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-thin dense set; but if jXj < !, X!1
has no <!1-thin dense set. Using similar proof techniques, we address the question
of the existence of very thin dense sets and slim dense sets of minimal cardinalities.
Examples are constructed showing that a separable space with a dense set that is
slim or very thin need not have a countable dense set of that type. These examples
are subspaces of 2c, where c is the cardinality of the continuum. We also prove (under
MA+:CH) an extension of a theorem of Schr oder [11] giving conditions on X which
guarantee the existence of countable very thin dense set in Xc.
1
Chapter 4 examines a weakened form of slim which was proposed by Gruenhage,
originally as an alternate de nition of slim. We show that this property, called codi-
mension 1 slim, is indeed weaker than slimness (in products of more than two factors)
by constructing a space which has a codimension 1 slim dense set but no slim dense
set. We then consider the property superslim (which, like \slim" and \codimension
1 slim", is based on cross-sections), and show an equivalence to a version of (NC), a
criterion related to the existence slim dense sets.
(NC) is studied more extensively in Chapter 5, along with the related criterion
(GC). (GC) and (NC) are speci c conditions on the factor spaces which imply the
existence of a slim dense set in a product. These are introduced in [6]. Both deal
with a space having a certain type of collection of pairwise disjoint nowhere dense
sets. We give some examples concerning the circumstances under which a collection
satisfying (GC) can consist of  nite or countable sets. We also discuss the possibility
of such a collection being uncountable, giving results in the direction of impossibility.
In particular, in any in nite T2 space, one cannot have a collection witnessing (GC)
which is uncountable and consists of  nite sets.
In further characterizing spaces which satisfy (GC) or (NC), we prove that if
X has a dense metrizable subspace, X satis es (GC). This leads to some results on
when a linearly ordered space (LOTS) satis es (GC), as well as a characterization of
generalized ordered spaces (GO-spaces) which satisfy (GC) as those which have a  -
disjoint  -base. We also construct a LOTS which does not satisfy (NC). The failure
occurs because the space is the topological sum of two spaces with very di erent
open sets; so we next consider when a sum X Y may satisfy (NC). We show that
if points have neighborhoods which are uniform in cardinality, the space will satisfy
(NC). This is applied to a linearly ordered space to give a characterization of (NC)
based on the intervals of the space. Also, (NC) satis ed by a sum of two strongly
2
irresolvable spaces is characterized by the relationship of related ultra lters under the
Rudin-Keisler order on  !.
Finally, we consider how a space satisfying (GC) witnessed by a collection of
 nite sets is related to the space possessing the property of GN-separability, a selective
separability property.
3
Chapter 2
Background
The concepts of \thin" and \very thin" sets were de ned by Piotrowski [10].
\Slim" was de ned by Gruenhage in [6].
De nition 2.1. Let X = Q < X be a product space, and let D X.
1. D is thin if whenever x;y2D with x6= y, then jf < : x 6= y gj> 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<! but X! does have
a slim dense set.
 Under (CH), a countable regular space X such that X2 has a very thin dense
subset, X3 has a thin dense set, and X3 has no slim or very thin dense subset.
Another useful result on the existence of slim dense sets in in nite powers is the
following, based on a sequence of dense sets in X.
Proposition 2.3. [6] Suppose X admits a decreasing sequence Dn of dense sets such
that Tn<!Dn =;. Then for any in nite cardinal  , X has a dense set which is both
slim and thin.
This is applied to show that under any of the following conditions on X, X has
a dense slim and thin set for any in nite cardinal  :
 X is !-resolvable;
 X has a meager dense subset;
 X is a separable Hausdor space with no isolated points.
The last condition can be generalized to products of di erent spaces. However, Propo-
sition 2.3 cannot be generalized to products of di erent spaces; an example is given
to show that there is a sequence of countably many spaces with no isolated points,
each having such a decreasing sequence of dense sets, whose product has no slim or
thin dense set. The authors also show that, under V = L, every in nite power of a
dense-in-itself space has a slim and thin dense set.
6
Gruenhage and Natkaniec [6] also introduce the criteria (GC) and (NC) on a
space X, which imply that certain products have slim dense sets.
(NC) There is a pairwise disjoint collectionN of nowhere dense sets in X such that,
given any  nite collectionU of nonempty open sets in X, there is some N2N
which meets every U2U.
(GC) There is a pairwise disjoint collectionN of nowhere dense sets in X such that
every nonempty open set U meets all but  nitely many N2N.
A weaker version of (NC) is also de ned: for each k < !, (NCk) is the statement
obtained by requiring the collection U in the de nition of (NC) to have cardinality
 k.
(NC) and (GC) are studied in [6], with the following results. If X satis es (NC),
then every power of X has a slim dense set. If X , satis es (GC) for all  < , then
there is a slim dense set in Q < X . Several theorems about types of spaces which
satisfy (NC) or (GC) are proved, and several spaces are constructed whose products
do not have special dense sets of some kind. In particular, every metrizable dense-in-
itself space, and every separable space with  -weight !, satis es (GC). While (GC)
implies (NC), there is a space which satis es (NC) but not (GC). (GC) is preserved
by taking a topological sum, while (NC) is not. However, both (NC) and (GC) are
productive.
It is not known whether there is a consistent example of a dense-in-itself space X
such that X! does not have a slim dense set. However, such an example, if it exists,
is shown in [6] to have a subspace which is strongly irresolvable and Baire. In seeking
candidates for such an example, the authors show that a strongly irresolvable space
X for which the ideal N of nowhere dense sets is selective cannot satisfy (NC). It is
also shown that if a strongly irresolvable space satis es (NC) by a collectionN, then
the family of all M N such that M also witnesses (NC) is an ultra lter on N.
7
There is a connection between a stronger version of (GC) and a selective sepa-
rability property GN-separable. Selective separability properties were studied in [1]
by Bella, Bonanzinga, and Matveev. They discuss several stronger versions of sep-
arability, along with related covering properties, giving many implications involving
the  -weight of a space and tightness properties, especially for subspaces of 2 .
Bella, Bonanzinga, and Matveev prove that X is GN-separable i X is R-
separable and every countable dense subset of X contains a groupable subset. Gru-
enhage observed that one could use techniques similar to those used in [6] for slim
spaces to answer questions about selective separability [4]; there are also connections
between the criterion (GC) and GN-separability, which we will look into.
Most of the results in the literature, and in this dissertation, involve X being
dense-in-itself (having no isolated points); isolated points inhibit the formation of
\nice" dense sets. This is especially true in the last section on (GC) and (NC).
We will denote the  th coordinate of a point x2
Y
 < 
X by x or by x( ). All
spaces are assumed to be Hausdor unless otherwise noted.  is a cardinal number,
and  + is the least cardinal greater than  .
8
Chapter 3
Thin-type Properties
Two questions naturally arise when considering thin and very thin dense sets
in product spaces. One is (if the product has very many factors at all) the gap
between the two concepts. Distinct points in thin sets need only di er at more than
one coordinate, while in very thin sets all coordinates must be di erent. There are
obvious proposals for an intermediate de nition, one of which we will consider here
under the name < -thin. Another natural question is whether a product space which
is separable, and has a special dense set, must have a special dense set witnessing the
separability. We construct an example to show that this does not have to be the case.
3.1 < -thin Dense Sets
De nition 3.1. Let D be a subset of Q < X and let  be a cardinal less than or
equal to  . D is < -thin if for any x;y2D, x6= y,
jf < : x = y gj< :
In the case that  =  is regular, this is equivalent to: D is < -thin if for any
x;y2D, x6= y, there is an   < such that x 6= y for all  >  .
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 <!1-thin dense
set.
In the case of small (relative to the power) factor spaces, the situation relates
directly to the existence of a very thin dense set.
