I-weight, Special Base Properties and Related Covering Properties
Except where reference is made to the work of others, the work described in this
dissertation is my own or was done in collaboration with my advisory
committee. This dissertation does not include proprietary or
classified information.
Bradley S. Bailey
Certificate of Approval:
Phillip Zenor
Professor
Department of Mathematics
Gary Gruenhage, Chair
Professor
Department of Mathematics
Stewart Baldwin
Professor
Department of Mathematics
Wenxian Shen
Associate Professor
Department of Mathematics
Stephen McFarland
Dean
Graduate School
I-weight, Special Base Properties and Related Covering Properties
Bradley S. Bailey
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulfillment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
16 December 2005
I-weight, Special Base Properties and Related Covering Properties
Bradley S. Bailey
Permission is granted to Auburn University to make copies of this dissertation at its
discretion, upon the request of individuals or institutions and at their expense.
The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
Brad Bailey was born in Savannah, Georgia in 1977. He entered Armstrong Atlantic
State University in 1995, where he met his future wife, Elaine LeCreurer. Each gradu-
ating from Armstrong in May of 2000, Brad and Elaine were married that same month
and began graduate studies at Auburn University that fall. Brad received his Masters
degree in 2002 and his PhD in 2005, both at Auburn and in the field of topology. At the
time this is being written, Brad and Elaine are happily anticipating the birth of their
child, who is expected to arrive in the summer of 2006.
iv
Dissertation Abstract
I-weight, Special Base Properties and Related Covering Properties
Bradley S. Bailey
Doctor of Philosophy, 16 December 2005
(M.S., Auburn University, 2002)
(B.S., Armstrong Atlantic State University, 2000)
80 Typed Pages
Directed by Gary Gruenhage
Alleche, Arhangel?ski?? and Calbrix defined the notion of a sharp base and posed the
question: Is there a regular space with a sharp base whose product with [0, 1] does not
have a sharp base? Chapter 2 contains an example of a space P with a sharp base whose
product with [0, 1] does not have a sharp base. The example in Chapter 2 also answers
the following 3 questions found in the literature: Is every pseudocompact Tychonoff space
with a sharp base metrizable? Is there a pseudocompact space X with a G?-diagonal
and a point-countable base such that X is not developable? Is every ?Cech-complete
pseudocompact space with a point-countable base metrizable? The space we construct
is pseudocompact, ?Cech-complete, has a G?-diagonal, a sharp base and a point-countable
base, but is not metrizable nor developable.
In Chapter 3, we study open-in-finite (OIF) bases and introduce the notion of a
?-open-in-finite (?-OIF) base. Each ?-OIF base is also OIF. We show that a base B for
the space X is ?-OIF if and only if for each dense subset Y of X, B harpoonupright Y is OIF. We
also define OIF-metacompact, ?-OIF-metacompact, (n,?)-metacompact, and (< ?,?)-
metacompact and show that for generalized order spaces and ? = ? these properties
v
are equivalent. The (< ?,?)-metacompact property is corresponds to the < ?-weakly
uniform base property. We show that for Moore spaces X, the space X has an OIF base
(resp. ?-OIF base, < ?-weakly uniform base) if and only if the space is OIF-metacompact
(resp. ?-OIF-metacompact, (< ?,?)-metacompact).
In the final chapter, we prove that for the class of linearly ordered compact spaces,
i-weight reflects all cardinals. We find necessary and sufficient conditions for i-weight
to reflect cardinal ? in the class of locally compact linearly ordered spaces. In the last
section we calculate the i-weight of paracompact spaces in terms of the local i-weight
and extent of the space. This result determines that for compact spaces i-weight and
local i-weight are the same.
vi
Acknowledgments
I am obviously greatly indebted to my advisor, Dr. Gruenhage, for his guidance
and support. I am also thankful for not only the members of my committee, Drs.
Baldwin, Shen and Zenor, but all the mathematics faculty at both Auburn University
and Armstrong Atlantic State University for all their help and for instilling in me their
passion for mathematics through skillful instruction and conversation. I also appreciate
Dr. Dong taking the time to be my outside reader.
I am grateful to the Fitzpatrick family. The endowment they generously established
will continue to support and encourage students with an interest in topology for many
years.
I would like to thank my family for their ongoing encouragement and my wife, Elaine
for being so supportive through the years.
vii
Style manual or journal used Transactions of the American Mathematical Society
(together with the style known as ?auphd?). Bibliograpy follows van Leunen?s A
Handbook for Scholars.
Computer software used The document preparation package TEX (specifically
LATEX) together with the departmental style-file auphd.sty.
viii
Table of Contents
List of Figures x
1 Introduction and Background 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Definitions and Background Results . . . . . . . . . . . . . . . . . . . . . 2
2 An Example of a Space with a Sharp Base 5
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.2 The Construction of Space P . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.3 Verifying Properties of P . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3 Open-in-finite bases 17
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.2 OIF Spaces and Cardinal Functions . . . . . . . . . . . . . . . . . . . . . 18
3.3 Stronger Base Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.4 The maltesecross Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.5 Covering Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.6 Set Theoretic and Combinatorial Conditions . . . . . . . . . . . . . . . . . 43
4 I-weight and separating weight 46
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
4.2 Compact Linearly Ordered Spaces . . . . . . . . . . . . . . . . . . . . . . 47
4.3 Locally Compact Linearly Ordered Spaces. . . . . . . . . . . . . . . . . . 54
4.4 Paracompact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
Bibliography 69
ix
List of Figures
2.1 A typical S? with root ??. . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.2 An open set [??] and the sequence of disjoint open sets {[T?(i)]}i<?. . . . 8
3.1 The tangent-disk space and some open sets . . . . . . . . . . . . . . . . . 29
4.1 The tree T(X). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
x
Chapter 1
Introduction and Background
1.1 Introduction
This dissertation focuses on three different research topics. In Chapter 2 we con-
struct a space P that answers several questions found in the literature. The space P has
what is known as a sharp base, yet it?s product with the space [0, 1] does not have a
sharp base.
Chapter 3 is about open-in-finite (OIF) bases and ?-open-in-finite (?-OIF) bases. In
[4] OIF bases are introduced and the question is posed: Must each dense subspace of a
space with an OIF base have an OIF base? That question is still open. However, we
show that a base B for a space X is a ?-OIF base if and only if for each dense subset Y
of X, the base BharpoonuprightY is an OIF base.
We demonstrate that for OIF spaces the cardinal functions weight and pi-weight
coincide, and that the same is true of character and pi-character. Further, we prove that
if X is OIF, then for each dense subset Y of X, then the weights Xand Y are the same.
Several covering properties related to OIF, ?-OIF and < ?-weakly uniform bases are
discussed in the final sections of Chapter 3. We prove that for the class of Moore spaces
these covering properties are equivalent to the corresponding base properties. We also
show that for generalized order spaces (GO spaces), these properties are all equivalent
to each other and to paracompactness.
The last chapter contains reflection theorems for i-weight and formulas for the i-
weight of linearly ordered compact and locally compact spaces, and for paracompact
1
spaces. We show that i-weight reflects for linearly ordered compact. Then we find condi-
tions under which i-weight reflects for linearly ordered locally compact spaces. We define
local i-weight, denoted liw(X), and the last section provides that for a paracompact space
X, the i-weight of X is max{log(e(X)),liw(X)}.
1.2 Definitions and Background Results
This section contains definitions from general topology and set theory that are used
throughout this work. All the material in this chapter may be found in one of [7], [12]
or [14].
First we define several cardinal functions.
Definition. The weight of space X, denoted w(X) is the minimum cardinality plus ?
of a base for the topology on X.
Definition. A pi-base for X is a collection V of nonempty open sets in X so that if U is
any nonempty open set in X, then V ? U for some V ? V. The pi-weight of X, denoted
piw(X), is the minimum cardinality plus ? of a pi-base for X.
Definition. A local base at x is a collection V of open sets each containing x, so that if
U is an open set containing x, V ? U for some V ? V. The character of X at x, denoted
?(x,X), is the minimum cardinality of a local base at x. Then ?(X) = sup{?(x,X) :
x ? X}+?.
Definition. A local pi-base at x is a collection V of open sets, so that if U is an open set
containing x, then V ? U for some V ? V. Then pi-character at x , denoted pi?(x,X),
is the minimum cardinality of a local pi-base at x, and pi?(X) = sup{pi?(x,X) : x ? X}.
2
We call a cardinal function ? monotone if ?(Y) ? ?(X) for each subspace Y of X.
Theorem 1.1 Weight is monotone.
Proof. For a space X let B be a base for X so that |B| = w(X) and let Y be any
subspace of X. Then BharpoonuprightY = {B ?Y : B ? B} is a base for Y and |BharpoonuprightY| ? |B|. square
Definition. A subset C of a space X is called discrete if for each c ? C there is
an open set Uc so that Uc ? C = {c}. We define the extent of X, denoted e(X) by
e(X) = sup{|C| : C is closed and discrete subset of X}+?.
Definition. Let A be a set and ? be a cardinal less than ? ? |A|Then define [A]? = {C ?
P(A) : |C| = ?}, [A]?? = {C ? P(A) : |C| ? ?} and [A]<? = {C ? P(A) : |C| < ?}.
Definition. A tree is a partial order , ?T,??, so that for each x ? T, {y ? T : y < x}
is well ordered by <. If x ? T, then we call x a node and the height of x in T, denoted
ht(x,T), is type({y ? T : y < x}). For each ordinal ?, the ?th level is {x ? T :
ht(x,T) = ?}. The height of T is the least ? so that ?th level is empty. A chain in T is
a set C ? T which is totally ordered by <, and an antichain is a set A ? T, such that
?x,y ? A(xnegationslash=y ? (x negationslash? y ?y negationslash? x)). A branch of a tree is a maximal chain. If a node t
is contained in a branch b, we write t ? b. If a node t is of height ?, then we say that t
is ?-branching, where ? is a cardinal, if |{x ? T : ht(x,T) = ?+ 1 and t < x}| = ?.
There are two specific types of trees that we discuss.
Definition. For any regular ?, a ?-Aronszajn tree is a tree such that every chain and
level is of cardinality < ?. A ?-Suslin tree is a tree T so that |T| = ? and every chain
and antichain of T has cardinality < ?.
3
Let T be a tree so that the 0th level has cardinality 1. If each node is either 0 or 2-
branching, we can identify the nodes t of T with functions ?t : ? ? 2 where ? = ht(t,T),
as we describe. Let the single node at the 0th level correspond to the empty sequence
and the two nodes of level 1 correspond to (0) and (1). Above those are (0,0), (0,1),
(1,0) and (1,1), et cetera. So a node t at the ?th level corresponds to an element ?t
of ?2. For each node t ? T that is 2-branching, let ?slurabovet (0) and ?slurabovet (1) correspond to
the two nodes above t in the tree ordering. Similar to nodes, a branch b corresponds to
a sequence from ht(b)2. We write b(?) = 0 (respectively 1) if the node of b on level ?
corresponds to a sequence from ?2 ending with 0 (respectively 1).
Let X(T) denote the set of branches of T. We are able to define two different
topologies on X(T). The first topology is a linear order topology. For branches c,d ?
X(T) so that cnegationslash=d, there is a minimal level n so that c(n)negationslash=d(n). We write c < d if and
only if c(n) = 0 and d(n) = 1. Then < orders the points of X(T).
For the second topology, let t ? T and define [t] = {b ? X(T) : t ? b}. Then
{[t] : t ? T} defines a clopen base for a topology on X(T).
These topologies are used on X(T) in different sections of this work. We rely on the
different notation for the basic open sets to make it clear to the reader which topology
is under discussion.
4
Chapter 2
An Example of a Space with a Sharp Base
2.1 Introduction
Definition. A sharp base is a base B such that whenever (Bi)i<? is an injective sequence
from B with x ?
intersectiondisplay
i<?
Bi, then {
intersectiondisplay
i?n
Bi : n < ?} is a base at x.
In [1], Alleche, Arhangel?ski?? and Calbrix introduced and studied sharp bases and
asked if there is a regular space with a sharp base whose product with [0, 1] does
not have a sharp base. Good, Knight and Mohamad [8] claimed to have a Tychonoff
counterexample, but it turns out that their space is not regular. It is not regular because
they added a closed discrete set L to the Baire metric space ?c, in such a way to make
the new space P pseudocompact. Such P cannot be regular: for if it is, one may find
a neighborhood of p ? ?c whose closure misses L. That neighborhood can be assumed
to come from a clopen basis for ?c, and would then be homeomorphic to ?c and be
pseudocompact, a contradiction.
In this chapter we give a modification of the Good, Knight and Mohamad space
which makes the space Tychonoff; instead of taking a union of a closed discrete set with
?c, we add a ?-discrete set to ?c so that the added set is dense in the union. The space
we construct is pseudocompact but not compact, hence not metrizable; we also show
it is not developable. Our space has no isolated points and a sharp base, and for T1
spaces a sharp base is always weakly uniform. Since Heath and Lindgren show that a
T2 space with a weakly uniform base has a G?-diagonal [11], our space has one also. In
[3], it is shown that a pseudocompact space with a G?-diagonal is ?Cech-complete, and
5
that if a space with not more than ?1 isolated points has a sharp base, then it has a
point countable base. Therefore, the space we construct also answers these three other
questions in the negative:
Is every pseudocompact Tychonoff space with a sharp base metrizable? [3]
Is every pseudocompact space X with a G?-diagonal and with a point-countable base
developable? [2]
Is every ?Cech-complete pseudocompact space with a point-countable base metrizable?
[2]
2.2 The Construction of Space P
Let B = ?c and for ? ? <?c define [?] = {g ? B : ? ? g}. We also denote
? ? n+1c by (?0,?1,???,?n), where ?(i) = ?i. By ?1 ? ?2 we mean that ?1 and ?2 are
incompatible (i.e. the two finite partial functions disagree at a point in both domains).
a1
a1
a1
a1
S?(0)
a0
a0
a0
a0
S?(1)
a24a24a24a24
S?(2)
a88a88a88a88
a88 S?(3)a64a64
a64 S?(4)
a112a112a112
??
Figure 2.1: A typical S? with root ??.
Define S to be the collection of elements of ?(<?c) subject to these two conditions:
1. For all S ? S there exists a ks < ? and a ?s ? <?c such that whenever ? ? S,
?harpoonuprightks = ?s. This ?s will be called the root of S.
6
2. Whenever ?1 and ?2 are distinct elements of S, ?1(ks)negationslash=?2(ks).
Let S = {S? : ? < c}, and let the root of S? be ??. Define T? ? ?(<?c) so that
T = {T? : ? < c} has these three properties:
(i) for i negationslash= j, T?(i) ? T?(j)
(ii) if ?,? < c, ? negationslash= ?, with T? and T? defined, then ranT? ?ranT? = ?, and
(iii) for ?,? < c, ? negationslash= ?, T? and T? defined, if T?(i) ? T?(j), then whenever jprimenegationslash=j,
T?(iprime) ? T?(jprime) for all iprime < ?.
Assume for ? < ? we have either constructed a T? ? ?(<?c) subject to the conditions
above or we have not constructed a T? at all. Now we define T?. Choose a ? ? c not in
uniondisplay
{ranT?(j) : ? < ?,j ? ?}. Then for each i ? ? let Sprime?(i) = S?(i)slurabove(?). The sequence
(T?(i))i<? will be a subsequence of (Sprime?(i))i<?, so the fact that no previous T? contains a
finite partial function with ? in the range will yield property (ii) for T?. In addition, the
fact that the elements of Sprime? are pairwise incompatible will make the elements of T? also
incompatible, satisfying property (i). So we need only concern ourselves with property
(iii).
