fp ο = fq ο _ 0Λ. The fp ο for ο 2 b p b q is just 4

5 any enumeration of all functions of the form f ο q _ g where g 2 Q f0λ g. It is easy to see that p as above works. LEMMA 8. For any ο 2! 2, the set E ο = fp 2 P : ο < sup(b p )g is dense in P. PROOF: Fix q 2 P. We can add to b q an interval [ο 0 ;ο 0 +!) with ο 0 > ο as in the second part of the proof of lemma 7, to obtain p 2 E ο. Given a generic set G for P, we define B = S fb p : p 2 Gg and f ο = S ff ο p : p 2 Gg. Thus, by the previous lemmas, X = ff ο : ο 2 Bg is a family 2 (in V, but below we will show 2 is preserved) functions in 2! 2. Of course the f ο 's depend on G; nevertheless we will skip the subscript G since we work with just one G. We will consider X as a subspace of (2! 2 ; T ) where T is the countable box topology described in the previous section. As a subspace of a regular space, X is regular. We need to prove more topological properties of X in the generic extension. LEMMA 9. X is Lindelöf in V P. PROOF: It will suffice to show that any cover of X by basic open sets [ff] of (2! 2 ; T ) includes a countable subcollection covering X. As the forcing is countably closed, these functions ff belong to the ground model; thus we can construct a decreasing sequence of conditions (p n : n 2!) such that for each ο 2 b pn the condition p n+1 decides at least one basic open set [ff ο ] from the original cover which contains f _ ο. We may without loss of generality assume by lemmas 6 and 7 that dom(ff ο ) a pn+1. Thus a p 2 P below each p n as in the proof of the lemma 6, has the following properties: a) 8ο 2 b p 9ff ο p k _ f ο 2 [»ff ο ] 2 _U. b) dom(ff ο ) a p. Now using 6) of the definition, of extension in P and the compatibility of the conditions of a generic set we conclude that p forces that f[ff ο ] : ο 2 b p g is a cover of X. As this collection is countable, this establishes the Lindelöfness of X. Actually, this proof is similar to the proof that a Kurepa tree constructed in [T] has no Aronszajn subtrees. According to fact 3 from the previous section we are left with showing that X has no! 1 -convergent sequence. For this, as well as for 2 -chain condition proof, the possibility of performing certain amalgamations is needed. The following technical lemma takes care of this. First we need the following: DEFINITION 10. We say that p; q 2 P are isomorphic if and only if a p a q [ b p b q < a p a q [ b p b q < a q a p [ b q b p and moreover there are order preserving bijections i : a p! a q and j : b p! b q such that i is the identity on a p a q, j is the identity on b p b q and for all ο 2 b p and ff 2 a p. fq j(ο) (i(ff)) = fp ο (ff) Although this use of isomorphic" is reasonably standard, it should be noted that the relation is not transitive. 5

6 LEMMA 11. a) Suppose p; q 2 P are isomorphic and there are ff 0 ;ο 0 2! 2 such that i) a p [ b p < ff 0 < a q a p [ b q b p < ο 0 and ii) a q a p 6= ;: Then there is an s» p; q. b) Suppose in addition that iii) r» p, iv) a r ;b r ff 0, v) ffi 2 b p b q, and that vi) there are ο 1 2 b r b p and 1 2 b q b p such that f ο 1 r ja p = f ffi p ja p and f 1 q j(a p a q ) = f ffi p j(a p a q ): Then s» r;q may be chosen so that f ο 1 s ja q = f 1 q. PROOF: We prove part b) only, as the proof of part a) can be obtained by a simplification of the proof of part b). Let = a p a q and = b p b q. We define the following amalgamation s = (a s ;b s ;F s ): a s = a r [ fff 0 g [ a q. b s = b r [ b q [ [ο 0 ;ο 0 +!). F s = Fr Λ [ Fq Λ [ F [ F where Fr Λ = ffs ο : ο 2 b r b q g, Fq Λ = ffs ο : ο 2 b q b r g, F = ffs ο : ο 2 g, F = ffs ο : ο 2 [ο 0 ;ο 0 +!)g. There is not much freedom in defining F ; for ο 2 we choose fs ο = fr ο [ f< ff 0 ; 0 >g [ fq ο: By the isomorphism of p and q, the defined sets are functions. Now let us define Fr Λ. Let ff : F r! F q be any function such that a) ff(f)j f, b) if fj = gj, then ff(f) = ff(g), c) ff(f