Properties of Nonmetric Hereditarily Indecomposable Subcontinua of
Finite Products of Lexicographic Arcs
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 classi ed information.
Regina Greiwe Jackson
Certi cate of Approval:
Stewart Baldwin
Professor
Mathematics and Statistics
Michel Smith, Chair
Professor
Mathematics and Statistics
Gary Gruenhage
Professor
Mathematics and Statistics
Jo Heath
Professor Emerita
Mathematics and Statistics
George Flowers
Dean
Graduate School
Properties of Nonmetric Hereditarily Indecomposable Subcontinua of
Finite Products of Lexicographic Arcs
Regina Greiwe Jackson
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Ful llment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
December 18, 2009
Properties of Nonmetric Hereditarily Indecomposable Subcontinua of
Finite Products of Lexicographic Arcs
Regina Greiwe Jackson
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
Regina Marie Greiwe, daughter of Michael Greiwe and Karen Broome, was born Febru-
ary 4, 1982. She attended Tri-County High School in Buena Vista, Georgia where she
graduated fourth in her class in 1999. She then entered Columbus State University under
an Honors scholarship where she graduated Magna cum Laude in June, 2003. She was
awarded with a Bachelor of Science degree in Applied Mathematics. While at Columbus
State University, she was president of the local chapter of the MAA and was awarded the
Mathematics Award in 2003. The following year she entered Auburn University to pursue
a Master of Science degree in Mathematics. While at Auburn University, she served as
president of the local chapter of SIAM and has been involved in Science Olympiad and
mathematics education. In 2006, she was awarded a Master of Science in Mathematics.
She is the wife of Raymond Jackson, Jr., and she is the mother of Laura Jeanette Amilia
Greiwe, and Caroline Elizabeth Jackson.
iv
Dissertation Abstract
Properties of Nonmetric Hereditarily Indecomposable Subcontinua of
Finite Products of Lexicographic Arcs
Regina Greiwe Jackson
Doctor of Philosophy, December 18, 2009
(M.A., Auburn University, 2006)
(B.S., Columbus State University, 2003)
41 Typed Pages
Directed by Michel Smith
Nonmetric hereditarily indecomposable subcontinua of  nite products of Lexicographic
arcs are examined. It is shown that these subcontinua cannot intersect certain subsets of the
products. Then nonmetric hereditarily indecomposable subcontinua of these same products
cross the Hilbert cubes are examined and are shown to not intersect certain subsets. In
conclusion, it is shown that all hereditarily indecomposable subcontinua of the product of
three Lexicographic arcs are metric.
v
Acknowledgments
I would like to thank my family for helping me take care of lifes little bumps so that
I could concentrate on my work. I would also like to thank my topology professors for
providing me with inspiration along the way. I would like to thank Jo Heath for pushing me
to work with Dr. Michel Smith. She knew, better than I did, where I would excel. Thank
you Dr. Smith for being patient with me, providing me with inspiration (and an interesting
problem), and showing so much excitement for what I?ve done. I would also like to thank
you, the reader, for taking the time to look at my thoughts and work.
vi
Style manual or journal used Journal of Approximation Theory (together with the style
known as \aums"). Bibliography follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (speci cally LATEX)
together with the departmental style- le aums.sty.
vii
Table of Contents
List of Figures ix
1 Introduction and Background 1
2 Nonmetric Hereditarily Indecomposable Subcontinua of Finite Prod-
ucts of Lexicographic Arcs 6
3 Nonmetric Hereditarily Indecomposable Subcontinua of Finite Prod-
ucts of Lexicographic Arcs and Hilbert Cubes 16
4 Conclusions for the Product of Three Lexicographic Arcs and the
Product of Two Lexicographic Arcs with the Hilbert Cube 27
Bibliography 30
Appendices 31
A Notes on the style-file project 32
viii
List of Figures
1.1 The Lexicographic Arc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 L2 and L3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.1 Proposition 2.1: A, B, and C . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Proposition 2.1: The Component I . . . . . . . . . . . . . . . . . . . . . . . 8
2.3 Proposition 2.2: Convergent Sequence in P10 . . . . . . . . . . . . . . . . . . 9
2.4 Proposition 2.2: Convergent Sequence and the component I . . . . . . . . . 10
2.5 Proposition 3.3: K is metric. . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.6 Proposition 3.3: Case 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.7 Proposition 3.3: Case 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.8 Proposition 3.3: Case 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.1 Theorem 2.1: A , B , and C . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.2 Theorem 2.1:    1   2  14 . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.3 Proposition 3.1: A, B, and C . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.4 Proposition 3.1: The Component I . . . . . . . . . . . . . . . . . . . . . . . 21
3.5 Proposition 3.2: Convergent Sequence in P10 . . . . . . . . . . . . . . . . . . 22
3.6 Proposition 3.2: Convergent Sequence and the Component I . . . . . . . . 23
3.7 Proposition 2.3: K is metric. . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.8 Proposition 2.3: Case 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.9 Proposition 2.3: Case 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.10 Proposition 2.3: Case 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
ix
Chapter 1
Introduction and Background
A continuum is traditionally de ned as a compact, connected, metric space. One of the
simplest examples is an arc, which is a space homeomorphic to the unit interval. Basically
it is a continuum with two endpoints, one of which is mapped to 0, and the other which
is mapped to 1. In this paper we are concerned with one of many nonmetric analogs to
the arc, the Lexicographic arc. Throughout we will de ne a continuum to be a compact,
connected Hausdor space. If a continuum is metric, it will be stated.
In 1951, Bing showed that higher dimensional metric continua must contain non-
degenerate hereditarily indecomposable subcontinua [1]. He also is responsible for showing
that the set of all pseudo-arcs is a dense G set in Rn [2]. In a sense this means that the set of
all hereditarily indecomposable subcontinua of Euclidean space is large. Recently, Michel
Smith has been studying how hereditarily indecomposable continua sit inside nonmetric
continua [6] [7] [8] [9] [10] [11]. Surprisingly, it seems that metrizability and hereditary
indecomposability could possibly be linked. He has shown that all hereditarily indecompos-
able subcontinua of the inverse limit of both Souslin arcs and Lexicographic arcs are metric
[8] [9]. Also, he has shown that the product of two Souslin arcs contain only metric hered-
itarily indecomposable continua [10] [11]. The purpose of this paper is to investigate the
existence of hereditarily indecomposable continua in  nite products of Lexicographic arcs.
