The Stone- Cech Compacti cation and the Number of Pairwise
Nonhomeomorphic Subcontinua of  [0;1)n[0;1)
by
David Lipham
A thesis submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Master of Science
Auburn, Alabama
May 4, 2014
Keywords: Stone- Cech, Subcontinua, Ultrapower
Copyright 2014 by David Lipham
Approved by
Michel Smith, Chair, Professor of Mathematics
Stewart Baldwin, Professor of Mathematics
Gary Gruenhage, Professor of Mathematics
Piotr Minc, Professor of Mathematics
Abstract
An introduction to the Stone- Cech compacti cation  X of a normal topological space X
is given. The method of invariantly embedding linear orders into ultrapowers is used to  nd
2c pairwise nonhomeomorphic continua in  R, under the assumption that the Continuum
Hypothesis fails.
ii
Acknowledgments
Thanks  rst and foremost to my adviser, Dr. Michel Smith, for his invaluable guidance
and patience. Thanks to Dr. Stewart Baldwin and Dr. Piotr Minc for their inspiring
graduate courses in Set Theory and Topology, respectively, and to Dr. Gary Gruenhage for
letting me present some of this work in his seminar. I would also like to thank Dr. Baldwin
for his help clarifying many of the combinatorial arguments needed in the  nal chapter.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Topology and Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.1 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.2 Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3 The Stone- Cech Compacti cation . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.1 Filters and Normal Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.2 Construction of  X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.3 Properties and Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.4 Additional Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.5.1  ! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.5.2  R and  H . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.5.3  !1 and  L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
4 Ultrapowers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.1 De nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.2 CH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
5 The Continua Iu (u2! ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
5.1 De nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
5.2 Decompositions and the Proof of Theorem 1.2 . . . . . . . . . . . . . . . . . 30
6 Constructing Ultra lters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
iv
6.1 Invariant Embeddings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
6.2 The linear orders J ( < 2c) . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.3 A Quick Proof of Theorem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . 47
6.4 The Proof of Theorem 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
6.5 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
v
List of Figures
5.1 Theorem 5.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
6.1 IX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.2 Theorem 6.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
6.3 Jv . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
6.4 Theorem 6.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
6.5 [L1 ;L2 ] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
vi
Chapter 1
Introduction
A continuum is a compact and connected topological space. The Stone- Cech remainder
 HnH of the half-line H = [0;1) is a continuum, and every free ultra lter u on ! generates
a subcontinuum Iu of  HnH. In [1], A. Dow  nds a family of 2c free ultra lters on ! such
that the corresponding Iu?s are pairwise nonhomeomorphic. Thus, he proves the following.
Theorem 1.1 (: CH). There exist 2c pairwise nonhomeomorphic subcontinua of  HnH.
The main result of [1] is achieved by  rst noting that each Iu is closely related to
the linearly ordered ultrapower R!=u. The following theorem says that in order to  nd 2c
pairwise nonhomeomorphic Iu?s, it su ces to  nd 2c pairwise nonisomorphic completions
R!=u of ultrapowers R!=u.
Theorem 1.2. If u and v are free ultra lters on ! and Iu?Iv, then R!=u R!=v.
Finding 2c nonisomorphic R!=u?s is no trivial matter. It was  rst established in [9] that
all ultrapowers R!=u are isomorphic if CH holds. Prior to [3], only c were known to exist
when CH fails. The authors of [3] indeed show that there are 2c nonisomorphic ultrapowers
when CH fails. A. Dow was able to modify some of their arguments to prove the following.
Theorem 1.3 (: CH). There exists a family fD :  < 2cg of free ultra lters on ! such
that R!=D 6 R!=D for any  < < 2c.
The goal of this paper is to develop the tools needed for proving the theorems stated
above. In the next chapter, we will state some relevant de nitions and theorems from
introductory topology and set theory. The Stone- Cech compacti cation is the subject of
Chapter 3, wherein we prove existence and uniqueness results and look at some examples.
1
In Chapter 4 we introduce the ultrapowers R!=u and prove the aforementioned CH result.
In Chapter 5 we examine the spaces Iu and prove Theorem 1.2. In Chapter 6 we work
through a series of results from [3] which lead to the proof of Theorem 1.3. As indicated
above, Theorems 1.2 and 1.3 yield a proof of Theorem 1.1. We o er an alternative proof of
Theorem 1.1 in Chapter 6 as well - one that does not require the full strength of Theorem
1.3.
2
Chapter 2
Topology and Set Theory
2.1 Topology
We refer the reader to Topology by J. Munkres [6] for the basics.
Theorem 2.1. Every closed subspace of a compact space is compact.
Theorem 2.2. Every compact subspace of a Hausdor space is closed.
Theorem 2.3. The continuous image of a compact space is compact.
Theorem 2.4. A continuous bijection f : X!Y is a homeomorphism if X is compact and
Y is Hausdor .
Suppose f : X !Y is a function. For A2P(X), let f(A) = ff(x) : x2Ag denote
the image of A under f. For B 2P(Y), let f 1(B) = fx 2 X : f(x) 2 Bg denote the
inverse image of B under f. For added clarity we will sometimes write f[A] instead of f(A)
(respectively, f 1[B] instead of f 1(B)). If f is invertible, f 1 may also denote the inverse
of f.
Theorem 2.5. f is continuous i f(clXA) clYf(A) for all A2P(X).
A collection C of subsets of X is said to have the  nite intersection property if every
 nite subcollection of C has nonempty intersection.
Theorem 2.6. X is compact i for every collection C of closed sets in X having the  nite
intersection property, the intersection TC2CC of elements of C is nonempty.
Theorem 2.7 (Tychono ?s Theorem). A product of compact spaces is compact in the product
topology.
3
A compacti cation of X is a compact Hausdor space containing a dense copy of X.
Every locally compact Hausdor space has a one point compacti cation, de ned to be the
set  X = X[f1g with the following topology: U  X is open if
(i) U is open in X, or
(ii) 12U and  XnU is compact in X.
Theorem 2.8. If A is a connected subspace of X then the closure of A in X is connected.
A continuum is a compact and connected topological space.
Theorem 2.9. The intersection of a family of continua with the  nite intersection property
is a continuum. That is, if C is a collection of compact connected subspaces of X with the
 nite intersection property, then TC2CC is compact and connected (as a subspace of X).
Theorem 2.10. Every metrizable space is normal.
Theorem 2.11. Every compact Hausdor space is normal.
Theorem 2.12. Every F subspace of a normal space is normal.
Proof. Suppose L is a countable union of closed sets in X, and A and B are relatively closed
disjoint subsets of L. We may write A and B as countable unions of closed subsets of X;
A = Si2!Ai and B = Si2!Bi.
Separate A0 and B0 with disjoint open sets U0 and V0 such that clXU0 \B = ? =
A\clXV0. (To see that the last condition may be met, note, for instance, that B = F\L
for some closed F  X such that F\A0 = ?.)
Separate A1 and B1 with disjoint open sets U0 and V0 such that clXU0\B = ? =
A\clXV0. Let U00 = U0\(XnclXV0) and V00 = V0\(XnclXU1). These disjoint open
sets still contain A1 and B1, respectively. Now let U1 = U00[U0 and V1 = V00[V0. We
have disjoint open sets containing A1 and B1, clXU1\B = ? = A\clXV1, and U0  U1
and V0  V1. Continuing in this manner, Si2!Ui and Si2!Vi will be disjoint open sets in
X containing A and B, respectively.
4
Theorem 2.13 (Urysohn?s Lemma). If X is normal and A and B are disjoint closed subsets
of X, then there exists a continuous f : X![0;1] such that f(A) =f0gand f(B) =f1g
Theorem 2.14 (Tietze?s Extension Theorem). If X is normal and A is a closed subspace
of X, then any continuous map from A into R may be extended to a continuous map from
X into R.
Let C(X) be the ring of continuous real-valued functions on X. Let C (X) be the
subring consisting of the bounded members of C(X). Suppose A;B;Z  X. A is C -
embedded if every function in C (A) can be extended to a function in C(X). E.g., Tietze?s
Extension Theorem says closed subsets of normal spaces are C -embedded. A and B are
completely separated if there exists f2C(X) such that f(A) =f0g and f(B) =f1g. E.g.,
Urysohn?s Lemma says disjoint closed sets in a normal space are completely separated. X
is completely regular if for each closed set A X and p2XnA there exists f2C(X) with
f(A) = f0g and f(p) = 1. If X is completely regular and T1, then distinct singletons are
completely separated. By Urysohn?s Lemma, every T1 normal space is completely regular. Z
is a zero-set if there is a continuous function f : X!R with Z = f 1(f0g). If f : X!R,
we let Z(f) = f 1(f0g) denote the zero set of f. Let Z(X) denote the collection of all zero
sets of X.
Theorem 2.15. Two subsets of X are completely separated i they are contained in disjoint
zero sets.
Proof. Suppose A and B are completely separated by f, with A  f 1(f0g) and B  
f 1(f1g). Then A Z(f), B Z(f 1), and Z(f)\Z(f 1) = ?. Conversely, suppose
A and B are contained in disjoint zero sets. Let f1;f2 2C(X) s.t. Z(f1) A, Z(f2) B,
and Z(f1)\Z(f2) = ?. De ne f = jf1jjf1j+jf2j. Then f is continuous, f 1(f0g) = Z(f1) A,
and f 1(f1g) = Z(f2) B.
Theorem 2.16. In a metric space, closed sets and zero sets are equivalent.
5
A linear order is a pair (L;<), where L is a set and < is a binary relation on L such
that for all a;b;c2L:
(i) Either a = b, a<b, or b<a. (comparability, antisymmetry)
(ii) a a. (irre exivity)
(iii) If a<b and b<c, then a<c. (transitivity)
We may refer to a linear order simply by its underlying set when no confusion will
arise. Suppose L is a linear order. L is dense if for all l1 < l2 2L there exists l3 2L with
l1 <l3 <l2. L is complete if every subset of L has a least upper bound. If L is a dense linear
order, (M; ) is a completion of L if
(i) M is complete,
(ii) L M and  extends the ordering < on L, and
(iii) L is dense in M, i.e., for all m1 m22M there exists l2L with m1 l m2.
Theorem 2.17. Every dense linear order has a unique completion (up to isomorphism).
Theorem 2.18. If L is dense and compact in the order topology, then L is complete.
Theorem 2.19 (Intermediate Value Theorem). Suppose L and L0 are linear ordered topo-
logical spaces, L is complete, and f : L!L0 is continuous. If a;b2L and r is a point of L0
lying between f(a) and f(b), then there exists a point c of L lying between a and b such that
f(c) = r.
The Intermediate Value Theorem may be used to prove the following.
Theorem 2.20. Suppose L and L0 are complete linearly ordered topological spaces and h :
L!L0 is a homeomorphism. Then h is either order preserving or order reversing.