Case 1. Suppose there exists some ? < ? for which T? was defined, such that for
infinitely many j there is some i ? ? with S?(i) ? T?(j). If this is the case, do not define
T?.
Case 2. If for each ? < ? there are at most finitely many j for which S?(i) ? T?(j)
for some i, we will define a T?.
Suppose that for i ? k we have already selected a sequence of natural numbers
0 = n0 < n1 < ??? < nk and defined T?(i) = Sprime?(ni).
7
There are at most finitely many different finite partial functions f such that f ?
T?(i) for some i ? k. The second induction condition implies that there are at most
finitely many ? < ? with such an f in the range of T?. List these as ?(0),...,?(m).
We have assumed that for each ? < ?, there are at most finitely many j for which Sprime?(i)
extends T?(j) for some i. Using this fact, we see that for each ?(p) there is a jp such that
for all j ? jp, Sprime?(i) does not extend any T?(p)(j). Then define nk+1 = max{jp : p ? m}
and T?(k+1) = Sprime?(nk+1). To check property (iii), suppose that ? < ? and T?(k) ? T?(j)
for some j,k < ?. Assume that k is the least possible for which there exists such a j.
Then ? = ?(p) for some p ? m in the above construction. Since nk+1,nk+2,... are all
greater than jp, T?(i) cannot extend T?(jprime) for any jprimenegationslash=j and any i, so we have property
(iii). Note that T?(j) could not extend T?(i) because ? ? ranT?(i)\ranT?(j). Indeed,
from this and from (iii) with conditions 1 and 2 of S ? S it is easy to see that we have
the following.
(iv) If ?? = ??, then T?(j) and T?(i) are compatible for at most one pair (i,j) in
? ??.
[??]
...{[S?(i)]}i<?
...{[T?(i)]}i<?
Figure 2.2: An open set [??] and the sequence of disjoint open sets {[T?(i)]}i<?.
8
Choose L disjoint from B such that L = {s? : T? is defined}. Let the root of s?
refer to ??. Let P = B ?L.
For ? ? <?c, let B(?) = [?]?{s? : ?? ? ?} and let Bn(s?) = {s?}?
uniondisplay
m?n
([T?(m)]?
{s? : ?? ? T?(m)}). These will be the basic open sets for P, and call the collection of
them B.
2.3 Verifying Properties of P
First, we will observe some properties of B.
Theorem 2.1 The collection B has the following properties.
(a) For ?1,?2 ? <?c, ?1 ? ?2 iff B(?1) ? B(?2) = ? and if ?? ? ?? then Bn(s?) ?
Bm(s?) = ?.
(b) ?1 ? ?2 iff B(?2) ? B(?1), and if ?? ? ?, then for each n < ?, Bn(s?) ? B(?).
(c) Suppose B(?) ? Bn(s?)negationslash=?. Then ? ? ?? or ?? ? ?. If ? ? ?? then B(?) ?
Bn(s?) = Bn(s?). If ? supersetnoteql ??, then the intersection is either B(?) or B(T?(m))
for some m ? n. Finally, if B(?) ? Bn(s?) then for some m ? n we have
B(?) ? B(T?(m)).
(d) If Bn(s?) ?Bnprime(s?prime)negationslash=? and ??prime ? ??, then the intersection is either Bn(s?) or a
set of form B(?), for some ? ? {T?(m),T?prime(mprime) : m ? n,mprime ? nprime}. In particular,
the latter holds if ??prime = ??.
Proof. (a) Suppose that ?1 ? ?2; then there is no point of B nor any finite partial
function that could extend both ?1 and ?2. If s? ? L is in B(?1) ? B(?2) then ??
9
extends both, contradiction. Suppose for the reverse, that B(?1)?B(?2) = ?; then since
[?1]?[?2] is contained in this set, it is clear that ?1 ? ?2.
Now if the roots of s? and s? are incompatible then each pair of extensions of the
roots will be incompatible, hence B(T?(nprime))?B(T?(mprime)) = ? for each nprime ? n and mprime ? m.
Further, s? ? Bm(s?) implies that ?? extends ??, which has been assumed to be not the
case. So Bn(s?)?Bm(s?) = ?.
(b) Clear from the definition of B(?) and s?.
(c) Suppose that B(?)?Bn(s?)negationslash=?. Since Bn(s?) ? B(??) we have ? negationslash? ??, by (a).
If ? ? ??, then for each m ? n, ? ? T?(m) and s? ? B(?), so Bn(s?) ? B(?).
Suppose ? negationslash? ??; then for some m ? n, B(?)?B(T?(m))negationslash=?, while property (i) of T
implies that B(?)?B(T?(k)) = ? for knegationslash=m. By (a) and (b), one of B(?) and B(T?(m))
is contained in the other, and the intersection is simply the contained set. This implies
the last sentence of (c).
(d) Suppose Bn(s?)?Bnprime(s?prime)negationslash=?, where s?negationslash=s?prime. If ??prime subsetnoteql ?? then s?primenegationslash?Bn(s?) and
[T?prime(j)]?[??]negationslash=? for at most one j ? ?. Therefore, Bn(s?)?Bnprime(s?prime) = B(T?prime(j))?Bn(s?)
for some j ? nprime. Now the rest follows from (c).
If ?? = ??prime, then the conclusion follows from condition (iv). square
Proposition 2.2 The base B is a clopen base for P.
Proof. Notice that the properties show immediately that B is a base. To see that
Bn(s?) is closed consider s? ? L\Bn(s?). Suppose that Bj(s?) meets Bn(s?), where j
is sufficiently large that s?negationslash?Bj(s?). Then by c) the intersection is one of B(T?(nprime)) for
some nprime ? n, B(T?(jprime)) for some jprime ? j, Bj(s?) or Bn(s?).
10
Since s?negationslash?Bn(s?) and s?negationslash?Bj(s?), we know that the intersection cannot be Bj(s?) or
Bn(s?). If the intersection is B(T?(jprime)) then Bjprime+1(s?) misses Bn(s?). So the intersection
has to be some B(T?(nprime)). We have that ?? negationslash? T?(nprime), so there exist jprimeprime < ? so that for
i > jprimeprime we have T?(i) negationslash? T?(nprime). So B(T?(nprime)) ? B(T?(jprime)) or Bn(s?) ? Bj(s?). So then
Bjprimeprime+1(s?) misses Bn(s?).
To see that each limit point of Bn(s?) in B is in Bn(s?), suppose that p is such a
limit point of Bn(s?) not in Bn(s?). Choose k < ? so that pharpoonuprightk negationslash? ??. Then by property
(d), B(pharpoonuprightk)?Bn(s?) = B(T?(m)) for some m ? n. Then for kprime < ? so that kprime > |T?(m)|
we have B(pharpoonuprightkprime)?Bn(s?) = ?.
Lastly, we observe that B(?) is clopen. Since B is dense and the subspace base is
clopen, we only need to turn our attention to limit points of B(?) in L. Suppose then
that s? ? L is a limit point of B(?), not in B(?); then for all n < ?, Bn(s?) meets
B(?). If ?? ? ?, then clearly B(?)?Bn(s?) = ?. If ?? ? ?, then s? is in B(?) which is
contrary to our assumptions. So assume that ? supersetnoteql ?a, then there is at most one T?(m)
that extends ? or is extended by ?. Then Bm+1(s?)?B(?) = ?. square
Proposition 2.3 The base B is sharp.
Proof. Let the injective sequence (B(?i))i<? come from B. If p ? B is contained in every
B(?i) then p ? ?i, so since |?i| must be unbounded, it is clear that {
intersectiondisplay
i?n
B(?i) : n < ?}
is a base at p. If s? is in every B(?i), then ?? extends every ?i, but since |??| is finite,
this is not possible.
Now consider an injective sequence (Bni(s?i))i<?, with nonempty intersection. If
there is an infinite subset J of ? such that the ??i, i ? J, are distinct, then it is
11
easy to see that {B(??i) : i ? J} is a base for a unique point p ? B. Hence, so is
{
intersectiondisplay
i?j
Bni(s?i) : j < ?}, since for each i ? J we have Bni(s?i) ? B(??i).
Next, suppose that s?i = s? for all i in an infinite subset J of ?. Then {Bni(s?i) :
i < ?} is a base at s?, therefore {
intersectiondisplay
i?j
Bni(s?i) : j < ?} is a base at s? too.
The final case, without loss of generality, is when the s?i?s are distinct, but ??i = ?
for all i < ?. Then by (d), pairwise intersections have the form B(?) for some ? in the
range of the corresponding pair from T . By property (ii) of T , {Bni(s?i)?Bni+1(s?i+1) :
i is even, i < ?} consists of distinct B(?)?s. Therefore, this must be a base at some
p ? B, and {
intersectiondisplay
i?j
Bni(s?i) : j < ?} is as well. square
Proposition 2.4 The space P is not compact.
Proof. Consider C0 = {s? ? L : ?? = ?}. Note that P \C0 =
uniondisplay
?<c
B((?)). We intend to
show that the closed set C0 is infinite and discrete. To see that this is a discrete set, notice
that for s? ? C0, the set B1(s?)?C0 can only contain s?. Examine {(??i)i<? : ? < c},
where ??i = ??primeiprime iff both i = iprime and ? = ?prime. Call this collection S0; then this is a subset
of S. Note, that for each S? ? S0 and i < ?, we have that the length of S?(i) is exactly
one. Each T?(j) is constructed to have length at least 2. Therefore, during the induction
that defined T , for each S? ? S0, Case 1 does not hold. Hence, a corresponding T? is
constructed for each S? ? S0. This implies that {s? ? L : ?? = ?} has size at least c. square
A space is called perfect if every open set is F?. A development for a space X is
a sequence (Gi)i<? of open covers so that for each point x ? X the set {G ? Gi : x ?
G,i < ?} is a local base for x. A space with such a development is called developable
and each regular developable space is perfect.
Proposition 2.5 The space P is not perfect, hence not developable.
12
Proof. Let U = P \ C0. We show that U is not F?, and hence P is not developable.
Suppose that {Fj}j<? is a collection of closed sets so that
uniondisplay
j<?
Fj = U. By the Baire
property of B, each [(?)] is Baire. So for all ? < c there is an n? and an [??] =
[(?,?1,???,?n?)] ? Fn?. Choose n0 so that {? : [??] ? Fn0} is infinite. Order {?i}i<? ?
{? : [??] ? Fn0}, then S = (( ??i))i<? ? S, and has the empty set as its root. So an s ? L
was defined as a limit point of S, and ? the root of s is also the empty set. Therefore,
s is a limit point of a closed set, Fn0, which implies that s ?
uniondisplay
?<c
B((?)). So there is a
? < c so that ? ? (?), a contradiction. square
Recall that a Tychonoff space is pseudocompact if every continuous real valued
function is bounded.
Proposition 2.6 The space P is pseudocompact.
Proof. Suppose that ? is an unbounded continuous real valued function on P. Since B
is dense, for each n ? ? there is an xn such that ?(xn) > n. Let D = {xn : n ? ?} and
let?s note that D is closed discrete, hence not compact. If p were a cluster point of D,
then every open neighborhood of p contains infinitely many elements of D. This implies
that ? increases unboundedly over every neighborhood of p, contradicting the continuity
of ?.
Since D is closed and not compact we can find a k < ? such that {xnharpoonuprightk : xn ? D}
is infinite. Choose the minimum such k. Then there is a ? ? <?c and an infinite subset
A of ?, such that xnharpoonupright(k?1) = ? for n ? A, and xn(k?1) is different for these infinitely
many n ? A.
Let D? = {xn : n ? A}. Since ?(xn) > n by continuity of ? there exists jn > k so
that ?(B(xnharpoonuprightjn)) > n. Then for some ? < c, {xnharpoonuprightjn : xn ? D?} is S? and ?? = ?. If s?
13
was not defined then for some ? < ?, T?(j) ? S?(n) = xnharpoonuprightk for infinitely many n. Then
each basic open neighborhood of s? contains infinitely many of the sets B(xnharpoonuprightk). So ?
takes on arbitrarily large values over every neighborhood of s? contradicting continuity.
If s? was defined, then T?(i) was chosen so that T?(i) ? xniharpoonuprightjni for each i ? ?, so
B(T?(i)) ? B(xniharpoonuprightjni). So again, ? takes on large values over every open set containing
s?, contradicting the continuity of ?. square
For a metric space, compact and pseudocompact are equivalent, so P clearly cannot
be metrizable.
The following definition can be found in [5].
Definition. An n-weakly uniform base B for the space X is a base so that given any
subset A of X, the set {B ? B : A ? B} is finite. A < ?-weakly uniform base B is
a base so that given any infinite subset A of X there is a finite subset F of A so that
{B ? B : F ? B} is finite.
The notion of a weakly uniform base, which is due to [11], corresponds to a 2-weakly
uniform base. For n < w < ? it is clear that n-weakly uniform base are m-weakly
uniform, and that each n-weakly uniform base is < ?-weakly uniform. Also, for any T1
space a sharp base is a weakly uniform base.
Lemma 2.7 Let X be a Tychonoff, pseudocompact, non-compact space with no isolated
points which partitions into B?L, and has an n-weakly uniform base B (resp. < ?-weakly
uniform base B). If
(a) B = B1 ?B2 where B1 is a ?-point finite base for B
14
(b) for all x ? L there is a local base {Bn(x) : n < ?} so that n < m implies Bm(x) subsetnoteql
Bn(x) and B2 = {Bn(x) : n < ?,x ? L}
(c) for xnegationslash=y ? L, n, m ? ?, Bn(x)negationslash=Bm(y).
Then X ?[0,1] does not have an n-weakly uniform base (resp. < ?-weakly uniform
base).
Proof. Assume, by way of contradiction, that W is an n-weakly uniform base for
X?[0,1]. Let C be a countable base for [0,1]. For each x ? L, choose Wxn ? W, Bxn ? B
and Cxn ? C so that (x, 12) ? Bxn ? Cxn ? Wxn ? Bn(x) ? [0,1]. Let BC = {B ? B : for
some n ? ? and x ? L, B = Bxn and C = Cxn}.
We claim that BC is point-finite. Suppose not; then there exists an infinite collection
(Bj)j<? from BC that has nonempty intersection. Let y ?
intersectiondisplay
j??
Bj; then there are xj ? L
and nj ? ? so that Bj = Bxjnj and C = Cxjnj. Then {y}?C ?
intersectiondisplay
j??
(Bxjnj ?Cxjnj) ?
intersectiondisplay
j??
Wxjnj .
If xjnegationslash=xk then Bn(xj)negationslash=Bnprime(xk).
There are two cases to consider.
Case 1. There is an infinite J ? ? so that xjnegationslash=xk whenever jnegationslash=k with j,k ? J. Then
{Wxjnj : j ? J} is infinite. Suppose not; then some open W is contained in infinitely
many different Bnj(xj)?[0,1]. That B is n-weakly uniform implies that
intersectiondisplay
j<?
Bnj(xj) is
finite. Let it be F, implying W ? F ?[0,1], which is impossible since X has no isolated
points. Hence {Wxjnj : j ? J} is infinite, and so {y} ? C ?
intersectiondisplay
j??
Wxjnj is a finite set, a
contradiction.
Case 2. There is an infinite K ? ? so that xj = xk = x for j,k ? K. Then the set
{nk : k ? K} is infinite, since the Bxknk are distinct. Again, {y}?C ?
intersectiondisplay
k?K
(Bxknk ?Cxknk) ?