Michel Smith, in conjunction with Jennifer Stone, showed that every hereditarily indecom-
posable subcontinuum of the product of two Lexicographic arcs is metric [11]. The author
will develop a technique to restrict the existence of nonmetric hereditarily indecomposable
subcontinua in  nite products, and in  nite products cross the Hilbert cube. Most of the
techniques used can be extended to a countable product of Lexicographic arcs.
1
The goal is to show that any hereditarily indecomposable subcontinuum of a  nite
product of Lexicographic arcs is metrizable. In this paper, the author will prove it for the
product of three of these arcs.
Let us start by describing the Lexicographic arc and by introducing some notation
that will be used throughout the paper. The Lexicographic arc is the set [0;1] [0;1] given
an order topology based on the following order. We say that (a;b) < (c;d) provided that
either a < c, or a = c and b < d. This produces a compact, connected, Hausdor space
with endpoints (0;0) and (1;1). We will denote the Lexicographic arc with L, and we
will introduce a notation of a point in [0;1] [0;1] to simplify the interval notation and
points in higher dimensional products. Let the point (x;y) = xy. Further research into the
Lexicographic arc can be found in the author?s Masters Thesis [13].
One way to think of the lexicographic arc is to think of the unit square as an arc which
has each point replaced with another arc. Refer to Figure 1.1.
Figure 1.1: The Lexicographic Arc
In turn, products of n Lexicographic arcs can be thought of as the unit n-cube with
each point replaced with an n-cube. Figure 1.2 depicts the product of two and three said
arcs. We will denote the product of n Lexicographic arcs as Qni=1L = Ln.
During our exploration we will have need of two more spaces, the Hilbert cube and the
compact Double Arrow Space. The Hilbert cube is de ned as the countable product of unit
intervals, and it will be denoted by [0;1]1 = Q1i=1[0;1] with the usual topology on [0;1].
2
Figure 1.2: L2 and L3
The Double Arrow space is the order topology induced by the lexicographic order described
at the beginning of this chapter on the set [0;1] f0;1g.
The following de nitions and theorems will be used extensively throughout the paper.
De nition 1.1. A space is said to be separable if it contains a countable, dense subset.
De nition 1.2. A space is said to be completely separable given it has a countable basis.
De nition 1.3. A continuum is indecomposable if it cannot be written as the union of
two proper subcontinuum.
De nition 1.4. A continuum is hereditarily indecomposable if every subcontinuum is
indecomposable.
De nition 1.5. Let A be a subset of a space X. The component of x2A is the union
of all connected subsets of A containing x.
De nition 1.6. Let A and B be subsets of a space X. X can be separated over A and
B provided that there exist disjoint open sets, U and V, such that U[V = X, A U, and
B V.
De nition 1.7. A point x2X is a cut point of X if X fxg is not connected.
De nition 1.8. An arc is a continuum having exactly two non-cut points.
3
Further exploration of the properties of a metric arc was done by Nadler [4].
De nition 1.9. A subset of a space is nowhere dense in the space if it has empty interior.
De nition 1.10. Let X be a topological space. Then the hyperspace of X denoted by 2X
is the space of nonempty compact subsets of X. Let fU1;U2;U3;:::;Ung be a  nite collection
of open subsets of X. Then the collection of sets of the form fK22XjK2Sni=1Ui and for
each 1 i n, Ui\K6=;g constitutes a basis for 2X.
De nition 1.11. Let X be a topological space. Then we denote the subspace of 2X known
as the hyperspace of continua as C(X) =fK22xjK is a continuumg.
For a more in depth look at hyperspaces refer to [5].
De nition 1.12. Let x be a point in a continuum X. De ne a partial order on C(X) using
inclusion. Then an order arc of x in X is an arc in C(X) with endpoints fxg and X.
Theorem 1.1. If A is a proper subcontinuum of an indecomposable continuum X, then A
is nowhere dense in X.
Theorem 1.2. If a space X is compact and completely separable, then X is metrizable.
Theorem 1.3. If an arc is separable, then it is completely separable.
Corollary 1.1. If an arc is separable, then it is metrizable.
Theorem 1.4. Let X be a compact Hausdor space, and let A and B be disjoint closed
subsets of X such that no component intersects both A and B. Then X can be separated
over A and B.
Theorem 1.5. Let X be a hereditarily indecomposable continuum. If E and F are disjoint,
closed subsets of X contained in open sets U and V respectively, then there exist closed sets
A, B, and C such that X = A[B[C, E A, F  C, A\B V  F, B\C U E,
and A\C =;.
A proof of the metric case can be found in [3].
4
Theorem 1.6. The order arc generated by a hereditarily indecomposable subcontinuum is
unique.
Theorem 1.7. Let X be a hereditarily indecomposable continuum, and let  (p) denote the
order arc of some point p2X. If there exists an open neighborhood, U of p, containing a
countable collection of open subsets, fGig1i=1, having the properties that for each H K2
 (p) such that H 6= K, there is an i <1 such that Gi\(K H) 6= ; and  Gi\H = ;,
then  (p) is separable.
Proof. Let X be a hereditarily indecomposable continuum, and choose p2X. Let U be
an open neighborhood, U of p, containing a countable collection of open subsets, fGig1i=1,
having the properties that for each H  K 2 (p) such that H 6= K, there is an i <1
such that Gi\(K H)6=;and  Gi\H =;. Let Xi be the irreducible subcontinuum of X
between p and  Gi. Then fXig1i=1 is countable.
We claim that fXig1i=1 is dense in  (p). Let H K2 (p) such that H6= K. Then
we must show that there exists i < 1 such that H  Xi  K and H 6= Xi 6= K. By
the hypothesis, there exists i <1 such that Gi\(K H) 6= ; and  Gi\H = ;. Since
 Gi\H =;, we have that H Xi and H6= Xi. Since Xi\Gi =;, by virtue of irreducibility,
and Gi\(K H) 6= ;, Xi  K and Xi 6= K. Thus fXig1i=1 is dense in  (p). Therefore
 (p) is separable.
