Thin-Type Dense Sets and Related Properties
by
Jennifer Diane Hutchison
A dissertation submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Doctor of Philosophy
Auburn, Alabama
May 14, 2010
Keywords: dense, thin, slim, superslim, (GC), (NC)
Copyright 2010 by Jennifer Diane Hutchison
Approved by:
Gary Gruenhage, Chair, Professor of Mathematics and Statistics
Stewart Baldwin, Professor of Mathematics and Statistics
Wlodzimierz Kuperberg, Professor of Mathematics and Statistics
Abstract
Dense sets in topological spaces may be thought of as those which are ubiquitous.
We discuss dense sets in product spaces which also have a thin-type property, making
them in some sense rare or spread out. Thin-type properties include the previously
studied properties thin, very thin, and slim. We construct examples showing that
even in a separable space, there may be no countable very thin or slim dense set. We
also de ne and discuss the properties < -thin, codimension 1 slim, and superslim.
The de nition of < -thin is between those of thin and very thin; the de nition of
superslim is between very thin and slim. Codimension 1 slim is slightly weaker than
slim, in that since only some of the cross-sections are required to be nowhere dense,
it is possible for a space to have a codimension 1 slim dense set but no slim dense set.
We give some results about the existence of a < -thin dense set, in one case relating
this to the existence of a very thin dense set. We show that a superslim dense set in a
nite power of X is related to the existence of a certain type of collection of nowhere
dense subsets of X.
The criteria (GC) and (NC), relating to collections of nowhere dense sets, are
discussed. These were shown by Gruenhage, Natkaniec, and Piotrowski to imply the
existence of a slim dense set in certain products. We consider when a space can
satisfy (GC) with a collection of nite sets, and show that a collection witnessing
(GC) cannot be uncountable if X is rst countable and separable. We particularly
consider ordered spaces, and characterize the linearly ordered and generalized ordered
spaces which satisfy (GC), along with the linearly ordered spaces which satisfy (NC).
The latter is connected to properties of ultra lters. We also introduce a connection
between a stronger version of (GC) and GN-separability.
ii
Acknowledgments
I would like to thank my family for their support and encouragement throughout
my education, and particularly my husband Aaron for his wisdom and encourage-
ment during the last phases of my research. I am also grateful to my advisor, Dr.
Gary Gruenhage, for his patience and for teaching me to do mathematical research.
Finally, I would like to thank my mathematics professors at Cedarville University and
Miami University, for preparing me for further study and giving me the con dence to
continue.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3 Thin-type Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.1 < -thin Dense Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.2 Bounds on the Cardinalities of Special Dense Sets . . . . . . . . . . . 17
4 Slim-Like Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.1 Codimension 1 Slim . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.2 Superslim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
5 (GC) and (NC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
5.1 Cardinalities of Collections Witnessing (GC) and (NC) . . . . . . . . 32
5.2 Ordered Spaces and (GC) . . . . . . . . . . . . . . . . . . . . . . . . 37
5.3 Ordered Spaces and (NC) . . . . . . . . . . . . . . . . . . . . . . . . 42
5.4 (GC) and GN-separability . . . . . . . . . . . . . . . . . . . . . . . . 50
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
iv
Chapter 1
Introduction
The concept of a dense set in a topological space is an elementary one: a set which
meets every open set. Dense sets may be thought of as those that are ubiquitous in
a space. In this dissertation, we look at the situations in which dense sets in product
spaces may be well spread out. Thin, very thin, and so on are terms describing how
sparsely a subset of a product space is distributed through the space. We shall be
discussing the situations under which such special kinds of dense sets exist in product
spaces, along with the characteristics of these sets. We will also consider some criteria
which imply the existence of slim dense sets.
In Chapter 3, we will look at a new kind of thin dense set: a subset of a product
of -many factors in which any two distinct points agree on less than coordinates.
This is called < -thin, and the de nition is between the de nitions of thin and very
thin. We will see that if each of + factor spaces has a dense set of size , there is a
< +-thin dense set in the product; whereas if each factor space has small cardinality
relative to , the situation reduces to the existence of a very thin dense set. In
particular, if X is separable, X!1 has a 1.
