An Irrational Problem

FUNDAMENTA

MATHEMATICAE

* (200*)

An irrational problem

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.

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

\Long Cantor Set", then X = X .

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

X \ M generated by fU \ M : U 2T \M g. Then X is de ned to b e X \ 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-

able space , then X = X .

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

power 2 would do. It followed that ! M , whence by relativization we

deduced that X had no uncountable left- or right-separated subspaces. This

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

to Q.

The problem that naturally arises then is:

Problem. If X is homeomorphic to R Q, is X also? In particular ,

does X = X ?

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

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-

able spaceofcardinality continuum (e.g. an uncountable separable completely

metrizable space ). Then X = X .

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 .

As noted byWelch [W], this also follows from the assumption that 2 is

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-

rizable space. Then X = X if any of the fol lowing conditions hold :

(a) X includes a perfect pre-image of a compact rst countable uncount-

able T space ,

(b) X is (aproduct of spaces ) of pointwise countable type , e.g. X is rst

countable or local ly compact or Cech-complete ,

. (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

acounterexample. (This conjunction do es yield a non-metrizable X such

An irrational problem 3

that X is a separable metrizable space of size 2 [T].) Note that (a) and

(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.

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

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].

(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

onto [0; 1] =[0; 1] \ M . (The equalityisby rst countability [JT].) Closed

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-

quence fG (n)g , each G (n) a collection of op en sets such that G (n)is

dense (wecall G (n) an \op en almost-cover"), and such that whenever F is a

: 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

of op en almost-covers. Let B b e a basis for X .For each n

x 2 G , pick a basic op en G (x) 2B containing x and included in some

0 0

member of G (n). Then, letting G (n)= fG (x):x 2 G g, claim X is

almost Cech-complete with resp ect to the sequence fG (n)g .For let D be

0 0

a centered collection of op en sets meeting each G (n), say D 2D\G . Then

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

b b

the basis B .Given B 2B,let B = f (Y f (X B )). Then B is op en

and included in B .Furthermore, by (2) (irreducibili ty), B 6= ;. Indeed,

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 )

implies x 2 f (Y f (X B )). Thus B \ U 6= ;.

Let H(n)=ff (G): G 2G(n)g.Claim H(n) is an op en almost-cover

of Y . Certainly each f (G) is op en; given U op en in Y ,some G 2G(n) meets

1 1

b b b

f (U ), hence so do es G, hence f (G) meets f (U ), hence so do es G, hence

f (G) meets U .Now supp ose D is a centered family of op en subsets of Y

meeting each H(n). Then C = ff (D ): D 2Dgis a centered family of op en

T T

: C 2Cg6=0.Ifx 2 fC : C 2Cg, subsets of X meeting each G (n), so fC

then f (x) 2 fD : D 2Dg.SoY is almost Cech-complete with resp ect to

fH(n)g .

Continuing with the pro of of Theorem 4(a), with the aid of Lemmas 7

and8we can prove:

Lemma 9. Suppose X includes a perfect pre-image of an uncountable

compact rst countable T space Y and X is Cech-complete. Then M

2 M

[0; 1].

Proof. First of all, as in the classical pro of that uncountable compact

(see e.g. [H, 4.5]), we notice rst countable T spaces have cardinality2

that each such space Y includes a compact subspace without isolated p oints,

and hence a Cantor tree of closed sets, which yields a map of the subspace

onto the Cantor Set (see e.g. [J , 3.18]) and hence onto [0; 1]. By the Tietze

Extension Theorem, this map extends over the whole compact space Y to

give us a p erfect map of it onto [0; 1]. Since the comp osition of p erfect maps

is p erfect, the p erfect pre-image of a compact space is compact, and p erfect

maps from a compact space are closed irreducible on a closed subspace, we

An irrational problem 5

conclude that there is a closed L X and a closed irreducible map from L

onto [0; 1]. By elementarity, there is an F 2 M such that:

(1) F \ M is closedin X ,

(2) thereisacontinuous surjection g from F \ M (as a subspaceof X )

to [0; 1] \ M such that if H 2 M is a closed subset of X , then

(a) g (F \ H \ M ) is closedin[0; 1] \ M ,

(b) if g (F \ H \ M )= [0; 1] \ M , then F \ H \ M = F \ M .