Theorem 3.4. If  is an in nite regular cardinal and jXj+ < , any < -thin set in
X has cardinality  jXj.
Proof. Suppose E is a < -thin set in X with jEj>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 X .
Proof. Let A !1 be such thatjX j= k for all  2A, where k is some  nite number.
Then, a <!1-thin dense set in Q <!1 X will give a <!1-thin dense set in Q 2AX 
by restricting the coordinates. Q 2AX is the same as k!1, and by Theorem 3.4, a
<!1-thin set in k!1 has size  k, and thus cannot be dense. So there is no <!1-thin
dense set in Q <!1 X .
Theorem 3.4?s e ect on the existence of a < -thin dense set may now be clearly
seen.
Corollary 3.6. If  is an in nite regular cardinal and jXj+ < , X has a < -thin
dense set if and only if X has a very thin dense set.
Proof. (() Clearly, a very thin set is < -thin.
()) Let D be a < -thin dense set in X . Then, the construction in Theorem
3.4 gives a very thin set D0 X consisting of the tails of points of D. Since D was
dense, so is D0.
Corollary 3.7. If X is countably in nite, then Xc+ has no <c+-thin dense set.
Proof. Any <c+-thin set in Xc+ gives a very thin set in Xc+, which must then be
countable; but it is well-known that a product of c+ Hausdor spaces cannot be
separable.
Let us consider <c-thin dense sets in Xc. Schr oder proved that a product of c-
many spaces, each of which has a countable weak  -base, has a countable dense very
thin subset. In Proposition 3.8 (below), we extend Schr oder?s result using Martin?s
Axiom to get the same result for Xc, where X is separable and has a  -base of size
< c.
12
Proposition 3.8. (MA+:CH) If a separable space X with  (X) ! has a  -base
of size  < c, then there is a countable very thin dense set in Xc.
Before proving Proposition 3.8, we borrow some notation from [15]. In construct-
ing a dense set in Xc from a dense set D in X, for a collection I of disjoint closed
intervals with rational endpoints, where jIj= n, and a set fd1;:::;dng of elements of
D, we will denote by p(I;d1;:::;dn) the point with  th coordinate di if  is in the ith
element of I, and  th coordinate d0 otherwise, where d0 is a  xed element of D.
We also need to note that in Schr oder?s construction of a countable very thin
dense set in a product of continuum-many spaces with countable weak  -bases, he
constructs a weak  -base in the product from the weak  -bases on the factors. We
will make this more explicit in our modi cation of Schr oder?s construction.
Proof of Proposition 3.8. Since X is separable, so is Xc; let E be a countable dense
set in Xc constructed from the countable dense set D in X using disjoint closed
intervals with rational endpoints (as referenced above).
We will construct a weak  -base in Xc from the  -base B =fB :  < g. Let
I = f[r;s]  [0;1] : r;s2Qg. Enumerate all  nite pairwise disjoint collections of
subsets of I as fFn : n<!g. For each Fn, de ne a family of sets
Vn =VFn =
 Y
 <!1
B ( )   : [0;1]! ;  ( ) = 0 if  62[r;s];
for some [r;s]2Fn;  ( ) =  [r;s] if  2[r;s]2Fn; [r;s]2 
 
:
Since each Fn is  nite, and there are only  many choices for each  [r;s], each Vn has
size  . So SfVn : n2!g also has size  . Furthermore, each basic open set in Xc
contains a member of one of the Vn?s: indeed, let U = Q <cU , where each U is
open in X and U = X for all but  nitely many  , be a basic open set in Xc. Say
U 6= X for  =  1; 2;:::; n. Since U i is open for each i = 1;:::;n; it contains an
13
element B i of B. Separate  1;:::; n by the disjoint intervals [r1;s1];:::;[rn;sn]2I.
De ne  : [0;1] ! by  ([ri;si]) =  i and  ( ) = 0 if  62[ni=1[ri;si]. Then  is
the type of function that will generate a member of Vn, and that member of Vn will
be contained in U. Thus, SfVn : n 2 !g is a weak  -base for Xc of size  . Let
P =fP :  < g= SfVn : n2!g.
Let P = ffjf is a function from nf 2! to E and ff(i) : i < nfgis very thing.
Partially order P by function extension. Since E is countable, there are only countably
many functions from elements of ! to E; so P is ccc.
For each k < !, de ne D1k = ff 2 Pjk 2 dom(f)g. D1k is dense for each k,
since if k  nf = dom(f), we may extend f as follows: Let f0(i) = f(i) for each
i < nf. For each i with nf  i  k, choose by induction a point f0(i) 2 E with
  (f0(i)) 6=   (f0(j)) for any j < i. This is possible, since at each step only  nitely
many f0(j)?s have been chosen, and   (f0(j)) 2   (E), which, being dense in X,
must be in nite. Then we have a function f0 : k + 1!E, so f0 is an extension of f
which is in D1k.
Now, for each  <  , de ne D2 = ff 2Pj9i < nf (f(i) 2P )g. We will show
that every f2P has an extension in D2 . For any f2P,ff(i) : i<nfgis  nite, and
we claim that   (E\P ) is in nite for each  < c:
Let I = f[a0;a1];:::;[an 1;an]g be the intervals which were used to de ne P .
This means that for each k < n, there is a  k such that  x(P ) = B k whenever
x2[ak 1;ak]. Consider the pointsfe2Eje = p(I;d1;:::;dn) for some d1;::;dn2Dg.
Since D is dense in X, B k \D is in nite for all  k, k = 1;:::;n. Thus, there are
in nitely many points of the form p(I;d1;:::;dn) where dk 2B k\D. These points
will be in P \E, so this intersection is in nite.
Thus, we will be able to  nd a point x in E\P so that ff(i) : i < nfg[fxg
is very thin. Then the function f0 = f[f(nf;x)g is in D2 and extends f, so D2 is
dense.
14
Let G be a  lter in P which meets each of the  dense sets in D := fD1k : k <
!g[fD2 :  <  g. Since G is a  lter and meets each D1k, [G := fG is a function
from ! to E. Clearly, F := ffG(i) : i < !g is countable. To see that F is very
thin, suppose fG(i);fG(j)2F are distinct. Without loss of generality, suppose i<j.
Because G meets D1j+1, there is a function h2G with domain nh j+1, and h fG.
So fG(i) = h(i) and fG(j) = h(j). Since the range of h is very thin, h(i) and h(j)
cannot agree at any coordinate. Thus, F is very thin.
F is also dense in Xc: Suppose U  Xc is open. Then U contains an element
P of the weak   base P. G meets D2 , so there is f  fG with f(i)2P for some
i<nf; that is, fG(i)2P  U. So F\U6=;.
Suppose we have a space X which satis es the conditions of Proposition 3.8.
Then Xc has a countable very thin dense set. IfjXj= ! and 2! >!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 <c-thin dense set will be
countable.
We have established:
Corollary 3.9. (MA+:CH) Let X be a countable space with a countable weak  -
base for each  < c. Then Q <cX has a countable <c-thin dense set.
We do not know if the converse is true.
Question 3.10. If Xc has a countable <c-thin dense set, must X have a countable
weak  -base?
In an e ort to answer this question, the following result may turn out to be
useful:
Lemma 3.11. If X has a countable weak  -base, and D is a countable dense subset of
X, there is a countable very thin dense set E in X!1 with the property that D   (E)
for all  <!1.
15
Proof. This is a modi cation of the construction in [11], also used in Theorem 3.8, of
a countable very thin dense set from a countable weak  base.
Let W = fWn : n < !g be a countable weak  -base for X, where W0 = D.
(If D is not already in W, we may add it and W will still be a weak  -base.) Say
D =fdn : n<!g.
Identify !1 with the closed unit interval and let I = f[r;s]  [0;1] : r;s2Qg.