ο 1 r ) = f 1 q. ff can be defined because for any ο 2 b r there is ο 0 2 b p such that fr ο ja p = fp ο0 (by 6) for the extension r» p) and because for any ο 0 2 b p there is 2 b r such that f q j = fp ο0 j (by the isomorphism of p and q). Also c) follows from the hypothesis about ο 1 and 1. Now, for ο 2 b r b q define fs ο = fr ο [ f< ff 0 ; 1 >g [ ff(fr ο ). To define Fq Λ we need the following CLAIM: For every ο 2 b q b r there is at most one (or none) 2 such that f q j = f ο q j. PROOF OF THE CLAIM: Suppose there are two such distinct 1 ; 2 ; Then by 4) of the definition of forcing (say, for q), there is an ff» max( 1 ; 2 ) + 1 such that f 1 q (ff) 6= f 2 q (ff). But, since < (a p [a q ) (see definition 10, of isomorphic conditions), we have that ff 2, a contradiction. Now we define ρ : F q! F r such that a) ρ(f)j f, b) if fj = gj, then ρ(f) = ρ(g), c) if there is 2 such that f q j = fj, then ρ(f) = f r, d) there is ο 2 b p such that ρ(f) = f ο r. 6

7 First, let us see that ρ as above can be defined. If the hypothesis of c) is satisfied, take ρ(f) as in c) (by the claim it is uniquely defined), then d) is satisfied, a) is clear and b) follows from the claim. If the hypothesis of c) is not satisfied and f = fq ο we pick for example fp j(ο) where j is an isomorphism between b q and b p (by the assumption of the lemma) which does not affect the values on. Then define ρ(f) = fr j(ο) and do it for other functions with the same restriction on so that b) is satisfied. Now for ο 2 a q a r define fs ο = ρ(fq ο ) [ f< ff 0 ; 0 >g [ fq ο. Finally a definition of F is needed so that 3) of the definition of P is fulfilled: let F = ffr ο [ f< ff 0 ; 1 >g [ f q : f q j = fr ο j ;ο 2 b r ; 2 b q g Fr Λ. and assume that it is somehow numbered by elements of [ο 0 ;ο 0 +!). This completes the definition of s. Let us now prove that s 2 P and that s» r;q. Clearly 1), 2), 5), 6) are satisfied by the construction. Now let us verify 7), for both of the extensions s» r and s» q, as it will be used in the proofs of 3) and 4). Let ο; ; ff be as in 7). Case 1: ο; 2 b r ;ff2 a r. As a s a r > a r, there is nothing to prove. Case 2: ο; 2 b q b r ;ff2 a q. As f ο q j = f q j, we have ρ(f ο q ) = ρ(f q ), thus 7) is satisfied. Case 3: ο; 2 b q b r ;ff2. By the construction ρ(fq ο ) = fr ο0, ρ(f q ) = f r 0, fs ο jff = fp ο0 jff and f s jff = f p 0 jff for some ο 0 ; 0 2 p (as defined in the definition of ρ, clause d)). Since r» p, by 7) for this extension we have that fr ο0 and f r 0 agree on a r ff and so fs ο and f s agree on a r ff as well. Case 4: 2, ο 2 b q and ff 2. Similar to case 3). Case 5: ο; 2 and ff 2 a q. By 4) for q or p, this is impossible. Case 6: 2, ο 2 b q and ff 2 a q. Since f p j = f ο q j and 2, by clause c) of the definition of ρ we have that and by the definition of F we have that f ο s = f r [ f< ff 0 ; 0 >g [ f ο q ; f s = f r [ f< ff 0 ; 0 >g [ f q : As f ο q j(a q ff) = f q j(a q ff) we obtain that f ο s j(a s ff) = f s j(a q ff) as required. This completes the proof of 7) To prove 3) let us consider various cases. Note that we will frequently use 7) without mentioning it. Case 1: ο 2, ff 2 a r or ff 2 a q. 3) follows from 3) for r or q respectively (using 7)). Case 2: ο 2, ff = ff 0. Since ff 0 < a q a p 6= ;, pick any ff 2 a q a r and apply 3) for a q. 7