15
intersectiondisplay
k?K
Wxknk =
intersectiondisplay
k?K
Wxnk. Once again, this must be finite, so we have the same contradiction
as in Case 1.
Therefore, BC is point finite. Let Bprime =
uniondisplay
C?C
BC; then B1 ?Bprime is a ?-point finite base
for X. All pseudocompact spaces with ?-point finite bases are metrizable [17]. However,
all metrizable pseudocompact spaces are also compact, contradiction. square
Theorem 2.8 The product P ?[0,1] does not have a sharp base.
Proof. We use Lemma 2.7. Let B1 =
uniondisplay
n<?
{B(?) : |?| = n} and B2 = {Bn(s?) : s? ?
L,n < ?}. Since for T1 spaces sharp implies weakly uniform, this means that P ?[0,1]
has no sharp base. square
16
Chapter 3
Open-in-finite bases
3.1 Introduction
Definition. A base for a topological space X is open-in-finite (OIF) if every nonempty
open set is contained in at most finitely many elements of the base. If space X has an
OIF base then we say X is an OIF space.
In [4] the base property OIF was introduced and the following question was asked:
Is every dense subspace of a regular OIF space OIF? This work came out of an attempt
to answer that question. In the next section, we examine the relationship between the
cardinal functions of dense subsets of an OIF space and the cardinal functions of the OIF
spaces. In section 4 we find properties that a space must have if it is densely contained
in an OIF space.
We show that every left or right separated dense subset of an OIF space must be
OIF. We also give a necessary and sufficient condition for when dense subsets of OIF
spaces are themselves OIF.
In the last section of this chapter we introduce some covering properties that are
related to the base properties OIF, ?-OIF and < ?-weakly uniform. We show that for
GO spaces these properties are equivalent to each other and to paracompact. We also
show that for Moore spaces the covering properties are equivalent to the corresponding
base property.
17
3.2 OIF Spaces and Cardinal Functions
Next, we present some results regarding cardinal functions and OIF spaces. The
definitions of these cardinal functions can be found in Chapter 1.
Theorem 3.1 If X is OIF, then w(X) = piw(X) and ?(X) = pi?(X).
Proof. For any space X, we know w(X) ? piw(X), because any base is a pi-base.
Suppose w(X) > piw(X). Let B be any base of X, and let A be a pi-base of size piw(X).
Then if each A ? A is in finitely many members of B then |B| = |A|. So some A ? A
is in infinitely many members of B and since B was arbitrary, X is not OIF. Hence
w(X) = piw(X) for any OIF space.
Let p ? X, where X is an OIF space. Suppose V is a local base for p where
the elements of V are taken from an OIF base. Then let U be a local pi-base for p of
cardinality pi?(p,X). Any local base is a local pi-base so pi?(p,X) ? ?(p,X). Suppose
that |U| = pi?(p,X) < ?(X) = |V|. Then each element of V must contain an element of
U. For each V ? V assign UV ? U so that UV ? V. Then some U ? U is assigned to
infinitely many V ? V, which means that V is not an OIF collection. So for each p ? X
we have pi?(p,X) = ?(p,X). Therefore, pi?(X) = ?(X). square
Theorem 3.2 If X is a regular OIF space with OIF dense subspace Y, then piw(Y) =
w(Y) = piw(X) = w(X).
Proof. Suppose that A is a pi-base of Y. Then let Aprime = {(A)? : A ? A}. We show that
Aprime is a pi-base for X. Let U be an open set in X; since Y is dense, U ? Y negationslash= ?. Let
x ? U ?Y; then by regularity there is a V open in X so that x ? V ? V ? U. Also,
V ?Y is open in Y, so there an A ? A so that A ? V. So (A)? ? V ? U. We know that
18
(A)? negationslash= ?. So Aprime is a pi-base of X of the same cardinality as A. Thus piw(X) ? piw(Y).
We already know that piw(Y) = w(Y) ? w(X) = piw(X), so equality holds. square
In Section 3.4 we will see that if X is an OIF space then all the dense subspaces of
X have the same weight as X.
Examples. The space ?? is not OIF because the weight of ?? is c and the weight
of ? is ?. Also, ?R is not OIF because the weight of R is ? while the weight of ?R is c.
Corollary 3.3 If X is a completely regular space and C is an infinite closed discrete
subspace of X such that 2|C| > w(X), then ?X is not OIF.
Proof. If C is an infinite closed discrete space, then ?C has weight 2|C| and w(?C) ?
w(?X). Then w(?X) negationslash= w(X), so ?(X) is not OIF. square
This means that if ?X has an OIF base and w(X) < 2?, then X is countably
compact. Recall that ?X is only defined for completely regular X, so if X is also second
countable, then X is metrizable. For metrizable spaces, countably compact is equivalent
to compact. Therefore if X is a completely regular second countable space for which ?X
is OIF, then X is compact. So if ?X and X are distinct OIF spaces, it must be the case
that w(X) > ?.
In [4] the authors noted that if X and Y are OIF then X ? Y is OIF. Also, the
question was raised : If X ? X is OIF does that imply that X is OIF? In connection
with that question we explore the relationship between the cardinal functions of X and
X ?X.
Proposition 3.4 If X ?X is OIF, then piw(X) = w(X) = w(X ?X) = piw(X ?X),
and pi?(X) = ?(X) = ?(X ?X) = pi?(X ?X)
19
Proof. Assume X ?X is OIF and let x ? X. Then piw(X) ? w(X) = w(X ?{x}) ?
w(X ?X) = piw(X ?X). If A is a pi-base of X, then A?A is a pi-base of X ?X, so
piw(X ?X) ? piw(X). So the claimed equality holds.
Suppose (x1,x2) is a point in X ?X. Let A1 and A2 be open neighborhood bases
at x1 and x2 respectively, with each Ai having size pi?(X). Then A1 ? A2 is a local
base at (x1,x2). It follows that pi?(X ? X) ? pi?(X), while it is already known that
pi?(X) ? ?(X) ? ?(X ?X) = pi?(X ?X). square
This last result is an easy observation.
Proposition 3.5 Let X be a space with OIF base B and let the character of X be ?.
Then B is a point-? ? base.
Proof. Suppose that x is a point in X contained in more than ? many elements of
an OIF base B. There is a local base at x of size cardinality less than or equal to ?.
Each of the ? ?+ sets in B containing x must contain some set from the local base.
Therefore some element of the local base is contained in infinitely many elements of B,
contradiction. square
Corollary 3.6 Suppose X is a regular, first countable, countably compact space. If X
is OIF, then X is a compact metrizable space.
3.3 Stronger Base Properties
Here we primarily discuss the property called ?-OIF, but we shall also refer to a
generalization of weakly uniform bases. The ?-OIF property is of interest because ?-OIF
implies OIF, each example of an OIF space in [4] is also a ?-OIF space, and every dense
subspace of a ?-OIF space is ?-OIF.
20
Definition. A base B for space X is called a ?-OIF base if every infinite intersection
from B is nowhere dense, and X is called a ?-OIF space if X has a ?-OIF base.
Proposition 3.7 If B is a ?-OIF base then B is an OIF base.
Proof. Suppose B is a base as above. Let {Bi : i < ?} be a subset of B. Since
intersectiondisplay
i<?
Bi
is nowhere dense,
intersectiondisplay
i<?
Bi must have empty interior. So it must be that every open set is
contained in at most finitely many members of B. square
Definition. A sequence of sets {Vi : i < ?} is called strongly decreasing if Vi+1 ? Vi for
each i < ?.
Lemma 3.8 Any strongly decreasing sequence {Vi : i < ?} of open sets from an OIF
base is a ?-OIF collection.
Proof. Suppose that {Vi : i < ?} is strongly decreasing and a subset of some OIF base.
Then consider C =
intersectiondisplay
i<?
Vi. If x ? C, either x ?
intersectiondisplay
i<?
Vi or x is a limit point of this set. In
the second case, for each i < ?, x ? Vi+1. Therefore, x ? Vi for each i < ?. So C ? Vi
for each i. It follows now that C must have empty interior. square
Definition. A collection U of open sets is called regular if for each U ? U there is a
Uprime ? U so that Uprime ? U.
Example. There is a regular OIF collection of open sets in R that is not ?-OIF. Let
Un =
parenleftbigg
2? 1n,2 + 1n
parenrightbigg
for each n ?N. Then U = {Un : n ?N} is a regular OIF collection
of open sets in R. Let An =braceleftbig k3n : 1 ? k < 3nbracerightbig. Then define U?n = (Un ?(0,1))\An.
21
Then
parenleftBiggintersectiondisplay
n?N
U?n
parenrightBigg?
= ?, since it misses the set
uniondisplay
n?N
An which is dense in (0, 1).
Therefore U? = U ? {U?n : n ? N} is OIF. Also, Un+1 ? Un ? U?n, so U? is a regular
collection. However,
parenleftBiggintersectiondisplay
n?N
U?n
parenrightBigg?
= (0,1), and so U? is not ?-OIF.
Theorem 3.9 An OIF base B of X is a ?-OIF base if and only if for every dense subset
Y of X, the trace of B on Y is an OIF base for Y.
Proof. For the forward direction, let X be a space with a ?-OIF base B. Let Y be a
dense subset of X and let BharpoonuprightY = {B ? Y : B ? B}. Suppose that Ci ? BharpoonuprightY for each
i ? ? and let G =
intersectiondisplay
i<?
Ci. Suppose G? negationslash= ? in Y. Then G? negationslash= ? in X, and G? ?
intersectiondisplay
i<?
Ci
?
,
which contradicts our assumption that B is ?-OIF. Thus, BharpoonuprightY is ?-OIF.
For the reverse direction, let B be an OIF base so that for each dense subset Y of
X, BharpoonuprightY is OIF. Suppose that B is not ?-OIF, and let Bn ? B for each n ? ? so that
W =
intersectiondisplay
n<?
Bn
?
negationslash=?.
Consider Y prime =
intersectiondisplay
n<?
Bn?(X\
intersectiondisplay
n<?
Bn). This Y prime is dense, for if U is an open set of X,
then either U ?
intersectiondisplay
n<?
Bn is empty or it isn?t. If it is empty then U ? (X \
intersectiondisplay
n<?
Bn) ? Y prime.
If the intersection is not empty then U meets Y prime. Next we add points to Y prime to form Y
so that {Bn ?Y : n < ?} is infinite.
Choose two disjoint open sets in W, say A1 and A2. Next for each {m,n} ? [?]2
choose xm,n ? (Bm \Bn) ? (Bn \Bm). Then let P1 = {{m,n} : xm,nnegationslash?A1} and P2 =
[?]2 \P1. Ramsey?s Theorem states that if we partition [?]2 into P1 and P2, then there
is an infinite P ? ? such that [P]2 is contained in one partition. Let Y = Y prime ?{xm,n :
{m,n} ? P}. We show that {Bn ?Y : n < ?} is infinite. Consider Bn ?Y and Bm ?Y
22
with {m,n} ? P. The point xm,n is contained in Y and is contained in only one of Bn
and Bm, therefore Bn ?Ynegationslash=Bm ?Y.
If P ? P1 then let U = A1 and let U = A2 otherwise. Note that U ? W \{xm,n :
{m,n} ? P} and W ?(X\
intersectiondisplay
i<?
Bi) = ?, and therefore, U?Y ?
intersectiondisplay
i<?
Bi?Y =
intersectiondisplay
i<?
(Bi?Y),
contradiction. square
Corollary 3.10 If X has a ?-OIF base then every dense subset of X has a ?-OIF base.
It is known that every metacompact Moore space has a OIF base; it is in fact true
that every metacompact Moore space has a ?-OIF base.
Theorem 3.11 Every metacompact Moore space has a ?-OIF base.
Proof. Let G = (Gi)n<? be a development for metacompact Moore space X, so that
Gn+1 refines Gn, each Gn is point-finite. Then consider {Vi : i < ?} ? uniontexti??G; with
intersectiondisplay
i<?
Vinegationslash=?. If some Gn contains infinitely many sets from {Vi : i < ?}, then because Gn
is point finite,
intersectiondisplay
i<?
Vi = ?, contradiction. Therefore, we have that {n : there exists i < ?
so that Vi ? Gn} is cofinal in ?. Therefore, {Vi : i < ?} either has empty intersection or
contains a base for some point of
intersectiondisplay
i<?
Vi. This means that |
intersectiondisplay
i<?
Vi| < 2, and therefore is
nowhere dense. square
In [4], the authors show that if a space has an OIF base of regular open sets, then
each dense subset of that space is OIF. The OIF property is also known to be productive
and the third theorem below shows the same is true for ?-OIF.
Theorem 3.12 An OIF base of regular open sets is a ?-OIF base.
23
Proof. Suppose B is a regular open OIF base for space X.
Consider
intersectiondisplay
i<?
Bi
?
?
parenleftBiggintersectiondisplay
i<?
Bi
parenrightBigg?
?
intersectiondisplay
i<?
Bi? =
intersectiondisplay
i<?
Bi. So each G? set from B is nowhere
dense, hence B is ?-OIF. square
Theorem 3.13 Suppose X is a space with an OIF base B. Then X has a ?-OIF base
if and only if for all B ? B there is an collection CB = {Vi : i ? IB} ? B so that
1. Vi? ? B for each i ? IB,
2. B =
uniondisplay
i?IB
Vi, and
3. for each infinite F ? IB, we have
intersectiondisplay
F
Vi
?
= ?.
Proof. Take X, B and CB as above and let C =
uniondisplay
B?B
CB. Then this C is a base for X.
Suppose that {On : n < ?} ? C and consider
intersectiondisplay
n<?
On. For each n < ?, let Bn be an
element of B so that On ? CBn. If {Bn : n < ?} is finite then infinitely many On are from
the same CB and therefore the intersection of them is nowhere dense. If {Bn : n < ?}
is infinite, then observe that
intersectiondisplay
n<?
On
?
?
parenleftBiggintersectiondisplay
n<?
On
parenrightBigg?
?
parenleftBiggintersectiondisplay
n<?
Bn
parenrightBigg?
= ?. Therefore, C is
?-OIF.
Now suppose that X has a ?-OIF base C and let B be a base for X. Then for each
B ? B let CB be any collection from C of sets contained in B whose union is B. square
Corollary 3.14 If X is a regular OIF space that is hereditarily metacompact, then X
has a ?-OIF base.
Proof. Let X be such a space and let B be an OIF base for X. For each B ? B and
each x ? B let Vx,B be a set from B so that x ? Vx,B ? Vx,B ? B. Then for each B the
24
collection VB = {Vx,B : x ? B} covers B. Find a point finite open refinement CB of each
VB. Then CB satisfies the conditions of the previous theorem. square
Theorem 3.15 If for each ? < ? the space X? is ?-OIF, then
productdisplay
?<?
X? is ?-OIF.
Proof. Let B? be a ?-OIF base for X?. For F ? [?]<?, a basic open set in the
product would be
intersectiondisplay
??F
pi?1? (A?) where A? ? B?. Suppose that the open set G =
intersectiondisplay
i<?
?
?intersectiondisplay
??Fi
pi?1? (A?,i)
?
?
?
where
??
?
intersectiondisplay
??Fi
pi?1? (A?,i)
??
?
i??
is an infinite collection of basic open
sets and each A?,inegationslash=X?. Let Gprime be a basic open set contained in G. Then for all but
finitely many ?, pi?(Gprime) = X?, and therefore pi?(G) = X? for all but finitely many ?.