Wethus haveaB -map from a closed (hence Cech-complete) subspace

of X onto [0; 1] \ M . Therefore [0; 1] \ M includes a dense Cech-complete

subspace. Such a subspace is then G in [0; 1] \ M . It follows that (0; 1) \ M

also includes a dense G . But (0; 1) \ M is a top ological subgroup of (0; 1)

and wehaveshown it is non-meager. It has the prop erty of Baire since

(0; 1) [(0; 1) \ M ] is meager. Wenow apply:

Lemma 10 [Ke, 0.11]. A subgroup of a topological group which is non-

meager and has the property of Baire is clopen.

But (0; 1) is connected, so (0; 1) \ M =(0; 1), so M includes (0; 1) and

hence [0; 1] and wehave established Lemma 9.

Theorem 4(a) can now easily b e established in the manner of Theorem 1.

Precisely:

Lemma 11. If [0; 1] M and X is T , hereditarily Lindelof and hered-

M 2

itarily separable , then X = X .

The pro of is outlined following Corollary 2 and is given in full in [T].

Theorem 4(c) is proved in a completely di erent fashion from 5(a), using

the following unpublished result of W. H. Wo o din, included with his kind

p ermission.

Theorem 12. If R \ M includes a perfect set , then R \ M = R.

The following pro of, due to E. T. Eisworth and included with his kind

p ermission, is much more elementary than Wo o din's pro of.

Proof of Theorem 12. By a routine mo di cation of the construction of a

disjoint Bernstein sets of R, i.e. fB g such Bernstein set, construct 2

r r 2R

that neither B nor R B includes a Cantor Set. De ne f : R ! R by

r r

B . Then there are such Bernstein sending B to r , f arbitrary outside

r r

r 2R

sets and suchanf 2 M .If P R \ M = R is p erfect, it includes a Cantor

Set and so meets each B . Therefore f P = R M .

It follows immediately that:

Corollary 13. If R \ M is analytic and uncountable , then R \ M = R.

6 F. D. Tall

Thus, to prove (c) we need only show R \ M is analytic, which follows

from the following two lemmas. We rst quote:

Lemma 14 [E, 4.3.20]. If Y is a completely metrizable space , then every

continuous map f : A ! Y from a dense subset A of a topological space X

to the space Y is extendible to a continuous map F : B ! Y de nedona

G set B X including A.

We also need:

Lemma 15 [E, Exercise 6.2.A.(e)]. Every uncountable separable com-

pletely metrizable space without isolatedpoints includes a dense subspace

homeomorphic to R Q.

Since the p erfect kernel of a separable completely metrizable space is

completely metrizable (as only countably manypoints havebeenremoved

and a G subspace of a complete metric space is completely metrizable), we

see that every uncountable separable completely metrizable space X has the

form Y [ I , where Y has a dense G subspace homeomorphic to R Q (G

since R Q is completely metrizable) and I is a countable set of isolated

points. Applying this to X , let L X b e homeomorphic to R Q and

M M

R Q. Then L is G in dense in the non-isolated part of X ,say h : L

X \ M for X , X for Y X . Apply Lemma 14 now, taking X \ M for A,

M M

and the identity function for f . Since f is onto, so will b e F . Then F (L)

is G in a G subset of a closed subset of X , and h F maps F (L)onto

R Q. Relativizing to M ,we see that R Q is a continuous image of a G

subset of X and hence is analytic. Then so is R \ M , and we are done.

Theorem 4(d) follows quickly from (c), for if jX j =2 , then jX \ M j =

@ @ @

0 0 0

j2 \ M j.ButjX \ M j =2 ,so2 \ M is uncountable.

jX j =2 will o ccur if X is hereditarily Lindelof; this is interesting

b ecause there is a consistent example|due to G. Gruenhage|of an X which

is hereditarily separable, hereditarily Lindelof, non-metrizable, but for which

X is a separable metrizable space [Ju]. Such an example thus cannot have

X complete.

We will also use Lemma 9 to answer a question of Gruenhage. In [PT],

we proved:

Theorem 16. (a) If ! M and X is homeomorphic to ! , then

1 M 1

X = X .

(b) Chang's Conjecture implies thereisanM such that (! ) is hom-

2 M

eomorphic to ! .

