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 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 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 ? ? 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 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 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
?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 ?, 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
? 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 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
? 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 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 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 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 a for all
a ? A\{?2}. So (?,?2]?A = {?2}; hence A is discrete.
68
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