8 Case 3: ο 2 b r b q. If ff 2 a r, we are done by 3) for r. If ff 2 a q a r, consider ff(f ο r ). By 3) for q there is 2 b q ff such that ff(f ο r )jff a q = f q jff a q and ff(f ο r ) 6= f q. The first part of these equalities implies that in particular f q j = f ο r j. Now note that by the construction f ο s = f ο r [ f< ff 0 ; 1 >g [ ff(f ο r ): Also the following holds: f ο r [ f< ff 0 ; 1 >g [ f q 2 F F s ; as f ο r [ f< ff 0 ; 1 >g [ f q cannot be in F Λ r because only f ο s is in F Λ r among the extensions of f ο r. If ff = ff 0 we can use the above case for ff's in a q a r, which completes the proof of this case. Case 4: ο 2 b q b p. We may without loss of generality assume that ff 2 a q a p. Then by 3) for q we are done. Case 5: ο 2 [ο 0 ;ο 0 +!) We have f ο s = f ο0 r [ f< ff 0 ; 1 >g [ f 0 q ; where f q 0 j = fr ο0 j and f q 0 6= ff(fr ο0 ). Again we may without loss of generality assume that ff 2 a q a r. By 3) and 4) for q we get the required 6= 0, > ff. Thus, we have completed the proof of 3), so now let us proceed to the proof of 4). Again we consider several cases. Case 1: fο; g 2 [b r ] 2 [ [b q ] 2. This follows from 4) for r or q respectively. Case 2: ο 2 (b r b q ) [ [ο 0 ;ο 0 +!), 2 b q b r. ff = ff 0 works by the construction. Case 3: ο 2 b r b q ; 2 [ο 0 ;ο 0 +!). As f s ja r = f r 0 2 F r, we can consider two subcases. First, if 0 6=, then the required ff is in a r (by 4) for r) and since a r < [ο 0 ;ο 0 +!) we are done. Secondly if = 0, then since f s 62 Fr Λ, that means that f s 6= f s 0, which is enough as all elements of [ο 0 ;ο 0 +!) are above a s. This completes the proof of 4) and the fact that s» r;q. LEMMA 12. (CH) P satisfies 2 -chain condition. Let A be a family of distinct conditions of P of 2. We need to find two compatible conditions. Using CH we can choose two isomorphic conditions of A, say p; q such that there are ff 0 ;ο 0 2! 2 such that a p ;b p < ff 0 < ff 0 +! < a q a p ;b q b p < ο 0. We have two cases. If a p a q 6= ;, then we can apply lemma 11 for r = p to obtain the required s» p; q. Otherwise a p = a q. The amalgamation s is obtained by putting a s = a p [ [ff 0 ;ff 0 +!), b s = b p [ b q [ [ο 0 ;ο 0 +!) and by defining fs ο = fp ο _ 0Λ for ο 2 b p, fs ο = fq ο _ 1Λ for ο 2 b q, and choosing f ο 's for ο 2 [ο 0 ;ο 0 +!) so that ff ο : ο 2 [ο 0 ;ο 0 +!)g = ff ο_ f : ο 2 b p ; f 2 Q f0 Λ gg where i Λ is the function constantly equal to i and defined on [ff 0 ;ff 0 +!) for i = 0; 1 and Q is the set of all functions defined on [ff 0 ;ff 0 +!) with at most finitely many values different than 0. A simple argument shows that s works. 8

9 LEMMA 13. (CH) P forces that X has no convergent! 1 -sequences in V P. PROOF: Suppose there is an! 1 -sequence converging to a point f ffi 2 X for some ffi 2 B. Let p 0 be a condition of P which forces it. Let f ο _ : <! 1 g be P -names for distinct indexes of points of the sequence. Let M be an elementary submodel of H(! 3 ) of 1 which is closed under countable subsets (i.e., [M]! M; here we are using CH) and contains all relevant objects, in particular p 0, ffi, f ο _ : <! 1 g. Note that M! 2 is an ordinal in! 2 of uncountable cofinality. Using the density lemmas 7, 8 and 3) and 4) of the definition 5 of the forcing P we can find a condition p 1» p 0, and an ordinal ff 1 2 a p1 such that ff 1 > M! 2 and an ordinal 1 2 b p1 M such that f 1 p 1 j(a p ff 1 ) = f ffi p 1 j(a p ff 1 ); f 1 p 1 (ff 1 ) 6= f ffi p 1 (ff 1 ): Since p 1» p 0 forces that ff ο _ : <!1 g is a sequence converging to f ffi, there is a q» p 1 and a 0 <! 1 such that q forces that for > 0 we have Λ) f _ ο (ff 1 ) = f ffi (ff 1 ): Let us see that the amalgamation lemma 11 implies that this is impossible. CLAIM: There is a condition p in M P which is isomorphic to q and such that a q (M! 2 ) = a q a p and b q (M! 2 ) = b q b p. Since ffi 2 a q (M! 2 ), it will follow that ffi 2 a p. Proof of the claim: This can be arranged using the elementarity of M and the closure of M under countable subsets. Now, work in M. As p forces that [fp ffi ja p ] is a basic neighbourhood of f ffi and ff ο _ : <!1 g is an! 1 -sequence converging to f ffi, p forces that there is a ο 2! 