Enumerate the collection of all  nite pairwise disjoint collections fromI asfFn : n<
!g, in such a way that F2k =   12 12k; 12 + 12k  for k = 1;2;3;::: and F3k has the
property that 12 62[F3k. For each Fn, de ne a family of sets
Vn =VFn =
 Y
 <!1
W ( )j : [0;1]!!; ( ) = 0 if  62[r;s] for some [r;s]2Fn;
 ( ) = c[r;s] if  2[r;s]2Fn; for some c[r;s]2!
 
:
Since eachFn is  nite, and there are only countably many choices for each c[r;s], each
Vn is countable. So SfVn : n2!g is also countable. Furthermore, each basic open
set in X!1 contains a member of one of the Vn?s: Indeed, let U = Q <!1 U , where
each U is open in X and U = X for all but  nitely many  , be a basic open set
in X!1. Say U 6= X for  =  1; 2;:::; n. For each i = 1;2;:::;n, U i is open, so
each U i contains an element Wni of W. Separate  1;:::; n by the disjoint intervals
[r1;s1];:::;[rn;sn] 2I. De ne  : [0;1] ! ! by  ([ri;si]) = ni and  ( ) = 0 if
 62[ni=1[ri;si]. Then  is of the type of function that will generate a member of Vn,
and that member of Vn will be contained in U. Thus, SfVn : n2!g is a countable
weak  -base for X!1.
We will de ne a very thin dense set in X!1 by carefully choosing one point from
each Vn. Choose x0 2 V0 arbitrarily. If x0;x1;:::;xn 1 have been selected, choose
xn2Vn such that:
16
(i)   (xn)6=   (xj) 80 j n
This is possible because there are only  nitely many previously chosen points
in each projection, and will make the set fxn : n2!g very thin.
(ii) If n = 2k for some k2N (so Vn = Vf[1
2 
1
2k;
1
2+
1
2k]g
), and  62 12 12k; 12 + 12k ,
choose   (xn) to be dj, where j = minfm<! : dm62  (xl); 0 l<ng.
(iii) If n = 3k, choose  1=2(xn) to be dj, where j = minfm<! : dm62 1=2(xl); 0 
l<ng.
Since for every  6= 1=2, there exists a k such that  62F2k for all k > 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  <!1 such that U is open in
(X;  )2 for all  > . 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<!g.
For each n2N, de ne the sets An as follows:
An =f(x;y;z)2Q3 : (y;z) = (qn;rn)2E;x2Dng
Note that each An is a dense subset of a line parallel to the x axis. Let A =[1n=1An.
Claim 1: A is dense in Q3.
Let U V W be open in Q3. Then, since E is dense in the dense subset D02 of
Q2, V W must contain a point of E; say (qm;rm). Because Dm is dense in Q, there
is an xm2U\Dm. Then (xm;qm;rm)2Am\(U V W) A\(U V W). So
A is dense. 4
Claim 2: A is codimension 1 slim.
First, we consider a cross-section obtained by  xing the  rst coordinate atx02Q.
C(x0) is homeomorphic to Q2. C(x0)\An = f(x;y;z) : x = x0 2 Dn; (y;z) =
(qn;rn) 2 Eg. Since the Dm?s are disjoint, x0 can only be in one Dm; if it is in
Dm for some m 2 N, then this determines a unique point (qm;rm) for (y;z). So
23
C(x0)\An is either empty or one point; and it is nonempty for only a single n. Thus,
C(x0)\([n2NAn) is a single point.
Now,  x a point y0 in the second coordinate and consider An\C(y0) =f(x;y0;z) :
(y0;z) = (qn;rn)2E; x2Dng. Since E is very thin, there is at most one n for which
y0 = qn, so the intersection C(x0)\([n2NAn) is either empty or is the dense subset
Dn fqng frng of the line f(x;qn;rn) : x2Qg.
Similarly, if we  x the third coordinate to be z0, C(z0)\([n2NAn) is either empty
or is a dense subset of the line f(x;qn;z0) : x2Qg, speci cally the dense subset for
which qn and z0 = rn are  xed and x is any element of Dn.
Thus, we see that C(a)\A is either empty, one point, or a dense subset of a line;
this is nowhere dense in Q2. So A is codimension 1 slim in Q3. It should be noted
that A fails to be slim; as we have seen, the intersection of A with certain lines is
dense in those lines. This proves the claim. 4
We will now construct a  ner topology on Q3, making all possible slim dense sets
fail to be dense, while keeping A dense. This construction is patterned after Example
3.5 in [6].
Begin by letting  0 be the usual topology on Q. Enumerate all subsets S of Q3
with the property that each s2S has distinct coordinates as fS : 0 <  < !1g,
with each appearing !1 times. We will add open sets U and V = QnU at each
stage  > 0 to construct a series of topologies f  :  <!1g with the properties:
1. (Q;  ) is regular and has no isolated points.
2. Dn is dense in (Q;  ) for each n<!.
3. A is dense in (Q;  )3.
4. If S is slim dense in (Q;  )3, S \U 3 =;.
We observe that  0 has these properties.
24
If  has been de ned for < satisfying (1)-(4), let  be the topology generated
by S <   . If S is not slim dense in   , we will take U = ;. Suppose S is slim
and dense in (Q;  )3.
Since there is a countable base for  0, and thus we have only added countably
many new open sets at each stage, there is a countable baseBfor   3. LethBn;inibe
an enumeration of all pairs of open sets B2B and i<! so that each hB;ii occurs
in nitely many times at an index which is k mod 8, for each k = 0;1;:::;7. We will
de ne  nite sets Fn;Gn Q and put U = SFn and SGn QnU. Let F0 = G0 =;.
Suppose Fm;Gm are de ned for m n 1. Look athBn;ini, where Bn = C0 C1 C2.
Let Jn =f(q;r)2E : [(fq;rg[Fn 1[Gn 1)3n(Fn 1[Gn 1)3]\S 6=;g.
Recall that for each (x0;x1;x2) 2 S , x0 6= x1 6= x2, so that any point in
[(fq;rg[Fn 1[Gn 1)3n(Fn 1[Gn 1)3]\S has a coordinate in Fn 1[Gn 1. Fixing
one such coordinate gives a cross-section of S , which is slim; so any group of points
(a;b) in Jn which belong to a particular cross-section will be nowhere dense in Q2.
Since there are only  nitely many choices for a point of Fn 1[Gn 1 and a coordinate
in which to put it, we see that Jn is a  nite union of nowhere dense sets. Thus Jn is
nowhere dense; butE is dense inC1 C2, so we can pick a (qn;rn)2(C1 C2)\E which
is not in Jn. An\Bn will be a somewhere dense subset of the linef(x;qn;rn) : x2Qg,
because Dn is dense in (Q;  ) and C0 is open. Consider the set
Hn =fm : [(fm;qn;rng[Fn 1[Gn 1)3n(Fn 1[Gn 1)3]\S 6=;g
This set is nowhere dense, because if we  x two elements offqn;rng[Fn 1[Gn 1, we
have a cross-section; and S is slim, so it will have nowhere dense intersection with
the cross-section. Since there are only  nitely many cross-sections to consider, the
set Hn is nowhere dense. If in = 0, choose a point xn from
fq :9r such that (r;q)2E or (q;r)2Egn(Hn[([m<n(Fm[Gm))):
25
Otherwise, choose a point xn 2Dinn(Hn[([m<n(Fm[Gm))): This is possible be-
cause E and Din are dense, Hn is nowhere dense, and Fm;Gm are  nite.