5
Chapter 2
Nonmetric Hereditarily Indecomposable Subcontinua
of Finite Products of Lexicographic Arcs
In this chapter we will restrict the subsets that a nonmetric hereditarily indecomposable
subcontinuum of Ln can intersect. We will start with a metric subset and add in a dimension
of L at a time. We will de ne some notation for the subsets we will be using. Fix ai2[0;1]
for each 1 i n.
1. Let S = Qni=1(ai0;ai1). We will refer to this set as the interior of a metric cube.
2. Let Sj0 = Q1 i<j(ai0;ai1) faj0g Qj<i n(ai0;ai1) be known as the \jth" lower metric
face.
3. Let Sj1 = Q1 i<j(ai0;ai1) faj1g Qj<i n(ai0;ai1) be known as the \jth" upper metric
face.
4. Let Pj = Q1 i<j(ai0;ai1) L Qj<i n(ai0;ai1). We will refer to this set as the \jth"
tube.
5. Let Pj0 = Q1 i<j(ai0;ai1) [00;aj0) Qj<i n(ai0;ai1) be the \jth" lower tube.
6. Let Pj1 = Q1 i<j(ai0;ai1) (aj1;11] Qj<i n(ai0;ai1) be known as the \jth" upper
tube.
First we will show that if M is a nonmetric hereditarily indecomposable subcontinuum
of Ln, then it cannot intersect the interior of a metric cube. Please note that the following
theorems are also true for a countable product of lexicographic arcs.
For the following propositions  x ai2[0;1] for each 1 i n.
6
Proposition 2.1. If M is a nonmetric hereditarily indecomposable subcontinuum of Ln,
then M\S =;.
Proof. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln. Since M is
nonmetric, we may assume wlog that  1(M) = L. Suppose M\S6=;. Then there exists
a point, ~a2M\S. Notice that the boundary of S is homeomorphic to a sphere which is
metrizable. Let B be a countable basis for bd(S), and let fDjg1j=1 be the collection of all
 nite unions of elements of B.
Let us assume that a1 < 1, and let K be an irreducible subcontinuum of M from ~a to
f11g Qni=2L. Then we will letfK g 2 denote the components of K in Ln S indexed by
the set  . Then for each  2 , let ^K = K \bd(S). Notice that ^K is a nonempty closed
subset of bd(S). For each j <1 let Kj = S ^K  Dj K . Finally, let xj = lubf 1(Kj)g. So
we have used K to construct fxjgj<1, a countable subset of L.
We will now use Theorem 1.5 and the metrizability of bd(S) to show that fxjgj<1 is
uncountable, contradicting its construction. Let b2(a1;1). We claim that there exists j <
1such that xj2(b0;b1). Let E = M\([b1;11] Qni=2L), F =f~ag, U = (b0;11] Qni=2L,
and V = S. Then by Theorem 1.5, there exists closed subsets A, B, and C of Ln such that
M = A[B[C, E A, F  C, A\B (V F), B\C (U E), and A\C =;. Refer
to Figure 2.1.
Figure 2.1: Proposition 2.1: A, B, and C
7
Let ^A = A\bd(S), ^B = B\bd(S), and ^C = C\bd(S). Then ^B and ^A[ ^C are disjoint
closed subsets of bd(S). Hence by normality there exists disjoint open subsets of bd(S), O
and W, such that ^B O and ^A[ ^C W. By compactness, there exists j <1 such that
^B Dj  O. Then ^B Dj and ( ^A[ ^C)\Dj = ;. By the Theorem 1.4, there exists a
component, I, of B intersecting both A\B and B\C.
Figure 2.2: Proposition 2.1: The Component I
Since I B, I\bd(S) ^B Dj. So I S is a subset of Kj. Now B\C (U E) =
(b0;b1) Qni=2L, implying that xj >b0. Also since ^A\Dj = ;, we have ^A\Kj = ; and
A\Kj = ;. So E\Kj = ; which implies that xj < b1. So for each b2 (a1;1) there is a
j <1 such that xj 2 (b0;b1). Thus fxjgj<1 is uncountable, a contradiction. Therefore
M\S =;.
Next we will note that M cannot intersect an isolated metric face. Recall that a metric
face is de ned as Sj0 = Qi<j(ai0;ai1) faj0g Qi>j(ai0;ai1) or Sj1 = Qi<j(ai0;ai1) faj1g 
Q
i>j(ai0;ai1). The proof follows a similar argument to the proof of Proposition 2.1.
Proposition 2.2. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln,
and choose 1  j  n such that aj 62f0;1g. If ~a2M\Sj0, then there is a sequence in
8
M\Pj0 converging to ~a. Similarly if ~a2M\Sj1, then there exists a sequence in M\Pj1
converging to ~a.
Proof. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln, and wlog
choose j = 1 to simplify notation. Suppose that a1 62f0;1g and that  1(M) = L. Let
~a2M\S10. Suppose that there is no sequence in M\P10 converging to ~a. Then there is
an open neighborhood of ~a, O P10 such that M\O S10. Hence O\M is metrizable.
LetB be a countable basis for bd(S), and letfDkg1k=1 be the collection of  nite unions
of elements of B. Notice that a1 6= 1. Let K be an irreducible subcontinuum of M from ~a
tof11g Qni=2L. LetfK g 2 denote the components of K in Ln O indexed by  . Then
for each  > 0, let ^K = K \bd(S). Notice that ^K is a nonempty closed subset of bd(S).
Using compactness, for each k<1, let Kk = S ^K  DkK . Finally, let xk = lubf 1(Kk)g.
We will now use Theorem 1.5 and the metrizability of bd(S) to show that fxkgj<1
is uncountable, a contradiction. Let b2 (a1;1). We claim that there exists j <1 such
that xk 2 (b0;b1). Let E = M\([b1;11] Qni=2L), F = f~ag, U = (b0;11] Qni=2L, and
V = O. Then by Theorem 1.5, there exists A, B, and C, closed subsets of Ln such that
M = A[B[C, E A, F  C, A\B (V F), B\C (U E), and A\C =;. Notice
that A\V, B\V, and C\V are each subsets of S10. Let ^A = A\bd(S), ^B = B\bd(S),
and ^C = C\bd(S). Refer to Figure 2.3.