Suppose L and L0 are linear orders and f : L!L0. f maps L co nally if for all l02L0
there exists l 2 L such that l0 < f(l). f maps L coinitially if for all l0 2 L0 there exists
l2L such that f(l) <l0. The co nality of L, denoted cf(L), is the least ordinal  such that
there is a map f :  !L co nally into L. The coinitiality of L, denoted coi(L), is the least
ordinal  such that there is a map f :  !L coinitially into L.
6
2.2 Set Theory
We refer the reader to Set Theory by T. Jech [5] for the basics.
Suppose  is a cardinal. The co nality of  , denoted cf( ), is the least  such that there
is an unbounded map from  into  .  is regular if cf( ) =  .  is singular if cf( ) < .
Theorem 2.21. Suppose  is regular. If  <  and fX :  <  g is a collection with
jX j< for each  < , then
  
 S < X 
  
 < .
Theorem 2.22. For every in nite cardinal  there exists an increasing sequence f  :  <
cf( )g such that  = sup <cf( )  and j  j< for each  < cf( ).
Theorem 2.23. If  is a limit cardinal, then 2 = (2 )cf( ).
Theorem 2.24. If  is an in nite cardinal, and h i : i< i is a nondecreasing sequence of
nonzero cardinals, then Qi<  i = (supi<  i) .
Theorem 2.25. Suppose  is singular. There exists a setf i : i< cf( )gof regular cardinals,
each  i >!1, such that
supi<cf( ) i =  and
Y
i<cf( )
2 i = 2 :
Proof. As a limit cardinal,  is the sup of cf( ) regular cardinals  i. We may assume each
 i >!1. By Theorems 2.23 and 2.24,
Y
i<cf( )
2 i = (supi<cf( )2 i)cf( ) = (2 )cf( ) = 2 :
Let  be a regular uncountable cardinal. A set C   is a closed unbounded subset
of  if C is unbounded in  and C contains all limit ordinals less than  . A set S   is
stationary if S\C = ? for every closed unbounded subset C of  .
7
Theorem 2.26. The intersection of fewer than  closed unbounded subsets of  is closed
unbounded.
Theorem 2.27 (Pressing Down Lemma). If  is regular uncountable, S is a stationary
subset of  , and f : S! such that f( ) < for all  2S, then there is a stationary set
S0 S and  < such that f( ) =  for all  2S0.
Theorem 2.28. Suppose  is regular uncountable and  <  is regular. Then S = f <
 : cf( ) =  g is stationary in  , and may be partitioned into  pairwise disjoint stationary
sets.
Proof. If C is closed unbounded in  , then the  -th element of C has co nality  , thus
S\C6= ?. So S is stationary in  . For each  2S, let (  ) < be an increasing sequence
in  with sup <   =  . For each  < and  < let
S ; =f 2S :     g:
Claim: There exists  <  such that S ; is stationary in  for all  <  . Well, otherwise
for all  <  there exists   <  and a closed unbounded C such that C \S  ; = ?, so
that each element  2C \S has   <   . Then C = T < C is closed unbounded and
 = sup <    sup <   < for each  2C\S. But C\S is stationary in  ; in particular
it is unbounded in  . Contradiction.
Let  <  be given by the claim and de ne f( ) =   for each  2S. Then for each
 < , f  S is a regressive function on the stationary set S . For each  < the Pressing
Down Lemma implies there exists a stationary S0  S and     with f( ) =   for all
 2S0 . Then   6=   0 implies S0 \S0 06= ?. In particular,   fS0 :  < g  =jf n :  < gj.
Since the   are unbounded in  and  is regular, this set has cardinality  . While it may
not be true that S = S < S0 , we could simply add the de cit SnS < S0 to one of the
S0 .
8
Theorem 2.29 ( -system Lemma). Suppose C is an uncountable collection of  nite sets.
Then there exists an uncountable S C and a set r (the root of the  -system) such that
A\B = r for any A;B2S.
Theorem 2.30 (Ramsey?s Theorem). Suppose f is an n-place function on ! with  nite
range. Then there is an in nite W  ! such that f is constant on all increasing n-tuples in
Wn.
9
Chapter 3
The Stone- Cech Compacti cation
In this chapter we will show that every T1 normal topological space X has a compact-
i cation  X which is unique with respect to certain properties. One of these properties is
that every continuous map f : X ! Y from X into a compact Hausdor space Y has a
unique continuous extension  f :  X !Y. We will prove some useful results concerning
the extensions  f. Then we will look at some examples of  X, showing in particular that
 HnH is a continuum, and that j !j= 2c.
3.1 Filters and Normal Bases
Suppose L is a collection of sets that is closed under  nite intersections. An L- lter is
a subcollection D of L such that
(i) ? =2D
(ii) If A;B2D, then A\B2D
(iii) If A2D and A B2L, then B2D.
We omit reference to L when no confusion will arise. By properties (i) and (ii), a  lter
is a collection of subsets of X with the  nite intersection property. Given any subcollection
E L with the  nite intersection property, de ne
(E) =fL2L: L is a superset of a  nite intersection of members of Eg:
Then (E) is an  lter containingE, called the  lter generated by E. A  lter u is an ultra lter
if no other  lter properly includes u.
Theorem 3.1 (Ultra lter Lemma). Every  lter may be extended to an ultra lter.
10
Proof. Suppose D is a  lter. Consider the set P of all  lters containing D, partially ordered
by inclusion. The union of a chain of  lters containing D is a  lter containing D, so every
chain in P has an upper bound. By Zorn?s Lemma P has a maximal element, an ultra lter
containing D which no other  lter properly includes.
Theorem 3.2. u is an ultra lter i u is a  lter and every set in L which intersects each
member of u is in u.
Proof. Suppose u is an ultra lter and A2L intersects every element of u. Then u[fAg
has the  nite intersection property, so we may consider (u[fAg). By maximality of u,
(u[fAg) = u. So A2u. Conversely suppose u is a  lter and A2L. If A =2u then A does
not intersect every set in u, so it cannot be added to u to generate a larger  lter.
Theorem 3.3. If u is an ultra lter, A1;A22L, and A1[A22u, then A12u or A22u.
Proof. Suppose neither is inu. Then there existA0;A002uwithA1\A0 = ?andA2\A00 = ?.
Then A0\A002u but (A1[A2)\(A0\A00) = ?, contradicting A1[A22u.
A  lter is principal if it consists of all members ofLwhich contain a particular element
of X. That is, a principal  lter is a  lter of the form fL2L : x2Lg. A principal  lter
may be an ultra lter, depending on the collection L.
Suppose X is a set. We refer to aP(X)- lter (P(X)-ultra lter) as simply a  lter on X
(ultra lter on X). An ultra lter on X which is not principal is said to be free.
Theorem 3.4. u is an ultra lter on X i u is a  lter on X and for all A X exactly one
of A, XnA belongs to u.
Proof. Both A and XnA cannot belong since their intersection is empty. One of the two
has to intersect all sets in the  lter, otherwise take one  lter set outside A and one outside
XnA; their intersection would be empty. Conversely, suppose u is a  lter. If A2P(X)nu
then XnA2u, so A may not be added to u to generate a larger  lter.
The  lter on ! consisting of the co nite subsets of ! is called the Fr echet  lter.
11
Theorem 3.5. An ultra lter on ! is free i it contains the Fr echet  lter.
Proof. Suppose u is a free ultra lter on !. Let A  ! such that !nA  nite. For a
contradiction suppose A =2 u. Then !nA 2 u. For each n 2 !nA there exists xn 2 u
such that n =2xn. So (!nA)\Tn2!nAxn = ?, contradicting the  nite intersection property
of u. Conversely, suppose u is an ultra lter containing the Fr echet  lter. If n 2 !, then
!nfng2u, so that fng=2u.
Suppose X is a T1 topological space and L(X) is a closed lattice base for X. That
is, L(X) is a collection of closed subsets of X that is closed under  nite unions and  nite
intersections, such that every closed subset of X is an intersection of members ofL(X). We
assume ?2L(X). L(X) is a normal base for X if, additionally,
(i) for any closed subset A and x2XnA there exists a member of L(X) containing x
missing A, and
(ii) disjoint members of L(X) are contained in disjoint complements of members of
L(X).
Theorem 3.6.
(1) If X is completely regular, then Z(X) is a normal base for X.
(2) If X is normal, then the collection of closed subsets of X is a normal base for X.
(3) If X is compact Hausdor , then any closed lattice base is a normal base for X.
Proof of (1). Z(X) is a lattice since Z(f1)\Z(f2) = Z(jf1j+jf2j) and Z(f1)[Z(f2) =
Z(f1 f2). Now let A be closed in X. Under the assumption X is completely regular, for
each x2XnA let fx : X !R be a continuous function with f(x) = 1 and f(A) = f0g.
Then A = Tx2XnAf 1x (f0g). ThusZ(X) is a closed lattice base for X. Next we establish the
normal base properties. (i) A andfxgare completely separated by zero sets Z1 and Z2. The
set Z2 is as desired. (ii) Suppose Z(f1);Z(f2)2Z(X) are disjoint. By Theorem 2.15, Z(f1)
and Z(f2) are completely separated. That is, there exists f 2C(X) such that f 1(f0g) =
Z(f1) and f 1(f1g) = Z(f2). Then Z(f1)  f 1(( 1; 12)) and Z(f2)  f 1((12;1)). As
12
[12;1) and ( 1; 12] are zero sets in R, f 1([12;1)) and f 1(( 1; 12]) are zero sets in X. We
have Z(f1)  f 1(( 1; 12)) = Xnf 1([12;1)), Z(f2)  f 1((12;1)) = Xnf 1(( 1; 12]),
so that Z(f1) and Z(f2) are contained in disjoint complements of zero sets.
Proof of (2). Trivial.
Proof of (3). (i) Suppose A closed in X and x2XnA. LetfL g=fL2L: x2Lg. Since
fxg is closed, fxg= TL . There exists  s.t. A\L = ?. Otherwise, since the collection
fL gis closed under  nite intersections,fL g[fAghas  nite intersection property, and we
have TL \A6= ?. (ii) Suppose A, B are disjoint sets in L. Since X is normal, A and B
are contained in disjoint open sets U and V. Let fL g and fL0 g be subcollections of L so
that XnU = TL and XnV = TL0 . Then U = XnTL = SXnL . Since A is compact,
 nitely many of the sets XnL cover A. Similarly,  nitely many sets XnL0 cover B.
That is, the complement of Tni=1L i 2L contains A, and the complement of Tmi=1L0 i 2L
contains B. Also, XnTL i\XnTL0 i = ? because XnTL i  U, XnTL0 i  V, and
U\V = ?.
Note that if X is metric, the bases in (1) and (2) are identical.