Let (xn;qn;rn) = (a;b;c). We will add the points a;b;c to Fn 1 and Gn 1 to
obtain Fn and Gn, respectively, according to the value of n (mod 8):
n (mod 8) Fn Gn
0 a;b;c
1 a;b c
2 a;c b
3 b;c a
4 a b;c
5 b a;c
6 c a;b
7 a;b;c
Note that if Fn 13\S = ;, so does Fn3: for, if (x;y;z) 2 (Fn3nFn 13)\S ,
at least one coordinate of (x;y;z) is in fxn;qn;rng. But, we de ned Jn, Hn and
(xn;qn;rn) in such a way that no point with at least one of those three coordinates,
and the others possibly in Fn, could be in S .
Now, let U = Sn<!Fn. De ne   +1 to be the topology generated by   along
with U and QnU . Since at each stage we have added a set and its complement, the
topology is regular, and we have that U and QnU have in nite intersection with
every open set because we chose (qn;rn)2C1 C2, and C1;C2 ran over all elements
of a basis. So there are no isolated points. We need to check that the Dn?s are dense
in   +1, that A is dense in (Q;  +1)3, and that (U )3 misses S .
First, to see that Dn is dense, consider U \B, where B is open in   . Then
  10 (B) is open in   3 and contains one of the Bm?s, at a step where m 0 mod 8
and im = n. Then the a chosen at that step is in  0(Bm)\Dn, and was put into
26
Fm U . So U \B meets Dn. A similar argument with m 7 mod 8 shows that
QnU \B meets Dn for any B2  .
Let U0 = U , U1 = QnU . Fix a basic open set B = B0 B1 B2 in Q3, and
consider B\A\(Uk0 Uk1 Uk2), where ki < 2. B appeared as the  rst coordinate of
a pair hBn;ini in nitely times at an n which is congruent to any k mod 8. So there
is a point (in fact, in nitely many points) of A\B for which the  rst coordinate is
in Uk0, the second is in Uk1, and the third in Uk2. Thus, A is dense in (Q;  +1)3.
Now, consider U 3\S . Suppose (a;b;c) 2U 3. Then, (a;b;c) 2Fn3\S for
some n, since each of a;b;c must have been added at some stage, but this is impossible
by the construction of Fn. So U 3\S =;.
Once we have constructed the   ?s as described, let  = [ <!1  , and let X =
(Q; ). A will be dense in X3, since it is dense at each stage. Suppose S  X3 is
slim and dense. Without loss of generality, each s2S has distinct coordinates, and
since there are only countably many cross-sections to consider, S will appear as a
slim dense S at some stage  . But then U +1 misses S. So X3 has no slim dense
set.
Example 4.3. A codimension 1 slim dense set in Q3, which fails to be slim in more
ways.
Let fDn : n<!g be a partition of Q into dense sets. For i = 0;1;2, let Si be a
very thin dense subset of Di2, such that for each (q;r)2Si, q6= r. Enumerate Si by
f(qin;rin) : n<!g.
For each i = 0;1;2, n<!, de ne the sets Ain as follows:
A0n =f(x;y;z)2Q3 : (y;z) = (q0n;r0n);x2D3n+3g
A1n =f(x;y;z)2Q3 : (x;z) = (q1n;r1n);x2D3n+4g
A2n =f(x;y;z)2Q3 : (x;y) = (q2n;r2n);x2D3n+5g
27
Note that each Ain is a dense subset of a line parallel to a coordinate axis. Let
A =
1[
n=0
(A0n[A1n[A2n)
A is dense in Q3 and is codimension 1 slim, essentially because A is the union of three
sets which are the same as the A in the preceding example.
4.2 Superslim
Now, we will consider another cross-section based property which is, in a sense,
between slim and very thin.
De nition 4.4. A subset S of Q < X is superslim i every cross-section of S is
 nite.
Suppose S is superslim; since fs 2 S : s( ) = xg is a cross-section for each
 <  , x2X, each point x2X can appear at most  nitely many times in each
coordinate. SojSj jXj. On the other hand, suppose T  X is such that T(x; ) =
ft2T : t( ) = xgis  nite for each  < ; t2T. Since any cross-section will involve
 xing one or more coordinates, and thus be contained in a T(x; ), this condition
implies that T is superslim. So superslim is equivalent to \each point of X appears
only  nitely many times in each coordinate." This is a di erence with slim; in that
case, since \nowhere dense" can be di erent in di erent dimensions, we had a space
which had a codimension 1 slim dense set but not a slim dense set. Here, the key
property is  niteness, which does not change when considering di erent powers of X.
We  nd that for a  nite power, the existence of a superslim dense set is related to
satisfying a strengthened version of the property (NCk), which (as discussed above)
is related to the existence of a slim dense set in Xk.
Proposition 4.5. Let k < !. X satis es (NCk) witnessed by a collection of  nite
sets i Xk has a countable superslim dense set.
28
Proof. Both directions are modeled after proofs in [6]; the  rst after Proposition
4.1(2), and the second after Proposition 4.10.
()) Suppose X satis es (NCk), witnessed by the collection N =fN :  < g,
where for each  , jN j<!. Let D =
[
 < 
N k. We will see that this set is superslim
and dense.
Fix an element d0 2D, and consider the set C(d0; ) =fd2Djd( ) = d0( )g.
Since the members of N are pairwise disjoint, there is a unique  <  such that
d0( )2N . But then, by construction of D, any d2C(d0; ) must be in N k, which
is a  nite set. Thus, any point of X will appear only  nitely many times in the  th
coordinate; so D is superslim.
To see that D is dense, let
Y
i<k
Ui be a basic open set in Xk. Since N satis es
(NCk), there is an N 2 N which meets each member of fU0;:::;Uk 1g. Thus,
N k\Qi<kUi6=;; so D is dense in Xk.
(() Conversely, suppose that Xk has a countable superslim dense set D. For
each d2D, let c(d) be the set of coordinates of D, and let c(D) =[fc(d) : d2Dg=
fxn : n2!g. Observe that c(D) is indeed countable because D is, and each c(d) is
 nite. Also, c(D) is dense in X, so X is separable.
De ne by induction a sequence of disjoint  nite setshHn : n<!i: LetH0 =fx0g.
If Hn has been de ned, let kn be the least k 2 ! such that xk =2 [i nHi. Let
Hn+1 =fxkng[fc(d)jd2D^c(d)\Hn6=;gn[i nHi. That is, we take xkn plus the
remaining unused coordinates of each point with a coordinate in Hn.
It is clear from the construction that Hn+1 is disjoint from Hi for i n. Also,
Hn+1 is  nite: indeed, suppose x2Hn+1. Then fd2D : d(i) = xg is  nite for each
i = 0;1;:::;n 1, because D is superslim. So when we consider the coordinates of all
d in [i nfd2D : d(i) = xg, we still have a  nite set. So, as long as Hn is  nite, so
is Hn+1; and we see that H0 is  nite.
29
Now, for each in nite subset A of !, enumerate A in an increasing fashion by
fa0;a1;:::g. De ne
NA =f[i a0Hi; [a1i=a0+1Hi; [a2i=a1+1Hi;:::g
We claim that for some A !, NA witnesses (NCk).
Suppose not. Then, for every in nite A !, we can  nd nonempty open sets
U(A;0);U(A;1),...,U(A;k 1) with the property that no one member of NA meets
all of them. Let A be an uncountable almost disjoint family of subsets of !. Since
X is separable, X is (ccc); the collection fU(A;0) : A2Ag is uncountable, so there
must be an uncountable subcollection A0 for which U(A;0)\U(B;0) 6= ; for any
A;B2A0. Then consider fU(A;1) : A2A0g; in the same way, we  nd that there is
an uncountable subcollection A00 A0 with U(A;1)\U(B;1)6=; for all A;B2A00.
Continuing this process, we  nd that there must be sets A;B2A such that A6= B
but U(A;i)\U(B;i)6=; for each i = 0;1;:::;k 1.