Figure 2.3: Proposition 2.2: Convergent Sequence in P10
9
Notice that ^C and ^A[ ^B are disjoint closed subsets of bd(S). By normality, there exists
disjoint open sets W1 and W2 such that ^C W1 and ( ^A[^B) W2. By compactness, we may
assume that there exists k<1such that W1 = Dk. Then ^C Dk and ( ^A[ ^B) Dk =;.
By Theorem 1.4, there exists a component of C intersecting bothf~agand B\C as depicted
in Figure 2.4.
Figure 2.4: Proposition 2.2: Convergent Sequence and the component I
Since I  C, I\(Ln O) is a subset of Kk. Since B\C  (b0;b1) Qni=2L, we
have that xk > b0. Now Kk\ ^A = ;, implying that Kk\E = ;. Thus xk < b1. Thus
xk2(b0;b1). So for each b2(a1;1) there is a k<1such that xk2(b0;b1). Thusfxkgk<1
is uncountable, a contradiction. Therefore there exists a sequence in M\P10 converging to
~a.
Similarly, if ~a2M\S11, then we can use K irreducible from ~a to f00g Qni=2L to
show that (0;a1) is countable, a contradiction. Therefore there exists a sequence in M\P11
converging to ~a.
Proposition 2.1 states that if ~z 2 M, then there is at least one coordinate, say xiyi,
for some 1  i n such that yi 2f0;1g. In other words, M is restricted to metric faces.
Proposition 2.2 states that these faces cannot be isolated.
10
For the following corollary, note that the subspace of L restricted to xy2L such that
y2f0;1g is homeomorphic to the Double Arrow Space, Z.
Corollary 2.1. If M is a nonmetric hereditarily indecomposable subcontinuum of Ln, then
M\Pj is embeddable in Z Qni=2(0;1).
Proof. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln, and choose
j = 1 to simplify notation. We claim that f : (M\P1) ! (Z Qni=2(0;1)) de ned by
f(a1y1;a2y2;   ;anyn) = (a1y1;y2;   ;yn) is an embedding. Since P1 = L Qni=2(ai0;ai1), we
have that f is an embedding if it is well-de ned. By Proposition 2.1, since yi62f0;1g for
1 <i n, y1 2f0;1g. Hence a1y1 2Z, and f is well-de ned.
We will now use Proposition 2.1 along with Theorem 1.4 to produce a separable order
arc in C(M). We will then use this order arc to show that M cannot intersect a tube.
Recall that a tube is Pj = Qi<j(ai0;ai1) L Qi>j(ai0;ai1).
Proposition 2.3. If M is a nonmetric hereditarily indecomposable subcontinuum of Ln,
and p2M\Pj, then  (p) is separable.
Proof. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln, and choose
j = 1 to simplify notation. Choose p2M\P1. Let U = M\P1. Then U is an open
neighborhood of p in M. LetBbe a countable basis for Qni=2(ai0;ai1),B0 be a countable basis
for [00;01) Qni=2(ai0;ai1), and B1 be a countable basis for (10;11] Qni=2(ai0;ai1). De ne
G = f(q0;r1) Bjq < r, q;r2 Q and B2Bg[fL BjB2Bg[B0[B1. Notice that G
is a countable collection of open subsets of U. To use Theorem 1.7, we will need to show
that for each H K2 (p) such that H6= K, there is a G2G such that  G\H =; and
G\(K H)6=;.
Let H K2 (p) such that H6= K. Now H and K are subcontinua of M implying
that they are each hereditarily indecomposable. Hence H is nowhere dense in K. Thus
there exists k2U\(K H) and there exists V, an open neighborhood of k in U, such that
V\H =;. By Proposition 2.1, k cannot be contained in a metric cube, thus k is contained
in a face. Now we will use a basis for P1 to  nd a G2G satisfying the desired properties.
11
Notice that either K is metric or nonmetric. If K is metric, then by Proposition 2.1
K cannot intersect any metric cube. So K fxyg Qni=2(ai0;ai1) where y2f0;1g. Hence
there exists B2B such that k2B fxyg  B fxyg V. Let G = B L. Then G2G,
k2G\(K H), and  G\H =;.
Figure 2.5: Proposition 3.3: K is metric.
Now suppose that K is nonmetric. Again, by Proposition 2.1, k2fxyg Qni=2(ai0;ai1)
for x2[0;1] and y2f0;1g. This gives rise to three cases:
1. k2f00g Qni=2(ai0;ai1),
2. k2f11g Qni=2(ai0;ai1), or
3. k2[01;10] Qni=2(ai0;ai1).
Case 1
Suppose k2f00g Qni=2(ai0;ai1).
Then k2V \[00;01) Qni=2(ai0;ai1). So there exists B2B0 such that k2B  B 
V \[00;01) Qni=2(ai0;ai1). Let G = B. Then G2G, k2G\(K H), and  G\H =;.
Case 2
Suppose k2f11g Qni=2(ai0;ai1).
12
Figure 2.6: Proposition 3.3: Case 1
Similar to Case 1, there is a B2B1 such that k2B  B V\(10;11] Qni=2(ai0;ai1).
Let G = B. Then G2G, k2G\(K H), and  G\H =;.
Figure 2.7: Proposition 3.3: Case 2
Case 3
Suppose k2[01;10] Qni=2(ai0;ai1).
Then k is in a metric face, and V is a nonmetric open subset of U. Since P1 =
L Qni=2(ai0;ai1), we have ab, cd 2 L such that a < c and W open in Qni=2(ai0;ai1) such
that k2 (ab;cd) W  V. By propositions 1 and 2, k is on an unisolated face. So there
exists ^k2(a1;c0) W  V. Now there exists B2B and q, r2Q such that  B W and
a<q<r<c. Let G = (q0;r1) B. Then G2G, ^k2G\(K H), and by proposition 3
 G = [q1;r0]  B = (q0;r1)  B V. Hence  G\H =;.
13
Figure 2.8: Proposition 3.3: Case 3
Hence we have satis ed the hypothesis for Theorem 1.7. Therefore  (p) is separable.
We will now show thatM cannot intersect a tube, by using Proposition 2.3 to contradict
the uncountability of an interval in L.
Proposition 2.4. If M is a nonmetric hereditarily indecomposable subcontinuum of Ln,
then M\Pj =; for each 1 j n.