Hence, there is a finite set F so that Fi ? F for each i ? ?. To verify this claim, aiming
for a contradiction, suppose that
uniondisplay
i<?
Fi is infinite. Then let ? < ? so that ? ?
uniondisplay
i<?
Fi and
pi?(G) = X?. Then let B? ? B? be a set that misses A?. Then
productdisplay
?<?
X??B? ?
productdisplay
?<?<?
X?
is an open set that misses
intersectiondisplay
i<?
?
?intersectiondisplay
??Fi
pi?1? (A?,i)
?
?, contradiction.
There is a ? ? ? so that ? ? Fi for infinitely many i < ? and {A?,i : ? ? Fi}
is infinite; else
??
?
intersectiondisplay
?i?Fi
pi?1a (A?i)
??
?
i??
is a finite collection of open sets. But notice that
{A?i : i ? I} ? B?.
Then we have pi?(G) ? pi?
?
??intersectiondisplay
i<?
?
?intersectiondisplay
??Fi
pi?1? (A?,i)
?
?
??
?? ?
parenleftBiggintersectiondisplay
i?I
A?,i
parenrightBigg?
. This contra-
dicts that B? is a ?-OIF base. square
Recall from Chapter 2 that an n-weakly uniform base B for a space X is a base such
that given any subset A of X with |A| = n, the collection {B ? B : A ? B} is finite.
Call B < ?-weakly uniform if, given any infinite set A, there is a finite subset F ? A
with {B ? B : F ? B} finite.
25
Theorem 3.16 Suppose X has a < ?-weakly uniform base that is point finite at every
isolated point in X. Then X is ?-OIF.
Proof. Let B be a base for X, as above. Suppose {Bi : i < ?} ? B.
Consider
intersectiondisplay
i<?
Bi. If this intersection is finite, then it is closed and does not contain
an open set because any isolated point is in only finitely many elements of the base.
Therefore, it is nowhere dense.
If
intersectiondisplay
i<?
Bi is infinite, then there exists a finite subset F of
intersectiondisplay
i<?
Bi so that {B ? B :
F ? B} is finite, contradiction. square
Theorem 3.17 There is a space that has a ?-OIF base but does not have a < ?-weakly
uniform base.
Proof. The product P?[0, 1], where P is the space from Chapter 2, is such a space. The
space P ?[0,1] does not have a < ?-weakly uniform base by Lemma 2.7, but Theorems
3.16 and 3.15 shows it does have a ?-OIF base. square
Example. If ? is an uncountable cardinal, then [0, 1]? will be ?-OIF but not
metrizable.
Definition. A space is neighborhood OIF at a point x if there is a local base for x that
is an OIF collection. We call a space neighborhood OIF if it is neighborhood OIF at each
point x ? X.
A space X is neighborhood ?-OIF at point x if there is a local base for x that is a
?-OIF collection. We call X a neighborhood ?-OIF space if it is neighborhood ?-OIF at
each point x ? X.
26
Naturally, each ?-OIF space is neighborhood ?-OIF, and each neighborhood ?-OIF
space is neighborhood OIF.
Proposition 3.18 If X is a regular space that is neighborhood OIF at x, then X is
neighborhood ?-OIF at x.
Proof. Let U be a OIF local base at x, and order U = {Ui : i < ?}. For each i < ?,
if possible, let Uprimei be a member of U so that x ? Uprimei, Uprimei ? Ui and Uprimejnegationslash=Uprimei for each
j < i. If for some i < ? we are unable to find a Uprimei that has not already been assigned
to a previous member of U, then do not define Uprimei. Then Uprime = {Uprimei : Uprimei defined and
i < ?} is also a local OIF base at x. It is a ?-OIF collection, for J ? [?]? and consider
intersectiondisplay
j?J
Uprimej
?
?
?
?intersectiondisplay
j?J
Uprimej
?
?
?
?
?
?intersectiondisplay
j?J
Uj
?
?
?
= ?. square
Corollary 3.19 All regular neighborhood OIF spaces are neighborhood ?-OIF spaces.
Corollary 3.20 If Z is a regular neighborhood ?-OIF space, then each dense subspace
of Z is a neighborhood ?-OIF space. In fact, each dense subspace of a regular OIF space
is neighborhood ?-OIF.
Proof. By the proof of Theorem 3.9, we know that each ?-OIF collection in Z will have
an OIF trace in any dense subspace. By Proposition 3.18, each of these OIF local bases
in the dense subset will contain a ?-OIF local base. square
Corollary 3.21 No non-trivial regular P-space can be embedded in a regular OIF space.
Proof. Suppose that X is a non-trivial regular P-space, and let x ? X be any non-
isolated point. Then let Bx be any local base for x in X. Let {Bi : i < ?} ? Bx;
27
then x ?
intersectiondisplay
i<?
Bi. So
intersectiondisplay
i<?
Bi is nonempty, and as a G? subset of a P-space it is also
open. Therefore Bx is not an OIF collection. Thus X is not neighborhood OIF at any
non-isolated point. square
3.4 The maltesecross Property
Recall that the question that motivates this research is: Is every dense subspace
of a regular OIF space OIF? While trying to answer this question we considered the
possibility of creating a counterexample by embedding a space that is known not to be
OIF in a space that is regular and OIF. Naturally, this cannot be done with an arbitrary
space. This section contains conditions under which a space cannot be densely embedded
in a regular OIF space. In this section we prove that every left (or right) separated space
that is dense in an OIF space is OIF, and every space contains left separated dense
subspaces. Therefore, if X is a space that is not OIF but can be densely embedded in
a regular OIF space, then each dense left separated subspace of X is OIF. Later in this
section we examine some of the properties that must be enjoyed in order for space Z
and some dense subspace X of Z to both be OIF. This section ends with a necessary
and sufficient condition for a dense subset X of an OIF space Z to be OIF.
The following maltesecross property is necessary for a space to be a dense subspace of a regular
OIF space.
maltesecross: There exists a base B so that for each point-selection p : B ? X with p(B) ? B there
is a gp : B ? B so that p(B) ? gp(B) ? gp(B) ? B and ran(gp) is OIF.
Proposition 3.22 If X is a dense subset of a regular OIF space then X has the maltesecross
property.
28
Proof. Suppose that X is a dense subset of a regular OIF space, Z. Let B be an OIF
base for Z, then BharpoonuprightX is a base for X. Let p : BharpoonuprightX ? X be a point-selection. For each
p(B?X) find Bprime ? B so that p(B?X) ? Bprime ? Bprime ? B, where the closure in taken in Z;
this is possible since Z is regular. Then define gp : BharpoonuprightX ? BharpoonuprightX by gp(B?X) = Bprime?X.
Aiming for a contradiction suppose that ran(gp) is not OIF. Then there is a V ? B
and {Bi : i < ?} so that V ?X ? gp(Bi ?X) = Bprimei ?X for each i < ?. Since Bprimei ? Bi,
we see that V ? Bi for i < ?. This contradicts the assumption that B is OIF. square
The maltesecross property arose from an examination of the proofs that the following two
examples cannot be densely embedded in a regular OIF space.
Example 1. Tangent-Disk space. The underlying set is X = {(x,y) ? R2 : y ? 0}.
By Bn(x,y) we denote the Euclidean ball of radius 1n centered at (x,y). For each
(x,0) we denote Dn(x,0) = {(x,0)} ? Bn
parenleftbigg
x, 1n
parenrightbigg
. Then B = {Bn(x,y) : y > 0 and
n ?N}?{Dn(x,0) : n ?N} is a base for the Tangent-Disk topology.
a45a27
a54
a114
a112 a112
a112 a112
a112a112
a112a112a114
a112a112
a112a112
a112 a112
a112 a112
a112
a112
a112
a112
a112
a112
a112
a112
Figure 3.1: The tangent-disk space and some open sets
Let B be any base for X and define p : B ? X so that if there exists a unique
(x,0) ? B then p(B) = (x,0). This is possible since for each Dn(x,0) there is an element
of B that contains (x,0) and is contained in Dn(x,0). For other B ? B, let p(B) be
arbitrary. Then for any gp : B ? B so that p(B) ? gp(B) ? gp(B) ? B we intend to
29
show that ran(gp) is not OIF. Since the cardinality of the real line is c, the range of
gp has cardinality c. Therefore there is at least one open set from a countable base for
R?(0,?) that is contained in infinitely many members of ran(gp).
Proposition 3.23 If X has maltesecross property then X is neighborhood OIF.
Proof. Suppose that X has the maltesecross property. Let B be the base guaranteed by maltesecross.
We intend to show that X is neighborhood OIF at each point. Pick x ? X and let
Bx = {B ? B : x ? B}. Then let p : B ? X be defined so that p(B) = x for B ? Bx
and let p(B) be arbitrary for Bnegationslash?Bx. Then there is a gp : B ? B so that ran(gp) is OIF.
Therefore, gp(Bx) is OIF and a local base at x in X. square
Example 2. Sequential Fan on ?1 many sequences. Suppose X is the space of
consisting of a point, ?, and ?1 many sequences converging to ?. The points of each
sequence are isolated, and each neighborhood of ? is made up of a tail from each
sequence. Then ? is a point whose local base must always fail to be OIF.
Since all first countable spaces are neighborhood OIF, the Tangent-Disk space de-
scribed above serves as an example of a space that is neighborhood OIF but not OIF.
We observed that if X has the maltesecross property, then X is neighborhood OIF. Further-
more, this implies that each dense subset of a regular OIF space must be neighborhood
OIF. For left or right separated dense subsets we can say more.
Proposition 3.24 Suppose that a regular space X is left or right separated, and has the
maltesecross property. Then X is OIF.
Proof. We will assume that X is a regular left separated space, and our proof will work
analogously for a right separated space.
30
Since the space is left separated, there is a well-ordering, say ?, under which each
{y : y ? x} is closed. Let B be the base from maltesecross and for each x ? X choose Bx ? B so
that x ? Bx ? [x,?). This assignment is one-to-one. For each x ? X let Bx = {B ? B :
x ? B ? Bx}. Then we see that Bx is a local base at x and Bx?Bynegationslash=? if and only if x = y.
Choose an arbitrary point x0 and define p : B ? X by p(B) = x if B ? Bx and p(B) = x0
otherwise. Then by maltesecross, there is a gp : B ? B so that p(B) ? gp(B) ? gp(B) ? B and
ran(gp) is OIF. Therefore, ran(gp) is an OIF base for X. square
The following lemma and its proof are both well known.
Lemma 3.25 Every space contains a left separated dense subset.
Proof. The subset is created recursively. At stage ?, choose x? from X \{x? : ? < ?},
if possible. If X \{x? : ? < ?} = ?, then L = {x? : ? < ?}. This method assures that
each [x,?) in L is open in L, and that L is dense. Clearly this does terminate at some
stage less than or equal to |X|. square
Corollary 3.26 If Z is a regular OIF space, then any dense subspace of Z will have the
same weight as Z.
Proof. Suppose X is a dense subspace of Z. We use Lemma 3.25 to find a left separated
dense subspace Xprime of X, which will also be dense in Z. Then by Proposition 3.24 Xprime is
OIF, and by Theorem 3.1 w(Xprime) = w(Z). Since weight is monotonic, w(X) = w(Z). square
Next, we present a condition that does not depend upon the points of the space,
just the open sets.
maltesecrossmaltesecross: There is B and a g : B ? B satisfying ?negationslash=g(B) ? g(B) ? B and {g(B) : B ? B} is
OIF.
31
We note that maltesecross implies maltesecrossmaltesecross. The next example shows that maltesecrossmaltesecross does not imply maltesecross.
Proposition 3.27 There is a space with the maltesecrossmaltesecross property that does not have the maltesecross
property.
Proof. LetLdenote the set of limit ordinals less than?1. DefineX = (?1+1)\Lwith the
topology inherited from the order topology. Let B = {[?,?1] : ? ? X}?{{?} : ? ? X}
be a base for X. Define g : B ? X by g([?,?]) = {?} and g({?}) = {?}. Then for each
B ? B we have g(B) ? B and g(B) is OIF.
Also, X is not neighborhood OIF at the point ?1, since any local base for ?1 in X
is would contain ?1-many different open sets. Therefore for any local base for ?1 some
isolated point is contained in infinitely many different sets from the local base. Since X
is not neighborhood OIF, X cannot have the maltesecross property. square
Proposition 3.28 If X is a dense subspace of a regular OIF space then X has the maltesecrossmaltesecross
property.
Proof. Since maltesecross implies maltesecrossmaltesecross, this follows from Proposition 3.22 square
Lemma 3.29 If X is a space with uncountable pi-weight and a pi-base A which is an
?1-Suslin tree under reverse inclusion, then X cannot be densely embedded in a regular
OIF space.
Proof. For contradiction, suppose that X and A are as above and is dense in an OIF
space. Then there is a base B for X and g : B ? B so that g(B) ? B and g(B) is an OIF
collection. Define g? : B ? A by g?(B) ? g(B). Because g(B) is OIF, the tree (g?(B),?)
has height ? and |g?(B)| = |g(B)| > ?. Therefore, (g(B),?) has an uncountable level ,
which is an uncountable antichain in (A,?), contradiction. square
32
Theorem 3.30 A Suslin line cannot be densely embedded in a regular OIF space.
Proof. If X if an arbitrary Suslin line we intend to show that there is a collection of
open sets in X that is an ?1-Suslin tree ordered by reverse inclusion. If X is any Suslin
line by Theorem II.4.4 in [14], there is an L which is dense in X, dense in itself and has
no separable open subset. To form L, we define equivalence classes in X by letting x ? y
if (x,y) or (y,x) is a separable subset of X. Then L is the set of ? equivalence classes.
Since X is ccc, only countably many equivalence classes are more than just one point.
For the countably many non-trivial separable intervals, there is a countable collection
of open intervals that is a pi-base for each interval. If the rest of X also has a pi-base
that is an ?1-Suslin tree under reverse inclusion, then the union of these countably many
countable trees with the ?1-Suslin tree is still an ?1-Suslin tree. Therefore, we work with
the line L.
In [14] Theorem II.5.13 describes the construction of an ?1-Suslin tree from a Suslin
line L which is dense in itself and has no nonempty open subset which is separable. In
the construction, the nodes of the tree are open intervals from the line, and the order is
reverse inclusion. Kunen let J denote the collection of all the nonempty open intervals
of L. Then for each ? < ?1 defined J? so that for each ?,
1. the elements of J? are pairwise disjoint,
2.
uniondisplay
J? is dense in L,
3. if ? < ?, I ? J? and J ? J? then either,
a. I ?J = ?, or
b. J ? I and I \Jnegationslash=?.
33
4. if ? < ? for each J ? J? there exists I ? J? so that J ? I.
These conditions ensure that
?
? uniondisplay
?<?1
J?,?
?
? is an ?1-Suslin tree. Now we intend to
show that
uniondisplay
?<?1
J? is a pi-base for L. Suppose that (a,b) is a nonempty open interval of L
and that no element of
uniondisplay
?<?1
J? is contained in (a,b). Then by properties 1 and 2, for each
? < ?1 there are c?,d? and e? so that d? < a < c? < b < e?, and (d?,c?),(c?,e?) ? J?.
Then {(d?,c?) : ? < ?1} is chain of cardinality ?1 in
?
? uniondisplay
?<?1
J?,?
?
?, which is not
possible. Therefore, for some ? < ?1 there is a U ? J? so that U ? (a,b). square
Recall that the space X(T) generated by the tree T has as its points the maximal
chains in the tree and in the following theorem the basic open sets are [?] = {c : ? ? c},
where ? ? T.