2 b p and > 0 such that f ο ja p = fp ffi ja p. So let r» p be a condition which decides this ο and such that r 2 M and ο 2 b r. Since the cofinality of M! 2 is! 1 we can choose ff 0 2 M such that a r ;b r ff 0. Now we check that we can apply lemma 11 for ο 1 = ο. Since f 1 p 1 (ff 1 a p1 ) = fp ffi 1 (ff 1 a p1 ), by 7) of the definition of P, f 1 q (ff 1 a p1 ) = fq ffi (ff 1 a p1 ), but a p a q = a q (M! 2 ) ff 1. >From lemma 11 then, we obtain s» r;q such that f ο s ja q = f 1 q. Then s forces that which contradicts *). f _ ο (ff 1 ) = f 1 q (ff 1 ) 6= f ffi q (ff 1 ) = f ffi (ff 1 ); Combining lemmas 6, 9, 12, 13 and fact 3 we obtain the following. THEOREM 14. It is consistent that there is an uncountable Lindelöf regular space with no Lindelöf subspace of Remarks There are a number of difficult long-outstanding open problems concerning Lindelöfness, for example Arhangel'ski» 's problem concerning the existence of a Lindelöf space with points G ffi of size bigger than 0 (see [A]). A breakthrough there was achieved by Shelah [S], who constructed a consistent example of 2 = 0 )+ by countably closed forcing. Later I. Gorelic (in [G]) found an essential 9

10 simplification of this construction which generalizes to any cardinal» = 2 = 0 )+. However no better ZFC bound for cardinalities of Lindelöf spaces with points G ffi than a measurable cardinal is known. Previous unsuccessful attempts to produce a Lindelöf space with no Lindelöf subspace of 1 may perhaps have been too influenced by Shelah's space, in that they carried the excess baggage of points G ffi ". We go to the opposite extreme here by having G ffi 's open, which simplifies the construction. Nonetheless, our method of obtaining the Lindelöfness of the whole space is similar to that of [S], although we originally extracted it while looking at an independent argument from [JW]. We do not know how to construct a consistent example of a Lindelöf space with points G ffi which does not include a Lindelöf subspace of 1. In [BT], it is shown consistent with ZFC - modulo a huge cardinal - that no such first countable space of 2 = 0 )+ exists. This is non-trivial in the T 1 case - it is not known whether the cardinality of Lindelöf, first countable T 1 spaces is bounded by 0. Allowing :CH, it is consistent from a weakly compact cardinal that no T 2 counterexample of character less than 0 exists ([BT]). The question remains of whether an uncountable Lindelöf space with no Lindelöf subspace of 1 exists in ZFC, and in particular whether there is such a Lindelöf P -space. References: [A] A. Arhangel'ski» ; On the cardinality of bicompacta satisfying the first axiom of countability; Soviet Math. Dokl. 10, 1969, pp [BT] J. E. Baumgartner, F. D. Tall; Reflecting Lindelöfness; Top. Appl. To appear. [vd] E. Van Douwen; The Integers and Topology; in [KV] pp [E] R. Engelking; General Topology; Helderman Verlag, Berlin [G] I. Gorelic; The Baire category and forcing large Lindelöf spaces with points G ffi ; Proc. Amer. Math. Soc. 118, 1993, No 2. pp [HJ] A. Hajnal, I. Juhász; Remarks on the cardinality of compact spaces; Proc. Amer. Math. Soc. 59, 1976, pp [JW] I. Juhász, W. Weiss; On a problem of Sikorski; Fund. Math. 100, 1978, pp [JuW] W. Just, M. Weese; Discovering Modern Set-Theory. Set-theoretic tools for every mathematician. Vol II. Graduate Studies in Mathematics. Vol 18. American Mathematical Society, providence, [K] K.Kunen; Set Theory; An Introduction to Independence Proofs; North Holland, [KV] K. Kunen, J. Vaughan eds. Handbook of set-theoretic topology; North Holland, [R] M. E. Rudin; Lectures on Set-Theoretic Topology; American Mathematical Society, Providence, [S] S. Shelah; On some problems in general topology pp , in Set Theory; Boise, Idaho; ; American Mathematical Society; Providence [T] S.Todorcevic; Trees, subtrees and order types; Ann. Math. Logic 20, 1981, pp [W] W. Weiss; Versions of Martin's Axiom; in [KV] pp Authors' addresses: Departamento de Matemática, Universidade de S~ao Paulo, Caixa Postal: 66281, S~ao Paulo, SP, CEP: , Brasil. piotr@ime.usp.br Department of Mathematics, University of Toronto, Toronto, Ontario, Canada M5S 3G3. tall@math.toronto.edu 10