THE COMPACT OPEN TOPOLOGY FOR A SPACE
OF RELATIONS AND CERTAIN MONOTONE
RELATIONS WHICH PRESERVE ARCS,
PSEUDOCIRCLES AND TREES
By
JANE MAXWELL DAY
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
April, 1964
ACKNOWLEDGMENTS
The author is especially indebted to her chairman, Dr. W. L.
Strother, for his instruction, constant encouragement and guidance
throughout her doctoral study, as well as during the writing of this
paper; she also wishes to thank Dr. J. E. Maxfield for his patience
and encouragement during her doctoral work, and Dr. A. D. Wallace for
his time and suggestions given during the preparation of this paper.
She is deeply grateful to her husband, Walter, for his under
standing, encouragement, generosity and the extra measure of patience
he continually offered. Without his support, her graduate work could
not have been done.
TABLE OF CONTENTS
AC.OIOWLI
INTRODUCE
PART I
Page
ii
1
2
2
EDGMEIITS . . . . . . . . . . .
CTIO1 . . . . . . . . . . .
SECTION 1. Background and definitions . . .
SECTION 2. The compact open topology and joint
continuity for a relation space .
SECTIOII 3. A relation space on which the compact
open topology is metrizable . .
SECTIOII 4. A relation space with the compact open
topology which is a semialgebra .
SECTION 5. An isomorphism theorem for semialgebras
of relations . . . . . .
SECTION 1. Background and definitions . . .
SECTIOII 2. Arcs and metric arcs . . . . .
SECTION 3. Pseudocircles and simple closed curves
SECTION 4. Trees and dendrites . . . . .
CES. . . . . . . .
DADT TT
REFEREII
INTRODUCTION
The work in Part I is a study of relation spaces with the
compact open topology. One result is: when X is Hausdorff and R
is the real numbers, a certain space of relations, each in X x R,
forms a semialgebra.
In Part II are theorems which generalize to monotone relations
various known theorems about monotone singlevalued functions. For
example, if X is Hausdorff and I is an arc, sufficient conditions are
given on a monotone relation in I x X to imply that X is also an arc.
This theorem has as a special case the wellknown result that a con
tinuous, monotone image of an arc in a Hausdorff space is again an arc.
PART I
SECTION 1. Background and definitions.
We will say f:X Y is a function from X to Y iff for each x
in X, f(x) is a single point of Y. If X is a set and is a collec
tion of subsets of X, J will be called a topology for X iff the empty
set and X are in c/ and any union or finite intersection of elements
of J is in .'. A topological space, or just space, is a pair (X,J),
where X is a nonnull set X and / is a topology for X. We will often
say X is a space without specifying a topology for X, if no confusion
can arise. If X is a space and J is any collection of subsets of X,
it is simply a logical exercise to prove that the collection J7 of all
unions and finite intersections of elements of J is a topology for X.
We will say J generates /, or J is a subbasis for ..
Let Y_ be the set of all functions from X to Y, and let
B(K,W) = {f Y I f(K)CWj. Let X and Y be spaces and let Z be the
topology for YX generated by fB(K,W) I K is compact in X and W is open
in Y]. 6 is known as the compact open topology for y Let FCY,
let F have topology J/ and let KCX; define p:F x K  Y by
p(f,x) f(x). p is known as the evaluation function, and J' is said
to be jointly continuous on K iff p is continuous.
The compact open topology for Y o, is characterized by these
facts: each topology for X which is jointly continuous on compact
3
is larger than 6; and when X is Hausdorff or regular and F is a subset
of Y containing only functions continuous on the compact of X, then
(e F is jointly continuous on compact. These theorems can be found
in [8 ]. 4 IF denotes IC n F Ce ;}, which is a topology for F.
Before we can define the compact open topology for a space of
relations, we need several more definitions and a lemma, 1.1.
A relation from X to Y is defined to be a set RCX x Y. For
each x in X,xR is defined to be n2(R n ( [x x Y)), and for each
y in Y, Ry is defined to be the set nl(R n (Xx fy})). For subsets
A of X and B of Y, we define AR to be U aR and RB to be U Rb.
a eA beB
Notice that nl(R) = RY and n2(R) = XR, and, if 0 represents the null
set, OR and RO are 3.
All authors of theorems about multivalued functions seem to
agree to define F:XY to be a multivalued function from X to Y iff
F(x) is a nonnull subset of Y for each x in X. Using this definition,
a relation RCX x Y such that X = RY is the graph of a multivalued
function F:X Y defined by F(x) = xR, and conversely if F:XY is
a multivalued function, the graph of F is a relation Re X x Y such that
X = RY.