Theorem 3.31 The space generated by an ?1-Suslin tree cannot be embedded in a reg-
ular OIF space.
Proof. Let S be an ?1-Suslin tree. Then the base B = {[v] : v ? S} for X(S) is clearly
an ?1-Suslin tree under reverse inclusion. square
Proposition 3.27 gives an example of a regular space that has maltesecrossmaltesecross property and not
the maltesecross property; hence the space is not OIF. This leaves us with the open question: Is
there a regular space that has the maltesecross and is not OIF?
3.5 Covering Properties
The base properties that we have studied have natural associations with covering
properties, which are just as intertwined. We define OIF metacompact, and (n,?)-
metacompact, determine that they are the same for GO spaces but establish that the
34
properties do not necessarily coincide even if the space is required to be monotonically
normal. We also define ?-OIF metacompact and find some conditions for which these
covering properties imply the base property.
The following definition can be found in [5].
Definition. A space X is n-metacompact if every open cover U has an open refinement
V so that for each A ? X such that |A| = n, then |{V ? V : A ? V}| < ?. A space is
< ?-metacompact if every open cover U has open refinement V so that for each A ? X
such that |A| = ? there is a finite subset B of A so that |{V ? V : B ? V}| < ?.
Thus the well-known property ?metacompact? is the same as 1-metacompact. The
following definitions allow us to generalize these metacompact properties to cardinals
larger than ?.
Definition. Let n < ? and ? be an infinite cardinal. A space X is (n,?)-metacompact if
every open cover U has an open refinement V so that for each A ? X such that |A| = n,
then |{V ? V : A ? V}| < ?. A space X is (< ?,?)-metacompact if every open cover U
has an open refinement V so that for every infinite set A ? X, there is a finite set B so
that |{V ? V : B ? V}| < ?.
Therefore, (n,?)-metacompact is n-metacompact and (< ?,?)-metacompact is < ?-
metacompact. Also, (1,?1)-metacompact is the same as meta-Lindel?of. Fix ? and let
n < m < ?. Clearly, (n,?)-metacompact implies (m,?)-metacompact which implies
(< ?,?)-metacompact.
Proposition 3.32 If a space has a < ?-weakly uniform base then it is < ?-metacompact.
35
Proof. For any open cover, any refinement consisting of members of the < ?-weakly
uniform base will witness < ?-metacompactness. square
Definition. A spaceX is OIF-metacompact if every open coverU has an open refinement
V that is an open-in-finite collection. A space X is ?-OIF-metacompact if V is a ?-open-
in-finite collection.
Since each ?-OIF collection is OIF, we know that ?-OIF-metacompact implies OIF-
metacompact. Also, clearly any OIF space (resp. ?-OIF space) will be OIF-metacompact
(resp. ?-OIF-metacompact).
Proposition 3.33 If X is an OIF-metacompact space with character ?, then X is
(1,?+)-metacompact.
Proof. Let U be an arbitrary open cover, and let V be the OIF-metacompact refinement
of U. Then since the character of X is ?, we know that each x ? X is contained in not
more than ? many elements of V, because if x is contained in ?+ many members of V,
then there is some member of the local base for x that is contained in infinitely many
members of V. square
Corollary 3.34 Every space that is OIF-metacompact and first countable is metaLin-
del?of.
Definition. A linearly ordered space is a pair (X,?) so that ? is a linear order on the
set X and {(a,b) : a,b ? X,a ? b}?{(??,a) : a ? X}?{(b,?) : b ? X} is a base for a
topology on X. This topology is referred to as the order topology. A generalized ordered
space, or GO space, is a subspace of a linearly ordered space.
36
Proposition 3.35 For GO spaces OIF-metacompact implies paracompact.
Proof. Let U be an open cover of X an OIF-metacompact space, and let V be an
OIF-refinement of U. We may assume that V consists of convex subsets of X. Then by
Bennet and Lutzer [6] we may assume that V is a ?-point-finite collection. In fact, they
show that if V0 = {V ? V : |V| = 1} and for n ? 0, Vn+1 is the collection of maximal
subsets of V \
uniondisplay
0?k?n
Vk, then each Vn is star-finite (e.g. each member of Vn meets only
finitely many other members of Vn). Then V0 ?V1 covers the same set as all of V and is
locally finite. square
Proposition 3.36 For GO spaces (< ?,?)-metacompact implies (1,?+)-metacompact.
Proof. Let X be a (< ?,?)-metacompact GO space, and let ? be the order on X.
Let U be an open cover of X. Then let V be a refinement of U witnessing (< ?,?)-
metacompact. We may assume that the elements of V are convex. We define V0 and V1,
V0 = {V ? V : |V| = 1} and V1 = V \V0. Obviously, V0 is point-finite.
We wish to show that each point of X is contained in only ? many elements of V1.
For contradiction, assume b ? X and that B = {V ? V1 : b ? V} has cardinality ?+.
Then the sets {inf(V) : V ? B} and {sup(V) : V ? B} cannot both have cardinality
less than or equal to ?. However, all the supremums are ?-greater than b and all the
infimums are ?-less than b.
Assume that the set of supremums is of size ?+, call this set S. Then we will show
that there is a x ? X so that the cardinality of the set of supremums in S ?-less than x
is at least ?.
For each x ? S, let L(x) = {y ? S : y ? x} and R(x) = {y ? S : x ? y}. These
are the ?left? and ?right? sides of x. Notice that L(x) ?R(x) ?{x} = S, therefore for
37
each x we have max{|L(x)|,|R(x)|} = ?+. Next, define L = {x ? S : |L(x)| ? ?} and
R = {x ? S : |R(x)| ? ?}; then clearly L?R = S. Finally, we note that |S \L| ? ?.
If |S \L| = ?+, then because ?+ is regular, no sequence of ? many elements of S \L is
cofinal in S \L. Therefore, for any ? sized subset of S \L, there is an upper bound z
in S \L. But then |L(z)| ? ?, contradiction. Similarly, S \R is also a set of cardinality
not more than ?. Hence, L?Rnegationslash=?.
Choose x ? L?R and let A = {V ? B : x ? sup(V)}. Then because the elements
of A are convex, all contain b and have supremums ?-greater than x. So for all V ? A
we have [b,x] ? V, contradicting (< ?,?)-metacompact.
If the infimums are uncountable, the proof is analogous.
Therefore, V1 is point-? ?. So V0 ?V1 is a (1,?+) open refinement of U. square
It is known that for GO spaces, meta-Lindel?of implies paracompact. Naturally,
metacompact implies both (< ?,?)-metacompact and ?-OIF-metacompact, so we have
the following corollary.
Corollary 3.37 For GO spaces, the following are equivalent:
a) X is paracompact,
b) X is ?-OIF-metacompact,
c) X is OIF-metacompact,
d) X is (< ?,?)-metacompact,
e) X is (n,?)-metacompact for each n < ?.
Proposition 3.38 The (n,?)-metacompact and (< ?,?)-metacompact properties are
hereditary for closed sets.
38
Proof. Let X be either (n,?)-metacompact or (< ?,?)-metacompact, and let C be a
closed subspace of X. Then let U be an open cover of the space C. For each U ? U,
there is an open set Uprime in X so that Uprime ?C = U. Let Uprime = {Uprime : U ? U}?{X\C} be an
open cover of X, then there is Vprime an open refinement of Uprime which is (n,?)-metacompact
or (< ?,?)-metacompact. So V = {V prime ?C : V prime ? Vprime} is an open refinement of U and is
either (n,?)-metacompact or (< ?,?)-metacompact, according to Vprime. square
Corollary 3.39 If X is a monotonically normal (< ?,?)-metacompact space, then X
is paracompact.
Proof. If X is not paracompact, then X contains a closed subspace C that is homeo-
morphic to a stationary subset of an uncountable cardinal. Then since C is closed, C is
(< ?,?)-metacompact. Also, C is a GO space therefore, C is paracompact, contradic-
tion. square
Corollary 3.40 For GO spaces, OIF-metacompact is hereditary for closed sets.
The following theorem can be found in [9].
Theorem 3.41 A space is monotonically normal if and only if for each open set U and
x ? U, one can assign an open set Ux containing x such that Ux ?Vynegationslash=? implies x ? V
or y ? U.
Theorem 3.42 Every space X is a closed subset of a ?-OIF space O(X) such that if X
is monotonically normal, then so is O(X).
Proof. We follow the construction found in [4]. The authors show that O(X) is OIF
and if X is T1, then X is a G? subset of O(X). They also state that O(X) has the same
separation axioms as X, but monotonically normal is not mentioned.
39
Let X be a space with base A and let F[A] be the set of all finite subsets of A. Let
Y = {?p,F? : p ? X and F ? F[A]}. For each U ? A, define f(U) = {?p,F? ? Y : p ?
U ? F} and let S(U) = U ?f(U). The space is O(X) = X ?Y where the topology is
generated by the subbase S = {S(U) : U ? A}?{{x} : x ? Y}.
First, we show that S generates a ?-OIF base. For any collection {Bi : i < ?} ? S
we may consider
intersectiondisplay
i<?
Bi =
intersectiondisplay
j<?
S(Uj) =
intersectiondisplay
j<?
(Uj ?f(Uj)) =
intersectiondisplay
j<?
Uj ?
intersectiondisplay
j<?
f(Uj). Note that
intersectiondisplay
j<?
f(Uj) = ?. Therefore,
intersectiondisplay
j<?
S(Uj) =
intersectiondisplay
j<?
Uj, and because the points of Y are isolated
Y ?
intersectiondisplay
j<?
Uj = ?. However, Y is dense in O(X), so
intersectiondisplay
j<?
Uj
?
= ?.
Now we show that if X is monotonically normal then O(X) is as well. Let O be
open in O(X), and let x ? O. There is a basic open set containing x and contained
in O, which has the form
intersectiondisplay
i<nx
S(Uxi ). If x ? Y, then let Ox = {x}. If x ? X, then
x ?
intersectiondisplay
i<nx
Uxi = Ox. Since X is monotonically normal, there is an assigned set Oxx as in
Theorem 3.41. In O(X) let Ox = S(Oxx) ? O. We need to check that if V is open in
O(X) and y ? V then Ox ?Vynegationslash=? implies x ? V or y ? O.
Assume that Ox ?Vynegationslash=?. There are essentially two cases; either both points x and
y are contained in X or not.
1. If y ? Y, then Ox ?Vynegationslash=? and Vy = {y} implies y ? Ox ? O. Similarly, if x ? Y,
then x ? Vy ? V.
2. If x,y ? X, then let Vy = S(Vyy )?V be assigned to V. Let
intersectiondisplay
i<ny
Vyi be the basic open
set containing y and used to define Vy. Therefore, Ox?Vy = S(Oxx)?S(Vyy )?O?V.
This means that Oxx ?Vyy negationslash=?, therefore y ?
intersectiondisplay
i<nx
Uxi or x ?
intersectiondisplay
i<ny
Vyi . Without loss
of generality, let y ?
intersectiondisplay
i<nx
Uxi . Then y ?
intersectiondisplay
i<nx
Uxi ?
intersectiondisplay
i<nx
S(Uxi ) ? O.
40
So we see that O(X) is monotonically normal. square
Corollary 3.43 There is a monotonically normal space that is OIF, hence OIF-metacompact,
but not paracompact.
Proof. Let X = ?1 with the order topology. Then X is monotonically normal, since it
is an ordered space, but not paracompact. Form O(X), as in Theorem 3.42, and then
O(X) is OIF, hence OIF-metacompact. This cannot be a paracompact space, because
all closed subspaces of paracompact spaces are paracompact, and X is such a closed
subspace. square
This is also an example of a space that is OIF-metacompact but not metacompact
or (< ?,?)-metacompact. For if it were (< ?,?)-metacompact, then X would be as
well, and therefore X would be paracompact.
It is already known that metacompact Moore spaces are OIF and have a n-weakly
uniform base for each n < ?, but we are able to be more precise in the next theorem.
We are able to say, informally, that for Moore spaces ?the base property is the same as
the covering property?.
Theorem 3.44 Suppose X is a Moore space. Then we have the following.
1. X has an OIF base if and only if X is OIF-metacompact.
2. X has an ?-OIF base if and only if X is ?-OIF-metacompact.
3. For n ? 2, X has an n-weakly uniform base if and only if X is (n,?)-metacompact.
Proof. For 1, suppose X has an OIF base B and let U be an open cover of X. Then
refine U with elements of B to find an OIF-refinement of U. This proves the forward
41
direction, we now prove the reverse. Suppose that X has development G = (Gi)i<? and
that X is OIF-metacompact. Then let Gprimeprime0 be an OIF refinement of G0. For Gprimeprimei defined,
let Gprimei+1 = {V ?U : V ? Gi+1,U ? Gprimeprimei }, then take Gprimeprimei+1 to be an OIF refinement of Gprimei+1.
Therefore, Gprimeprime = (Gprimeprimei )i<? is a development for X with each Gprimeprimei an OIF collection of open
sets, and each Gprimeprimei+1 a refinement of Gprimeprimei . Therefore
uniondisplay
i<?
Gprimeprimei is an base for X. To see that
uniondisplay
i<?
Gprimeprimei is OIF, suppose that V is an open set from
uniondisplay
i<?
Gprimeprimei . Then for some n < ? we have
that V ? Gprimeprimen and V is a subset of finitely many of the elements of each Gprimeprimei for i ? n. If
there are infinitely many sets from
uniondisplay
i>n
Gprimeprimei containing V, then we may assume that every
Gprimeprimei for i > n has an element containing V. For each i < ?, let Ui be an element of
Gprimeprimei containing V. Then (Ui)i<? is a base for any point of V. This implies that V is a
singleton, contradiction.
Based on the proof of 1, the proof for 2 should be clear.
For 3, the forward direction is analogous to Proposition 3.32, and the reverse direc-
tion is much the same as 1. For each i < ?, choose Gprimeprimei to be a (n,?) refinement. Suppose
that F is a subset of X having cardinality n, and that F is contained in infinitely many
sets from
uniondisplay
i<?
Gprimeprimei . Then we may assume that there is an Ui ? Gprimeprimei so that F ? Ui for each
i < ?. Yet, (Ui)i<? is a base for some point of F, implying |F| = 1, contradiction. square
There is an example of a space that has each of these covering properties but does
not have any of these base properties. Indeed, the example we present is paracompact.
Example. The sequential fan was described in Section 3.4. Suppose that X is the
sequential fan, and let U be any covering of X. Choose U? ? U so that ? ? U?. Then
let V = {U?}?{{x} : x ? X \U?}. Then V refines U and is locally finite.
42
3.6 Set Theoretic and Combinatorial Conditions
In [5] the axiom CECA was introduced and proven consistent with ZFC. The authors
show that CECA is equivalent to GCH plus a weakening of square? for singular ?. So in
particular, CECA holds in G?odels constructible universe. In the same paper, CECA is
used to prove the following result.
Theorem 3.45 Assume CECA. Suppose that ?, ? are regular infinite cardinals, and let
?A???<? be a sequence of sets such that for every I ? [?]? there is J ? [I]<? so that
|
intersectiondisplay
??J
A?| < ?. Then there exists ?Aprime???<? such that |Aprime?| ? ? for each ? < ? and the
sequence ?A?\Aprime???<? is point-< ?.
We have already established the next result for GO spaces without special set the-
oretic considerations; in [5] this result is given for ? = ?.