3.2 Construction of  X
Suppose X is a T1 topological space and L(X) is a normal base for X. De ne  X(L)
to be the set of all L(X)-ultra lters. Let  L(X) =fXnL : L2L(X)g. For O2 L(X), let
B(O) = BL(X)(O) =fu2 X(L) : (9L2u)(L O)g:
Theorem 3.7. The collection fB(O) : O2 L(X)g is a basis for a topology on  X(L).
Proof. The collection covers  X(L) since ?2L(X) and B(Xn?) = B(X) =  X(L). Now
suppose u2B(XnL1)\B(XnL2) for some L1;L2 2L(X). We  nd V 2 L such that
u2B(V)  B(XnL1)\B(XnL2). Well, there exist L3;L4 2u such that L3  XnL1
13
and L4  XnL2. Let V = XnL1\XnL2 = Xn(L1[L2) 2 0 and L = L3\L4. Then
L2u and L V, so u2B(V). If v2 (V) then there exists L2v with L V, so that
v2B(XnL1)\B(XnL2). Thus B(V) B(XnL1)\B(XnL2).
For L2L(X) de ne
F(L) = FL(X)(L) =fu2 X(L) : L2ug:
Theorem 3.8. F(L1)\F(L2) = F(L1\L2) for any L1;L22L(X).
Proof. ( ) holds by ultra lter property (ii). ( ) holds by ultra lter property (iii).
Theorem 3.9. F(L1)[F(L2) = F(L1[L2) for any L1;L22L(X).
Proof. ( ) holds by ultra lter property (iii). ( ) holds by Theorem 3.3.
Theorem 3.10. The collection fF(L) : L2L(X)g is a closed lattice base for  X.
Proof. By the two preceding theorems, it su ces to show every closed subset of  X is an
intersection of members of fF(L) : L2L(X)g. Well, for XnL2 0 we have
 X(L)nB(XnL) =fu2 X(L) : (:9L02u)(L0 XnL)g
=fu2 X(L) : (8L02u)(L0\L6= ?)g=fu2 X(L) : L2ug= F(L);
so that fF(L) : L2L(X)g is the set of complements of basic open sets.
Theorem 3.11.  X(L) is compact Hausdor .
Proof. To prove  X is Hausdor , suppose u1;u2 2  X(L) are distinct. Then there exist
L1 2u1; L2 2u2 with L1\L2 = ?. There exist L01 and L02 in L such that L1  XnL01,
L2 XnL02, and XnL01\XnL2 = ?. It follows that u12B(XnL01), u22B(XnL02), and
B(XnL01)\B(XnL02) = ?.
14
To prove  X is compact, we show that any collection of closed subsets of  X(L) with
the  nite intersection property has a nonempty intersection. Suppose C = fC :  2Ag is
a collection of closed subsets of  X with the  nite intersection property. Each C can be
written as an intersection of basis closed sets; C = T 2B F(L  ). Since C has the  nite
intersection property, so does fF(L  ) :  2A; 2B g, thus does fL  :  2A; 2B g.
Generate anL(X)- lter containing all of the sets L  , and then extend it to an ultra lter u
in  X(L). We have
u2
\
 2A
\
 2B 
F(L  ) =
\
 2A
C :
De ne e : X! X(L) by x7!fL2L(X) : x2Lg. That is, for x2X let e(x) be the
principal L(X)- lter on x. Normal base property (i) guarantees e(x) is an ultra lter.
Theorem 3.12. cl X(L)e[L] = F(L) for any L2L(X).
Proof. Let L2L. Since e[L]  F(L) and F(L) is closed in  X(L), we have cl X(L)e[L]  
F(L). To show F(L) cl X(L)e[L] it su ces to show F(L) is contained in every basic closed
set containing e[L]. To that end, suppose L02L with e[L] F(L0). Then L02e(l) for all
l2L, so L L0. By the superset property of ultra lters, we have F(L) F(L0).
Theorem 3.13. e is a dense embedding.
Proof. First we show e is injective. Let x1 6= x2 2 X. Then fx1g is closed in X and
x2 2Xnfx1g. So there exists L2L with x2 2L, and L\fx1g = ?. That is, x2 2L
and x1 =2 L. So L 2 e(x2) but L =2 e(x1). Thus e(x1) 6= e(x2), proving e is injective. e
is continuous since e 1[F(L)] = fx 2 X : L 2 e(x)g = fx 2 X : x 2 Lg = L. Since
e[L]\e[X] =fe(x)2 X(L) : x2Lg= F(L)\e[X], it is clear that the inverse of e (de ned
on e[X]) is continuous. Finally cl X(L)e[X] = F(X) =  X(L) by Theorem 3.12, so that e[X]
is dense in  X(L).
15
We will frequently identify X with its copy e[X] in  X(L), viewing points of X as prin-
cipalL(X)-ultra lters. For instance, under this identi cation Theorem 3.12 says cl X(L)L =
F(L) for any L2L(X). Theorems 3.11 and 3.13 yield the following.
Theorem 3.14.  X(L) is a compacti cation of X.
Corollary 3.15. If X is compact, then X? X(L).
Proof. Suppose X is compact. Since  X(L) is Hausdor , X is closed in  X(L). So X =
cl X(L)X =  X(L)
Theorem 3.16. cl X(L)(L1\L2) = cl X(L)L1\cl X(L)L2 for any L1;L22L(X).
Proof. Using Theorems 3.8 and 3.12 , cl X(L)L1\cl X(L)L2 = F(L1)\F(L2) = F(L1\L2) =
cl X(L)(L1\L2).
3.3 Properties and Uniqueness
Suppose X is a T1 normal space. LetL(X) be the collection of closed subsets of X and
let  X =  X(L). Consider the following statements.
(1) cl X(L1)\cl X(L2) = cl X(L1\L2) for any L1;L22L(X).
(2) Disjoint closed sets in X have disjoint closures in  X.
(3) Every continuous function from X into a compact Hausdor space has a unique
continuous extension to  X.
(4) X is C -embedded in  X.
(5) If  X is any compacti cation of X, then there is a unique continuous surjection
f :  X! X which is the identity on X.
If  X is a topological space containing X, say  X satis es (i) if statement (i) holds when
all instances of  are replaced by  .
Theorem 3.17. (1))(2).
Proof. Trivial.
16
Theorem 3.18.  X satis es (1);(2).
Proof. Theorems 3.16 and 3.17.
Theorem 3.19.  X satis es (3).
Proof. Suppose f is a continuous function from X into a compact Hausdor space Y. De ne
 f :  X!Y by choosing
 f(u)2
\
L2u
clYf[L] (3.1)
for each u2 X. For each u2 X, fclYf[L] : L2ug has the  nite intersection property
since u does. Compactness of Y guarantees the intersection in (3.1) is indeed nonempty for
each u2 X. When x2X it is clear that we may choose  f(x) = f(x), so that  f extends
f. To prove  f is continuous, we show  f[cl XA] clYf[A] for any A2P(X). To that end
let A2P(X) and suppose y2 f[cl XA] =  f[cl XclXA]. There exists u2cl XclXA such
that  f(u) = y. We have clXA2u, so by de nition of  f and continuity of f,
y =  f(u)2clYf[clXA] clY clYf[A] = clYf[A]
Uniqueness of the continuous extension follows from the facts Y is Hausdor and X is dense
in  X, and implies TL2u clYf[L] (u2 X) is a singleton.
Theorem 3.20. (3))(4);(5) and (4))(2).
Proof. (4) ) (2): Suppose L1;L2 are disjoint closed sets in X. By Urysohn?s Lemma
there exists f 2 C (X) such that f(L1) = f0g and f(L2) = f1g. Then  f(cl X(L1))  
f(L1) = f0g. Similarly,  f(cl X(L2))  f1g. So cl X(L1)\cl X(L2) = ?. (3) ) (4):
Let Y = clRf[X] and apply the assumption. (3) ) (5): The continuous extension of the
inclusion i : X ,! X is a surjection since the image of  X is closed, contains X, and X is
dense in  X.
Corollary 3.21.  X satis es (4);(5).
17
Theorem 3.22. Suppose  X and  X are compacti cations of X. If  X satis es 5 and
 X satis es 2, then  X? X.
Proof. Let h: X! X be a continuous function which is the identity on X. Then h is closed
and surjective, so we just need to show that h is injective to prove h is a homeomorphism.
To that end, let p;q be distinct points in  X. There exists f 2C( X) s.t. f(p) = 0 and
f(q) = 1. Let L1 = fx2X : f(x)  13g and L2 = fx2X : f(x)  23g. Suppose V is an
open set  X containing p. Then V\f 1(( 1; 13)) is a nonempty open set in  X (it contains
p). Since X is dense in  X, there exists x2V\f 1(( 1; 13))\X. Then x2V\L1. Thus
every open set in  X containing p contains a point of L1. That is, p2 cl XL1. Similarly,
q2cl XL2. By continuity of h we have h(p)2h(cl XL1) cl Xh(L1) = cl XL1. Similarly,
h(q)2cl XL2. By hypothesis cl XL1\cl XL2 = ?, thus h(p)6= h(q).
Theorem 3.23. If  X is a compacti cation of X satisfying any of (1)-(5), then  X? X.
Proof. Suppose  X satis es one of (1)-(5). By Theorems 3.17 and 3.20,  X satis es 2 or 5.
 X satis es 2 and 5, so by the previous theorem  X? X.
Thus  X is the unique compacti cation of X with the properties (1)-(5). We call  X
the Stone-  Cech compacti cation of X.
Theorem 3.24. (i)  X? X(Z), and (ii) if X is compact Hausdor and C(X) is a closed
lattice base for X, then X? X(C).
Proof. By Theorem 3.6, Z(X) is a normal base for X. Applying Theorems 2.15 and 3.16,
we see that  X(Z) is a compacti cation of X satisfying (2). This proves (i). Suppose X
is compact Hausdor and C(X) is a closed lattice base for X. By Theorem 3.6, C(X) is a
normal base for X, so  X(C) is a compacti cation of X. Therefore X? X(C).
3.4 Additional Properties
Theorem 3.25. If A is a closed subset of X, then there is an embedding  i :  A! X
such that  i[cl AL] = cl XL for all L2L(A).
18
Proof. A closed subspace of a normal space is normal, so  A exists. Let  i :  A! X be
the continuous extension of the inclusion i : A ,!X.  i[cl AL] = cl XL for all L2L(A)
follows from the fact that  i is continuous and closed. To prove  i is an embedding, we just
need to show it is injective. Well, since A is closed in X,L(A) L(X). By the de nition of
 i (3.1) we see that  i(p) is the ultra lter in  X extending p. Thus distinct points in  A
map to distinct points in  X.
Note that image of  i is  i[ A] =  i[cl AA] = cl XA. So we have the following.
Corollary 3.26.  A?cl XA for each A2L(X).
Theorem 3.27. If f : X!Y is a homeomorphism, then  f :  X! Y is a homeomor-
phism.