We will not write in multifunction terminology, but we will
restrict our attention to relations "defined on all of X," that is,
relations RCX x Y such that X = RY. Such a relation might be called
a functional relation since there is a unique multifunction associated
with it.
Since many readers may be familiar with multifunction rather
than relation theory terminology, we will attempt, when defining a
property for relations, to mention by what other names it is known in
the literature.
If X and Y are spaces, we will let S(X.Y) fRCX x Y I X = RY).
The set of all nonnull subsets of a set Z will be denoted by
2_ If Y is a set and UCY, let 2U = fA 2 I 0 U / 0 1.
When Y is a space, f2U and 2U I U open in Y) generates a topol
ogy for 2 and throughout this paper when we speak of the space 2 ,
we mean 2 with this topology. It appears to have originated with
Vietoris in [17), who named it the finite topology for 2 Michael
uses this name in [11]. Choquet defined it essentially the same way
as Vietoris in [3 ]. Their definitions are easily shown to be equiv
alent to Frink's neighborhood topology, defined in [5 ]. This is the
name Strother uses for it, in [15] and other papers.
If fe(2 Y), let us call U fx x f(x) I xeX], where f(x) is
considered as a subset of Y, the graph of f in X x Y.
SECTION 2. The compact open topology and joint continuity for a
relation space.
The following lemma is due to many authors.
(1.1) Lemma. If R S(X,Y), there is a unique
that R is the graph of f in X x Y. Conversely, if fe
of f in X x Y is unique and is an element of S(X,Y).
f (2Y)x such
(2Y ), the graph
Proof: Let ReS(X,Y). By definition then, X = RY: i.e., for
each xe X, there is some y eY such that x Ry. Hence xR / 0 for any
xeX. Define f: X2 Yby f(x) = xR for each xeX. xR / O so xR e 2 ,
hence fe(2 ) The graph in X x Y of f is U ((x} x f(x) I xeX) which
is by definition, U ((xJ x xR xe X, which is just R. Clearly, differ
ent functions have different graphs, so f is unique.
Conversely, let fe(2 ) and define RCX x Y thus:
R = U [(x) x f(x) xeX), where f(x) is considered as a subset of Y.
Since f(x) e 2Y, f(x) / 0 for any xeX, so xR = f(x) / 0 for any xeX.
Hence X = RY, so ReS(X,Y). By definition R is the graph of f in X x Y,
and it is clearly unique.
Define i: (2 Y) S(X,Y) by letting i(f) equal the graph of
f in X x Y. In view of 1.1, i is 11 and onto.
Let C? be the compact open topology for (2 ) : that is, < is
generated by (B(K,W) I K compact in X and W open in 2 Y. Let )0 be the
topology for S(X,Y) such that i is a homeomorphism onto (S(X,Y),9 ).
We will call A the compact open topology for the relation space S(X,Y).
An obvious problem is how to describe RZ directly, and 1.2
gives such a description, though it is not very satisfactory. There may
be a simpler description for 2Z which we have overlooked. Define
A(K,U) = fR e S(X,Y) IR KC U) and
Z(K,U) (R eS(X,Y) I KcRU).
Let M(x,U) denote a set of type A(x,U) or Z(x,U), and let f#
be the collection of all sets of the form
na
n ( u nI M(x,u a)]}
x C K aeD i 1 a
where K is compact in X, each Uai is open in Y, and D is any indexing
set.
(1.2) Theorem. ', the compact open topology for S(X,Y),
is generated by 9.
Proof: Let d be the compact open topology for (2 )", let
B(K,W) be an arbitrary subbase element of ((2 )X, ), and let
i: (2Y)X S(X,Y) be the function identifying each f with its graph
in X x Y. By definition of 17, i(B(K,W))is an arbitrary subbase
element of (S(X,Y),0).
Since W is an arbitrary open set of 2 W = U N where
a D a
each N is a finite intersection of subbase elements of 2Y
The following qualities are easily established:
The following qualities are easily established:
B(K, U N) = n B(x, U N )
a D a xeK aeD a
= U B(x,N ).
xeK aeD a
m
Fix a in D; then N = l V., where each V is of the form
a j 3 j
U.
2 or 2U for some U. open in Y. The following qualities are
easily established:
m m
B(x, Jn ) = n B(x,Vj),
ii j j=1 1
i(B(x,2U)) = A(x,U), and
i(B(x,2U)) = Z(x,U).