2. D is very thin if whenever x;y2D with x6= y, x 6= y for all < .
3. D is slim if for every non-empty proper subset K and 2Q 2KX , the
set D\C( ) is nowhere dense in C( ), where C( ) =fx2X : x K = g is
the cross-section of X at . We will call D\C( ) the cross-section of D at .
The de nition we have given for very thin is stated in such a way as to show
the connection with thin; very thin was de ned in Piotrowski?s paper [10] in the
equivalent formulation: 8 < and p2X , jfx2D : x = pgj 1.
Geometrically, in R3, no two points of a thin set can lie on a line parallel to an
axis. For a very thin set in R3, no two points can be on a line or in a plane parallel
to an axis. It is clear that for X2, the notions of thin and very thin coincide.
Piotrowski [10] showed that the product of 2! separable spaces has a countable
thin dense subset, and that the product of 2! spaces, each of which has a countable
-base, has a very thin dense set. (Recall that a -base is a collection B of non-
empty open subsets of X such that each open set in X contains a member of B.)
He asked whether the second result could be weakened to \separable" instead of
\countable -base". Schr oder [11] and Szeptycki [12] each constructed a separable
4
space with no (very) thin dense set in its square. Schr oder, in fact, constructed a
class of counterexamples, and also discussed the number of topologies on a countable
set which do not allow the square to have a thin dense set. He also weakened the
-base requirement slightly by only requiring a countable weak -base; that is, the
sets in the collection B are only required to be in nite, not open.
In [6], Gruenhage, Natkaniec, and Piotrowski prove a variety of results about
the existence and nonexistence of thin, very thin, and slim dense sets in products of
di erent spaces and in powers of a space.
The authors begin by presenting a generalization of Schr oder?s result about
countable -bases, plus an implication of the existence of a very thin dense set. This
is based on some cardinal functions of X: (X) is the least cardinal of a non-empty
open subset of X; d(X) is the least cardinal of a dense subset of X; and w(X) is
the least cardinal of a -base for X.
Proposition 2.2. [6] Assume X = Q < X , where all X are dense in themselves.
Consider the following statements:
(i) = inf < (X ) sup
<
w(X ) and 2 .
(ii) There is a very thin dense set in X.
(iii) (X ) d(X ) for any ; < ; 6= .
Then (i))(ii))(iii).
This is useful for demonstrating whether or not X has a very thin dense set. The
authors also construct spaces which do not have certain types of dense sets, including:
A metrizable space X for which Xn has no thin dense set for any n2N, and
X! has no very thin dense set.
Countably many dense-in-themselves spaces whose product does not have a thin
dense set.
5
Under (CH), a countable regular space X such that X2 has a thin dense subset,
but X3 does not.
Under(CH), a countable regular space X such that X2 has no slim dense subset.
A space X for which Xn has no slim dense set for any n .
We will consider when products of -many spaces have < -thin dense sets, for
an in nite cardinal . Note that if jXj , then X has a (more than) < -thin
dense set by Theorem 2.4 in [6]. The dense set constructed there has the property
that any distinct points di er on all but nitely many coordinates. We will now show
9
that we also get a < +-thin dense set in a product of + di erent spaces, provided
each factor space has a dense subset of size .
First, we have a lemma, which has been established previously (see the reference
to this fact in, for instance, [2], and in [16] for + = !1), but is proved here for the
convenience of the reader.
Lemma 3.2. If is an in nite cardinal, there is a family ff : < +g of functions
from + to , with the property that if < < +, jf < +jf ( ) = f ( )gj .
Such a family is called almost disjoint or eventually di erent.
Proof. For each in nite ordinal < +, let g : ! be a 1-1 function. Since
j j for any < +, this is possible. Then, for each ordinal with ! < +,
let f be de ned by
f ( ) =
8
><
>:
0
g ( ) >
It is clear that ff : ! < +g is a family of functions from + to .
Suppose ! < < < +. For any > , f ( ) = f ( ) implies that g ( ) =
g ( ), which is impossible since g is one-to-one and 6= . So
jf < +jf ( ) = f ( )gj j j :
This shows rst that the functions are distinct for distinct < +; and also that the
family is almost disjoint.
We can now prove:
Theorem 3.3. For each < +, let X have a dense subset of size . Then
Q
< + X has a <
+-thin dense set.