Theorem 3.46 Assume CECA. Fix a regular infinite cardinal ?. If X is (< ?,?)-
metacompact, then X is (1,?+)-metacompact.
Proof. Let U be an open cover of X and let V be an (< ?,?)-metacompact refinement
of U. List X = {x? : ? < ?} and for each ? < ? let A? = {V ? V : x? ? V}. For
each infinite collection of points, there is some finite subcollection that is contained in
less than ?-many elements of V. Hence for each I ? [?]? there is some finite subset J
of I so that |
intersectiondisplay
??J
A?| < ?. We apply Theorem 3.45 with ? = ? and ? = ?. Therefore
for each ? < ? there is a Aprime? ? [A?]?? such that ?A?\Aprime? : ? < ?? is point-finite on the
set V. For each V ? V let I(V) = {x? ? V : V ? A?\Aprime?}. Each I(V) is finite and
if x? ? V\I(V), then V ? Aprime?. For each ? < ? choose some V? ? V so that x? ? V?.
Then for each V ? V, define V ? = (V\I(V)) ?{x? ? I(V) : V = V?}, note that each
43
V ? ? V. Therefore V? = {V ? : V ? V} is an open refinement of U. Such a V? is a
(1,?+)-metacompact refinement, because if x? ? V, then V ? Aprime? ?{V?}, and this is a
set of size ? ?. square
Now the question is raised whether the statement ?(< ?,?)-metacompact implies
(1,?+)-metacompact? is independent of ZFC. If the following combinatorial principal
holds (and is consistent with MA + ?3 ? 2?), then there is a space that is (< ?,?)-
metacompact and not (1,?1)-metacompact.
(*) If X is a set and |X| = ?3, then there exists a collection H of subsets of X
and a partition {Hn : n < ?} of H such that : (1) if H1,H2 ? H and H1negationslash=H2, then
|H1?H2| < ?; and (2) if Y ? X and |Y| = ?3, then for each n ? ?, there exists H ? Hn
such that |Y ?H| = ?2.
Theorem 3.47 (MA + ?3 ? 2? + *) There is a space with a weakly uniform base that
is not (1,?2)-metacompact.
Proof. Assume MA + ?3 ? 2?. This construction is essentially due to [18]. Let S be
a subset of the x-axis of size ?3 and let K be a countable dense subset of the upper
half plane of R2. Denote K = {pn : n < ?}. Then we let X = S ?K. For x ? S, the
neighborhoods are Bn(x) define to be an open disk in the upper half plane of radius 1/n
tangent to the axis at x intersected with K, together with {x}. The points of K are
isolated. Now let H =
uniondisplay
n<?
Hn be the collection and partition satisfying condition (*).
For each n < ? let Kprime(n) = {(pn,H) : H ? Hn} and define Kprime =
uniondisplay
n<?
Kprime(n). Now let
Xprime = S ?Kprime.
In Xprime we define the open neighborhoods of x ? S to be Bprimen(x) = {x}?{(pi,H) :
(pi,H) ? Kprime(i),pi ? Bn(x) and x ? H}. The points of Kprime are isolated. Let Bn =
44
{Bprimen(x) : x ? S}?{{q} : q ? Kprime}, then define B =
uniondisplay
n<?
Bprimen. We claim that this is a weakly
uniform base for Xprime. Suppose that x1 and x2 are both in Kprime, and let x1 = (pk,H1) and
x2 = (pj,H2). We have that |H1 ? H2| < ?; suppose that y0,y1,???,ym list H1 ? H2.
Then for each yi there is an ni so that x1 is not in Bprimeni(yi). Let n = max{ni : 0 ? i ? m};
then we have that {x1,x2} is contained in less than n?m many sets from B.
Next, we claim that Xprime has an open cover with no point ? ?1 open refinement.
Consider Bprime1 = {Bprime1(x) : x ? S}?{{q} : q ? Kprime}, which is an open cover of Xprime. Suppose
that V is an open point ? ?1 open refinement of Bprime1. Then for each x ? S, there is an
nx so that Bprimenx(x) is contained in some element of V; then Uprime = {Bprimenx(x) : x ? S} must
be point ? ?1. Let U = {Bnx(x) : x ? S}. Since K is dense in X, there is pi that is
contained in ?3 many sets from U, say {Bnx(x) : x ? Y} for some subset Y of X of
cardinality ?3. So there exists H ? Hi so that |H ?Y| = ?2 and the point (pi,H) is in
Kprime. For each x ? Y ?H we have (pi,H) ? Gprimenx(x). So (pi,H) is contained in ?2 many
elements of Uprime, contradiction. square
45
Chapter 4
I-weight and separating weight
4.1 Introduction
The study of reflection was started by Tkacenko in [16], and a systematic study was
made by Hodel and Vaughan in [13]. Hajnal and Juh?asz proved that weight reflects every
infinite cardinal [10]. Ram??rez-P?aramo proved that under GCH for the class of compact
Hausdorff spaces, i-weight reflects all infinite cardinals [15]. In the second section of this
chapter we prove that for compact linearly ordered spaces i-weight reflects all infinite
cardinals. We show that the point-separating weight must reflect, which implies that
i-weight must reflect.
In section three, we find necessary and sufficient conditions for i-weight to reflect
in the class of locally compact linearly ordered spaces. The lemmas used to determine
under what conditions i-weight will reflect for these spaces provide a means of calculating
the i-weight of an ordinal space.
We begin with some definitions which may be found in [15], [13].
Definition. A cardinal function ? is said to reflect cardinal ?, if when ?(X) ? ? there
is a subset Y of X so that |Y| ? ? and ?(Y) ? ?.
Definition. We say that X is condensed onto Z if there is a bijection from f : X ? Z
so that for each open subset U of Z, f?1(U) is open in X. Commonly, Z is regarded as
a copy of X, and the topology on Z is considered to be contained in the topology on X.
46
Definition. For a Tychonoff space X the i-weight of X is the minimum weight of a
Tychonoff space onto which X may be condensed.
So for example, the i-weight of Tychonoff space X is ? if and only if X has a weaker
separable metric topology.
Definition. We say that V ? P(X) is separating if for each pair (x,y) ? X2, xnegationslash=y, there
is an U ? V so that x ? U and ynegationslash?U. By separating weight, denoted sw(X), we mean
min{|V| : V is a separating open cover of X}.
4.2 Compact Linearly Ordered Spaces
In this section we show that for compact linearly ordered spaces, point-separating
weight and i-weight reflect all cardinals. For the reader?s comfort we now briefly outline
how we intend to show this.
A ?+-Aronsajn tree is a tree such that every chain and every level is of cardinality
< ?+. For a ?+-Aronsajn tree T we construct the space X(T) of chains in T, and
show that there is no point-separating open cover of cardinality less than ?+. For each
compact linearly ordered space we construct the tree T(X). If every subset of X that has
cardinality ?+ may be point-separated by ? or fewer open sets, then T(X) will contain
?+-Aronsajn tree A for which the points of X(A) may be point-separated by ? or fewer
open sets, which contradicts the earlier result.
Let A be a ?+-Aronszajn tree where every node is 2 or 0-branching, and level 0 has
only one node. For node t at level ?, node t corresponds to a sequence ?t ? 2? and, if t is
2-branching, the two nodes above t correspond to the sequences ?slurabovet (0) and ?slurabovet (1). Then
for each node t that is 2-branching let l(t) = ?slurabovet (0,0,0,???) and r(t) = ?slurabovet (1,1,1,???)
47
be 2 branches passing through node t. Let c and d be two maximal branches of A. Since
cnegationslash=d, we know that for some n < ?+ we have c(n)negationslash=d(n). Then define c < d if and only
if for the least n < ?+ for which ?c(n)negationslash=?d(n), ?c(n) = 0 and ?d(n) = 1. We give X(A),
the space of maximal branches of A, the topology generated by this order.
Lemma 4.1 Let A be a ?+-Aronszajn tree such that
1. A has one node at level 0.
2. A is 2-branching, or 0-branching at each node t
Then less than or equal to ? many open sets in X(A) cannot separate the pairs
{{l(t),r(t)} : t ? A}.
Proof. Let A be as above and let Aprime be the collection of nodes that are 2-branching.
Consider in X(A) the branches l(t) and r(t) for each t ? Aprime. Suppose that V is a
collection of ? ?-many open sets that point separate r(t) and l(t) for t ? Aprime. For each
t ? Aprime, let ?Vt,Wt? be a pair from V that separates l(t) and r(t) with l(t) ? Vt, r(t) ? Wt,
r(t)negationslash?Vt and l(t)negationslash?Wt. Each V ? V can be decomposed into disjoint convex subsets, so
for each t ? Aprime let (vt,vprimet) be the neighborhood of l(t) from the decomposition of Vt.
Similarly, define (wt,wprimet) so that r(t) ? (wt,wprimet) ? Wt.
For the set (vt,vprimet) to contain l(t) and not r(t), the branch vt must contain a node at
that is immediately to the left of a node below t and vprimet must be a branch that contains
node t.
Let S be a stationary subset of ?+, and for each t ? Aprime so that level(t) ? S, let
p(t) = at. Now we define a map f from S into ?+. First, for each s ? S, pick ts ? Aprime in
the sth level of A, then let f(s) = ? if and only if the level of (p(ts)) is ?. By the Pressing
48
Down Lemma, there is a level of the tree, ?, so that f?1{?} is a stationary subset of ?+.
Since each level has cardinality less than or equal to ? this means that there is a node a
in level ? so that |p?1{a}| = ?+. Consider {Vt : t ? p?1{a}} ? V. Since |V| ? ?, there is
a V so that V = Vt for ?+-many t ? p?1{a}. Since a is immediately to the left of a node
below each t ? p?1{a}, there is one convex set (v,vprime) from the disjoint decomposition of
V so that v = vt and vprime = vprimet for each t ? p?1{a}. Let ? be the height of branch vprime, and
note that |?| ? ?. Then the number of nodes t so that l(t) ? (v,vprime) is less than |?| ? ?,
yet p?1({a}) ? (v,vprime), contradiction. square
Lemma 4.2 For a compact Hausdorff space X, iw(X) = w(X) = nw(X) = sw(X).
Proof. By [12], we know that sw(X) ? nw(X), and for compact Hausdorff spaces,
w(X) = nw(X) = psw(X). Also, psw(X) ? iw(X) ? w(X), therefore, iw(X) = w(X)
if X is compact Hausdorff. We intend to show that sw(X) ? nw(X) if X is compact
Hausdorff. Suppose V is a separating open cover of X. Then let N = {X \ W : W is
a finite union of elements of V}. We claim that N is a net for X. Suppose that U is
an open set of X, then pick p ? U. For each q ? X \ U, find Vq ? V so that q ? Vq
and pnegationslash?Vq. Then {U}?{Vq : qnegationslash?U} covers X, hence there is a finite subcover, Vprime. Let
W = uniontext{V ? Vprime : Vnegationslash=U}, and N = X \ W. Then N ? N and clearly, p ? N ? U.
Therefore, N is a net with the same cardinality as V. square
The following lemma and its proof are found in [13].
Lemma 4.3 If ? is a monotone cardinal function that reflects successor cardinals, then
? reflects all infinite cardinals.
49
Next, we need to observe that separating weight is monotone. We combine this with
a similar lemma for i-weight.
Lemma 4.4 I-weight is monotone, and for compact spaces point-separating weight is
monotone.
Proof. Let X and Y be topological spaces, with Y ? X. If B is a base for X, then
{U ?Y : U ? B} is a base for Y. So the minimum cardinality of a base for X is not less
than the minimum cardinality of a base for Y.
For separating weight, let X and Y have the same relationship as above. Then, if
V is a separating open cover of X, VharpoonuprightY is a separating open cover of Y with cardinality
not more than |V|. square
Theorem 4.5 For a compact linearly ordered space, separating weight reflects ?+. Hence,
for compact linearly ordered spaces, separating weight reflects all infinite cardinals.
Proof. Suppose X is a compact linearly ordered space and sw(X) ? ?+, but every
subset Y of X such that |Y| ? ?+ has separating weight ?.
We construct a tree T(X) from the space X.
I?
I?0? I?1?
I?0,0? I?0,1? I?1,0? I?1,1?
a112a112
a112
I? a112 a112 a112
Level 0
Level 1
Level 2
a112a112
a112
Level ?
a112a112
a112
Figure 4.1: The tree T(X).
50
Let X = I?, then divide the space X into two closed intervals with at most one point
in common, I?0? and I?1?, so that every point of I?0? is less than or equal to every point
of I?1?. For a reason that will only be important near the end of this construction, we
choose to split the interval at a non-isolated point, if the interval contains a non-isolated
point. Next, form I?0,0? , I?0,1?, I?1,0? and I?1,1? by dividing each of I?0? and I?1? into two
parts, again at a non-isolated point, if possible. This process is done at each successor
level of the tree. At the level ?, consider ? : w ? 2. If
intersectiondisplay
n<?
I?harpoonuprightn is a non-degenerate
interval, then I? is a node at the ? level. Likewise at each limit level, nodes appear only
if the intersection of preceding intervals is a non-degenerate interval. If at some successor
level an interval should have only one point in it, then that node does not branch. The
ordering on the tree is that I? ? I? if and only if ? ? ?.
Suppose that at some level less than ?+, there are at least ?+ many non-degenerate
intervals (if necessary, choose ?+ of them). Let ? be the least such level.
Since |?| ? ?, the collection of nodes that precede the ? level has size not more than
?. Let the ?+ many non-degenerate intervals at the level ? be {I? : ? < ?+} = {[l?,r?] :
? < ?+}. Assume that there is a collection V of ?-many open sets in X that T1 separate
points r? and l? for ? < ?+. Then for each ? < ?+ there is a pair ?V?,W?? ? V ?V so
that l? ? V?, r? ? W?, r?negationslash?V? and l?negationslash?W?. There is a pair ?V,W? and some J ? [?+]?+
so that ?V,W? = ?V?,W?? for ? ? J.
Next, decompose W into disjoint convex subsets, and for each ? ? J let (w?,wprime?) ?
W be the convex neighborhood of r? from this disjoint decomposition. The assign-
ment ? mapsto? (w?,wprime?) is one-to-one since if (w?,wprime?) = (w?,wprime?), then either l? or l? is in
(w?,wprime?) ? W. Now each r? is the limit of a sequence of right end points of the intervals
that precede I? in the ordering of the tree. Therefore, for each ? ? J, there is a right
51
end point rprime? of an interval preceding I? that is contained in (w?,wprime?). However, there are
only ?-many nodes below the ? level of this tree, and therefore only ?-many potential
rprime? for ?+-many disjoint (w?,wprime?), contradiction. Therefore, we conclude that every level
below the ?+ level has cardinality less than ?+.
Suppose this tree contains a branch of length ?+. Then either the sequence of left
endpoints or the sequence of right endpoints of intervals forming the branch in the tree
has cardinality ?+. Without loss of generality, the set of right end points {r? : ? < ?+}
has cardinality ?+. Then {r?}?<?+ is a non-increasing sequence, and must converge to
r ? sup{l? : ? < ?+}. For each ? < ?+ let U? ? V be the open set that contains r and
not r?. Then some U is U? for ?+-many ?. However, for some ? < ?+ we will have
r? ? U whenever ? > ?, contradiction.
We have now shown that every level of T(X) has cardinality not more than ? and
that every chain has length less than or equal to ?. So either T(X) is a ?+-Aronszajn
tree or the height of T(X) is less than ?+. It follows from Lemma 4.1 that T(X) cannot
be a ?+-Aronszajn tree. If T(X) were a ?+-Aronszajn tree, then consider X(T(X)) and
the points corresponding to r(t) and l(t) for the nodes t ? T(X). By our assumptions, in
X we are able to separate these ? ?+ points in X with ?-many open sets, contradiction.