Finally, since i is 11 it can be distributed through inter
sections, so
ma
i(B(K,W)) = n U n i(B(x,V ))
xeK aeD 1 l a
ma
n U n M(x,U )
xeKaeD I= 1 a
and this last set is in P by definition. This completes the proof.
Let 6 be the topology for (2 ) generated by fB(K,2 U) and
B(K,2U) I K compact in X and U open in Y}. We will call 4 the
subcompact open topology for (2Y ) and the topology for S(X,Y) which
is homeomorphic to 5 will be called the subcompact open topology for
S(X.Y).
Let be the topology for S(X,Y) generated by fA(K,U) and
Z(K,U) K compact in X and U open in Y}.
(1.3) Theorem. 9Q is the subcompact open topology for S(X,Y),
and the compact open topology for S(X,Y), R, contains ?F.
Proof: Let i be the function of 1.2. Then by definition the
subcompact open topology for S(X,Y) is generated by fi(B(K,2U )) and
i(B(K,2)) K compact in X and U open in Y}; i(B(K,2U)) A(K,U) and
i(B(K,2U)) Z(K,U), which completes the proof that / is the subcompact
open topology for S(X,Y).
Since B(K,2U) and B(K,2U) are in the compact open topology for
(2 ) their images under i will be in by definition. This completes
the proof that /7.
If X is a space whose only compact sets are finite, then of
course R 3 /O, hence A( = =/Z However, in general ?Z properly con
tains I, as the following example shows.
(1.4) Example. Let I be the unit interval, and define
RCI x I as follows:
0 if x <
xR=
1 if x > .
2
Let U = [0, ) and V = ( ,l]; note B(I,2U 2V) is open in
((2 )1, ) and R i(B(I,2 U 2V)), which is open in (S(I,I),?f) by
definition of 5 (i is the function used to define 9? ).
An arbitrary basic set of i is
n m
M = ( Ln (K )) n ( (
a 1 ( l i j.1
Z(L.V.))
j J
where each K. and L. is compact in I and each U. and V. is open in I.
Suppose ReM. We will prove M i(B(I,2U U 2 )).
1 "n
Case 1. U K Then there is some e > 0
C i1 1
open interval ( e,)CI \ U K..
S2 i. 1
such that the
1 x
Let x0 2 , and define
ScI x I thus:
xR if x / x0
xS =
(x0)R U 1 if x = x0.
It is easily seen that S eM but S/i(B(I,2U U 2 )) since if
fS = i1(S), fS(x0) 2UU U 2 so fS Q B(I,2U U 2V).
1 n
Case 2. 2 e U
2 iol
1
Ki. Then we may suppose 2
Ki. Again there is some e > 0 such that
1 1
( )CI\ U Ki.
i = p+l
Let and define S as before.
Let x Tt and define S as before.
Again it is easily seen that S cM but Si(B(I,2 U U 2V)) since
f,(x0) #2U U 2.
The following lemma is an easy consequence of the fact that
a compact Hausdorff space is regular. It appears in [9 ].
n
Su
i= p+l
p
n Ki and
i=l i
(1.5) Lemma. If C is compact in a Hausdorff space X and if
(U 3 n is a finite open cover of C, then there are closed sets
n
(Ci n such that C CU. for each i and C = U C..
i i= 1 i i i=l 1
Proof: fU C]i n is a collection of sets open in C.
C is regular so if x e U. n C, there is some V open in C such that
1 X
ve V CV *CU. ri C. IV I x & C] is an open cover of C, so there is a
X X 1 X
finite subcover, fV m Let C. = UfV I V CU. n C), for
xj j 1 1 Xj Xj
each i. Then C. is a finite union of closed sets, hence is closed;
1
C CU n CCU ; and finally, each Vj was chosen so that Vxj cU n
i i l Xj Xj i
n m
for some i, so U C U / = C.
i 1 j = xj
If S is a topology for a space S and TCS, I T will denote
C n T C IC .
The following lemma is known.
(1.6) Lemma. Let Z be Hausdorff, X a space and !> be the
compact open topology for Z". Let T = (f: X Z f is continuous
on every compact subset of X) and let V/ be a subbasis for Z. Then
r I T is generated by (B(K,V) n T I K compact in X and Ve 2/).
Proof: Let K be any compact set in X and W any open set of Z.