Consider the nonempty open subset of Xk given by U = Qi<k[U(A;i)\U(B;i)].
There is a d2D\U, since D is dense. By the construction of the Hn?s, there must
be an m such that c(d)\Hm6=;. Let n be the least such m.
By construction, c(d) Hn[Hn+1: since n is the  rst such that c(d)\Hn6=;,
there are no coordinates of d in any previous Hk; and when we constructed Hn+1, we
would have therefore have included all remaining coordinates of c(d).
If n62A, then some N 2NA contains both Hn and Hn+1. But then N meets
each U(A;i) (speci cally, at d(i)). This contradicts that no member of NA meets
each U(A;i); so n2A. Similarly, n2B.
Since A and B are members of an almost disjoint family, A\B is  nite. Each
Hn is also  nite; so[n2A\BHn is  nite. But we have just shown that each d2D\U
must have a coordinate in this  nite set. Since D is superslim, each coordinate can
30
appear only  nitely many times; so D\U is  nite; but this contradicts denseness of
D.
Therefore, some NA must witness (NCk) in X, and clearly the members of NA
are  nite.
Question 4.6. What is the relationship between \X! has a superslim dense set" and
\X satis es (NC) witnessed by a collection of  nite sets"?
This is more di cult than the question settled by Proposition 4.5; the above
proof does not extend because the sets Hn as de ned above will not be  nite. This in
turn is because c(d) is not  nite for most points d2X!. For the other direction, the
slim dense set constructed as in the proof will not be  nite if the power in question
is !; N! does not remain  nite even if N was.
We observe that there are (nice) spaces which have no superslim dense set in
their square.
Example 4.7. There is a metrizable space X such that X2 has no superslim dense
set.
Proof. Let X = c Q, where c has the discrete topology and Q has the usual topology.
In [6], it is shown that a dense set in X2 = c Q2 will have uncountable cross-sections
of the type f( ;q1;q2) : q1;q2 2Qg for some  xed (q1;q2)2Q2. The goal there is to
show that X2 has no very thin dense set; but since uncountable is more than  nite
as well as more than 1, this also shows that X2 has no superslim dense set.
Since this example also has no very thin dense set, it remains to ask:
Question 4.8. Is there a space such that X2 has a superslim dense set but X2 does
not have a very thin dense set?
31
Chapter 5
(GC) and (NC)
The properties (GC), (NC), and (NCk) de ned in [6] (and mentioned above, in
Chapter 2) were de ned and studied as criteria which guarantee the existence of slim
dense sets in certain products. However, they are interesting for reasons other than
the one which led to their establishment. We have already seen that (NCk) witnessed
by  nite sets is applicable (even equivalent) to a property related to slim; we will
eventually see that (GC) is related to a selective separability property on the factor
space. This application does not even involve product spaces. Thus, we wished to
study the criteria further; especially with the added condition that the sets in the
collection be  nite, we gain some interesting results.
5.1 Cardinalities of Collections Witnessing (GC) and (NC)
In [6], many of the results which show that certain types of spaces satisfy (GC)
or (NC) result in the collection N consisting of  nite sets. The next example shows
that it is possible for a separable space which satis es (GC) to have no collection of
 nite sets witnessing the property.
Example 5.1. Let X = D[F as in Fact 3.13. Then X is separable and satis es
(GC), but no collection of  nite sets will witness (GC).
Proof. A collection witnessing (GC) in X isN =fNk : k2!gwhere for each k<!,
Nk = fx2F : jf < c :   (x) = 1gj = kg. It is clear that the elements of N are
pairwise disjoint. To see that the Nk?s are nowhere dense, let x2XnNk. Then the
number of coordinates of x which are 1 is either less than or equal to k, or more than
k.
32
Suppose   i(x) = 1 for i = 1;2;:::;k + 1. Then Tk+1i=1   1 i (1) is an open set
separating x from Nk. If x has 1 in less than or equal to k coordinates, then x2Nk.
In fact, Nk = fx2X : jf < c :   (x) = 1g kg. However, any open set in X
may only be restricted on  nitely many points; thus any open set U has points in it
with arbitrarily many coordinates equal to 1 and cannot be contained in Nk. Thus,
Int(Nk) =; and the Nk?s are nowhere dense.
Finally, suppose U is a basic open set in X. Then U is the restriction to X of a
set of the form
n\
i=1
  1 i (f1g)\
m\
j=1
  1 j (f0g). Then, points in U have at least n ones,
so U misses Nk for all k <n. However, U contains points with any  nite number of
ones greater than or equal to n, so U\Nk 6= ; for all k  n. Thus, N witnesses
(GC).
Now, suppose that N is a collection of  nite sets which witnesses (GC). If
Nk, k < !, and Mk, k < ! are disjoint countable subcollections of N,
1[
k=1
Nk =
 1[
k=1
(Nk\F)
!
[
 1[
k=1
(Nk\D)
!
and
1[
k=1
Mk =
 1[
k=1
(Mk\F)
!
[
 1[
k=1
(Mk\D)
!
must both be dense in X, and are disjoint. However, since the topology on D is gen-
erated by a maximal independent family, D is irresolvable. So one of S(Mk\D) and
S(N
k\D) is not dense in D. Without loss of generality, suppose
S(M
k\D) := MD
is not dense in D. Then, there is an open set U in D which misses MD.
That means U  [Mk\F. However, since [Mk \F consists of countably
many points, each with ones in  nitely many coordinates, there is a  < c with
  (Mk\F) = 0 for all  > . 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<!g be a dense subset of X and let fN :  <!1g witness
(GC) in X. By Lemma 5.4, for each dn, there is an element  dn of !1 with the
property that dn2N for all  > dn. Let   = supf dn : n2!g<!1. Then for all
 >   , 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<! such that jNj= k for all N2N.
Proof. Suppose to the contrary thatN is a collection witnessing (GC) or (NC) in X,
and there is a k2N such that for every N2N,jNj= k. LetU =fU0;U1;:::;Ukgbe
a pairwise disjoint collection of k+1 open sets. IfN witnesses (GC), it also witnesses
(NC) (see [6], remarks after the de nitions of (GC) and (NC)). So there is an N2N
which meets every member of U; say xi2N\Ui for i2k + 1. Since U is a pairwise
disjoint collection, xi 6= xj for any i 6= j. This contradicts that jNj = k; so the
collection N cannot exist as described.
Corollary 5.8. No uncountable family of  nite sets can witness (GC).
Proof. If N were such an uncountable family, for some k 2 ! there would be a
countable subcollectionNk consisting of all elements of size k. This would still witness
(GC); but this contradicts Proposition 5.7. So, in any collection witnessing (GC) in
X, at most countably many sets can be  nite.
36
This raises the question:
Question 5.9. Can an uncountable family of closed sets witness (GC)?
5.2 Ordered Spaces and (GC)
We now turn our attention to examining the conditions under which ordered
spaces satisfy (GC). First, we look at linearly ordered spaces (LOTS), and then at
generalized ordered spaces (GO-spaces).
Lemma 5.10. If a space X contains a dense subspace Y which satis es (GC), then
X satis es (GC).
Proof. LetN be the collection witnessing (GC) in Y. We claim thatN also witnesses
(GC) in X.
Let N 2N. If there is an open set U  clX(N), then U\Y  clX(N)\Y =
clY (N). Since N is nowhere dense in Y, U\Y must be empty. But U\N U\Y =
;)U =;. Thus N is nowhere dense in X.
If U is an open set in X, then U\Y is nonempty (since Y is dense) and open
in Y, so U\Y meets all but  nitely many members of N. Thus, U meets all but
 nitely many members of N.
So,N is a pairwise disjoint collection of nowhere dense sets in X, and each open
set in X meets all but  nitely many members of N; that is, N witnesses (GC) in
X.