Proof. It su ces to show  f is injective. Well,  (f 1)  f :  X! X continuously extends
iX. By uniqueness of the extension,  (f 1)  f = i X, so  f must be injective.
Theorem 3.28. Suppose A2L(X), B2L(Y), and f : A!B is a homeomorphism. Then
f may be extended to a homeomorphism ^f : cl XA! cl YB such that ^f[cl XL] = cl Yf[L]
for all L2L(A).
Proof. Let  f :  A! B be the homeomorphism that extends f. Let  i :  A! X and
 j :  B ! Y be given by Theorem 3.25. Then  j  f ( i) 1 : cl XA! cl YB is a
homeomorphism and for each L2L(A) we have
 j  f ( i) 1[cl XL] =  j  f[cl AL] =  jcl Bf[L] = cl Yf[L]:
Theorem 3.29. If f is closed, then  f 1(fyg) = TB2y cl Xf 1(B) for any y2 Y.
19
Proof. Let y2 Y. For each x2 X we have
x2cl Xf 1(B) for all B2y , f 1(B)2x for all B2y
, f 1(B)\A6= ? for all B2y and A2x
, B\f(A)6= ? for all B2y and A2x
, f(A)2y for all A2x
,  f(x) = y , x2 f 1(fyg):
Theorem 3.30. If  X is a compacti cation of X, then the remainder  XnX is the con-
tinuous image of the Stone-  Cech remainder  XnX.
Proof. Let  iX :  X !  X be the surjective continuous extension of the identity on X.
It su ces to show there is no p 2  XnX such that  iX(p) 2 X. For a contradiction,
suppose there is such a p. Then p 6=  iX(p). Separate p and  iX(p) with disjoint open
sets Up;U iX(p)   X. There exists an open V   X such that V \X = U iX(p)\X. By
continuity of  iX there exists an open W   X containing p, mapping into V. There exists
x2Up\W\X. But x =2U iX(p)\X = V \X, so  iX(x) = x =2V. Contradiction.
We will usually denote the remainder  XnX by X .
Theorem 3.31. Suppose X is a locally compact countable union of compact spaces (in
addition to being T1 normal). If A is F in  XnX, then cl XnXA? A.
Proof. Since X is locally compact Hausdor it has a one point compacti cation X[f1g.
The continuous extension of the inclusion i : X ,!X[f1gsatis es  i 1(f1g) =  XnX,
thus  XnX is closed in  X. In particular, cl XnXA is a compacti cation of A. Now we
show A is C -embedded in cl XnXA.
Let f2C (A). The assumption that X is a countably union of compact spaces implies
X is F in  X. Since  XnX is closed in  X, X[A is F in  X and A is closed in
20
X[A. By Theorem 2.12, X[A is normal. Apply Tietze?s Extension Theorem to extend f
to ^f2C (X[A). Since X is dense in X[A,  ( ^f  X) must extend ^f.  ( ^f  X)  cl XnXA
is the desired extension of f.
3.5 Examples
3.5.1  !
 ! is the set of all ultra lters on !. The embedded copy of ! consists of the principal
ultra lters fA ! : n2Ag, n2!, while the remainder ! =  !n! consists of the free
ultra lters. The next theorem says there exists the maximum possible cardinality of (free)
ultra lters on !. For clarity of presentation we prove the theorem for  N instead of  !.
Theorem 3.32. j Nj= 22!.
Proof. Clearly j Nj 22!. We show j Nj 22! by giving a surjection from  N onto a set
of cardinality 22!. Let X = f0;1g2! with the product topology. X is a compact Hausdor 
space with cardinality 22!. If we  nd a map from N onto a dense subspace of X, then its
continuous extension to  N will be as desired.
Identify 2! with the product space f0;1g!, with countable basis B. Since f0;1g! is
Hausdor , we may separate any  nite number of distinct points in 2! with  nitely many
pairwise disjoint sets in B. For each  nite subset fU1;:::;Ung of B, consider
 f2f0;1g2! : (8i2f1;:::;ng)((f  U
i 0_f  Ui 1)^( f  (2!nU1[:::[Un) 0))
 :
LetDbe the union of all such collections (over the  nite subsets ofB). Claim: Dis dense
in X. To that end, letObe a nonempty basis open set off0;1g2!. WriteO= Tni=1  1 i (fpig),
where each pi2f0;1gand the  i22! are distinct. LetfU1;:::;Ungbe a collection of pairwise
disjoint sets in B such that  i2Ui for each i2f1;:::;ng. Take f2D with f  Ui pi for
each i. In particular f( i) = pi for each i, so f2D\O. This proves the claim.
21
Note that jDj= !. Let f : N!D be any surjection onto D. Then  f :  N!X is a
surjection onto X.
3.5.2  R and  H
Let H = [0;1) and H =  HnH.
Theorem 3.33. H ?Tn2! cl R[n;1).
Proof. By Corollary 3.26, cl RH? H. So there is a homeomorphism h :  H!cl RH which
is the identity on H. Thus the remainders H and cl RHnH are homeomorphic. Now we
show cl RHnH = Tn2! cl R[n;1) . Suppose p2 R. Then
p2cl RHnH ( ), (8A2p)(8n2!)(A\[n;1)6= ?)
, (8n2!)([n;1)2p)
, p2
\
n2!
cl R[n;1):
( ) Only the direction ()) needs proving. We prove the contrapositive. Suppose p2cl RH,
A02p and n2! s.t. A0\[n;1) = ?. Then A0\H2p is compact. SincefA0\H\A : A2pg
is a collection of closed subsets of A0\H with the  nite intersection property we have
T
A2pA =
T
A2p(A
0\H\A)6= ?, so that p is principal.
Using similar arguments to those in the preceding proof,
R ?
\
n2!
cl R( 1;n][
\
n2!
cl R[n;1):
Thus R is the union of two disjoint copies of H , and R is compact but not connected.
Theorem 3.34. H is a continuum.
Proof. For each n2!, cl R[n;1) is compact and connected by Theorem 2.8. So H is the
intersection of a nested collection of continua. By Theorem 2.9, H is a continuum.
22
3.5.3  !1 and  L
Let !1 be the  rst uncountable ordinal with the order topology. Let L = !1 [0;1) with
the lexicographic order topology (the Long Line). Each topological space is locally compact
Hausdor , and therefore has a one-point compacti cation  X.
Theorem 3.35. Every continuous real-valued function on !1 is eventually constant.
Proof. Let f 2C(!1). Let S be the stationary set f <!1 :  is a limit ordinalg. For each
n2N we de ne  n <!1 such that f varies by less than 1n on ( n;!1). Fix n2N. Using the
continuity of f, for each  2S let gn( ) < such that f varies by less than 1n on (gn( ); ].
By the Pressing Down Lemma, there exists a stationary set Sn S and  n <!1 such that
gn( ) =  n for all  2Sn. So f varies by less than 1n on ( n; ] for all  2Sn. Since Sn is
unbounded, f varies by less than 1n on ( n;!1). Since !1 is regular, supn2N n <!1. Clearly
f must be constant on the  nal segment (supn2N n;!1) of !1.
Theorem 3.35 implies !1 is C -embedded in its one-point compacti cation  !1 = !1+1.
Corollary 3.36.  !1 =  !1.
Theorem 3.37. Every continuous real-valued function on L is eventually constant.
Proof. Let f 2 C(L). Note that any subspace of L of the form fh ;x i :  < !1g is
homeomorphic to !1 in the order topology. By Theorem 3.35, f is eventually constant on
any set of the form fh ;x i :  < !1g. In particular, for each q 2Q\[0;1) there exists
 q <!1 and rq2R such that f(h ;qi) = rq for all  > q. Letting  = supq2Q\[0;1) q <!1,
we have f(h ;qi) = rq for each  > and q2Q\[0;1). Since Q\[0;1) is dense in [0;1),
we have f(f g [0;1)) = f(f g [0;1)) for all  ; > . Suppose for a contradiction that
f is not constant on S > f g [0;1). Since f g [0;1) is connected and f is continuous,
f(f g [0;1)) is connected in R. So there exists a nonempty interval (a;b) R such that
(a;b) f(f g [0;1)) for all  > . Asj(a;b)j !1, there existsfr :  < <!1g (a;b)
23
such that r 6= r for any  < < <!1. For each  > leth ;x i2f g [0;1) such that
f(h ;x i) = r . Then f is not eventually constant onfh ;x i:  > g, a contradiction.
Corollary 3.38.  L =  L.
24
Chapter 4
Ultrapowers
4.1 De nition
Suppose X is a set and u is an ultra lter on !. De ne a relation on X! by
f g,fn2! : f(n) = g(n)g2u:
It is easily checked that  is an equivalence relation: symmetry is obvious, re exivity
follows from the fact that X 2 u, and transitivity follows from the fact that u is closed
under  nite intersections and supersets. The ultrapower X!=u is the set of corresponding
equivalence classes f=u. If X is a linearly ordered set then we may de ne a relation on X!=u
by
f=u<g=u,fn2! : f(n) <g(n)g2u:
Theorem 4.1. X!=u is linearly ordered by <.
Proof. Irre exivity: fn2!f(n) <f(n)g= ? =2u)f=u6<f=u. Antisymmetry:
f=u<g=u)fn2! : f((n) <g(n)g2u)fn2! : f(n) g(n)g=2u)f=u g=u:
Transitivity: Suppose f=u <g=u and g=u <h=u. Then fn2! : f(n) < h(n)g fn2! :
f(n) <g(n)g\fn2! : g(n) <h(n)g2u, so f=u<h=u. Comparability: If f=u6= g=u then
fn2! : f(n) = g(n)g =2u. So the complement fn2! : f(n) < g(n)g[fn2! : f(n) >
g(n)g is in u. By Theorem 3.3, fn2! : f(n) <g(n)g2u or fn2! : f(n) >g(n)g2u, so
that f=u<g=u or g=u>f=u.
25
We will be primarily interested in the case X = R. We may view R as a linearly ordered
subset of R!=u by identifying c 2R with the equivalence class f=u, where f is given by
f(n) = c for all n2!. Under this identi cation, R!=u = R when u is a principal ultra lter.
If u is free then R!=u properly contains R, and is sometimes called hyper-real. In Chapter
6 we will assume:CH and  nd 2c free ultra lters u2! such that the corresponding R!=u
are pairwise nonisomorphic.
4.2 CH
A dense linear order L is countably saturated if for any countable subsets A and B with
A<B, there exists v2L such that A<v<B.
Theorem 4.2. R!=u is countably saturated for any u2! .
Proof. Let A and B be countable subsets of R!=u. Let (aj=u)j2! be an increasing co nal
sequence in A and (bk=u)k2! a decreasing coinitial sequence in B. Choose a representative
(a0i)i2! from a0=u. Assuming a representative has been chosen from aj=u, select a represen-
tative from aj+1=u such that aji  aj+1i for all i2!. Recursively select representatives from
each member of (bk=u)n2! in a similar manner so that bk+1i  bki for all k;i2!.