10
Proof. For each < +, let D = fd : < g be the dense set in X . Let
F = ff : < +g be the family of all functions f from a nite subset of +
into . Let fg : < +g be a family of + almost disjoint functions from +
to . De ne e 2 Q < + X as follows: e ( ) = d f ( ) for all 2 domf and
e ( ) = d g ( ) otherwise. Then the set E = fe : < +g is < +-thin dense: E
is dense because, given any basic open set
nT
i=1
1 i (Ui), there is a function f 2F
with domf = f 1;:::; ng and d if ( i) 2Ui. Given e ;e 2E, with 6= , for all
but nitely many 2 +, e ( ) is d g ( ) and e ( ) is d g ( ). Since g ;g are almost
disjoint functions on +, they agree on at most < + points of +. Thus, E is
< +-thin.
In particular, Theorem 3.3 says that if X is separable, X!1 has a jXj. Choose a set D consisting
of jXj+ distinct points from E; say D = fd : < jXj+g. For each ( ; ) in
jXj+ jXj+ with < , let ( ; ) be the least element of with the property
that d ( ) 6= d ( ) for all > ( ; ). This is possible because E is < -thin. Let
= supf ( ; ) : ; < jXj+; < g. Note that since jXj+ jXj+ < , and
is regular, < . Then, for all > and ; 2jXj+, d ( ) 6= d ( ). That is,
D0 = f(d )( > ) : d 2Dg is very thin. Thus, jD0j jXj, and jDj = jD0j. This
contradicts our choice of jDj>jXj, so X has no < -thin subset of size greater than
jXj.
A speci c consequence of this is:
11
Corollary 3.5. If jX j < ! for all < !1, there is no < !1-thin dense set in
Q
!1, then Corollary
3.6 shows that Xc has a < c-thin dense set i it has a very thin dense set. Since
the very thin dense set must be countable, we see that our k , we see
that (fxn : n2!g) will contain D for each n. Similarly, 1=2(fxn : n2!g) will
also contain D.
The set fxn : n2!g will be dense (as in the proof of Theorem 3.8).
If it is possible to choose the countable weak -base so that each member is
contained in D, you may construct the very thin dense set so that all the projections
are actually equal to D. For example, one may use this method to construct a very
thin dense set in Q!1 with the property that all coordinates are from the dyadic
rationals.
3.2 Bounds on the Cardinalities of Special Dense Sets
As mentioned above, another question concerning these special dense sets relates
to their cardinality. Must a product space with a very thin dense or slim dense set
have such a special dense set which also witnesses the density of the space? For
example, must a separable space which has a very thin dense set have a countable
very thin dense set? We construct examples showing that the answer is negative both
for very thin and for slim sets. We begin with a couple of facts:
17
Fact 3.12. Let be a topology on ! generated by a maximal independent family.
Then (!; ) embeds in 2c as a dense subset.
Proof. Let be a topology on ! generated by a maximal independent family fA :
< cg. Let f : ! ! 2c be given by f(n) = xn, where xn( ) = 1 if n 2 A and
xn( ) = 0 if x62A .
Following the notation in [6], let basic open sets in (!; ) be denoted by [ ] =
T
2dom A
( )
where is a function from a nite subset of c into 2, A1 = A , and
A0 = !nA . Denote basic open sets in 2c by U = T 2dom 1 (U ( ) ), where is as
above, U1 =f1gand U0 =f0g. Since the only other open sets in 2 are;and 2, which
may be avoided in writing out a nonempty basic open set, this notation describes any
proper nonempty subset of 2c. Then,
xn2U , xn( ) = ( ) 8 2dom
, ( ) = 1 i xn( ) = 1 8 2dom
, ( ) = 1 i n2A 8 2dom
, n2[ ]
So the open sets in (!; ) correspond exactly under f to the open sets in D = f(!)
2c.
To see that D is dense in 2c, let U be a nonempty basic open set in 2c. Then there
is a k2! in [ ] = Tni=1A ( i) i . The corresponding xk = f(k) will be in U \D.
We use D to get a subspace of 2c with a very thin dense set in its square.
Fact 3.13. Let F be the set of all points x2 2c for which the set f : x = 1g is
nite. Let D be a dense subset of 2c which is homeomorphic to (!; ), where (!; ) is
as in Fact 3.12. If X = D[F, then X2 has a very thin dense set.