M M

general ly , if X is local ly compact T , local ly hereditarily Lindelof , and

M 2

connected , then X = X .) M

An irrational problem 7

The pro of of (c) used connectedness in an essential way, so Gruenhage

(p ersonal communication) asked what happ ens if we for example have X

homeomorphic to the \Long Cantor Set". We can answer that X = X in

that case and indeed for more general ones.

Let us call a line space the linearly ordered top ological space obtained by

replacing each 2 ! by some non-empty compact subspace R of R. That

is, we are considering the order (and resulting ordered top ological space) on

fh ; r i : 2 ! and r 2 R g given by:

h ; r i < h ; si if either < ; or = and r

The Long Cantor Set is obtained by taking each R to b e a copyofthe

Cantor Set K .

We shall show:

Theorem 17. If X is homeomorphic to a line space , and some point

of X has no countable neighborhood , then X = X .

M M

Corollary 18. If X is homeomorphic to the Long Cantor Set , then

X = X .

Rather than proving these directly (which can b e done by an extension

of the argument for Theorem 16(a), we shall obtain them as corollaries to

more general results that|as in 16(c)|remove linearity from the picture.

Before doing so, however, we o er up the following examples which show

that nding the \right" generalization is not so easy.

Theorem 19. (a) 2 < @ implies thereisalocal ly compact T , lo-

! 2

cal ly hereditarily Lindelof , countably compact space X and an elementary

submodel M such that X 6= X , although X is local ly compact , T , lo-

M M 2

cal ly hereditarily Lindelof , countably compact , includes a Cantor Set , and

has size 2 .

(b) If 2 = @ , we may additional ly require that X is the union of an

1 M

increasing col lection of @ compact sets.

Proof. [JNW] construct lo cally compact T , lo cally countable, countably

compact spaces X of size @ ,each n 2 ! . Supp ose 2 = @ ;let X be

n n n

the disjoint sum of [0; 1] with X .Take a countably closed elementary

n+1

= @ containing X . Then X 6= X , but includes submo del M of size 2

n M

[0; 1]. By countable closure, X is a countably compact subspace of the rst

countable space X , and hence is a closed subspace and therefore is lo cally

compact. It also is lo cally hereditarily Lindelof.

To prove (b), consider the space X mentioned in the pro of of (a). In

fact|see [JNW]|this space has the prop erty that countable sets have com-

pact closure. Let X = X [0; 1]. Let M b e a countably closed elementary

submo del M of size 2 containing X . Then X 6= X , but X is lo cally

2 M M

8 F. D. Tall

compact T , lo cally hereditarily Lindelof, has no p oint with a countable

neighb orho o d, and countable sets have compact closure (since X is a closed

fx : < g. subspace of X ). Let X = fx :

M 1 M

Despite these examples, there are non-linear generalizations of Theo-

rem 17 available. First recall:

Definition. Let b e an ordinal. fx g is a freesequence (of p oints

fx : < g\ fx : g =0. in a space X ) if for every <,

We shall prove:

Theorem 20. Suppose X is local ly compact T , local ly hereditarily

M 2

Lindelof , contains a point with no countable neighborhood , and has no free

sequence of length > c . Then X = X .

Arkhangel'ski [A] proved that compact rst countable spaces haveno

uncountable free sequences, so line spaces cannot have free sequences of

length greater than ! , so Theorem 17 and Corollary 18 immediately follow.

After proving Theorem 20, we shall deduce some more corollaries.

We need to quote some results from previous pap ers that wewillneed

in order to prove Theorem 20.

Lemma 21 [PT]. X local ly hereditarily Lindelof and ! M imply X

M 1

is local ly hereditarily Lindelof.

Lemma 22 [T]. X local ly compact T implies X is local ly compact T .

M 2 2

Lemma 23 [JT]. X rst countable implies X is a subspaceof X .

Proof of Theorem 20. Since X is lo cally compact T ,byLemma6some

M 2

subspace of X maps p erfectly onto X and hence X includes a p erfect

pre-image of an uncountable compact rst countable T space. Thus by

Lemma 9, M [0; 1] and hence ! . Therefore X is lo cally hereditarily

Lindelof and lo cally compact T so it is rst countable, so X is a subspace

2 M

of it.

