FUNDAMENTA MATHEMATICAE * (200*) An irrational problem by Franklin D. Tall (Toronto) Abstract. Given a top ological space hX; Ti 2 M ,anelementary submo del of set theory,we de ne X to b e X \ M with top ology generated by fU \ M : U 2T \M g. M Supp ose X is homeomorphic to the irrationals; must X = X ?Wehave partial results. M M We also answer a question of Gruenhage byshowing that if X is homeomorphic to the M \Long Cantor Set", then X = X . M In [JT], we considered the elementary submodel topology , de ned as fol- lows. Let hX; Ti b e a top ological space which is an elementofM , an elemen- tary submo del of the universe of sets. (Actually, an elementary submo del of H ( ) for a suciently large regular cardinal. See [KT] or Chapter 24 of [JW] for discussion of this technical p oint.) Let T b e the top ology on M X \ M generated by fU \ M : U 2T \M g. Then X is de ned to b e X \ M M with top ology T .In[T]we raised the question of recovering X from X M M and proved for example: Theorem 1. If X isalocal ly compact uncountable separable metriz- M able space , then X = X . M In particular: Corollary 2. If X is homeomorphic to R, then X = X . M M The pro of pro ceeded byshowing that [0; 1] M |any de nable set of @ 0 power 2 would do. It followed that ! M , whence by relativization we 1 deduced that X had no uncountable left- or right-separated subspaces. This @ 0 and hence were included in implied b oth X and T had cardinality 2 M . This sort of pro of will recur frequently in what follows. 2000 Mathematics Subject Classi cation : Primary 03C62, 03E35, 54A35, 54B99; Secondary 03E15, 03E55, 54F65, 54G20, 54H05. Key words and phrases : elementary submo del, irrationals, reals, Cantor Set, Bernstein Set, Long Cantor Set. Research partially supp orted by NSERCGrant A-7354. [1] 2 F. D. Tall On the other hand, we noted in [JT]: Example.LethX; Ti be any regular space without isolated p oints, X; T2M ,acountable elementary submo del. Then X is homeomorphic M to Q. The problem that naturally arises then is: Problem. If X is homeomorphic to R Q, is X also? In particular , M does X = X ? M This is the problem of our title. Unde ned top ological terms can b e found in [E], set-theoretic ones in # [Ku], with the exception of 0 ,Jonsson cardinal, and Chang's Conjecture, for which consult [K]. However let us de ne two top ological notions which may not b e familiar to set theorists: Definition. A space is of pointwise countable type if each p oint is con- tained in a compact set K for which there exist op en fU g including K n n<! such that each op en set including K includes some U .Amapisperfect if n it is continuous, closed, and pre-images of p oints are compact. We obtained a partial result concerning the Irrational Problem in [T]: # Theorem 3. Suppose 0 does not exist and X is a separable metriz- M able spaceofcardinality continuum (e.g. an uncountable separable completely metrizable space ). Then X = X . M The only use of the set-theoretic hyp othesis (whichfollows for example @ @ 0 0 from V = L)was to conclude (via [KT]) that 2 M since jM j 2 . @ 0 As noted byWelch [W], this also follows from the assumption that 2 is @ 0 notaJonsson cardinal. Once one has 2 M , the pro of outlined ab ove for Theorem 1 applies in this situation. In this pap er we obtain several more partial results: Theorem 4. Suppose X is an uncountable separable completely met- M rizable space. Then X = X if any of the fol lowing conditions hold : M (a) X includes a perfect pre-image of a compact rst countable uncount- able T space , 2 (b) X is (aproduct of spaces ) of pointwise countable type , e.g. X is rst countable or local ly compact or Cech-complete , (c) jR \ M j is uncountable , @ 0 . (d) jX j =2 For all I know, the theorem is true without any additional hyp othesis, at least for R Q. Chang's Conjecture , which yields an uncountable M with M \ ! countable, is a natural hyp othesis to use, say with CH, to try to get 1 acounterexample. (This conjunction do es yield a non-metrizable X such An irrational problem 3 @ 0 that X is a separable metrizable space of size 2 [T].) Note that (a) and M (b) rule out a numb er of plausible attempts, such as the \Irrational Long Line" or the pro duct of @ copies of the countable discrete space. 