For each i2! let Mi =fm i : ami <bmi g. Claim: For each i2! there exists xi2R
such that ami < xi < bmi for each m2Mi. Fix i2!. Assume Mi 6= ?. Let  = max(Mi).
Then amr  a i <b i  bmi for all m2Mi, by choice of representatives. Let xi2(a i;b i).
De ne (xi)i2!2R! as indicated above, setting xi = 0 if Mi = ?. Now suppose j;k2!.
Let m = max(j;k). Let E 2 u such that ami < bmi for all i 2 E. For each i 2 E with
i m we have m2Mi. Thus ami < xi < bmi for all i2E\fn2! : n mg2u. So
aj=u am=u<x=u<bm=u bk=u.
Theorem 4.3. All countably saturated linear orders of cardinality !1 are isomorphic.
Proof. Suppose L and L0 are countably saturated linear orders of cardinality !1. Enumerate
L =fl :  <!1g and L0 =fl0 :  <!1g. We construct a bijection ? : !1 !!1 inductively
26
so that the induced mapping l 7! l0?( ) is an order preserving isomorphism from L onto
L0. Let ?0 = f(0;0)g. Suppose  <!1, and for each  < a bijection ? has been de ned
between subsets of !1 so that
(i)   dom(? )\ran(? ), j? j !, ?i ?j for all i<j  , and
(ii) the induced mapping l 7!l0? ( ) is an order preserving isomorphism from fl :  2
dom(? )g onto fl0 :  2ran(? )g.
De ne ? as follows. Let  = S < ? . Let  1 <!1 be the least ordinal not in dom( ).
Let
A =f 2dom( ) : l <l 1g and B =f 2dom( ) : l 1 <l g:
There exists  2 <!1 such thatfl0 ( ) :  2Ag<l0 2 <fl0 ( ) :  2Bg. Extend  by de ning
 ( 1) =  2. Now let  3 <!1 be the least ordinal not in ran( ). Let
A0 =f 2ran( ) : l0 <l0 3g and B0 =f 2ran( ) : l0 3 <l0 g:
There exists  4 <!1 such that fl  1( ) :  2A0g<l 4 <fl  1( ) :  2B0g. Extend  again
by de ning  ( 4) =  3. Let ? =  .
Since jR!=uj= c, we have the following.
Corollary 4.4 (CH). R!=u?R!=v for all u;v2! .
27
Chapter 5
The Continua Iu (u2! )
5.1 De nition
Let I = [0;1], M = ! I, and In =fng I. For u2! , let
Iu =
\
A2u
cl M
[
n2A
In:
That is, Iu is the subspace of ultra lters in  M which contain the sets Sn2AIn, A2u.
Theorem 5.1. Iu?TA2u cl HSn2A[n;n+ 1].
Proof. Let E denote either the set of evens or the set of odds, whichever set u contains. The
map  : E I!Sn2E[n;n + 1] de ned by  (hn;xi) = n + x is a homeomorphism. By
Theorem 3.28, there exists a homeomorphism ^ : cl M(E I) ! cl HSn2E[n;n + 1] such
that ^ [cl ML] = cl H [L] for all L2L(E I). We have
\
A2u
cl M
[
n2A
In =
\
A2u
cl M
[
n2A\E
In? ^ 
\
A2u
cl M
[
n2A\E
In =
\
A2u
^ cl M
[
n2A\E
In
=
\
A2u
cl H 
[
n2A\E
In =
\
A2u
cl H
[
n2A\E
[n;n+ 1] =
\
A2u
cl H
[
n2A
[n;n+ 1]:
Every free ultra lter on ! contains the  lter of co nite sets, so the preceding theorem
implies Iu is a subspace of  HnH. Clearly Iu is compact. Now de ne  : M ! ! by
28
hn;xi 7! n. Note that   1(n) = In. Thus  is continuous, and the  bers   1(n) are
connected. Clearly  is also surjective. Let   be the continuous extension of  , from  M
onto  !. Theorem 3.29 implies
Iu =
\
A2u
cl M
[
n2A
In =
\
A2u
cl M  1(A) =    1(u):
We use this characterization of Iu to prove the following.
Theorem 5.2. Iu is connected.
Proof. Suppose A and B are disjoint closed sets in  M and    1(u) = A[B. Separate
A and B with open sets U and V. Let O =  !n  ( Mn(U[V)): Then u 2 O, so
   1(u)    1(O): And    1(O) U[V: Let U0 = U\   1(O) and V0 = V \   1(O),
so that U0 and V0 are disjoint open sets containing A and B respectively, and
   1(O) = U0[V0:
Claim U0\M =   1( (U0\M)). ( ) always holds. ( ). Suppose n2 (U0\M). Then
n2O, so   1(n) U0[V0. Since   1(n) is connected, we must have   1(n) U0\M.
Similarly, V0\M =   1( (V0\M)). So  (U0\M)\ (V0\M) = ?. Since ! is discrete
we have cl ! (U0\M)\cl ! (V0\M) = ?. That is,
  cl M(U0\M)\  cl M(V0\M) = ?:
Claim U0 cl M(U0\M). Well, if U0ncl M(U0\M)6= ? then, as a nonempty open set
in  M, there exists x2M\(U0ncl M(U0\M)), which is absurd.
Similarly, V0 cl M(V0\M). So we have A cl M(U0\M) and B cl M(V0\M).
Either A or B must be empty, otherwise u2  cl M(U0\M)\  cl M(V0\M).
Thus Iu is a continuum - a subcontinuum of  HnH.
29
5.2 Decompositions and the Proof of Theorem 1.2
In this section we will identify R with the interval (0;1). For each x=u2R!=u, choose
a representative (xn)2R! from x=u and let
xu =fL2L(M) : (9A2u)(L fhn;xni: n2Ag)g;
i.e., xu is the  lter of closed subsets of M generated by
ffhn;xni: n2!gg[ Sn2AIn : A2u :
Theorem 5.3.
(i) xu is an ultra lter in Iu, and
(ii) xu = yu i fn2! : xn = yng2u.
Proof. If xu is an ultra lter then clearly it is in Iu. Suppose L 2L(M) intersects every
set in xu. Then A = fn2! : hn;xni2Lg2u, otherwise !nA2u and L\(fhn;xni :
n 2 !g\Sn2!nAIn) = ?. So L  fhn;xni : n 2 Ag, hence L 2 xu. This proves (i).
Now suppose A = fn 2 ! : xn = yng2 u. Let L 2 xu. There exists B 2 u such that
L  fhn;xni : n 2 Bg. Then A\B 2 u and L  fhn;xni : n 2 A\Bg = fhn;yni :
n 2 A\Bg, so L 2 yu. This proves xu  yu. The reverse inclusion follows similarly.
Conversely, if xu = yu then A = fn 2 ! : xn = yng 2 u. Otherwise !nA 2 u and
(fhn;xni: n2!g\fhn;yni: n2!g)\Sn2!nAIn = ?.
Thus element in R!=u corresponds to a unique element in Iu. However, not every
element in Iu is of the form xu. Consider, for example, the collection
 S
n2AIn : A2u
 [fMnG : G M is open and  (G\I
n) < 1=n for each n2!g:
30
Each set in u is in nite, so sets of the  rst type contain In for arbitrarily high n2!. For
su ciently large n, no  nite number of the sets G can cover In, so the collection has the
 nite intersection property. The ultra lter generated by this collection can contain no set of
the form fhn;xni: n2!g - we could cover fhn;xni: n2!g with one of the sets G.
Theorem 5.4. R!=u is dense in Iu.
Proof. Let B(O)\Iu be a nonempty basic open subset of Iu. Then A =fn2! : O\In6=
?g2u. For each n2A choose xn2O\In with xn6= 0n;1n. Then xu2B(O)\R!=u.
For each au <bu2R!=u, de ne an interval in Iu by
[au;bu] =
\
A2u
cl M
[
n2A
fng [an;bn]
It is easily checked that [au;bu] is well-de ned. De ne a relation on Iu by x y i every
interval containing x contains y.
Theorem 5.5.  is an equivalence relation.
Proof. Transitivity and re exivity are obvious. To prove  is symmetric, suppose x  y.
Then there is an interval [au;bu] containing x but not y. There exists L 2 y such that
L\Sn2!fng [an;bn] = ?. For each n 2 ! let cn = supfc 2 L\In : c < ang and
dn = inffd2L\In : bn <dg. Then y2[0u;cu][[du;1u]. Without loss of generality, assume
y2[0u;cu]. Since [0u;cu]\[au;bu] = ?, we have x =2[0u;cu]. Thus y x.
For each x2Iu de ne the layer Lx = x= to be the equivalence class of x.
Theorem 5.6. Lxu =fxug for all xu2R!=u.
Proof. Clearly xu 2Lxu. Now suppose y2Iunfxug. There exists A2y and B2u such
that A\fhn;xni: n2Bg= ?. For each n2B we may de ne a subinterval fng [an;bn]
of In containing hn;xni, missing A\In. Then xu2[au;bu] but y =2[au;bu]. So y =2Lxu.
31
A partial ordering on the intervals in Iu is given by [au;bu] < [cu;du] i bu < cu. Now
de ne a relation on the set of layers of Iu by Lx Ly if there are intervals I1 and I2 with
x2I1, y2I2, and I1 <I2.
Theorem 5.7.  linearly orders Iu= , and (Iu= ; ) contains a dense copy of R!=u.
Proof. Transitivity and irre exivity of are obvious. To show any two elements of (Iu= ; )
are comparable, suppose Lx 6= Ly. By Theorem 5.5, x and y are contained in disjoint
intervals. One interval must be less than the other, so Ly  Lx or Lx Ly. Now we show
 is antisymmetric. Suppose for a contradiction that Lx  Ly and Ly  Lx. Then there
exist I1;I22x and I3;I42y with I1 <I3 and I4 <I2. But I1\I2 and I3\I4 are nonempty
disjoint intervals with I1\I2 <I3\I4 and I3\I4 <I1\I2, a contradiction.
By Theorem 5.6 we may identify each xu2R!=u withfxug2Iu= . If au <bu2R!=u
then it is an easy matter to select intervals [cu;du] and [eu;fu] such that au 2 [cu;du],
bu 2 [eu;fu], and [cu;du] < [eu;fu]. Thus  extends the ordering on R!=u. That R!=u is
dense in this ordering follows from Theorem 5.4 and the fact that (Iu= ; ) is the continuous
image of Iu (shown below).
De ne [0u;Lx) = SLy LxLy and [0u;Lx] = [0u;Lx)[Lx:
Theorem 5.8. [0u;Lx] is closed in Iu.