18
Proof. Let X = D[F. Since X 2c, w(X) w(2c) = c. (Recall that w(X) is the
minimum size of a base for X.) To see that (X) c, note that any open set in X
contains a basic open set of the form
U = (D[F)\U ;
where U is basic open in 2c. Then the points x , where
x ( ) =
8
>>>>
<
>>>>
:
1 ( ) = 1
1 =
0 otherwise
are in U for each < c for which ( ) 6= 0; that is, all but nitely many ?s. So
jUj c; that is, (X) c.
So, (X) c w(X) w(X), which implies that X2 has a very thin set by
Proposition 2.1 in [6].
Now we will show that we can construct X in such a way that D cannot make a
signi cant contribution to a special dense set.
Fact 3.14. Let X be as in Fact 3.13. If P is a hereditary property (such as very
thin), and any subset of D2 with property P is nowhere dense in D2, then X2 cannot
have a countable dense set with property P.
Proof. Suppose that E X2 is a countable dense set in X2 that has the propertyP.
Then, by hypothesis, E\D2 is nowhere dense in D2. Since D2 is dense in X2, that
means that E\D2 is nowhere dense in X2.
Since E is countable, the set EF := ( 1(E)\F)[( 2(E)\F) is a countable subset
of F, so there is an < c with the property that for all x2EF, x( ) = 0 for all > .
(Speci cally, we may take to be 1+supf < c : x( ) = 1 for some x2EFg.) Then
19
the open set V = 1 (f1g) 1 (f1g) is open and nonempty in X2 and does not
contain any points with a coordinate in EF. Thus, V does not meet the closure of
the set of points from E which have at least one coordinate in F. But every point of
E has either at least one coordinate in F or both coordinates in D. So V E\D2,
which contradicts that E\D2 must be nowhere dense.
We will now apply this to the properties P = very thin and P = slim. Observe
that these properties are hereditary, in the sense that a subset of a very thin (resp.
slim) set is also very thin (resp. slim).
Example 3.15. There is a separable space X such that X2 has a very thin dense
subset, but X2 does not have a countable very thin dense subset.
Proof. Let X = D[F as in Fact 3.13. Then X2 has a countable dense set (D2)
and a very thin dense set. Since the topology on D is homeomorphic to a topology
generated by a maximal independent family, D is strongly irresolvable. A very thin
dense set in X2 results in disjoint dense sets in X; for a strongly irresolvable space,
this is impossible. In fact, if X is strongly irresolvable, any thin set in X2 is nowhere
dense. This follows from a result in [11] which says that if a space Y2 has a thin dense
set, there is a one-to-one map : Y !Y with the property that for all nonempty
U Y, (U) is dense in Y. Clearly, if A X2 were somewhere dense, then we could
nd disjoint open sets U1;U2 Int(A), and apply jA to get disjoint dense sets in
(Int(A)), contradicting that X is strongly irresolvable.
Thus, if E is a very thin subset of X2, E\D2 is also very thin, and so E\D2 is
nowhere dense in D2. By Fact 3.14 withP = very thin, X2 does not have a countable
very thin dense set.
Example 3.16. (CH) There is a separable space X such that X2 has a very thin
dense subset, but X2 does not have a countable slim dense subset.
20
Proof. Consider the set X = D[F where D is the space constructed in Example
3.4 in [6]. That is, D is the space (!; ) where is the topology generated by a
particular maximal independent family. The family is constructed by induction in
such a way that at stage + 1, a potentially slim dense subset E of !2 is made not
dense by adding to the independent family a set T +1 that meets every open set from
the preceding stage, but misses E . The potential slim dense subsets each appear
!1-many times.
One result of this construction is that any slim set in D2 is nowhere dense. To
see this, consider an open set U D2. There is some . If E is slim dense in D2, then E was slim dense in some stage
of the construction. Thus, E would appear as E for some > in the construction.
So, when +1 was de ned, U was open, and thus contains one of the basic open sets
which T +12 was required to meet. However, T +12 was constructed to miss E ; so
T +12\E =;. Thus, U6 E . So E is nowhere dense in D2.