1 We will prove (a) and (b) rst. Their pro ofs are natural extensions of the pro of of Theorem 1. (b) in fact follows immediately from (a) by using Lemmas 5 and 6 b elow together with the classical top ological fact that an uncountable separable completely metrizable space includes a copyofthe Cantor Set. Lemma 5 [JT]. If X is T , i 3, so is X . M i Lemma 6. If X is T and a product of spaces of pointwise countable 3 type , then X isaperfect image of a subspaceof X [P].For X compact T , M 2 X is a perfect image of X itself [Ju]. M (A pro of of Lemma 6 for spaces of p ointwise countable typ e may b e found in [JT]. Alternatively, one can omit Lemma 6 and divide into 2 cases: nite pro ducts of spaces of p ointwise countable typ e havepointwise countable typ e, so [JT] applies; in nite pro ducts include copies of the Cantor Set.) The idea of the pro of of (a) is to use the p erfect map plus another classical fact, namely that the Cantor Set maps onto [0; 1], to obtain via elementarity a closed map from a closed|hence complete|subspace of X M onto [0; 1] =[0; 1] \ M . (The equalityisby rst countability [JT].) Closed M metrizable images of completely metrizable spaces are completely metrizable (see e.g. [E, 4.4.17]), hence absolute G .If[0; 1] \ M is a G in [0; 1], it is in fact all of [0; 1] (see b elow), so [0; 1] M , which, by the pro of of Theorem 1, is enough to get the desired result. Unfortunately, elementaritydoesnot quite give us a closed map, so wehavetowork a bit harder. Now for the details. We rst need de nitions and a technical lemma which enable us to make do with an elementary version of a closed map. Definition. A regular space is almost Cech-complete if there is a se- S quence fG (n)g , each G (n) a collection of op en sets such that G (n)is n<! dense (wecall G (n) an \op en almost-cover"), and such that whenever F is a T : F 2Fg6=0. centered collection of op en sets meeting each G (n), then fF If we require the G (n)'s to b e covers, wegetCech-completeness. Lemma 7[F]. Acompletely regular spaceisalmost Cech-complete if and only if it includes a dense G Cech-complete subspace. Definition. Let B b e a basis for a space X such that B is closed under nite intersections. f : X ! Y is a B -map if: (1) for all B 2B, f (X B ) is closed, (2) f is onto, but for no B 2B is f j(X B )onto Y . 4 F. D. Tall Lemma 8. A B -image of an almost Cech-complete space is almost Cech- complete. Proof. We mo del our pro of on that for closed irreducible images in [AL]. Supp ose X is almost Cech-complete with resp ect to the sequence fG (n)g n<! of op en almost-covers. Let B b e a basis for X .For each n<! and each S 0 x 2 G , pick a basic op en G (x) 2B containing x and included in some n n S 0 0 member of G (n). Then, letting G (n)= fG (x):x 2 G g, claim X is n n 0 almost Cech-complete with resp ect to the sequence fG (n)g .For let D be n<! 0 0 a centered collection of op en sets meeting each G (n), say D 2D\G . Then n n D is included in some D 2G by construction. Then D = D[fD g n n n<! n n T T : D 2D g 6=0,so fD : D 2Dg6=0. is centered, so fD So without loss of generality,wemay assume each G is included in n 1 b b the basis B .Given B 2B,let B = f (Y f (X B )). Then B is op en b and included in B .Furthermore, by (2) (irreducibili ty), B 6= ;. Indeed, b B is dense in B .To see this, let U 2Bbe suchthatU \ B 6= ;. Then d d b d ; 6= U \ B U \ B , but also U \ B B ,sincex 2 U \ B implies f (x) 2 Y f (X (U \ B )) implies f (x) 62 f (X B ) implies f (x) 2 Y f (X B ) 1 b implies x 2 f (Y f (X B )). Thus B \ U 6= ;. b Let H(n)=ff (G): G 2G(n)g.Claim H(n) is an op en almost-cover b of Y .