Proof. Suppose y2Iun[0u;Lx]. Then there are intervals [au;bu] containing x and [cu;du]
containing y such that [au;bu] < [cu;du]. Then B(Sn2!fng (bn;1n]) is an open set in Iu
containing y and no points in [0u;Lx].
Thus, the decomposition map Iu!Iu= is continuous if Iu= has the order topology
induced by  . In particular, Iu= is compact in the order topology.
Theorem 5.9. (Iu= ; ) is the completion of R!=u.
Proof. Theorems 2.18 and 5.7.
32
Theorem 5.10. Countable co nality layers in Iu contain copies of ! .
Proof. Let (anu)n2! be any strictly increasing sequence of elements in R!=u. Let A = fanu :
n 2 !g. There exists Lx 2 Iu=  such that Lx = supIu= A = (clIu= A) nA. Then
Lx =   1[(clIu= A)nA]  (clIuA)nA (where  : Iu !Iu= is the decomposition map).
As A is relatively discrete in Iu, Theorem 3.31 says clIuA = cl MnMA? A? !. We have
Lx (clIuA)nA?! .
Theorem 5.11. Each interval is a continuum.
Proof. Suppose [au;bu] is an interval. We may assume an < bn for all n2!. The natural
homeomorphism between M and Sn2!fng [an;bn] induces a homeomorphism between Iu
and [au;bu]. Alternatively, view [au;bu] as    1(u), where   :  Sn2!fng [an;bn] ! !
is the continuous extension of the  rst coordinate projection  .
Theorem 5.12. Each layer is a continuum.
Proof. Notice Lx = Tx2[au;bu][au;bu]. Apply Theorems 5.11 and 2.9.
Theorem 5.13. If u;v 2 ! , then a homeomorphism Iu ! Iv induces an isomorphism
(Iu= ; )!(Iv= ; ) which is either order preserving or order reversing.
Proof. Suppose h : Iu !Iv is a homeomorphism. De ne ? : (Iu= ; ) ! (Iv= ; ) by
?(Lx) = Lh(x). We show ? is well de ned and injective, i.e., that the following diagram is
commutative.
Iu h! Iv
# yx #
Iu= ?! Iv= 
Figure 5.1: Theorem 5.13
Suppose x2IunR!=u and h[Lx] intersects two di erent layers Ly0 and Ly00 in Iv. Then
(Ly0;Ly00) is a nonempty open subset of h[Lx] (it contains a point of R!=v). So Lx contains
33
a nonempty open subset of Iu, thus Lx contains a point xu 2R!=u. This contradicts our
assumption about x. Thus, since h(x)2h[Lx]\Lh(x), we have h[Lx] Lh(x) (*). Similarly, if
h(x)2IvnR!=v then h 1[Lh(x)] Lh 1h(x) = Lx, i.e., h[Lx] Lh(x) (**). We may now prove
h[Lx] = Lh(x) for all x2Iu. This is clear if x2R!=u and h(x)2R!=v. If x2IunR!=u and
h(x)2R!=v, apply (*). If x2R!=u and h(x)2IvnR!=v, apply (**). If x2IunR!=u and
h(x)2IvnR!=v, apply (*) and (**). Thus,
?(Lx) = ?(Ly),h[Lx] = h[Ly],Lx = Ly,x y:
So ? is well-de ned and injective. Moreover, ? is a homeomorphism since the decomposition
maps are continuous and closed. By Theorem 2.20, ? must be order preserving or order
reversing.
The preceding theorem will be used to  nd nonhomeomorphic Iu when CH fails. This
approach would fail badly under CH (see Corollary 4.4). In fact, if CH is assumed, A. Dow
showed that any two have isomorphic closed lattice bases, hence all Iu are homeomorphic
(see Theorem 3.24(ii)).
34
Chapter 6
Constructing Ultra lters
6.1 Invariant Embeddings
Suppose  ; >! are regular cardinals. (A;B) is a ( ; )-cut of L if A<B, L = A[B,
cf(A) =  , and coi(B) =  . An order preserving map ? : L!L0 is an invariant embedding
if every ( ; )-cut (A;B) of L maps to a cut of L0, in the sense that there is no x2L0 with
?[A] <x<?[B]. The main result of this section is that every linear order of cardinality c
admits an invariant embedding into some ultrapower !!=u.
Suppose D is a  lter over !. Let ID =fX ! : !nX2Dgbe the corresponding dual
ideal. Then ID contains ? but not X, and is closed under  nite unions and subsets. De ne
A B mod D,AnB2ID
A = B mod D,A B2ID:
Suppose G !! is a family of surjective functions. G is independent mod D if for all
distinct g1;:::;gl2G and j1;:::;jl2! (not necessarily distinct), we have
fn2! : gk(n) = jk for all k lg6= ? mod D:
Note that \A6= ? mod D" means A is not a subset of a complement of a set in D, i.e., A
intersects every set in D.
Let
FI(G) =fh : h is a function; dom(h) is a  nite subset of G; ran(h) !g
35
be the set of all  nite partial functions from G to !. For each h2FI(G) let
Ah =fn2! : g(n) = h(g) for all g2dom(h)g,
and
FIs(G) =fAh : h2FI(G)g:
Theorem 6.1. (i) G is independent mod D i Ah 6= ? mod D for all h 2 FI(G). (ii)
If G is independent mod D, then there exists a maximal  lter D  D modulo which G is
independent.
Proof of (i). ()) Let h 2 FI(G). Enumerate dom(h) = fg1;:::;gkg. For each i  k let
ji = h(gi). Then
Ah = fn2! : h(gi) = gi(n) for all i kg
= fn2! : gi(n) = ji for all i kg6= ? mod D:
(() Let g1;:::;gl 2 G distinct and j1;:::;jl 2 !. De ne h 2 FI(Gg by dom(h) =
fg1;:::;glg and h(gi) = ji for each i l. Then fn2! : gi(n) = ji for all i lg = Ah 6=
? mod D.
Proof of (ii). Let P = fP  P(!) : P is a  lter, D  P; and G is independent mod Pg,
partially ordered by inclusion. Suppose (P ) < is a chain of  lters (P  P for    < )
in P. It is easily seen that S < P 2P, thus every chain in P has an upper bound. The
existence of D follows from Zorn?s Lemma.
A P(!) is a partition mod D if
(i) A6= ? mod D for all A2A
(ii) if A;A02A with A6= A0, then A\A0 = ? mod D
(iii) for all B2P(!) with B6= ? mod D, there exists A2A s.t. A\B6= ? mod D.
36
Suppose A is a partition mod D and B 2P(!). B is based on A mod D if for all
A2A, either A B mod D or A\B = ? mod D. Suppose A  P(!). B is supported by
A mod D if it is based on some partition A A, mod D.
Theorem 6.2. Suppose D is a maximal  lter over ! modulo which G is independent. For
every B2P(!) there exists a countableG0 G such that B is supported by FIs(G0) mod D.
Proof. Claim: For every X 2P(!) with X 6= ? mod D, there exists Ah 2 FIs(G) such
that Ah X mod D. Well, for a contradiction suppose X2P(!), X6= ? mod D, and (*)
Ah\(!nX)6= ? mod D for all Ah2FIs(G). Then !nX6= ? mod D, so !nX (which is
not in D) intersects every set in D. The  lter generated by D[f!nXg properly extends
D, and G is independent mod (D[f!nXg) by (*) and Theorem 6.1(i). This contradicts
the maximality of D.
Now let B2P(!). Enumerate P(!) =fX :  < 2!g. Using the claim we may de ne
a collection fA :  < 2!g inductively so that for each  < 2! either
(i) A = ? if X \B = ? mod D or (X \B)\A 6= ? mod D for some  < , or
(ii) A 2FI(G) with A  (X \B) mod D.
De nefA0 :  < 2!gsimilarly for !nB. ThenA= (fA :  < 2!g[fA0 :  < 2!g)nf?g
is a partition on which B is based mod D. We now show A is countable. Suppose not, and
assume jAj = !1. Enumerate A = fAh :  < !1g. Then fdom(h ) :  < !1g is an
uncountable collection of  nite sets. By the  -system Lemma there exists an uncountable
S  !1 and a  nite set r  G such that dom(h )\dom(h ) = r whenever  6=  2 S.
Enumerate r =fg1;:::;gng. There are only countably many tuples (h (g1);:::;h (gn)),  2S,
so there must exist  6=  2 S such that h and h agree on the common part of their
domains. But then Ah \Ah 6= ? mod D, contradicting partition property (ii). So A is
countable, which impliesG0 =fg2G : (9h2FI(G))(Ah2A and g2dom(h)gis countable.
Clearly A FIs(G0), so B is supported by FIs(G0) mod D.
37
Theorem 6.3. There is a family of c surjective functions in !! which is independent modulo
the Fr echet  lter F.
Proof. Consider the triples (A;hCk : k<ni;hjk : k<ni), where A is a  nite subset of !,
n2!, the sets Ck are distinct subsets of A, and jk2! for each k<n. The collection of all
such triples is countable. Let f(Ai;hCik : k<nii;hjik : k<nii) : i2!g be an enumeration.
For each B ! de ne a function fB : !!! by
fB(i) =
8
>><
>>:
jik if B\Ai = Cik
0 otherwise
:
Claim ffB : B !g is as desired. Let fBk : k<ng be a  nite collection of distinct subsets
of ! and fjk : k<ng a  nite set of values.
First we show there exists i2! such that fBk(i) = jk for each k. For each l6= m<n
let alm2Bl Bm. Let A =falm : l6= m<ng. Then A is a  nite subset of ! such that the
Bk\A are distinct . Let i2! such that ni = n, Ai = A, and for each k, Cik = Bk\Ai and
jik = jk. Then fBk(i) = jk for each k.
jffB : B !gj = 2!: If B1 6= B2  ! and j1 6= j2 2 !, then there exists i 2 ! s.t.
fB1(i) = j1 and fB2(i) = j2, so that fB1 6= fB2. It is also clear that the functions fB are
surjective (i.e. for any j2! there exists i2! such that fB(i) = j).
Now we show fi 2 ! : fBk(i) = jk for each k  ng 6= ? mod F. Note that \6=
? mod F" simply means \in nite." Suppose not. Enumerate the setfik : n+1 k n+mg
and letfBk : n+ 1 k n+mgbe collection of distinct subsets of !, distinct from the Bk
(1 k n). For each k2fn+ 1;:::;n+mg let jk2! with jk6= fBk(ik). Then
fi2! : fBk(i) = jk for each k n+mg= ?,
a contradiction.
38
Theorem 6.4. If I is a linear order with jIj= c, then there exists a free ultra lter u on !
such that I admits an invariant embedding into !!=u.