Then W = U where each U is a finite intersection of elements
a e D a a
of V/. By definition of the compact open topology, B(K,W) is an
arbitrary subbase element of & Let fe B(K,W) n T, which implies
f(K)C U U f in T implies f continuous on K, so f(K) is compact;
aeD a
hence there is a finite collection, fUai i=, which covers f(K).
ai i 1' which
n
f(K)C U1 Uai and f(K) compact in the Hausdorff space Z, so by 1.3
n
there are closed sets Ci such that f(K) = U C and CCUai for
fK 1 1 a1
each i = 1,..,n. Let K = K f (Ci) for each i and note that f
continuous on K and K compact imply that each Ki is compact. Note
n
also that f(Ki)cCiCUai for each i. Then i B(Ki,Ui) e i and
i ai i= 1 ai
n
fe l B(Ki,Ua) n T, which is clearly contained in B(K,W) r T.
m
To complete the proof, recall that for each U ., U = n V
al ai j .1 i
m
for some finite subcollection of 2/, fV j = 1, and note that
m m
B(L, n V ) n B(L,VJ) for any sets L and V .
J J1 j=1
Z
Let Z be a set and UCZ. We define 2_ to be the collection of
all nonnull closed subsets of Z, and 2U to be the collection of all
nonnull closed subsets of Z which have a nonnull intersection with U.
The following lemma is easily proved. It can be found in
[ll], [13] and [15].
Y
(1.7) Lemma. If Y is regular and if 2Y has the topology
generated by IfU and U U open in Y), then 2 is Hausdorff.
(1.6') Corollary. If X is a space, Y is regular, is the
compact open topology for (2Y) and if T = (f: X* I f is
continuous on the compact sets of Xj, then .I T is generated by
(B(Y.,U) n T and B(Y,EU ) T I K compact in X and U open in Y}.
Let X and Y be spaces, and let FcX x Y. R is said to be
upper semicontinuous on X iff for every x in X and neighborhood V of xR,
there is a neighborhood U of x such that URCV. We will use u.s.c. to
abbreviate upper semicontinuous. Notice that R is u.s.c. on X iff for
each V open in Y, [x c X I xRcV is open in X.
R is said to be lower semicontinuous on X iff for every V open
in Y, RV is open in X. We will use l.s.c. to abbreviate lower semi
continuous. Notice that R is 1.s.c. on X iff for each V open in Y,
(x e X xR n V / 0 } is open in X.
These names are those used by Kuratowski in [10] and by other
authors. In [15] and other papers, Strother called these forms of
continuity weak continuity and residual continuity, respectively.
The following remark will be useful in Section 4 of Part II.
If KCX, RCY. x Y and R is u.s.c. and l.s.c. on K, then RCX x Y but
R need not be u.s.c. or l.s.c. on X. First consider upper semicontin
uity: let V be open in Y and U = {x Y. I xRCV); U is open in K, but
U U(X\ K) = fxX I yxRCV] need not be open in X.
Similarly for lower semicontinuity: if V is open in Y, since RCK x Y,
RVCK; RV is open in K but need not be open in X.
If A and B are subsets of a set X, A\ B will denote
(aeA I ajB}.
The following lemma is due to many authors.
(1.8) Lemma. If f: X 2 is continuous, the graph of f in
X x Y is u.s.c. and l.s.c. on X. Conversely, if ReS(X,Y) and R is
u.s.c. and l.s.c. on X, then the function f: X 2Y whose graph in
X x Y is R is continuous.
Proof: Let f: X  2 be continuous and let R be the graph
of f in Xx Y. Then for any U open in Y, f1(2U) = fxeX I xRCU}
is open in X and f1(2U) RU is open in X, hence R is u.s.c. and
l.s.c. on X.
Conversely, if ReS(X,Y) and R is u.s.c. and l.s.c. on X,
define f: X 2Y by f(x) = xR for each x in X. f is continuous if
f1 (2U) and f (2U) are open for arbitrary U open in Y; R u.s.c. on
X implies (x&X I xRCUJ = f1(2U) is open and R l.s.c. on X implies
RU = f1(2U) is open. This completes the proof.
If RCX x Y, we will say R is point closed iff xR is closed
for each x in X.
Recall that A(K,U) fRCX x Y I X = RY and KRCU} and
Z(K,U) = fRCX x Y I X RY and KCRU}.
Since 2 is difficult to describe directly, the following
lemma is very useful.
(1.9) Theorem. Let X be a space, Y be regular and
F = (RcX x Y X = RY, R is point closed, and for each compact KCX,
R n (K x Y) is u.s.c. and l.s.c. on K). Then fL F, the compact open
topology for F, equals ZJ  F, the subcompact open topology for F:
that is, 1\ F is generated by fA(K,U) n F and Z(K,U) 0 F I K compact
in X and U open in Y).