Proposition 5.11. If X has a dense metrizable subspace, then X satis es (GC).
Proof. By Proposition 4.2 in [6], every metrizable dense-in-itself space satis es (GC),
so this follows directly from Lemma 5.10.
We now characterize the ordered spaces which satisfy (GC).
Lemma 5.12. If a LOTS satis es (GC), it has a  -disjoint  -base.
37
Proof. Suppose N = fNk : k < !g is a collection witnessing (GC) in a linearly
ordered space X. For each k < !, de ne Uk = Xn(N0[N1[:::[Nk). Uk is open,
so it may be written as a collection Ik of disjoint open intervals. Let B = [k<!Ik.
ObviouslyB is  -disjoint. We claim that each nonempty interval (a;b) X contains
a member of B:
Since (a;b) is open, it must meet all but  nitely many members of N. In par-
ticular, there are k;l < ! such that Nk and Nl both meet (a;b), say at ck and cl.
Without loss of generality, ck < cl. Then, for any m > 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 = Tk<!Uk is dense in the Baire
space X. We will show that Y is metrizable by constructing a  -locally  nite base
for Y.
Consider the  -disjoint  -base Sk<!Ik for X given by Lemma 5.12. Note that
for each k, SIk = Uk. Let B be the restriction of this  -base to Y. If x2Y, then
x2SIk for all k <!. Since Ik is a collection of disjoint open intervals, this means
that the interval J2Ik containing x is a neighborhood of x which does not meet any
other members of Ik. Thus, J\Y witnesses that fI\Y : I2Ikg is locally  nite.
To see that B forms a base for Y, let x2(a;b)\Y, where (a;b) is a basic open
set in X. Since x is in Y = Tk<!Uk = Tk<! (SIk), there is an Ik for each k < !
38
such that x 2 Ik  Ik. Suppose there is no k such that Ik  (a;b). Then either
(a;b) Ik for all k<!, or one of a;b is in Ik for all k<!.
If (a;b)  Ik for all k, then (a;b)  XnNk for all k, which contradicts that N
witnesses (GC).
If one of a;b, say a, is in Ik for all k, (a;x) Ik for all k because x2Ik and Ik is
an interval. This again gives a contradiction with N witnessing (GC), because then
(a;x) would miss in nitely many Nk?s.
So, for some k, Ik  (a;b), and thus Ik\Y  (a;b)\Y. This shows that B is
a  -locally  nite base for Y. Y is T3, being a subspace of a LOTS, and this means
that Y is metrizable.
Combining Proposition 5.11 and Theorem 5.13 shows that a Baire dense-in-itself
LOTS satis es (GC) i it has a dense metrizable subspace. For non-Baire spaces,
this is not true, as the following example shows.
Example 5.14. There is a LOTS satisfying (GC) which does not have a dense metriz-
able subspace.
Proof. The space is Gruenhage and Lutzer?s example in [5] of a LOTS that is Volterra
but not Baire. In their paper, they show that Volterra=Baire in any space with a
dense metrizable subspace; so this space clearly does not have such a subspace.
Let X be the set of all functions f from !1 to the integers with the property
that f( ) = 0 for all but  nitely many  ; give X the topology generated by the
lexicographic order. A neighborhood base at each f 2 X is fB(f; ) :  < !1g,
where B(f; ) =fg2X :8   ; g( ) = f( )g.
For each k < !, let Nk = ff 2X : jf < !1 : f( ) 6= 0gj = kg. The collection
N =fNk : k <!g is clearly pairwise disjoint. If f2X is nonzero for at least k + 1
coordinates, then there is  < !1 such that f(x) 6= 0 for at least k + 1 coordinates
less than  ; so B(f; )\Nk =;. Thus, Nk ff2X :jf <!1 : f( )6= 0gj kg.
39
This cannot contain any B(g; ), since these open sets contain points with arbitrarily
many nonzero coordinates. So each Nk is nowhere dense.
Now, let U  X be open and nonempty. Then there exist f and  such that
B(f; )  U. Let K = jf : f( ) 6= 0 and    gj. Then, since every g2B(f; )
must agree with f up to the  th coordinate, every g in B(f; ) has at least K
nonzero coordinates. In fact, for every n>K, 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<!Bn is a
 -disjoint  -base for X, such that each Bn is a disjoint collection. Without loss of
generality, we may assume that the elements of B are convex open sets.
For each n, consider SBn. If this is not dense in X, then XnSBn is a nonempty
open set; write it as a collection of disjoint convex open sets fU :  < Ag. Let
Cn = Bn[fU :  2Ag. Note that SCn is dense in X, and Cn is a collection of
disjoint convex open sets.
We will de ne In, n < !, by induction. Let I0 = C0. For each n > 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<!In is a  -base, since
S
n2!Cn contains a  -base and every member of Cn contains some member of an Ik.
Now, we will use I to de ne, by induction, the collection N witnessing (GC).
Pick an element xI0 2I from each I 2I0. Set N0 = fxI0jI 2I0g. If Nk has been
de ned, for each I 2Ik+1, consider the set FI = fxJnjI  J and J 2In; n kg.
Observe that there are only  nitely many In?s with n k. By the construction, the
members ofIn are pairwise disjoint; so there can only be one J in eachIn with I J.
So FI is a  nite set.
Now, for each I 2Ik+1, chose an element xIk+1 2InFI. Since X is a T2 space
without isolated points, and each I 2Ik+1 is open, I is in nite; so this is always
possible. Let Nk+1 =fxIk+1jI2Ik+1g.
Since X has no isolated points, InFI is open; it is also dense in I. Thus, since
each [Ik is open and dense in X, so is each Uk = SfInFI : I 2Ikg. Thus, XnUk
is nowhere dense; in particular, Nk 1  XnUk is nowhere dense. (Recall that FI
contains the points xIk 1.) Also, by construction, Nk\Nl =;if k6= l; so we see that
N =fNkjk<!g is a pairwise disjoint collection of nowhere dense sets.
It remains to see that N witnesses (GC) in X. Let U be a nonempty open
subset of X. Then, U contains a member of the  -base I. Say I0 2Ik is contained
in U. Then xI0k 2 I0  U implies that U\Nk 6= ;. Now, by the construction of
the In?s, there is an I1 2Ik+1, with I1  I0. Since xI1k+1 2I1\Nk+1, and I1  U,
U\Nk+1 6= ;. Continuing in this way, we can  nd In 2Ik+n which is contained in
U; the xInk+n chosen to be in In at stage k + n of the induction will be in Nk+n\U.
So U meets all but  nitely members of N.
This result leads easily to the following example:
Example 5.16. A Souslin line cannot satisfy (GC).
41
Proof. Recall that a Souslin line is a dense linearly ordered space which is (ccc) but
not separable. Let S be a Souslin line with no isolated points. (Note that spaces
with isolated points cannot satisfy (GC).) Suppose S satis es (GC); then there is a
 -disjoint  -base in S. But, S is ccc; so the  -base will be countable, and choosing
a point from each member of the  -base gives a countable dense set in S, which is
impossible. So S cannot satisfy (GC).
5.3 Ordered Spaces and (NC)
We consider next the conditions under which an ordered space satis es (NC).
Lemma 5.17. If a space X satis es (NC), then there is a collection of cardinality at
most  (X) which witnesses (NC).
Proof. Suppose N witnesses (NC) in X, and that U X is open, with jUj=  (X).
Let V = XnU. De ne N0 =fN2N : N\U6=;g. The members of N0 are clearly
pairwise disjoint, and since each meets U, there can be at mostjUj=  (X) of them.
We claim thatN0 witnesses (NC). Given a  nite pairwise disjoint collectionU of open
sets in X, either at least one of them meets U, or none of them do.