Thus, by Fact 3.14 withP = slim, X2 does not have a countable slim dense set.
However, D2 is countable and dense in X2, and X2 has a very thin (hence slim) dense
set by Fact 3.13.
21
Chapter 4
Slim-Like Properties
There are many ideas of smallness in topology: nite, < for some relevant
cardinal , nowhere dense, meager, etc. We have considered some of these already
in the restrictions we have placed upon coordinates of dense sets; we now turn our
attention to cross-sections of a set. The de nition of a slim set explicitly places a
restriction on the cross-sections; speci cally that they be nowhere dense in the cross-
section of the whole space. However, very thin also restricts cross-sections. One
might consider a very thin set to be one for which the cross-sections are singletons.
Indeed, suppose D Q < X is very thin, and consider the cross-section of D at
some 2Q 2KX . D\C( ) =fx2D : x K = g; but this means that we have
xed at least one coordinate, say ; so D\C( ) fx2Djx( ) = ( )g, which has
only one point. So, once again, we observe a gap between two de ned notions. We
will de ne a superslim set which falls into this gap.
We will also consider what happens when we only place conditions on D?s inter-
section with some cross-sections, rather than all; this will lead us to the concept of a
codimension 1 slim set.
4.1 Codimension 1 Slim
We de ne the property codimension 1 slim, suggested by Gruenhage as an alter-
native to the de nition of slim. We will show that it is slightly weaker than slim, in
the sense that a space may have a codimension 1 slim dense set but no slim dense
set.
22
De nition 4.1. Let D be a subset of Q < X . D is codimension 1 slim if for every
2 and x2X , D\C(x) is nowhere dense in C(x), where C(x) =ff2Q < X :
f( ) = xg.
The di erence between codimension 1 slim and slim is that we only require
the cross-sections of codimension 1 to be nowhere dense, not every cross-section.
Obviously for a product of two spaces, the notions would coincide.
Example 4.2. (CH) There is a space which has a codimension 1 slim dense set, but
no slim dense set.
Proof. We begin by constructing a codimension 1 slim dense set in Q3. Let fDn :
n < !g be a partition of Q into dense sets. Let E be a very thin dense subset of
D02, such that for each (q;r)2E, q6= r. Note that D02 has a very thin dense subset
because (D0) = w(D0) = !. Enumerate E by f(qn;rn) : n 0 to construct a series of topologies f : . Since for any point x with x( ) = 1 for some > ,
x2 1 (1) and 1 (1)\(Mk\F) =;, this means (U) = 0 for all > . Suppose
U = U \D, where U is a basic open set in X. U X corresponds to the set [ ] in
!. Because the topology on ! is generated by a maximal independent family, [ ]\A
is nonempty, in fact in nite, for any . So the n?s in [ ]\A will correspond under
the embedding map to points in U \D which have a one in the th coordinate, in
33
particular when > . This is a contradiction. So there is no collection N of nite
sets in X witnessing (GC).
We now see that in any non-separable space, no collection of nite or countable
sets can witness (GC).
Proposition 5.2. In any non-separable space satisfying (GC), the collection N wit-
nessing this must contain at most nitely many nite or countable sets.
Proof. Suppose N is a collection of pairwise disjoint nowhere dense sets in a non-
separable space X, and there is a countably in nite subcollection M of N for which
each N2Mis nite or countable. Any in nite subcollection ofN must still witness
(GC), so [M is a countable dense set in X. This is contradicts that X is not
separable.
Since any in nite subcollection of a collection witnessing (GC) also witnesses
(GC), we may assume that every member of a (GC) collection in a non-separable
space is uncountable. We may apply this speci cally to the lexicographic square,
which does satisfy (GC).
Example 5.3. There is a space satisfying (GC), witnessed by a countable collection,
but no collection of nite or countable sets can witness (GC).
Proof. Let X = ([0;1]2; ), where is the topology generated by the lexicographic
order.
Since X is not separable, Proposition 5.2 implies that a collection witnessing
(GC) cannot contain more than nitely many nite or countable sets.