Since the tree has height less than ?+, without loss of generality, assume the height
of the tree is ?. Then the left and right endpoints of the intervals contained as nodes in
the tree form a subset of X, call this collection Y and |Y| = ?. We claim that this subset
together with the isolated points is dense in X. Consider a nonempty open convex subset
(a,b) of X. Either (a,b) contains a left or right endpoint of some interval contained in
the tree, or (a,b) is contained in an interval from each level of the tree. Then let ?
be the height of the tree, and for each ? < ? let J? be the interval from level ? that
52
contains (a,b). Then (a,b) ?
intersectiondisplay
?<?
J?, while |
intersectiondisplay
?<?
J?| = 1. So nonempty (a,b) = {x}. So
Y together with the isolated points is a dense set in X.
We now claim that the set of isolated points has cardinality at most ?. Let {Jm :
m ? M} be the collection of minimal intervals contained in the tree which we were not
able to split at a non-isolated point. We claim each Jm can contain at most countably
many isolated points. Let [lm,rm] = Jm, then pick am ? Jm; we have [lm,am] is a
closed set, and is therefore compact. Pick any open neighborhood W of lm, and then
{W}?{{x} : x ? [lm,am]\W} covers [lm,am]. Therefore, all but finitely many of the
points of [lm,am] must be in W. Since W is arbitrary, this implies that[lm,am] contains
countably many isolated points. By symmetry, so does [am,rm]. Now it remains to note
that every isolated point of X is contained in Jm for some m ? M, and since |M| ? ?,
we have at most ? many isolated points.
We use Y to construct a base of cardinality ? for X. Let B = {(a,b) : a,b ? Y,a <
b}?{{x} : x is isolated }. Let Y prime = Y ?{x : x is isolated }.
Let x be a point in X and (u,uprime) be a convex neighborhood of x. Unless x is isolated,
we have that at least one of (u,x) and (x,uprime) is nonempty. Without loss of generality,
assume that (u,x) is nonempty, and choose a ? Y prime ?(u,x). If (x,uprime) is nonempty, then
choose b similarly, and x ? (a,b) ? (u,uprime). So assume instead that (x,uprime) is empty.
Consider an increasing sequence of nodes in the tree (which is a decreasing sequence
of intervals) that contain the point x, say {K? : ? < ?}. Then
intersectiondisplay
?<?
K? = {x}. Let
? be the least ordinal so that for r?, the right end point of K?, r? ? uprime. If r? = uprime,
then uprime ? Y, and the set we need is (a,uprime). If r? = x, then there is a second increasing
sequence of intervals in the tree that contain x, but for this sequence x is a left end
point. Then consider all the members of this second sequence that also contain uprime. The
53
intersection of them would be [x,uprime], and would be a node of the tree. Therefore, uprime ? Y,
and again the set we want is (a,uprime), for then x ? (a,uprime) ? (u,uprime). Therefore, X has
weight ?, and hence i-weight ?, contradiction. square
Corollary 4.6 For compact linearly ordered spaces, i-weight reflects all cardinals.
Proof. Since iw(X) = sw(X) for each compact Hausdorff space, if iw(X) ? ? then
sw(X) ? ?. Therefore, there is Y ? X so that |Y| ? ? yet sw(Y) ? ?. Any base for a
Tychonoff topology, would also be a separating open cover, therefore, iw(Y) ? sw(Y) ?
?. square
4.3 Locally Compact Linearly Ordered Spaces.
In this section we prove reflection theorems for locally compact linearly ordered
spaces and i-weight. We begin with several lemmas that build toward the main result.
We determine that the i-weight of an ordinal space is the cardinality of the ordinal.
Also, we determine the i-weights of subspaces of ordinal spaces. We find necessary and
sufficient conditions for i-weight to reflect cardinal ? in the class of locally compact
linearly ordered space. This section ends with the presentation of several examples.
We use the following definitions throughout the rest of this section.
Definition. For a locally compact linearly ordered space X and a,b ? X we write a ? b
if and only if either [a,b] or [b,a] is compact. Then ? is an equivalence relation. Let
?a = {b ? X : a ? b} denote the equivalence class of a. Then for a locally compact linearly
ordered space X, we define E(X) to be the number of distinct equivalence classes under
the ? relation, i.e., E(X) = |{?a : a ? X}|.
54
Lemma 4.7 For a locally compact linearly ordered space X, each ?a is an open subset of
X.
Proof. To see that ?a is open, let p ? ?a. We assume that a < p, and the proof for the
case with p < a is analogous. Then since X is linearly ordered and locally compact there
is an open interval (c,d) of X so that p ? (c,d) and (c,d) = [c,d] is compact. Then
either a < c or c ? a ? d. If a < c, then [a,d] = [a,p] ? [p,d]. Since [p,d] is a closed
subset of the compact space [c,d], [p,d] is compact and [a,d] is compact as a union of
finitely many compact sets. Also, for each b ? (c,d) the set [a,b] is compact. Therefore,
(c,d) ? ?a, which implies that ?a is open.
If c ? a ? d then for each b ? (c,d), either [a,b] or [b,a] is nonempty and compact
as a closed subset of [c,d], so in this case (c,d) ? ?a. square
Lemma 4.8 For a locally compact linearly ordered space X, if E(X) is infinite then
E(X) = e(X).
Proof. Notice first that e(X) ? E(X). We form a closed discrete set C of cardinality
E(X) by taking one point from each equivalence class. We have shown that for each
a ? X the set ?a is open, so C is discrete. Also, ?a\{a} is open, so C is closed.
Next, suppose that e(X) > E(X). Since E(X) is infinite, e(X) ? ?1. Then there
is at least one equivalence class, call it ?a, that contains at least ?1-many members of a
closed discrete set C. Choose a point pprime ? C??a. At least one of P = {c ? C??a : c < pprime}
and S = {c ? C ? ?a : c > pprime} is uncountable. We assume the set P is uncountable,
as the proof for S uncountable is analogous. We claim that we can find p < pprime ? ?a so
that |C ? [p,pprime]| ? ?. For c ? P, let Ac = {d ? P : c < d}. If Ac were finite for each
c ? P, then P would be an increasing union of sets which are all finite, and so |P| ? ?,
55
contradiction. Hence there is a p so that |Ap| ? ?, then [p,pprime] ? C is infinite, [p,pprime] is
compact and cannot contain an infinite closed discrete set, contradiction. square
Lemma 4.9 For spaces X and Y, iw(X ?Y) = max{iw(X), iw(Y)}.
Proof. Suppose that X and Y are topological spaces, and consider X ?Y. If BX and
BY are bases for X and Y, then BX ?BY is a base for X ?Y.
So w(X ? Y) ? |BX ?BY| = |BX||BY|. Therefore, iw(X ? Y) ? iw(X)iw(Y) =
max{iw(X),iw(Y)}.
Next, suppose that B is a base for a Tychonoff topology on X ?Y which is coarser
than the product topology. Fix y0 ? Y and consider UX = {U ?(X?{y0})negationslash=? : U ? B}.
Then pi1(UX) is a base for a Tychonoff topology on X.
The above argument is symmetric with respect to x and y, so the i-weights of X and
Y are not more than |B|. Therefore, i-weights of X and Y are not more than i-weight of
X ?Y. square
Theorem 4.10 Let ? be a regular cardinal, and let A be a stationary subset of ?. Then
A with the subspace topology inherited from the order topology on ? has i-weight ?.
Proof. Let ? be a regular cardinal and assume that A is a stationary subset of ?.
Suppose by way of a contradiction that B is a base for a Tychonoff topology on A so
that |B| < ?. For each U ? B, there is an open subset Uprime of ? so that Uprime ?A = U. Let
Bprime = {Uprime : U ? B}. Since A is stationary, A contains stationarily many limit ordinals.
Let S denote the limit ordinals contained in A.
For each s ? S, let ps be any element of A so that ps > s. Also, for each s ? S let
(Us,Vs) ? B2 be such that s ? Us, ps ? Vs and Us?Vs = ?. Since Uprimes must be open in the
56
order topology, we know that each Uprimes contains a convex segment containing s. Let g(s)
be an ordinal less than s so that (g(s),s] ? Uprimes. Then since S is stationary, there is a ?
so that g?1(?) is stationary. Because |B| < ? and |g?1(?)| = ?, there is (U?,V ?) so that
(U?,V ?) = (Us,Vs) for ?-many different s ? g?1(?). Then s ? (?,s] ? U? and psnegationslash?U?
for each s ? g?1(?). For any fixed s ? g?1(?), let p? = ps. We claim that p? is an upper
bound on the set g?1(?), else if there is a sprime so that p? ? sprime then p? ? (?,sprime],? U?. square
Corollary 4.11 For any cardinal ?, the i-weight of the ordinal space ? is ?.
Proof. If ? is not regular then ? must be a limit ordinal, since each successor ordinal
is regular. Each limit cardinal is the limit of the preceding regular cardinals. Therefore,
let L = {? < ? : ? is a regular cardinal}, and notice that ? is equal to
uniondisplay
??L
?. Then
|?| ? iw(?) ? sup{iw(?) : ? ? L} = ?. square
Corollary 4.12 The i-weight of any ordinal space ? is |?|.
Proof. Assume that ? is an ordinal but not a cardinal. Then, as ordinals, |?| < ?. By
monotonicity of i-weight we know that iw(|?|) ? iw(?). Also, iw(?) ? w(?) = |?| =
iw(|?|). square
Lemma 4.13 A Tychonoff space X with iw(X) ? ? can be condensed into I?.
Proof. Any Tychonoff space of weight m can be embedded in Im. So if a space X has
i-weight m ? ?, then X has a Tychonoff topology ? so that (X,?) is homeomorphic to
a subset of Im. Call the corresponding embedding f.
Then we embed each Im into I? by the defining h : Im ? I? as follows. Let x ? Im
be denoted as (xi)i<m; then h(x) = (xi)i<m slurabove(0)m?j<?. Let G : X ? I? be defined by
h?f. Clearly, G is one-to-one and continuous. square
57
Recall that D? denotes the discrete space of cardinality ?.
Theorem 4.14 Let ? be an ordinal, and C a club subset of ?. Suppose that ? \ C =
uniondisplay
i<?
(ai,bi) so that ? ? |?| and (ai,bi)?(aj,bj)negationslash=? if and only if i = j. Then, iw(?\C) =
max{iw(D?), sup{iw((ai,bi)) : i < ?}}.
Proof. Let ? and C be as above, i.e. ? \ C =
uniondisplay
i<?
(ai,bi). Then by monotonicity of
i-weight, we have iw(? \ C) ? iw((ai,bi)) for each i < ?. Also, since the (ai,bi) are
pairwise disjoint, and each is nonempty and open, we may choose xi ? (ai,bi) so that
X = {xi : i < ?} is a discrete space homeomorphic to D?. Then, invoking monotonicity
once again, iw(?\C) ? iw(X).
We prove that iw(?\C) ? max{iw(D?), sup{iw((ai,bi)) : i < ?}} by considering
two cases.
Suppose that there is ? < |?|so that ? = sup{iw((ai,bi)) : i < ?}. We condense each
(ai,bi) into I?. Then, ?\C can be condensed to a subset of D??I?. So by Lemma 4.9,
iw(D? ?I?) = max{iw(D?),?}. Therefore, iw(?\C) ? {iw(D?), sup{iw((ai,bi) : i <
?}}. Suppose on the other hand that sup{iw((ai,bi)) : i < ?} = |?|. Since |?| = iw(?) ?
iw(?\C) ? sup{iw((ai,bi)) : i < ?}, iw(?\C) = |?| = max{iw(D?), sup{iw((ai,bi)) :
i < ?}}. square
For the following corollary, recall that log(?) = min{? : 2? ? ?}.
Corollary 4.15 The i-weight of D? is log(?).
Proof. From Theorem 4.2 in [12], we know that for any Hausdorff topology on X,
|X| ? w(X)?(X) ? w(X)w(X) = 2w(X). This gives us a means of bounding the i-
weight of a space, in particular, ? ? 2iw(D?). So iw(D?) ? {? : 2? ? ?}. Notice
58
that for each ? so that 2? ? ?, we may consider ? to be a subset of 2?. Then under
the product topology on 2?, the weight of 2? is ?, which means that iw(X) ? ?. So
iw(X) = min{? : 2? ? ?} = log(?). square
Lemma 4.16 If X is a locally compact linearly ordered space so that for each pair
a,b ? X with a < b the set [a,b] is compact, then iw(X) = w(X). Moreover, for such a
space X, i-weight reflects all cardinals.
Proof. Let X be as above. Then pick a ? X. Either (??,a] or [a,?) has the same
weight as X. Without loss of generality, let w([a,?)) = w(X).
We intend to show that w([a,?)) = iw([a,?)) = max{cf([a,?)), sup{w([a,b]) :
b > a}}.
First, suppose that B is a base for a Tychonoff topology on [a,?) which is coarser
than the order topology. Since weight equals i-weight for compact Hausdorff spaces, we
know that iw([a,b]) = w([a,b]) and by monotonicity of weight, we know that w([a,?)) ?
sup{w([a,b]) : b > a}. Also, suppose that cf([a,?)) = ?. We construct a set C that is
homeomorphic to ?. Let c0 = a. Suppose that for j ? i each cj has been defined, and
pick ci+1 > ci. If ? is a limit ordinal so that for each j < ?, cj has been defined, define
c? = sup{cj : j < ?}. Since the cofinality of [a,?) is ?, ci is defined for each i < ?.
Let C = {ci : i < ?}. If i is a successor ordinal, (ci?1,ci+1)?C = {ci} and is open. If
? is a limit ordinal then (ci,c?] ?C = {cj : i < j ? ?} is open for i < ?. We map C
homeomorphically to ? by h(ci) = i. Then the i-weight of C is ?, the i-weight of ?. This
implies that iw([a,?)) ? ? = cf([a,?)).
Next, we observe that w([a,?)) ? max{cf([a,?)), sup{w([a,b]) : b > a}}. Let K
be a cofinal subset of [a,?) of cardinality cf([a,?)); so K = {ki : i < ?} and ki < kj
59
iff i < j. The set {[a,??) : ? < ?} is an open cover of [a,?). Also, w([a,??)) ?
w([a,??]). Let B? be a base for [a,??) under the subspace topology for the order
topology on [a,?). Then B =
uniondisplay
?<?
B? is a base for [a,?). The cardinality of B is
less than max{?, sup{w([a,k?)) : ? < ?}} ? max{?, sup{w([a,k?]) : ? < ?}} ?
max{?, sup{w([a,b]) : b > a}}.
So w([a,?)) = iw([a,?)).
Now we will show that i-weight reflects. Suppose that ? ? iw([a,?)). Then consider
several quick cases.
1. If ? ? cf([a,?)), then let C be the cofinal subset above in this proof. Then
Y = {ci : i < ?} is a subset of [a,?) that is homeomorphic to ?, hence Y has
i-weight ?.
2. If there is a b > a so that iw([a,b]) ? ?, then take Y to be a subset of [a,b] that
reflects ?.
3. Now assume that ? > cf([a,?)) and that iw([a,b]) < ? for each b > a. Then
let Yi ? [a,ki] so that |Yi| ? iw([a,ki]) = iw(Yi). Take Y =
uniondisplay
i<?