Proof. Let F be the Fr echet  lter on !. LetG !! be a family of c surjective function such
thatG is independent mod F (Theorem 6.3). Let D F be a maximal  lter modulo which
G is independent (Theorem 6.1 (ii)). Enumerate G = fft : t2Ig. For each pair s < t2I
let
Bs;t =fn2! : fs(n) <ft(n)g:
For each pair r<s2I and each function g2!! such that g 1(l) is supported by FIs(fft :
t2In[r;s]g) mod D for all l2!, let
Cg;r;s =fn2! : g(n) <fr(n) or fs(n) <g(n)g:
Claim. The sets Bs;t;Cg;r;s have the  nite intersection property (any  nite intersection
contains Ah (mod D) for some h2FI(G)).
Assuming the claim holds, there is an ultra lter u containing all of the sets Bs;t;Cg;r;s.
The map s7!fs=u is an invariant embedding from I into !!=U:
(i) Suppose s<t2I. Then Bs;t2u, so fs=u<ft=u.
(ii) Suppose (I1;I2) is a ( ; )-cut of I and g 2 !!. For each l 2 ! there exists a
countableGl G such that g 1(l) is supported by FIs(Gl) mod D. Then g 1(l) is supported
by Sl2!Gl for all l2!. Since  ; >! and   Sl2!Gl  = !, there exist r2I1 and s2I2 such
that g 1(l) is supported by FIs(fft : t2In[r;s]g) mod D for all l2!. Thus Cg;r;s2u, so
either g=u<fr=u or fs=u<g=u.
Proof of Claim. Consider a  nite intersection
n\
i=1
Bui;vi\
b\
k=1
Cgk;rk;sk:
39
Re-label the indices ui;vi;rk;sk from t0 to ta in increasing order. For each k b let  (k) a
such that t (k) = rk and  (k) a such that t (k) = sk. To show Tni=1Bui;vi\Tbk=1Cgk;rk;sk 6=
?, it su ces to show
\
i<j a
Bti;tj\
\
k b
Cgk;t (k);t (k) 6= ?:
LetT =ffti : i agand for k b letTk =ffti : i =2[ (k); (k)]g. We de ne a sequence
of functions hm2FI(G) so that
(1) hm hm+1
(2) dom(hm)\T = ?
(3) if h :T !! and k b, then for m su ciently large either
(i) there exists l2! such that Ahm[h jTk  g 1k (l) mod D, or
(ii) Ahm[h \g 1k (l) = ? mod D for all l2!.
Enumerate the countably many pairs (h ;k) where h :T !! and k b. Suppose m2!
and hm 1 has been de ned for (h m 1;km 1) (to de ne h0, follow the cases below and ignore
\hm 1"). We now de ne hm for (h m;km):
Case 1: There exists l2! such that Ahm 1[h m\g 1km(l)6= ? mod D. By assumption g 1km(l)
is supported by FIs(fft : t 2 In[ (km); (km)]g) mod D. So there exists ~h 2 FI(fft :
t 2 In[ (km); (km)]g) such that A~h  g 1km(l) mod D and A~h\Ahm 1[h m 6= ? mod D.
In particular, dom(~h)\(T nTkm) = ?, and we may assume hm 1[h m  Tkm  ~h. Let
hm = ~hn(h m jTkm). Then Ahm[h m Tkm = A~h  g 1km(l) mod D, dom(hm)\T = ?, and
hm 1 hm.
Case 2: Ahm 1[h m\g 1km(l) = ? mod D for all l2!. Let hm = hm 1.
40
Our goal now is to  nd h :T !! and M2! such that
AhM[h  
\
i<j a
Bti;tj\
\
k b
Cgk;t (k);t (k) mod D: (6.1)
For each in nite W  ! let W(T) denote the set of all increasing functions h :T !W. For
each k b de ne ?k : !(T)!f1;2;3;4g by
?k(h ) =
8>
>>>
>>>>
>><
>>>>
>>>
>>>
:
1 if (i) and h (ft (k)) l h (ft (k))
2 if (i) and l<h (ft (k))
3 if (i) and l>h (ft (k))
4 if (ii)
:
De ne  : !(T) !f1;2;3;4gf1;:::;bg by  (h ) = (?1(h );:::;?b(h )). Then  is an a-place
function on ! with  nite range. By Ramsey?s Theorem there exists an in nite W  !
such that   W(T) is constant, i.e., such that ?k  W(T) is constant for each k  b. Let
h 2W(T). For any m we have
Ahm[h  Ah  
\
i<j a
Bti;tj:
Now  x k b.
Case 1: ?k(h ) = 1. This case may be ruled out by our selection ofW: Supposing?k(h ) = 1,
we have ?k  W(T)  1. Using the fact that W is in nite, de ne h0 2 W(T) so that 
  fw2W : h0(f
t (k)) w h0(ft (k))g
  
 = 2ji2! :  (k) i  (k)j. Then ?k(h0) = 1. Let
l1 2! satisfying (i), such that h0(ft (k))  l1  h0(ft (k)). We may modify h0 on functions
fti, i2 [ (k); (k)], to get h002W(T) so that either h00(ft (k)) > l1 or l1 < h00(ft (k)). Again,
?k(h00) = 1. Let l2 2! satisfying (i), such that h00(ft (k)) l2  h00(ft (k)). We have l1 6= l2,
41
Ahm[h0 Tk  g 1k (l1) mod D, Ahm[h00jTk  g 1k (l2) mod D. But Ahm[h0 Tk = Ahm[h00 Tk. Con-
tradiction.
Case 2: ?k(h ) = 2. Let l2! such that Ahm[h  Tk  g 1k (l) mod D for m su ciently large
and l<h (ft (k)). Then
Ahm[h  Ahm[h  (Tk[fft (k)g) fn2! : gk(n) = l<h (ft (k))^h (ft (k)) = ft (k)(n)g mod D
(6.2)
Case 3: ?k(h ) = 3. Similar to Case 2. For m su ciently large,
Ahm[h  fn2! : ft (k)(n) <gk(n)g mod D: (6.3)
Case 4: ?k(h ) = 4. For m su ciently large, Ahm[h \g 1k (l) = ? mod D for all l2!. As
ID is closed under  nite unions,
Ahm[h \
[
l h (ft (k))
g 1k (l) = ? mod D:
That is, almost none of the points in Ahm[h map under gk to  h (ft (k)). Thus,
Ahm[h  fn2! : ft (k)(n) = h (ft (k))^h (ft (k)) <gk(n)g mod D: (6.4)
Each set in (6.2)-(6.4) containing Ahm[h (mod D) is contained in Cgk;t (k);t (k). For each
k b let mk be su ciently large for one of (6.2)-(6.4) to hold. Letting M = maxk b(mk),
we have (6.1).
Corollary 6.5. There exists u 2 ! such that !!=u has a ( ; )-cut for each pair of un-
countable regular  ;  c.
Corollary 6.6. If I is a linear order with jIj= c, then there exists a free ultra lter u on !
such that I admits an invariant embedding into R!=u.
42
Proof. We show the inclusion !!=u ,!R!=u is invariant. Suppose (A;B) is a ( ; )-cut of
!!=u. Let (f =u) < and (g =u) < be co nal and coinitial in A and B, respectively. For a
contradiction, suppose there exists h=u2R!=u with f =u<h=u<g =u for all  < and
 < . We may assume h(n) 0 for all n2!. Let h 2!! be de ned by h (n) =bh(n)c.
It must be the case that E1 = fn2! : h(n) 2Rn!g2u, so h =u < h=u. There exists
 0 < and E2 2u s.t. h (n) <f 0(n) <h(n) for all n2E1\E2 6= ?. This is impossible,
as h(n) h (n) < 1 for all n.
6.2 The linear orders J ( < 2c)
Theorem 6.7. If  >!1 is a regular cardinal, then there exists a set fI :  < 2 g of linear
orders satisfying
(i) cf(I ) =jI j=  .
(ii) If  6=  and ? : I !L, ? : I !L0 are co nal invariant embeddings, then L
and L0 have no isomorphic  nal segments.
Proof. There exists a partition fS :  < g of the stationary set S =f < : cf( ) = !1g
into  pairwise disjoint stationary subsets (Theorem 2.28). Fix X   . For each  <  ,
de ne
 X =
8
>><
>>:
!1 if  2S 2XS 
!2 otherwise
.
De ne IX = f( ; ) :  <  ; <  X g. Give IX the lexicographic ordering, with the order
reversed in the second factor.
( X0 ;0]( X1 ;0]            ( X ;0]               
Figure 6.1: IX
Suppose X6= Y   and ?X : IX !L, ?Y : IY !L0 are co nal invariant embeddings
into linear orders L;L0, respectively. For a contradiction, suppose  : M ! M0 is an
43
isomorphism between  nal segments M, M0 of L, L0, resp. For each  < , let
M =fm2M : m<?X( ;0) for some  < g
M0 =fm02M0 : m0<?Y ( ;0) for some  < g:
Let C = f <  :  [M ] = M0 g. Claim. C is closed unbounded in  . Assuming the
claim holds, let  2 X Y. Without loss of generality assume  2 XnY. There exists
 2C\S . We have  [M ] = M0 and  X = !1. Since the stationary sets S are pairwise
disjoint,  2S , and  2XnY, we have  =2S for any  2Y. Thus  Y = !2. Since ?X
maps co nally into L, M 6= ? for  su ciently large. As C\S is unbounded in  , we may
assume  was chosen so that M 6= ?.
Figure 6.2: Theorem 6.7
Cut IX and IY directly below (!1;0] and (!2;0] , respectively. Since cf( ) = !1, these
are (!1;!1) and (!1;!2) cuts of IX and IY , respectively. The assumption that ?X is an
invariant embedding implies M has no elements between M and ?X(!1;0] . Similarly, there
are no elements of M0 between M0 and ?Y (!2;0] . So coi(MnM ) = !1 and coi(M0nM0 ) =
!2. But  [M ] = M0 implies  [MnM ] = M0nM0 . Contradiction.
Proof of Claim. First we show that M = S < M (and M0 = S < M0 ) when  is a limit
ordinal. We just need to show M  S < M . To that end suppose m2M . Then there
exists  < such that m<?X( ;0). Since  is a limit ordinal there exists  < < . We
have m2M .
44
Now we can show C is unbounded in  . To that end, let  < . Since ?X and ?Y are
co nal and order preserving, we may choose a strictly increasing sequence ( i)i2! of elements
in  such that  <  0 and M0 i   (M j)  M0 k for all i < j < k2!. Let  = supi2! i.
Then
 [M ] =
[
i2!
 [M i] =
[
i2!
M0 i = M0 :
So  <  2C, proving C is unbounded in  . Now suppose  is a limit point of C (a limit
ordinal to which elements in C limit). Then
 [M ] =  [
[
 < 
M ] =  [
[
 2C\ 
M ] =
[
 2C\ 
 [M ] =
[
 2C\ 
M0 = M0 ;
so  2C. This proves C is closed.