Suppose no element of U meets U; then U[fUg is a pairwise disjoint,  nite
collection of open sets in X. So there is an N 2N which meets each; but this N
meets U, so it is in N0.
Now, suppose that at least one element of U meets U. Consider the re nement
of U given by
U0 = (fA\U : A2Ug[fA\V : A2Ug)nf;g
This is a  nite pairwise disjoint collection of nonempty open sets in X; so there is
an N 2N which meets each member of U0. Since there was an A 2U for which
A\U6=;, this N must meet U. So N2N0; but since each A2U must meet either
U or V, N meets each member of U (the original collection) as well.
42
Thus, N0 witnesses (NC).
Note that this shows also that whenever N witnesses (NC) in a space X, and
U  X is open, the subcollection NU = fN 2N : U\N 6= ;g also witnesses (NC)
in X.
Example 5.18. There is a LOTS which does not satisfy (NC).
Proof. The space is X = Q Y, where Q is the rationals with the usual topology
and Y is the set Z!1 with the lexicographic order topology.
We will show that no countable union of nowhere dense sets is dense in Y. This
will be su cient because the union of any collection witnessing (NC) must be dense in
X, and thus in Y. However,  (X) = ! (since Q is an open subspace), so Proposition
5.17 shows that if X satis es (NC), there is a countable collection witnessing (NC).
Now, let us see that, if fNk : k < !g is a collection of nowhere dense sets in
Y, [k<!Nk is not dense in Y. Let U be a nonempty open set in Y. Since N0 is
nowhere dense, there is a nonempty open set U0  UnN0. Since a LOTS is regular
and the intervals form a base, we may choose an interval I0 = (f0;g00) I0 U0. Let
 0 = minf <!1 : f0( )6= g00( )g. De ne
g0( ) =
8
>>>>
<
>>>>
:
g00( )  62f 0; 0 + 1g
f0( )  =  0
f0( ) + 2  =  0 + 1
Observe the following about g0:
 f0 <g0 because  0 + 1 is the  rst  at which f0( )6= g0( ), and f0( 0 + 1) <
f0( 0 + 1) + 2 = g0( 0 + 1).
 g0 <g00 because g0, f0, and g00 agree until  0, and then g0( 0) = f0( 0), which
is less than g00( 0) because f0 <g00.
43
Thus, (f0;g0) is an open interval which, being contained in I0, misses N0. Since Y
has no adjacent points, the interval is nonempty.
Now, suppose  k;(fk;gk) have been de ned for k<n, satisfying:
 (f0;g0) (f1;g1) ::: (fn 1;gn 1)
  0 < 1 <:::< n 1
 fk( ) = gk( ) for all    k
 gk( k + 1) = fk( k + 1) + 2
 gk( k 1 + 1) = fk( k 1 + 1) = fk 1( k 1 + 1) + 1
De ne  n and (fn;gn) as follows:
Since N0 [:::[Nn is nowhere dense, choose an interval (fn;g0n)  [fn;g0n]  
(fn 1;gn 1) which misses N0[:::[Nn, with the property that fn( n 1+1) = g0n( n 1+
1) = fn 1( n 1 + 1) + 1. This is possible because any function which is equal to fn 1
and gn 1 up to  n 1, with  n 1 +1st coordinate so de ned, is between fn 1 and gn 1,
since fn 1( n 1 + 1) < fn 1( n 1 + 1) + 1 < fn 1( n 1 + 1) + 2 = gn 1( n 1 + 1).
Choosing any two such functions to be endpoints of an open interval would give an
interval J properly contained in (fn 1;gn 1), with the property that every function
in J agrees with fn 1 and gn 1 up to the point where they split, and is directly
between them at the next coordinate. We may then take a subinterval of J which
misses N0 [:::[Nn. Let  n = minf < !1 : fn( ) 6= g0n( )g. Note that, since
fn 1 <fn <g0n <gn 1,  n > 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 <gn 1( n 1 + 1).
Consider Tn2NIn. Suppose there is an h2Z!1 such that h > fn for all n and
h<gn for all n; in this case, h is an upper bound forffngn2N; but we claim fn6!h.
To see this, for each n, we let  n <!1 be the least such that fn( n)6= h( n). Then
 = supf n : n 2 Ng < !1 because the set is countable. Let h be any function
which is identical to h for all    , and h ( + 1) = h( + 1) 1, and h+ be any
function which is greater than h. If = (h ;h+) is an interval containing h which
misses every element offfng, because at the point  n, fn( n) <h( n) = h ( n), and
for all  < n, f( ) = h( ) = h ( ).
Similarly, we may  nd an interval Ig = (p ;p+), where p is any function less
than h and p+ is a function that agrees with h until after the point at which the gn?s
have di ered from h. This interval contains h but misses every element offgng. Then,
(h ;p+) is a nonempty interval with endpoints that are less than every element of
fgng and greater than every element of ffng. So (h ;p+) \n2NIn. But, any point
in \n2NIn must miss Nk for every k<!; so [k<!Nk is not dense in Y.
45
We now show that there is indeed an h which is greater than fn and less than gn
for each n2N. De ne
h( ) =
8
>>>>
>>>
>>>
>>>>
>>>
>>>>
>>>
>>><
>>>
>>>>
>>>
>>>
>>>>
>>>
>>>>
>>>
:
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 <h<gn for all n2!.
By de nition of the functions fn;gn, for any k < n,  <  k, fn( ) = gn( ) =
fk( ) = gk( ). So minf : h( ) 6= fn( )g = minf : h( ) 6= gn( )g =  n + 1. And
h( n+1) = fn( n+1)+1, which is between fn( n+1) and gn( n+1) = fn( n+1)+2.
So fn <h<gn. Thus, h2\n<!(fn;gn).
This example failed to satisfy (NC) because it contained disjoint subspaces on
which the cardinality and structure of the open sets varied widely. We now examine
some conditions under which the topological sum X Y will satisfy (NC), given that
each of X and Y do.
Proposition 5.19. Let X be a space with the property that each point has a neighbor-
hood of cardinality  (X), and a local  -base of cardinality  (X). Then X satis es
(NC).
46
Proof. Fix a maximal pairwise disjoint collectionC =fC :  <jXjgof open subsets
of X, each of size  =  (X). For each  , there is a  -base B for C of cardinality
 . (For instance, we could take B to be the union of the local  -bases of size  for
each of the  -many points of C .) Index B as fB  :  <  g. Since  (X) is the
minimum cardinality of an open set, and each B  is contained in the open set C 
which has size  , jB  j =  for each  ; . Index the collection of all  nite subsets of
 as fF :  < g. For each  < , let A =fB  :  <jXj; 2F g. That is, A is
the collection of all  -base elements whose index is in F , across all the C ?s. Choose
sets N by induction so that:
 N contains one point from each element of A 
 N 0\N =; for all  0< 
This is possible because jB  j=  > 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  <jXj, in particular for  1;:::; n. Thus, N \Ui  N \B i i 6= ; for each
i = 1;2;:::;n.
In particular, we may apply this proposition to an ordered space. The fact that
the intervals are a base gives us the uniformity we need, as long as we have one
interval for each point that matches a standard.
Corollary 5.20. A LOTS X with the property that for every point, there is an
interval of cardinality  (X) containing that point, satis es (NC).
47
Proof. If I is such an interval, a local  -base of cardinality  (X) is given byf(a;b) :
a<b;a;b2Ig.
In [6], the authors show that if a collection N witnesses (NC2) in a strongly
irresolvable space X, the family F of all M N such that M also witnesses (NC2)
in X is an ultra lter on N. They also show that in a strongly irresolvable space, a
collectionN witnesses (NC2) if and only if it witnesses (NC). So in strongly irresolv-
able spaces, we have similar ultra lters on the collections which witness (NC). These
ultra lters may be used to characterize the conditions under which X Y satis es
(NC), given that X and Y are strongly irresolvable and satisfy (NC).