Proof. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln, and choose
j = 1 to simplify notation. Suppose that p 2 M\P1. Then by Proposition 2.3  (p)
is separable. Let D = fDig1i=1 be a countable dense subset of  (p), and wlog assume
 1(M) = L. So  1(p) = xy for some xy2L. By Proposition 2.1, y2f0;1g. We can assume
that x< 1.
Let yi = lubf 1(Di)gfor each i<1, and letY =fyig1i=1. We claim thatY is dense in
(xy;11]. We must show that for ab <cd2(xy;11] there exists i<1such that yi2(ab;cd).
Let ef, gh2(ab;cd) such that ef <gh. Let H be an irreducible subcontinuum of M from p
tofefg Qni=2L, and let K be an irreducible subcontinuum of M from p tofghg Qni=2L.
Then H  K 2  (p). Hence there exists i < 1 such that Di 2 (H;K)   (p). Since
Di K, we have that Di\(gh;11] Qni=2L =;. So yi gh. Since H Di, we have that
14
Di\[ef;11] Qni=2L6= ;. so yi  ef. Then yi 2 [ef;gh]  (ab;cd). Thus Y is dense in
(xy;11] implying that (xy;11] is separable, a contradiction. Therefore M\P1 =;.
15
Chapter 3
Nonmetric Hereditarily Indecomposable Subcontinua
of Finite Products of Lexicographic Arcs and Hilbert Cubes
Our investigation continues with an exploration of how hereditarily indecomposable
continua behave in products of Lexicographic arcs and Hilbert cubes. The arguments in
this chapter are generalizations of the arguments in the previous chapter. The  rst theorem
follows from Theorem 1.5.
Theorem 3.1. If M is a hereditarily indecomposable subcontinuum of L [0;1]1, then M
is metric.
Proof. Let M be a hereditarily indecomposable subcontinuum of L [0;1]1. Suppose that
M is nonmetric. Then it can be assumed that  1(M) = L. For each 0 < < 14, let
E = M\([00; 0] [0;1]1) (3.1)
F = M\([(1  )1;11] [0;1]1) (3.2)
U = [00; 1) (3.3)
V = ((1  )0;11] (3.4)
Then E , F are closed subsets of M, and U , V are open subsets of L [0;1]1
containing E and F respectively. By Theorem 1.5, there exists nonempty closed subsets,
A , B , and C such that
M = A [B [C (3.5)
16
E  A (3.6)
F  C (3.7)
A \C =; (3.8)
A \B  (V  F ) (3.9)
B \C  (U  E ) (3.10)
We will now focus on the subset of L [0;1]1, X = [12 0; 12 1] [0;1]1. This subset is
homeomorphic to the Hilbert cube, [0;1]1, and thus it is metrizable. Refer to Figure 3.1
Figure 3.1: Theorem 2.1: A , B , and C 
Recall that these sets exist for each 0 <  < 14. Let ^A = X\(T   <1
4
A ), and let
^C = X\(T < <1
4
A )\C . We claim that ^A and ^C are nonempty. Note that each
is an intersection of a collection of closed subsets of a compact space. We will show that
the intersection of a  nite collection of these sets is nonempty. By compactness, the  nite
intersection property applies. So let  < 1 <   < m < 14. For each 1  i<j m we
have that U i  E j  A j and that V i  F j  C j by (3.1), (3.4), (3.6), and (3.9). Now
we have that A i\A j 6=;, C \A i 6=;, A i\X6=;, and C i\X6=; as in Figure 3.2.
Thus X\(A 1 \   \A m)\A 6=; and X\(A 1 \   \A m)\C 6=;. Hence by
the  nite intersection property, ^A 6=; and ^C 6=;.
Now recall that X = [12 0; 12 1] [0;1]1 is homeomorphic to the Hilbert cube, and we
have shown that ^A and ^C are nonempty closed subsets of X. Notice since A and C are
17
Figure 3.2: Theorem 2.1:    1   2  14
disjoint, ^A and ^C are disjoint. Hence by normality, there exist disjoint open subsets, O ,
W , of X such that ^A  O and ^C  W . Let B be a countable basis of X. Let D be
the collection of all  nite unions of elements of B. By construction, D is countable. By
compactness, it may be assumed that O and W are elements of D. Thus we have that
O = O for uncountably many 0 < < 14 and 0 < < 14. Straight from this construction,
we get that there exist  and  2(0; 14) such that  < and O = O = O . Since ^A  ^A 
and ^C  ^A , we have that ^C  O \W . By normality, O \W =;, a contradiction.
Therefore M is metrizable.
Now that we have shown that L [0;1]1 does not contain a nonmetric continuum,
let?s explore the subsets of Ln [0;1]1. We will start with a metric subset of Ln [0;1]1,
same as in the previous chapter, and add in a dimension of L at a time. Notation for
the subsets we will be using is rede ned. Fix ai 2 [0;1] for each 1  i n. Recall that
[0;1]1 = Q1i=1[0;1], the space known as the Hilbert cube.
1. Let S = Qni=1(ai0;ai1) [0;1]1. We will refer to this set as the interior of a metric
cube.
18
2. Let Sj0 = Q1 i<j(ai0;ai1) faj0g Qj<i n(ai0;ai1) [0;1]1 be known as the \jth"
lower metric face.
3. Let Sj1 = Q1 i<j(ai0;ai1) faj1g Qj<i n(ai0;ai1) [0;1]1 be known as the \jth"
upper metric face.
4. Let Pj = Q1 i<j(ai0;ai1) L Qj<i n(ai0;ai1) [0;1]1. We will refer to this set as
the \jth" tube.
5. Let Pj0 = Q1 i<j(ai0;ai1) [00;aj0) Qj<i n(ai0;ai1) [0;1]1 be the \jth" lower tube.
6. Let Pj1 = Q1 i<j(ai0;ai1) (aj1;11] Qj<i n(ai0;ai1) [0;1]1 be known as the \jth"
upper tube.
For the following propositions,  x ai2[0;1] for 1 i n.
Proposition 3.1. If M is a nonmetric hereditarily indecomposable subcontinuum of Ln 
[0;1]1, then M\S =;.
Proof. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln [0;1]1.