Theorem 6.8. If  >!1 then there exists a set fJ :  < 2 g of linear orders satisfying
(i) jJ j=  
(ii) coi(J ) = cf( ) +!2
(iii) if  6=  and ? : I !L, ? : I !L0 are coinitial invariant embeddings, then L
and L0 have no isomorphic initial segments.
Proof. If  is regular, this follows from the previous theorem. Suppose  is singular. By
Theorem 2.25 there exists a setf i : i< cf( )gof regular cardinals, each  i >!1, such that
supi<cf( ) i =  and
Y
i<cf( )
2 i = 2 :
Let  = cf( ) + !2. There exists a partition fS :  < cf( )g of the stationary set
S =f < : cf( ) = !1g into cf( ) pairwise disjoint stationary subsets. For each  2S, let
h( )2cf( ) such that  2Sh( ), and for  2 nS, let h( ) = 0.
For each i< cf( ) let fIi; :  < 2 ig be the set of linear orders of cardinality  i given
by Theorem 6.7. For each v 2Qi<cf( ) 2 i, de ne Jv = f( ;x) :  <  ; x2Ih( );v(h( ))g
45
with the lexicographic order reversed in the  rst factor. Note: Jv needs at least one element
when  2 nS, to ensure its coinitiality is  instead of only cf( ).
            (Ih( );v(h( )))         (Ih(1);v(h(1)))(Ih(0);v(h(0)))
Figure 6.3: Jv
Suppose u;v 2Qi<cf( ) 2 i with u6= v and ?u : Ju !L, ?v : Jv !L0 are invariant
coinitial embeddings. For a contradiction suppose there is an isomorphism  : M ! M0
between initial segments M and M0 of L and L0, respectively. For each  < , let
M =fm2M : m>?u( ;x) for some  < and x2Ih( );u(h( ))g
M0 =fm02M0 : m0>?v( ;x) for some  < and x2Ih( );v(h( ))g:
There exists  2 S s.t. v(h( )) 6= u(h( )) (h  S maps onto cf( )). Choose  2
Sh( )\f < :  [M ] = M0 g with M 6= ?. Note that h( ) = h( ). Consider the cuts in L
and L0 below M and M0 respectively.
Figure 6.4: Theorem 6.8
Recall cf(Ih( ); ) =  h( ) and cf( ) = !1 are regular uncountable. By the invariant
property of ?X and ?Y , M nM and M0nM0 have Ih( );v(h( )) and Ih( );u(h( )) co nally
invariantly embedded. But  [M ] = M0 , so MnM  M0nM0 . This contradicts a property
of the linear orders fIh( ); :  < 2 h( )g.
46
We conclude this section by making a slight modi cation to Theorems 6.7 and 6.8.
Suppose L and L0 are linear orders. An order preserving map L ! L0 is an invariant-
1 embedding if the image of each ( ; )-cut of L is  lled by precisely one element of L0.
Theorems 6.7 and 6.8 hold if we replace \invariant" with \invariant-1." For instance, if in
the proof of Theorem 6.7 there are unique l2L and l02L0 such that M <l <?X(!1;0] 
and M0 <l0 <?Y (!2;0] , then  must map l to l0. The contradiction follows as before. In
the next sections we will apply Theorem 6.8 with this modi cation, when  = c and CH
fails. It is consistent that c is not regular, so the singular case provided by Theorem 6.8 is
of use.
6.3 A Quick Proof of Theorem 1.1
Let R!=u denote the completion of R!=u (i.e., R!=u = Iu= ).
Theorem 6.9 (: CH). There exists a family fD :  < 2cg of free ultra lters on ! and
a collection f[L1 ;L2 ] :  < 2cg of continua, [L1 ;L2 ]  ID for each  < 2c, such that
[L1 ;L2 ]6?[L1 ;L2 ] for any  < < 2c.
Proof. LetfJ :  < 2cgbe the family of linear orders of cardinality c given by Theorem 6.8.
By Corollary 6.6, for each  < 2c there exists D 2! such that J + !1 has an invariant
embedding ? into R!=D (an invariant-1 embedding into R!=D ). Let L1 = inf? [J +!1]
and L2 = sup? [J +!1]. Then [L1 ;L2 ], the union of the layers in ID between L1 and L2 ,
is a subcontinuum of ID ([L1 ;L2 ] = Tf[aD ;bD ] : aD <L1 and L2 <bD g).
Figure 6.5: [L1 ;L2 ]
47
Suppose  < < 2c and [L1 ;L2 ]?[L1 ;L2 ]. Arguing as in the proof of Theorem 5.13,
there is an order preserving or order reversing isomorphism between [L1 ;L2 ] and [L1 ;L2 ]
(their linearizations). Then (L1 ;L2 ) and (L1 ;L2 ) must be isomorphic via an order preserving
map, since their coinitialities are cf(c) + !2 and their co nalities are !1. This contradicts
a property of the linear orders J (J and J are coinitially invariantly-1 embedded into
(L1 ;L2 ) and (L1 ;L2 ), respectively).
Note also that J0 +!1 +J1 +!1 +:::+Jc +!1 may be invariantly embedded into some
R!=u, yielding a \chain" of c pairwise nonhomeomorphic subcontinua of Iu.
6.4 The Proof of Theorem 1.3
We require one additional result from [3].
Theorem 6.10. If  6=  , then the number of ( ; )-cuts of a linear order I is at most jIj.
Proof. Suppose I is a counterexample of minimal cardinality and letf(Ai;Bi) : i<jIj+gbe
a set of jIj+ distinct ( ; )-cuts of I. Let  = cf(jIj). Enumerate I = fi :  <jIjg. jIj is
the supremum of  many   withj  j<jIjfor each  < (Theorem 2.22). For each  < 
let I =fi :  <  g. Then jI j<jIj, I = S < I , and I  I for all  < < .
Case 1:  6=  ; . Claim: For each i<jIj+ there exists  i < such that Ai\I i is co nal
in Ai. Let i<jIj+. Suppose  < . Let (a ) < be co nal in Ai. Since  = S < f < :
a 2I g and  is regular, there exists  i <  such that jf < : a 2I igj =  (Theorem
2.21). Then Ai\I i is co nal in Ai. Suppose  < and there no  < such that Ai\I is
co nal in Ai. We may recursively de ne strictly increasing sequences (a ) < and (  ) < 
such that for each  < , a 2Ai\I  and Ai\I  <a for all  < . Then (a ) < is co nal
in Ai, a contradiction. This completes or proof of the claim. Similarly, for each i <jIj+
there exists  i <  such that Bi\I i is co nal in Bi. Thus, for each i <jIj+ there exists
 i < such that Ai\I i is co nal in Ai and Bi\I i is co nal in Bi. Because  <jIj+,jIj+
is regular, and jIj+ = S < fi<jIj+ :  i < g, there exists X jIj+ and  < such that
48
jXj=jIj+ and  i =  for all i2X (Theorem 2.21). But thenf(Ai\I ;Bi\I ) : i2Xgis
a set of jIj+ distinct  ; cuts of I , contradicting minimality of jIj.
Case 2:  =  . Then  6=  . Arguing as in the previous case, there exists a subset
X jIj+ of cardinality jIj+ and  < such that Bi\I is coinitial in Bi for each i2X.
We may assume there are less thanjIj+ many i?s for which Ai\I is co nal in Ai, otherwise
a contradiction follows as in the previous case. Therefore we may assume for each i2X,
Ai\I is not co nal in Ai. Then for each i 2 X there exists ai 2 AinI such that
Ai\I < ai < Bi. Suppose i 6= j 2 X. The cuts (Ai;Bi), (Aj;Bj) are distinct, so one
of Bi and Bj must be a proper subset of the other. Assume Bi  Bj, so that BjnBi is a
nonempty subset of Ai. Then there exists c2(BjnBi)\I  Ai\I . We have aj <c<ai,
thus ai6= aj. So fai : i2Xg is a collection of jIj+ distinct elements of I. The case  =  
yields a contradiction similarly.
Suppose I is a linear order,  ; > ! are regular cardinals, A;B I and x2I. Then
(A;x;B) is a ( ;1; )-cut of I if A<x<B, I = A[fxg[B, cf(A) =  , and coi(B) =  .
Proof of Theorem 1.3. LetfJ :  < 2cgbe the family of linear orders of cardinality c given
by Theorem 6.8. By Corollary 6.6, for each  < 2c there exists D 2! such that !1 + J 
invariantly embeds into R!=D (invariantly-1 embeds into R!=D ).
Fix  < 2c and let
E =f < 2 : !1 +J has an invariant-1 embedding into R!=D g:
We showjE j c. For each  2E let ? : !1 +J !R!=D be an invariant-1 embedding.
Let  = !1 and  = cf(c) + !2. For each  2E , the image under ? of the cut (!1;J )
of !1 + J produces a ( ;1; )-cut (A ;x ;B ) of R!=D . Each ( ;1; )-cut of R!=D 
corresponds to either a ( ; )-cut or a ( ;1; )-cut of R!=D , each type of which there
are only c many (Theorem 6.10). So if jE j > c, there exist  1 6=  2 2 E such that
(A 1;x 1;B 1) = (A 2;x 2;B 2). In particular B 1 = B 2. But ? 1  J 1 and ? 2  J 2
49
are coinitial invariant-1 embeddings of J 1 and J 2 into B 1 and B 2, respectively. This
contradicts a property of the orders J .
We may now recursively de ne X 2 ,jXj= 2 , such that !1+J admits an invariant-1
embedding intoR!=D but notR!=D for any  < 2X. Thus there is no order preserving
isomorphism between R!=D and R!=D , for  <  2X. For a  xed  2X, there is at
most one  2X such that there exists an order reversing isomorphism between R!=D and
R!=D , so there exists S  X, jSj = 2c, such that there is no order preserving or order
reversing isomorphism between R!=D and R!=D , for  < 2S.
By Theorem 5.13, we have the following.
Theorem 6.11 (: CH). There exists a family fD :  < 2cg of free ultra lters on ! such
that ID 6?ID for any  < < 2c.
6.5 Concluding Remarks
In [10], A. Dow proves there are also 2c subcontinua of  HnHwhen CH holds. Combined
with Theorem 1.1, we have the following theorem of ZFC.
Theorem 6.12. There exist 2c pairwise nonhomeomorphic subcontinua of  HnH.
Prior to this result approximately 20 subcontinua were found in the ZFC setting, many
by M. Smith in [7]. In [7] it is also shown that the layers of Iu are indecomposable continua
unlike Iu. Note that each ( ; )-cut of R!=u corresponds to a layer in Iu. A. Dow indicates
in [10] that the following question remains open: If CH fails, are there 2c pairwise nonhome-
omorphic indecomposable subcontinua of  HnH? In particular, it is not known if one can
produce 2c pairwise nonhomeomorphic layers by the method of invariantly embedding linear
orders into ultrapowers.
50
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51