Let us construct an N showing that X satis es (GC). For a prime p, let Np =
f(x;a=p) : x2R;a2N;1 a < pg. Suppose p1 6= p2; then if (x;y) 2Np1 \Np2,
y = a1p1 = a2p2 , which is impossible, since the fractions are both in lowest terms. Also,
Int(Np) = Int(Np[((0;1] f0g)[([0;1) f1g)) = ;, so the Np?s are nowhere
34
dense. If U X is open, there is a basic open interval contained in U of the form
((x;a);(x;b)). Then U must meet at least all Np for which 1=p< (b a); that is, all
p> 1b a or all but nitely many p. So N =fNp : p primeg witnesses (GC) in X.
We now turn our attention to the cardinality of the collection witnessing (GC).
The main question we will consider is whether such a collection may be uncountable.
In [6] it is noted that the collection may be assumed to be countable; we want to
know whether it is possible for it to be otherwise.
Our rst result is that for separable rst countable spaces, we cannot have an
uncountable collection which witnesses (GC). We begin with a lemma.
Lemma 5.4. If X is a rst countable space which satis es (GC) witnessed by an
uncountable family N = fN : < !1g, then for any x2X, there is an x < !1
such that x2N for all > x.
Proof. Let x2X, and let fUn(x) : n2!g be a countable neighborhood base at x,
consisting of open sets. Let N = fN : < !1g be a collection witnessing (GC) in
X. For each n, Un(x) meets all but nitely many elements of N; so let n be the
least element of !1 such that Un(x)\N 6= ; for all n. Let x = supf n :
n2Ng< !1. Let N0 = fN : x < < !1g. Then, for each with x < < !1,
N \Un(x)6=; for all n2N; that is, x2N for all N 2N0.
Note that N0, being an in nite subcollection of N, still witnesses (GC).
Proposition 5.5. A rst countable, separable space X cannot satisfy (GC) witnessed
by an uncountable collection of sets.
Proof. Let D =fdn : n dn. Let = supf dn : n2!g , n2!, dn 2N . That is, D N for all > . But this contradicts the
fact that N must be nowhere dense.
35
Suppose a space X satis es (GC), witnessed by a collection (of arbitrary size) of
which in nitely many of the members are countable or nite sets. Then, that space
is separable, since if we have a countably in nite collection of countable or nite
sets which witnesses (GC), their union is a countable dense set. So, Proposition 5.5
means that: if a rst countable space satis es (GC) witnessed by a collection N,
either all but nitely members of N are uncountable, or N is countable (or both).
This is a partial answer to the question of whether a collection witnessing (GC) may
be uncountable.
Question 5.6. Is there a rst countable space which satis es (GC), witnessed by an
uncountable collection (of which all but nitely members must be uncountable)?
Even in the general case, an uncountable family witnessing (GC) cannot sub-
stantially consist of nite sets.
Proposition 5.7. SupposejXj !. If a collectionN witnesses (GC) or (NC) in X,
and N is composed of nite sets, there is no k maxfk;lg, ck and cl are in
M := N0[N1[:::[Nm. M is nowhere dense in X, so it is nowhere dense in (ck;cl).
Thus, Im contains an interval which is contained in (ck;cl), say I. Then I (a;b)
and I2B. So B is a -disjoint -base for X.
If X is a Baire LOTS, we see that the converse of Proposition 5.11 is true.
Theorem 5.13. Let X be a linearly ordered space with no isolated points. If X is
Baire and satis es (GC), X has a dense metrizable subspace.
Proof. Let X be a Baire LOTS which satis es (GC), witnessed by a collection N =
fNk : k < !g. For each k < !, de ne Uk = Xn(N0[N1[:::[Nk). Since the Nk?s
are nowhere dense, each Uk is dense and open; so Y = TkK, there is a g2B(f; ) with n nonzero
coordinates. So B(f; )\Nk for all k>K.
Thus, N witnesses (GC) in X.
We will now build on our work with LOTS to obtain a more general result: the
characterization of all GO-spaces which satisfy (GC).
Theorem 5.15. A GO-space with no isolated points satis es (GC) i it has a -
disjoint -base.
Proof. ()) Suppose a GO-space X satis es (GC). X may be considered the dense
subspace of a LOTS L. By Lemma 5.10, L satis es (GC), which means that L has
a -disjoint -base (Lemma 5.12). Restricting the members of this -base to X will
give a -disjoint -base for X.