Since M is nonmetric, we may assume wlog that  1(M) = L. Suppose M\S6= ;. Then
there exists a point,~a2M\S. Notice that the boundary of S is homeomorphic to a sphere
which is metrizable. LetB be a countable basis for bd(S), and letfDjg1j=1 be the collection
of all  nite unions of elements of B.
Let us assume that a1 < 1, and let K be an irreducible subcontinuum of M from
~a to f11g Qni=2L [0;1]1. Then we will let fK g 2 denote the components of K in
(Ln [0;1]1) S indexed by the set  . Then for each  2 , let ^K = K \bd(S). Notice
that ^K is a nonempty closed subset of bd(S). By compactness, there is a j <1 such
that ^K  Dj . Let Kj = SfK jj = jg= S ^K  Dj K . Finally, let xj = lubf 1(Kj)g. So
we have used K to construct fxjgj<1, a countable subset of L.
We will now use Theorem 1.5 and the metrizability of bd(S) to show that fxjgj<1
is uncountable, contradicting its construction. Let b2 (a1;1). We claim that there exists
19
j < 1 such that xj 2 (b0;b1). Let E = M \([b1;11] Qni=2L [0;1]1), F = f~ag,
U = (b0;11] Qni=2L [0;1]1, and V = S. Then by Theorem 1.5, there exists closed subsets
A, B, and C of Ln [0;1]1 such that M = A[B[C, E A, F  C, A\B (V  F),
B\C (U E), and A\C =;. Refer to Figure 3.3.
Figure 3.3: Proposition 3.1: A, B, and C
Let ^A = A\bd(S), ^B = B\bd(S), and ^C = C\bd(S). Then ^B and ^A[ ^C are disjoint
closed subsets of bd(S). Hence by normality there exists disjoint open subsets of bd(S), O
and W, such that ^B  O and ^A[ ^C  W. By compactness, there exists j < 1 such
that ^B Dj  O. Then ^B Dj and ( ^A[ ^C)\Dj = ;. By Theorem 1.4, there exists a
component, I, of B intersecting both A\B and B\C.
Since I B, I\bd(S) ^B Dj. So I S is a subset of Kj. Now B\C (U E) =
(b0;b1) Qni=2L [0;1]1 implying that xj >b0. Also, since ^A\Dj =;, we have ^A\Kj =;
and A\Kj =;. So E\Kj =;, which implies that xj <b1. So for each b2(a1;1) there is
a j <1 such that xj 2(b0;b1). Thus fxjgj<1 is uncountable, a contradiction. Therefore
M\S =;.
Now we will show that if a point of M lies on a face, then there is a sequence of points
in M converging to it.
Proposition 3.2. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln,
and choose 1  j  n such that aj 62f0;1g. If ~a2M\Sj0, then there is a sequence in
20
Figure 3.4: Proposition 3.1: The Component I
M\Pj0 converging to ~a. Similarly if ~a2M\Sj1, then there exists a sequence in M\Pj1
converging to ~a.
Proof. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln [0;1]1,
and choose j = 1 to simplify notation. Suppose that a1 62f0;1g and that  1(M) = L. Let
~a2M\S10. Suppose that there is no sequence in M\P10 converging to ~a. Then there is
an open neighborhood of ~a, O P10 such that M\O S10. Hence O\M is metrizable.
LetB be a countable basis for bd(S), and letfDkg1k=1 be the collection of  nite unions
of elements of B. Notice that a1 6= 1. Let K be an irreducible subcontinuum of M from ~a
tof11g Qni=2L [0;1]1. LetfK g 2 denote the components of K in (Ln [0;1]1) O
indexed by  . Then for each  > 0, let ^K = K \bd(S). Notice that ^K is a nonempty
closed subset of bd(S). By compactness, there is a k < 1 such that ^K  Dk . Let
Kk = SfK jk = kg= S ^K  DkK . Finally, let xk = lubf 1(Kk)g.
We will now use Theorem 1.5 and the metrizability of bd(S) to show that fxkgj<1 is
uncountable, a contradiction. Let b2(a1;1). We claim that there exists j <1 such that
xk2(b0;b1). Let E = M\([b1;11] Qni=2L), F =f~ag, U = (b0;11] Qni=2L, and V = O.
Then by Theorem 1.5, there exists A, B, and C, closed subsets of Ln [0;1]1 such that
21
M = A[B[C, E A, F  C, A\B (V F), B\C (U E), and A\C =;. Notice
that A\V, B\V, and C\V are each subsets of S10. Let ^A = A\bd(S), ^B = B\bd(S),
and ^C = C\bd(S). Refer to Figure 3.5.
Figure 3.5: Proposition 3.2: Convergent Sequence in P10
Notice that ^C and ( ^A[^B) are disjoint closed subsets of bd(S). By normality there exists
disjoint open sets W1 and W2 such that ^C W1 and ( ^A[^B) W2. By compactness, we may
assume that there exists k<1such that W1 = Dk. Then ^C Dk and ( ^A[ ^B) Dk =;.
By Theorem 1.4, there exists a component of C intersecting bothf~agand B\C as depicted
in Figure 2.4.
Since I C, I\[(Ln [0;1]1) O] is a subset of Kk. Since B\C (b0;b1) Qni=2L,
we have that xk > b0. Now Kk\ ^A = ; implying that Kk\E = ;. Thus xk < b1. Thus
xk2(b0;b1). So for each b2(a1;1) there is a k<1such that xk2(b0;b1). Thusfxkgk<1
is uncountable, a contradiction. Therefore there exists a sequence in M\P10 converging to
~a.
Similarly, if~a2M\S11, then we can use K irreducible from~a tof00g Qni=2L [0;1]1
to show that (0;a1) is countable, a contradiction. Therefore there exists a sequence inM\P11
converging to ~a.
For the next proposition we will need a direct corollary of Proposition 3.1. Recall that
the \jth" tube is de ned as Pj = Q1 i<j(ai0;ai1) L Qj<i n(ai0;ai1) [0;1]1.
22
Figure 3.6: Proposition 3.2: Convergent Sequence and the Component I
Corollary 3.1. If M is a nonmetric hereditarily indecomposable subcontinuum of Ln 
[0;1]1, then M\Pj can be embedded in Z [0;1]1.