Yi. We claim
iw([a,?)) ? iw(Y) ? sup{w([a,ki]) : i < ?} = sup{w([a,b]) : b > a} = ?. It?s
clear that sup{w([a,ki]) : i < ?} = sup{w([a,b]) : b > a} since K is cofinal. To
verify that sup{w([a,b]) : b > a} = ? recall that w([a,b]) = iw([a,b]) and that
iw([a,b]) < ? ? iw([a,?)), so sup{w([a,b]) : b > a} = ?. So iw(Y) ? ?, because
iw(Y) = max{?,sup{iw([a,?i]) : i < ?}} = max{?,?} and ? ? ?.
Therefore, for [a,?), i-weight reflects all cardinals. Recall, thatw([a,?)) = w(X) ?
iw(X) ? iw([a,?)) = w([a,?)). So the i-weight of X is the i-weight of [a,?); therefore,
i-weight reflects for the space X. square
60
Theorem 4.17 Let X be a locally compact linearly ordered space. Then iw(X) =
max{iw(DE(X)), sup{iw(?a) : a ? X}} = max{log(e(X)), sup{iw(?a) : a ? X}}.
Proof. By monotonicity, we know that iw(X) ? max{iw(DE(X)), sup{iw(?a) : a ? X}}.
Now suppose that ? = sup{iw(?a : a ? X}. Then there is a condensation of X into
DE(X) ?I?, which has i-weight max{iw(DE(X)), ?}. So iw(X) = max{iw(DE(X)),
sup{iw(?a) : a ? X}}.
Also iw(DE(X)) = log|E(X)| = log(e(X)), therefore, iw(X) = max {log(e(X)),
sup{iw(?a) : a ? X}}. square
Theorem 4.18 Let X be a locally compact linearly ordered space. If iw(X) = iw(DE(X))
= log(e(X)), then i-weight reflects the cardinal ? if and only if either 2? < ? for all ? < ?
or ? ? sup{iw(?a) : a ? X}. Hence, if iw(X) = sup{iw(?a) : a ? X}, then i-weight reflects
all cardinals.
Proof. Suppose first that X is as above, and iw(X) = iw(E(X)). Assume that i-weight
reflects the cardinal ? and ? > sup{iw(?a) : a ? X}. There is a Y contained in X so that
|Y| ? ? and iw(Y) ? ?. Since iw(Y ? ?a) ? iw(?a) we have that iw(Y ? ?a) < ? for each
a ? X. Also, since |Y| ? ?, Y ? ?a is nonempty for only ?-many different equivalence
classes. So let {?ai : i < ?} = {?a : ?a ? Ynegationslash=?}. Then for we may condense Y into
??Isup{iw(?ai):i<?}, which has i-weight iw(k) since ? > sup{iw(?a) : a ? X}. Therefore,
? = iw(Y) ? iw(?) ? ?. So, for each ? < ?, we have 2? < ?; else, if 2? ? ? for some
? < ?, the i-weight of ? would be ?.
Next, assume that i-weight reflects ? and 2? ? ? for some ? < ?. Aiming for a
contradiction, further assume that ? > sup{iw(?a) : a ? X}. Since i-weight reflects
61
?, there is a set Y so that |Y| ? ? and iw(Y) ? ?. Then Y can be condensed into
??Isup{iw(?a):a?X} which has i-weight max{?, sup{iw(?a) : a ? X}} ? ?, contradiction.
Now we prove the reverse direction.
Assume that 2? < ? for all ? < ? and ? ? iw(X) = iw(E(X)) ? E(X). Pick
?-many different ai so that {?ai : i < ?} is a collection of pairwise disjoint open sets.
Then for each i < ?, pick yi ? ?ai and define Y = {yi : i < ?}. So Y is a discrete space
of cardinality ?, so the i-weight of Y is log(?) = ?.
If ? ? sup{iw(?a) : a ? X}, then consider two cases. First, if ? < sup{iw(?a) : a ?
X}, then let x ? X be so that iw(?x) ? ?. Then for ?x, i-weight reflects ?.
Suppose that ? > iw(?a) for each a ? X. Then for some ? ? ? let {ai : i < ?} be
a subset of X so that {iw(?ai) : i < ?} is cofinal in ?. Then for each i < ? pick Yi ? ?ai
so that |Y| ? iw(?ai) = iw(Yi). Let Y =
uniondisplay
i<?
Yi. Then |Y| ? ? and iw(Y) = sup{iw(?ai) :
i < ?} = ?. square
Next, we present some examples of linearly ordered spaces for which i-weight does
not reflect some cardinal because E(X) > iw(X). So the conditions on E(X) in the
preceding theorem may not be omitted. There is also an example of a locally compact
linearly ordered space for which E(X) > iw(X) and yet i-weight will reflect all cardinals
less than iw(X).
Lemma 4.19 Any infinite discrete space is homeomorphic to a linearly ordered space.
Proof. Suppose that X is a discrete space of cardinality ?. Let Xprime = ??Z and order Xprime
lexicographically. Then ((?,n?1),(?,n + 1)) = {(?,n)} for each point (?,n) ? ??Z,
so Xprime is a discrete space of cardinality ?. Then X is homeomorphic to Xprime. square
62
Theorem 4.20 (GCH). There is a locally compact linearly ordered space X with iw(X) =
?1, yet for X i-weight does not reflect ?1.
Proof. Consider X = D2?1. Then by [12], we know that 2?1 ? 2iw(X). Since the order
topology is coarser than the discrete topology, iw(X) = ?1.
Take any subset Y of X so that |Y| ? ?1. Since Y is discrete, Y may be condensed
onto a subset of the real line. Thus iw(Y) = ?. square
We may eliminate the need for GCH if we are willing to allow the i-weight to exceed
?1. The following is also an example of a paracompact space for which the i-weight does
not reflect.
Proposition 4.21 There is a locally compact linearly ordered space X that has iw(X) ?
?1, yet for X i-weight does not reflect ?1.
Proof. Take X = D22?1. Then iw(X) ? ?1; else if iw(X) = ? then |X| < 2? ?
2?1 < 22?1. Take Y to be a subset of cardinality not exceeding ?1 and just as above,
iw(Y) = ?. square
Theorem 4.22 For each ? there is an example of a linearly ordered locally compact
space X so that ? = iw(X) < E(X) and yet i-weight reflects all cardinals.
Proof. Suppose Xprime is any compact linearly ordered space so that iw(Xprime) = ?. Give
? = |2?| the order topology, and let L be the set of successor ordinals in ?. Notice that
L has cardinality ? and that iw(L) ? ?. Let X = L?Xprime have the topology induced by
the lexicographic order.
First, notice that E(X) = e(X) = ?. Next, we observe that iw(X) ? ?. We
know that the i-weight of X under the product topology is ?. We just need to show
63
that the product topology is weaker than the order topology. Suppose that U ? V is
a basic open set in the product topology. Then let (?,x) ? U ? V. So ? ? ? and
x ? Xprime. If ? is a successor ordinal then for any (a,b) so that x ? (a,b) ? V, the point
(?,x) ? ((?,a),(?,b)) ? U ?V. So iw(X) ? ? and by monotonicity, iw(X) ? ?; hence
iw(X) = ?.
Now suppose that iw(X) ? ?. Then iw({j} ? Xprime) ? ? for each j < ?, and by
Corollary 4.6 i-weight reflects cardinal ? for Xprime. Find Y prime ? Xprime so that |Y prime| ? ? and
iw(Y prime) ? ?. Then iw({j}?Y prime) ? ? and |Y prime| ? ?. square
4.4 Paracompact Spaces
In this section we calculate the i-weight of paracompact spaces. This gives us a
formula for the i-weight of a compact space as well. We begin with a definition.
Definition. For a Tychonoff space X and x ? X, let the local i-weight of x in X be
defined as liw(x,X) = min{iw(U) : U is an open neighborhood of x}. Then the local
i-weight of X is liw(X) = sup{liw(x,X) : x ? X}.
To see that the local i-weight of a space need not coincide with the i-weight, consider
the ordinal space ?1. The i-weight of ?1 has been shown to be ?1. However, the local
i-weight of each point in ?1 is ?, thus liw(?1) = ? < iw(?1). This example is not
compact, indeed it is not paracompact. If we consider the space X = ?1 ?{?1} we find
that the liw(?1,X) = ?1 = iw(X), so in this case local i-weight and i-weight are the
same. We will prove that this is true for all compact spaces.
Recall that, by monotonicity of i-weight, iw(X) ? max{log(e(X)),
liw(X)} for any space X.
64
Theorem 4.23 If X is a paracompact, Tychonoff space then iw(X) = max{log(e(X)),
liw(X)}.
Proof. Let X be a paracompact Tychonoff space, and cover X with open sets witnessing
local i-weight. We may accomplish this by taking for each x ? X, a neighborhood Nx
of x so that iw(Nx) = liw(x,X); then {Nx : x ? X} is the desired cover. Next, let
V be a ?-discrete open refinement of the cover; V =
uniondisplay
i<?
Vi and each Vi is a discrete
family. Then let Wi =
uniondisplay
{V ? Vi}, so that {Wi : i < ?} is a countable open cover of X.
Notice that since for each i < ?, the set Wi is the union of pairwise disjoint open sets,
iw(Wi) = max{log|Vi|,sup{iw(V) : V ? Vi}} ? max{log(e(X)),liw(X)}.
Now for each i < ? find a open set Wprimei so that Wprimei ? Wi, and {Wprimei : i < ?} is a cover
of X; we can do this by Remark 5.1.7 in [7]. Also, let R be a countable base of convex
sets for the interval [0, 1].
Denote the original topology on X by T. Let Bi be a base for a Tychonoff topology
?i on Wi which is weaker than the subspace topology inherited from T and |Bi| = iw(Wi).
Since we will be discussing several different topologies and subspaces we will denote the
closure of a set U as a subset of X under the topology T by clX,T(U). Let Bprimei = {B?Wprimei :
B ? Bi} and define Ai = {(Uprime,U) ? Bprimei ?Bi : clWi,?i(Uprime) ? U}.
We claim for each x ? Wprimei and ynegationslash=x there is a (Uprime,U) ? Ai with x ? Uprime and
ynegationslash?U. Take x ? Wprimei ? Wi, and suppose that ynegationslash=x. If y ? Wprimei then there is a pair
U,V ? Bi so that x ? U and y ? V and U ? V = ?. Next, find Uprimeprime ? Bi so that
x ? Uprimeprime ? clWi,?i(Uprimeprime) ? U and let Uprime = Uprimeprime ?Wprimei. Then x ? Uprime ? clWi,?i(Uprime) ? U and
Uprime ? Bprimei so (Uprime,U), so (Uprime,U) ? Ai.
65
Now assume that ynegationslash?Wprimei. Then find U ? Bi so that x ? U and ynegationslash?U, and find Uprime as
above.
We claim that if (Uprime,U) ? Ai, then clX,T(Uprime) ? clWi,?i(Uprime). Aiming for a contradic-
tion, suppose that y ? clX,T(Uprime) and ynegationslash?clWi,?i(Uprime). Since Uprime ? Wprimei and clX,T(Wprimei) ? Wi,
we have clX,T(Uprime) ? clX,T(Wprimei) ? Wi. Then there is a set V so that y ? V ? ?i and
V ?Uprime = ?. Yet, V is open in X, contradiction.
Next, since X is normal and clX,T(Uprime) ? U for each (Uprime,U) ? Ai , there is a map
fUprime,U : X ? [0, 1] so that fUprime,U(Uprime) = {0} and fUprime,U(Uc) = {1} for each (Uprime,U) ? Ai.
Let Fi = {fUprime,U : (Uprime,U) ? Ai} and let F =
uniondisplay
i<?
Fi. Suppose x,y ? X and xnegationslash=y. Choose
i < ? so that x ? Wprimei. We showed above that there is a (Uprime,U) ? Ai with x ? Uprime and
ynegationslash?U, so fUprime,U(x) = 0 and fUprime,U(y) = 1.
Then define F : X ? [0, 1]F by F(x) = (f(x))f?F. This map is continuous and
one-to-one. Consider F(X) ? [0, 1]F. The weight of the Tychonoff topology on F(X)
is less than or equal to |F|??. Let Tprime be the topology on F(X) and take F?1(Tprime) as a
topology on X that is Tychonoff and weaker than T.
Now note that |F| = max{|Fi| : i < ?} and that |Fi| = |Ai| = |Bi| = iw(Wi).
Therefore, this will generate a Tychonoff topology whose weight is less than or equal
to sup{iw(Wi) : i < ?} and recall, iw(Wi) ? max{log(e(X)),liw(X)}. Therefore,
iw(X) ? max{log(e(X)),liw(X)}. square
Corollary 4.24 In the class of compact spaces, i-weight equals local i-weight. Further-
more, for a compact space X, there is a point x ? X so that liw(x,X) = liw(X) =
iw(X).
66
Proof. Suppose that X is a compact space. For each point x ? X, choose open sets
Wx and Wprimex so that x ? Wprimex ? Wprimex ? Wx and iw(Wx) ? liw(x,X). Then {Wprimex : x ? X}
is an open cover of X. Choose a finite subcover and denote it {Wprimexi : i < N}. Then
{Wxi : i < N} is also an open cover of X. By the proof of Theorem 4.23 above, iw(X) =
sup{iw(Wxi) : i < N} = max{iw(Wxi) : i < N} ? liw(X). So iw(X) = liw(X), and
liw(xi,X) = liw(X) for some i < N. square
Example. There is a paracompact space X so that iw(X) = log(e(X)). Let Xprime be
a paracompact space for which i-weight and local i-weight coincide. Let iw(Xprime) = ? and
consider D22?. Define X = D22? ?Xprime. Then {?}?Xprime is clopen for each ? ? D22?, so
X is still paracompact. Also, liw(X) = ? while log(e(X)) = 2?, so iw(X) = log(e(X)).
Recall Theorems 4.17 and 4.18 for locally compact linearly ordered spaces. Theorem
4.23 somewhat parallels Theorem 4.17, so we considered the question: For the class of
paracompact Tychonoff spaces, if i-weight is determined by the local i-weight, does i-
weight reflect all cardinals? This next example shows that that is not necessarily the
case.
Example. Let L denote the set of limit ordinals strictly less than ?2. Then let
X = (?2?{?2})\L with the topology inherited from the order topology. We claim that
X is a paracompact GO space for which i-weight is ?2 and i-weight does not reflect the
cardinal ?1.
Aiming for a contradiction, suppose that the i-weight of X is ?1 and that B is a
base for a Tychonoff topology for X of cardinality ?1 or less. For each x ? X, let Ux ? B
be so that xnegationslash?Ux and ?2 ? Ux. Then for each x ? X let ?x be the least ordinal so that
(?x,?2] ? Ux. Since |B| ? ?1 at most ?1 different ?x are chosen; therefore, for some
67
? < ?2 the set {x ? X : ? = ?x} has cardinality ?2. However, |?| ? ?1. Therefore, for
some x ? X we have x ? (?x,?2] ? Ux, contradiction.
We now claim that any subset of X of cardinality ?1 is discrete and therefore has
i-weight ?. Note that if A is a subset of cardinality ?1 that does not contain ?2 then
A is clearly discrete. Next, suppose that A is a subset of X that contains the point ?2.
Observe that since ?2 is regular, no subset of cardinality ?1 is cofinal in ?2. Therefore,
each subset of ?2 with cardinality ?1 is bounded. We find ? < ?2 so that ? > a for all
a ? A\{?2}. So (?,?2]?A = {?2}; hence A is discrete.
68
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