(() Suppose X is a GO-space with no isolated points, and B = Sn 0, let In =
fU\V : U2Cn; V 2In 1g. Note thatIn is a disjoint collection of convex open sets:
If x2(U1\V1)\(U2\V2) for some U1\V1;U2\V22I0n, then x2U1\U2)U1 = U2,
40
since the elements ofCn are pairwise disjoint. Similarly, x2V1\V2 and V1;V22In 1
implies that V1 = V2. So U1\V1 = U2\V2.
Also note that SIn is still dense in X. Moreover,I = Sn>>>
<
>>>>
:
g00( ) 62f 0; 0 + 1g
f0( ) = 0
f0( ) + 2 = 0 + 1
Observe the following about g0:
f0 n 1. De ne
gn( ) =
8
>>>
><
>>>
>:
g0n( ) 62f n; n + 1g
fn( ) = n
fn( ) + 2 = n + 1
44
As before, (fn;gn) is a nonempty open interval contained in (fn;g0n), thus contained
in (fn 1;gn 1) and missing [i nNn.
Continuing in this way, we get a sequence of intervals (f0;g0) (f1;g1)
(f2;g2) ::: with the property that fn 1 < fn < gn < gn 1, fn( ) = gn( ) for
all n, fn( n + 1) + 2 = gn( n + 1), and fn( n 1 + 1) = gn( n 1 + 1) =
fn 1( n 1 + 1) + 1 fn for all n and
h>>>
>>>
>>>
>>>>
>>>
>>>>
>>>
>>><
>>>
>>>>
>>>
>>>
>>>>
>>>
>>>>
>>>
:
f0( ) < 0 + 1
f0( ) + 1 = f1( 0 + 1) = 0 + 1
f1( ) 0 + 1 < < 1 + 1
f1( ) + 1 = f2( 1 + 1) = 1 + 1
:::
fn( ) n 1 + 1 < < n + 1
fn( ) + 1 = fn+1( n + 1) = n + 1
fn+1( ) n + 1 < < n+1 + 1
:::
0 supf n + 1 : n2!g
with the understanding that a line is omitted if there is no in the prescribed range
(ie, if k = k 1 + 1).
It is clear that h : !1!Z, so h2Y. We claim that fn for all ; .
Note that since each N has nite intersection with each C , the N ?s are nowhere
dense.
To see that N =fN : < g witnesses (NC) in X, let U1;:::;Un be a pairwise
disjoint collection of open sets in X. For each Ui, choose a C i 2 C such that
Ui\C i 6=;. For each i, Ui\C i is an open subset of C i, so it contains an element
B i i 2B i. The collection of indices f 1; 2;:::; ng is a nite subset of , so it was
one of the F ?s. Then, N contains a point from B i for each i = 1;::;n and for
every !, it is clear that X cannot be GN-separable. For then, if Y is a dense
subset of X with no countable dense subset, the collectionhDn : n!)X is not R-separable. GN-separability implies
R-separability ([1] Proposition 72), so a non-R-separable space is not GN-separable.
However, it is (consistently) possible for a space with (X) >! to satisfy (GC)
witnessed by a collection of nite sets, as we show in the following example. Recall
that p is the least cardinal of a family of in nite subsets of ! with every nite inter-
section in nite, such that there is no in nite set almost contained in every member
of the collection.
Example 5.27. Assume p >!1. There is a space which satis es (GC), witnessed by
a collection of nite sets, which is separable but not GN-separable.
Proof. Let X = 2!1. By Proposition 4.5 in [6], if !1 < p, then 2!1 satis es (GC)
witnessed by a collection of nite sets. Also, 2!1 is separable.
However, X is not GN-separable. Consider the set
Y =fx22!1 :jf n. Let = supf n : n < !g< !1. Then 1 +1(f1g) is
open and misses Y0.
Therefore, it will be useful to turn our attention to spaces with (X) = !,
particularly countable spaces.
Question 5.28. Is there a countable space X which satis es (GC) witnessed by a
collection of nite sets, but is not GN-separable?
R-separability along with \every countable dense subset contains a groupable
dense subset" is equivalent to GN-separability, so we could answer this question by
answering:
Question 5.29. Is there a countable space X which satis es (GC) witnessed by a
collection of nite sets, but is not R-separable?
53
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