Proof. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln [0;1]1,
and choose j = 1 to simplify notation. We claim that f : (M\P1) ! (Z Qni=2(0;1))
de ned by f(a1y1;a2y2;   ;anyn;~x) = (a1y1;y2;   ;yn;~x) is an embedding. Since P1 = L 
Qn
i=2(ai0;ai1) [0;1]1, we have that f is an embedding if it is well-de ned. By Proposition
3.1, since yi 62 f0;1g for 1 < i  n, we have y1 2 f0;1g. Hence a1y1 2 Z, and f is
well-de ned.
Now we can use Proposition 3.2 and Corollary 3.1 to restrict nonmetric hereditarily
indecomposable subcontinua of Ln [0;1]1 even further.
Proposition 3.3. If M is a nonmetric hereditarily indecomposable subcontinuum of Ln 
[0;1]1, and p2M\Pj, then  (p) is separable.
Proof. Let M be a nonmetric hereditarily indecomposable subcontinuum of Ln [0;1]1,
and choose j = 1 to simplify notation. Choose p2M\P1. Let U = M\P1. Then U
is an open neighborhood of p in M. Let B be a countable basis for Qni=2(ai0;ai1) [0;1]1,
23
B0 be a countable basis for [00;01) Qni=2(ai0;ai1) [0;1]1, and B1 be a countable basis
for (10;11] Qni=2(ai0;ai1) [0;1]1. De ne G = f(q0;r1) Bjq < r, q;r 2 Q and B 2
Bg[fL BjB2Bg[B0[B1. Notice thatG is a countable collection of open subsets of U.
To use Theorem 1.7, we will need to show that for each H K2 (p) such that H6= K,
there is a G2G such that  G\H =; and G\(K H)6=;.
Let H K2 (p) such that H6= K. Now H and K are subcontinua of M implying
that they are each hereditarily indecomposable. Hence H is nowhere dense in K. Thus
there exists k2U\(K H) and there exists V, an open neighborhood of k in U, such that
V\H =;. By Proposition 3.1, k cannot be contained in a metric cube, thus k is contained
in a face. Now we will use a basis for P1 to  nd a G2G satisfying the desired properties.
Notice that either K is metric or nonmetric. If K is metric, then by Proposition 3.1 K
cannot intersect any metric cube. So K fxyg Qni=2(ai0;ai1) [0;1]1 where y2f0;1g.
Hence there exists B2B such that k2B fxyg  B fxyg V. Let G = B L. Then
G2G, k2G\(K H), and  G\H =;.
Figure 3.7: Proposition 2.3: K is metric.
Now suppose that K is nonmetric. Again, by Proposition 3.1, k2fxyg Qni=2(ai0;ai1) 
[0;1]1 for x2[0;1] and y2f0;1g. This gives rise to three cases:
1. k2f00g Qni=2(ai0;ai1) [0;1]1,
24
2. k2f11g Qni=2(ai0;ai1) [0;1]1, or
3. k2[01;10] Qni=2(ai0;ai1) [0;1]1.
Case 1
Suppose k2f00g Qni=2(ai0;ai1) [0;1]1.
Then k 2 V \([00;01) Qni=2(ai0;ai1) [0;1]1). So there exists B 2B0 such that
k2B  B V\([00;01) Qni=2(ai0;ai1) [0;1]1). LetG = B. ThenG2G, k2G\(K H),
and  G\H =;.
Figure 3.8: Proposition 2.3: Case 1
Case 2
Suppose k2f11g Qni=2(ai0;ai1) [0;1]1.
Similar to Case 1, there is a B2B1 such that k2B  B V\((10;11] Qni=2(ai0;ai1)).
Let G = B. Then G2G, k2G\(K H), and  G\H =;.
Case 3
Suppose k2[01;10] Qni=2(ai0;ai1) [0;1]1.
Then k is in a metric face, and V is a nonmetric open subset of U. Since P1 = L 
Qn
i=2(ai0;ai1) [0;1]1, we haveab, cd2Lsuch thata<candW open in
Qn
i=2(ai0;ai1) [0;1]1
such that k2(ab;cd) W  V. By Propositions 3.1 and 3.2, k is on an unisolated face. So
there exists ^k2(a1;c0) W  V. Now there exists B2B and q, r2Q such that  B W
25
Figure 3.9: Proposition 2.3: Case 2
and a<q<r<c. Let G = (q0;r1) B. Then G2G, ^k2G\(K H), and by Corollary
3.1  G = [q1;r0]  B = (q0;r1)  B V. Hence  G\H =;.
Figure 3.10: Proposition 2.3: Case 3
Hence we have satis ed the hypothesis for Theorem 1.7. Therefore  (p) is separable.
In other words, if M is a nonmetric hereditarily indecomposable subcontinuum of Ln 
[0;1]1, it must travel along the \edges" of a metric subspace, where an \edge" constitutes
the subspace of Ln [0;1]1 which restricts two of the Lexicographic coordinates to single
points of the form xy for y2f0;1g.
26
Chapter 4
Conclusions for the Product of Three Lexicographic Arcs
and the Product of Two Lexicographic Arcs
with the Hilbert Cube
Our  rst conclusion follows from the results of Chapter 2. There we discovered that
adding nonmetrizability restricts a hereditarily indecomposable subcontinuum of Ln from
intersecting metric cubes, or their faces. A nonmetric hereditarily indecomposable subcon-
tinuum must then travel along the edges of the metric cubes. In the case of L3, these edges
are arcs. This contradicts the hereditary indecomposability.
Theorem 4.1. If M is a hereditarily indecomposable subcontinuum of L3, then M is metriz-
able.
Proof. Let M be a hereditarily indecomposable subcontinuum of L3. Suppose that M is
nonmetric. We can assume that pi1(M) = L. Then by Theorem 1.4, there is an irreducible
subcontinuum of M fromf00g L L tof01g L L. We will denote this subcontinuum
as K0. Now either K0 is metric or nonmetric.