Theorem 5.21. Let X and Y be strongly irresolvable spaces which satisfy (NC).
X Y satis es (NC) if and only if there are collections NX, NY witnessing (NC) in
X and Y respectively, and a function f : NX !NY such that f(M) 2FY for all
M2FX and f 1(M) 2FX for all M2FY (where FX, FY are the ultra lters on
NX and NY respectively, discussed above).
Proof. (() Suppose we have such a function from NX to NY , and for N 2NY ,
consider f 1(N)  NX. Since FY is an ultra lter on N and fNg does not witness
(NC) and thus is not inFY ,NYnfNg2FY . So f 1(NYnfNg) =NXnf 1(N)2FX.
So SNXnf 1(N) is dense in X. Since X is strongly irresolvable, this means that
Sf 1(N) must be nowhere dense.
De ne NX Y to be fN[(Sf 1(N)) : N2NYg. Since f is a function, this is a
pairwise disjoint collection of sets in X Y. Each N[(Sf 1(N)) is nowhere dense,
since N2NY and we have shown that f 1(N) is nowhere dense.
To see that NX Y witnesses (NC), it is enough to show that if U1;::;Un is a
collection of pairwise disjoint open sets in X and V1;::;Vk is a collection of pair-
wise disjoint open sets in Y, there is an element of NX Y which meets each of
U1;:::;Un;V1;:::;Vk. For an open set U  X, de ne MU to be the set of elements
48
of NX which meet U. By the remark after Lemma 5.17, MU 2FX. Since FX is a
 lter, M:=MU1 \MU2 \:::\MUn 2FX. Consider f(M)2FY . f(M) witnesses
(NC), so there is an NY 2f(M) which meets every member of fV1;:::;Vkg. Then,
f 1(NY ) M, so[f 1(NY ) meets each of the U?s. Thus, NY[([f 1(NY ))2NX Y
and meets each of the U?s and V?s.
()) Suppose X Y satis es (NC), witnessed by a collection NX Y . Without
loss of generality, we may assume that each member of NX Y meets both X and Y.
Indeed, since X and Y are each open in X Y, MX = fN 2NX Y : X\N 6= ;g
and MY =fN 2NX Y : Y \N 6=;g are in FX Y . So MX\MY 2FX Y , and if
necessary we will use MX\MY in place of NX Y .
De ne
NX =fN\X : N2NX Yg
NY =fN\Y : N2NX Yg
f :NX !NY : N\X7!N\Y
It is clear that NX witnesses (NC) in X and NY witnesses (NC) in Y. Let FX, FY
be the ultra lters associated with NX and NY , respectively.
Claim: Let M2FX.  M:=fN2NX Y : N\X2Mg2FX Y
It is enough to show that[ Mis dense in X Y (see proof of Lemma 4.13 in [6]).
Suppose that there is an open set U X Y which does not meetS  M, and consider
MU =fN2NX Y : N\U6=;g2FX Y . MU\  M=;, since every member ofMU
meets U and no member of ~M meets U. Since these are both subcollections of the
pairwise disjoint collectionNX Y ,[MU and[ ~Mare disjoint. But[MU is dense in
X Y, hence in X; and
 S  
M
 
\X = SM is also dense in X. This contradicts
that X is strongly irresolvable. So S  M must be dense in X Y, and  M2FX Y .
49
Now, for M2FX,
f(M) = fY \N : N\X2Mg
= fY \N : N2  Mg
Since  M witnesses (NC) in X Y, f(M) witnesses (NC) in Y. So f(M)2FY .
Similarly, f 1(M)2FX for each M2FY .
WhenNX andNY are countable, the ultra ltersFX andFY may be regarded as
members of  !. In this case, the function condition described above is the statement
thatFX  FY , where is the Rudin-Keisler order on  !. Recall that when p;q2 !,
p q i there is a function f : !!! such that  f(q) = p, where  f is the Stone
extension of f. It is well-known (see, for instance, [13]) that  f(q) = p is equivalent
to 8Q2q(f(Q)2p), which is equivalent to 8P 2p(f 1(P)2q). Thus, we have:
Corollary 5.22. If X and Y are strongly irresolvable spaces which satisfy (NC),
witnessed by countable collections NX and NY , respectively, X Y satis es (NC) if
and only if FX is Rudin-Keisler equivalent to FY .
Proof. FX  FY ,X Y satis es (NC) ,Y X satis es (NC) ,FY  FX
5.4 (GC) and GN-separability
The de nitions of (GC) and GN-separable are very similar. We want to in-
vestigate the relationship between these two properties. We will use the following
de nitions from [1]:
De nition 5.23. A dense set D is groupable if it can be partitioned as D = SfAn :
n2!g where each An is nonempty and  nite, and every nonempty open set in X
intersects all but  nitely many An.
50
De nition 5.24. A space X is GN-separable if for every sequence hDn : n<!i of
dense subsets of X there are dn2Dn such that D =fdn : n<!g is groupable.
De nition 5.25. A space X is R-separable if for every sequence hDn : n < !i of
dense subsets of X, one can pick dn2Dn such that fdn : n<!g is dense in X.
It is clear from the de nitions that X has a groupable dense set i X satis es
(GC) witnessed by a collection of  nite sets. Therefore if X is GN-separable, since
one then has a groupable dense set, then X satis es (GC) witnessed by a collection
of  nite sets. We now wish to know if \(GC) witnessed by a collection of  nite sets"
implies GN-separable. We have the following partial result, which addresses one of
the key ideas: going from the existence of one groupable dense set to having every
dense set be groupable.
Fact 5.26. If X is countable and satis es (GC) witnessed by a collection of  nite
sets, then every dense set in X is groupable.
Proof. Let N =fNkjk<!g be a collection witnessing (GC) in X, where each Nk is
 nite. (N must be countable because X is.) Let X0 = XnSN. If X06=;, jX0j !;
enumerate X0 by fxk : k <jX0jg. Then N0 =fNk[fxkgjk <jX0jg[fNkjk jX0jg
is also a collection witnessing (GC) in X. N0 which covers X, and each member of
N0 is still  nite. (This step is based on the proof of Proposition 73 in [1].)
Let D X be dense. For each n < !, de ne An = fD\Nng. Since jDj = !
and the Nn?s are  nite and pairwise disjoint, countably many of the An?s will be
nonempty; take these nonempty An?s to form the collection which witnesses that D
is groupable.
This proof relied on the fact that X was countable. If X is merely separable,
the situation becomes more complicated. GN-separability relies in part on the ability
to pick a countable dense set from within any set that is dense in X; when X is
countable, any subset of X is already countable. If X is larger, though, it is possible
51
that although X may be separable, there will be a dense set without a countable
subset which is also dense. We can use the cardinal function  (X) to describe this
situation:
 (X) = maxfminfjYj: Y  D; Y is dense in Xg: D X denseg:
If  (X) > !, 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<!iwith Dn = Y
for all n has no dense set of the formfdn : n<!;dn2Dng, much less a groupable one.
In the terminology of [1],  (X) >!)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 <!1 : x = 1gj<!g:
Y is dense in X: if U  X is open, where U =ffjf : !1!2 ^ fjdom =  g for a
function  de ned on a  nite subset of !1,  ( ) = 1 for only  nitely many  ; letting
x =  ( ) if  2dom( ) and 0 otherwise, we see that x2U \Y.
52
But, no countable subset of Y is dense in X. Suppose to the contrary that
Y0 Y is countable. Say Y0 =fynjn2!g. For each n<!, there is a  n <!1 such
that yn( ) = 0 for all  >  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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54