In order to show that X = X , it will suce to show that jX j 2 .

is By lo cal compactness, x for each x 2 X an op en set U such that U

x x

compact. De ne a strictly increasing collection of closed subsets of X as

follows. Let F b e some U .Given F 6= X , take y 2 X F and let

0 x

S S

F = fU : x 2 F g[U .For limit with cf ( )=! ,if F 6= X ,

+1 x y

F and let take y 2 X

[

F = fU : x 2 some F ; < g[U :

x y

S S

If F = X ,let F = X .For limit with cf ( ) >!, let F = F .

< <

For such , F is then closed by rst countability. By construction, for

such 's F is op en as well. Since X is rst countable, and compact rst

An irrational problem 9

countable T spaces have cardinality 2 , it suces to prove that our

construction cannot continue up to say c + 1. If it did, by just lo oking at

indices of co nality c,wewould get a strictly increasing (c + 1)-sequence

of clop en sets. c + 1 is de nable and of size c and hence is in M ,soagain

by elementarity, there is an F 2 M , F : c +1 !T, such that for <

2 2

elementarity again, F ( ) \ M ( F ( ) \ M .Thus the F ( ) \ M 's form a

strictly increasing sequence of clop en subsets of X .For 2 c + 1, pick

f ( ) 2 (F ( +1)\ M ) (F ( ) \ M ). Then ff ( )g is a free sequence

2c +1

in X , contradicting our hyp othesis. Thus Theorem 20 is established.

Now for some more corollaries:

Corollary 24. Suppose X is a T , local ly hereditarily Lindelof , per-

M 2

fect pre-image of ! that contains a point with no countable neighborhood.

Then X = X .

Corollary 25. Suppose X is homeomorphic to the disjoint union of

! and R. Then X = X .

1 M

Proof. The second one is immediate; for the rst, observe that a p erfect

pre-image of a (lo cally) compact space is (lo cally) compact. Then X is

rst countable and so has no free sequence of length > c .

The cardinal arithmetic hyp othesis in Theorem 19(b) is necessary.A

pro of similar to that for Theorem 20 yields the following:

Theorem 26. Suppose 2 > @ and X is local ly compact T , local ly

n M 2

hereditarily Lindelof , contains a point with no countable neighborhood , and

is the union of @ compact sets. Then X = X .

n M

Note that the \lo cally hereditarily Lindelof " condition is essential in

these results. Otherwise, one could replace eachpointin! by the one-

@ +

point compacti cation of the disjoint sum of (2 ) copies of [0; 1]. Takea

countably closed elementary submo del M of size 2 containing this space X .

Then X will b e like X , except with only 2 copies of [0; 1].

Theorem 16(b) shows that the condition that X include a p oint with

no countable neighb orho o d is essential if wewantZFC results.

Iwould like to thank the referee for numerous useful suggestions, includ-

ing Theorem 19(b). Among other things, s/he noted that if one is mainly

interested in Theorem 20 and its corollaries, there is an easier way to obtain

M [0; 1]. Lemma 9 is dicult b ecause it assumes only Cech-completeness

rather than lo cal compactness. With lo cal compactness, wehave:

Lemma 27. If X includes an open set with uncountable compact rst

countable closure , then M [0; 1].

10 F. D. Tall

Proof. If there is suchanopensetU , claim there is one whichisofform

W \ M , W 2 M , W op en in X .To see this, rst claim each p ointinU is

contained in an op en set with compact closure, and hence a basic op en set

U U ,takeanopen V containing p. Then with compact closure. For if p 2

p 2 U \ V .Ifeach such basic op en set had countable closure, by compactness

U would b e countable, contradiction. Let therefore U = W \ M , W 2 M ,

2 M .NoteU =(W ) ; that (W ) = W \ M U is W op en in X . Then W

M M

were not in W , there would b e a V op en in X and containing x clear; if x 2 U

such that V \ U =0.Take suchaV 2 M . Then V \ M \ U =0,contradiction.

Since (W ) is compact, so is W [Ju] and so by Lemma 6 there is a p erfect

map from W onto [0; 1]. But then (W ) maps onto [0; 1] =[0; 1] \ M ,

M M

whence by compactness, [0; 1] M .

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Department of Mathematics

UniversityofToronto

Toronto, Ontario M5S 3G3, Canada

E-mail: [email protected]

Received22November 2001;

in revised form 1 September 2002 (145)