Suppose that K0 is metrizable. Then there exists a and b2 [0;1] such that K0 is a
subset of [00;01] [a0;a1] [b0;b1]. By Proposition 2.3 M\((00;01) (a0;a1) L) = ;,
M \((00;01) L (b0;b1)) = ;, and M \(L (a0;a1) (b0;b1)) = ;. This implies
that K0 \((00;01) (a0;a1) [b0;b1]) = ;, M\((00;01) [a0;a1] (b0;b1)) = ;, and
M\([00;01] (a0;a1) (b0;b1)) =;. In other words, each point of K0 has two coordinates
that are of the form x0 or x1. This implies, along with the fact that K0 is irreducible from
f00g L L tof01g L L, that   11 ([00;01])\K0 contains [00;01] fayg fbygwhere y,
z2f0;1g. Hence K0 contains an arc contradicting the hereditary indecomposability of M.
27
Therefore K0 is nonmetric. Thus it can be assumed that pi2(K0) = L. By Theorem 1.4
there exists an irreducible subcontinuum of K0 from [00;01] fa0g L to [00;01] fa1g L
for some  xed a 2 [0;1], say K1. So K1 is contained in [00;01] [a0;a1] L which by
Theorem 3.1 means that K1 is metrizable. Hence by the same argument as in the case
that K0 was metrizable, we have that K1 contains an arc of the form f0yg [a0;a1] fbzg
for some  xed b2 [0;1] and y, z 2f0;1g. This in turn again contradicts the hereditary
indecomposability of M. Therefore M is metrizable.
The generalization in Chapter 3 was that instead of Euclidean cubes, our points were
replaced with Hilbert cubes. This means that each point is replaced with a space for which
all metric continua are embedded. The same argument as in L3 generalizes with a few
changes to L2 [0;1]1.
Theorem 4.2. If M is a hereditarily indecomposable subcontinuum of L2  [0;1]1, then
M is metrizable.
Proof. Let M be a hereditarily indecomposable subcontinuum of L2  [0;1]1. Suppose
that M is nonmetric. We can assume that pi1(M) = L. Then by Theorem 1.4, there is
an irreducible subcontinuum of M from f12 0g L [0;1]1 to f12 1g L [0;1]1. We will
denote this subcontinuum as K0. Now either K0 is metric or nonmetric.
Suppose that K0 is metrizable. Then there exists a2 [0;1] such that K0 is a subset
of [12 0; 12 1] [a0;a1] [0;1]1. By Proposition 3.1 M \((12 0; 12 1) (a0;a1) [0;1]1) =
; which implies that K0 \((12 0; 12 1) (a0;a1) [0;1]1) = ;. By Proposition 3.3, M\
((12 0; 12 1)  L [0;1]1) = ; and M \ (L (a0;a1)  [0;1]1) = ;. This implies that
K0 \((12 0; 12 1) [a0;a1] [0;1]1) = ;, and K0 \([12 0; 12 1] (a0;a1) [0;1]1) = ;. This
implies, along with the fact that K0 is irreducible fromf12 0g L [0;1]1 tof12 1g L [0;1]1,
that   11 ([12 0; 12 1])\K0 contains [12 0; 12 1] fayg f~zgwhere y2f0;1gand ~z2[0;1]1. Hence
K0 contains an arc, contradicting the hereditary indecomposability of M.
Therefore K0 is nonmetric. Thus it can be assumed that pi2(K0) = L. By Theorem
1.4 there exists an irreducible subcontinuum of K0 from [12 0; 12 1] fa0g [0;1]1 to [12 0; 12 1] 
28
fa1g [0;1]1 for some  xed a2[0;1], say K1. So K1 is contained in [12 0; 12 1] [a0;a1] [0;1]1
which by Theorem 3.1 means that K1 is metrizable. Hence by the same argument as in the
case that K0 was metrizable, we have that K1 contains an arc of the formf12yg [a0;a1] f~zg
for some  xed ~z 2 [0;1]1 and y 2f0;1g. This in turn again contradicts the hereditary
indecomposability of M. Therefore M is metrizable.
Theorem 4.2 is the beginning of the next step to showing that the product of four
Lexicographic arcs contains only metric hereditarily indecomposable subcontinua. From
there the author is working on establishing a inductive argument to show the same for all
 nite products of Lexicographic arcs.
29
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[12] K. Kuratowski, \Topology," vII, Academic Press, New York, 1968.
[13] R. Greiwe, \A Brief Exploration of the Sorgenfrey Line and the Lexicographic Order,"
Masters Thesis, Auburn University (2006).
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Appendices
31
Appendix A
Notes on the style-file project
These style- les for use with LATEX are maintained by Darrel Hankerson1 and Ed
Slaminka2.
In 1990, department heads and other representatives met with Dean Doorenbos and
Judy Bush-Crofton (then responsible for manuscript approval). This meeting was prompted
by a memorandum3 from members of the mathematics departments concerning the Thesis
and Dissertation Guide and the approval process. There was wide agreement among the
participants (including Dean Doorenbos) to support the basic recommendations outlined in
the memorandum. The revised Guide re ected some (but not all) of the agreements of the
meeting.
Ms Bush-Crofton was supportive of the plan to obtain \o cial approval" of these style
 les.4 Unfortunately, Ms Bush-Crofton left the Graduate School before the process was
completed. In 1994, we were revisiting some of the same problems which were resolved at
the 1990 meeting.
In Summer 1994, I sent several memoranda to Ms Ilga Trend of the Graduate School,
reminding her of the agreements made at the 1990 meeting. Professors A. Scottedward
Hodel and Stan Reeves provided additional support. In short, it is essential that the Grad-
uate School honor its commitments of the 1990 meeting. It should be emphasized that Dean
Doorenbos is to thank for the success of that meeting.
Maintaining these LATEX  les has been more work than expected, in part due to contin-
uing changes in requirement by the graduate school. The Graduate School occasionally has
complete memory loss about the agreements of the 1990 meeting. If the Graduate School
rejects your manuscript based on items controlled by the style- les, ask your advisor to
contact the Graduate school (and copy to me) to urge cooperation.
Finally, there have been several requests for additions to the package (mostly formatting
changes for  gures, etc.). While such changes are not really part of the thesis-style package,
it could be bene cial to collect these options and distribute with the package (making
it easier on the next student). I?m especially interested in changes needed by various
departments.
1Mathematics and Statistics, 221 Parker Hall, 844-3641, hankedr@auburn.edu
2Mathematics and Statistics, 218 Parker, slamiee@auburn.edu
3Originally, the memorandum was presented to Professor Larry Wit. A copy is available on request.
4Followup memoranda gave a de nition of \o cial approval." Copies will be sent on request.
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