Proof: Let i: (2 ) S*(X,Y) be the function of 1.2, and
let T = ff: X  2 I f is continuous on each compact KCX). By 1.8,
i(T) = F, and if t is the compact open topology for (2Y)X, by defin
ition of /, i: (T, I T) (F, Al  F) is a homeomorphism.
By 1.6', el T is generated by fB(K, U) n T and B(K,IU) n T
K compact in X and U open in Y). It is easy to see that i(B(K,2j) =
A(K,U) and i(B(K,2U)) = Z(K,U), which completes the proof.
Let us recall several things.
Following Kelley in [8 ], if is a topology for Z and TCZ",
we define J I T to be jointly continuous on K iff KCX and the
function p: T x K Z defined by p(f,x) = f(x) is continuous.
6, the compact open topology for Z", is characterized by these
facts, proofs of which can be found in Chapter 7 of [8 ]:
(1.10) Theorem. If J1 is a topology for Z" which is jointly
continuous on compact, J is larger than .
(1.11) Theorem. When X is Hausdorff or regular and T is a
subset of ZX containing only functions continuous on the compact of X,
ST is jointly continuous on each compact KCX.
Letting Z = 2 and drawing the obvious parallels with S(X,Y),
we define an evaluation relation QC((S(X,Y) x X) x Y) thus: (R,x)Q =
xR for each (R,x) e S(X,Y) x X; and if J is a topology for S(X,Y) and
FcS(X,Y), we define JI F to be jointly continuous on K iff KCX and
Q 0 ((F x K) x Y) is u.s.c. and l.s.c. on F x K.
These definitions will give theorems for S(X,Y), 1.10' and
1.11', which parallel 1.10 and 1.11. The following lemma is useful
in the proofs of 1.10' and 1.11'.
(1.12) Lemma. Let S(X,Y) have topology Jand (2Y)X have
topology /. Let TC(2 ) and let F = i(T), where i(f) is the graph
of f in X x Y, for each f in T. Then if KCX and if i: (T, JI T) 
(F, JI F) is a homeomorphism, JI T is jointly continuous on K iff
SI F is jointly continuous on K.
Proof: We need to prove Q' = Q n ((F x K) x Y) u.s.c. and
l.s.c. on F x K iff p: T x K2 is continuous. By 1.7, Q' is u.s.c.
and l.s.c. on F x K iff q: F x K 2 is continuous, where q(R,x) =
(R,x)Q' = xR for each (R,x) eF x K. Since p(f,x) = q(i(f),x) and i is
a homeomorphism, it is clear that p is continuous iff q is. This
completes the proof.
(1.10') Theorem. If / is a topology for S(X,Y) which is
jointly continuous on compact, J is larger than t, the compact open
topology for S(X,Y).
Proof: Let i: (2Y)X  S(X,Y) be the function such that i(f)
is the graph of f in X x Y, for each fc (21) and let c be the
topology for (2Y ) such that i is a homeomorphism onto (S(X,Y),J ).
By hypothesis, if K is compact in X, Q n ((S(X,Y) x K) x Y) is u.s.c.
and l.s.c. on S(X,Y) x K; then 1.12 implies that p: (2Y)X x K 2Y
is continuous, when (2Y ) has topology Jf. Hence c/is jointly con
tinuous on K, which is an arbitrary compact set, so by 1.10, J .
Then by the homeomorphisms which define Z and /, 3 .
(1.11') Theorem. When X is Hausdorff or regular and F is a
subset of S(X,Y) containing only relations R such that for each compact
KCX, R n (K x Y) is u.s.c. and l.s.c. on K, then 1 I F, the compact
open topology for F, is jointly continuous on each compact KCX.
Proof: Let i: (2 ) *S(X,Y) be the function of 1.10' and let
T = i1(F). Then if g is the compact open topology for (2Y)X by
definition of 9, i: (T, &I T) (F, 1I\ F) is a homeomorphism.
By 1.8 and hypothesis, each f in T is continuous on each
compact KCX; and then by 1.11, (I T is jointly continuous on compact.
Therefore, by 1.12, i F is jointly continuous on compact.
SECTION 3. A relation space on which the compact open topology is
metrizable.
1.15 and the following definitions and remark are due to Arens.
(See [1 ].)
A space X will be called hemicompact iff X is the union of a
n
denumerable collection of compact sets, TH i) and for any compact
NK
KCX, there is some NK < such that KC U H This is a true
K i=l 1
generalization of compactness, since for example any locally compact
space with a denumerable basis is hemicompact.
(1.15) Theorem. Let X be hemicompact, Z be a metric space, and
let T = If: X Z I f is continuous). Then the compact open topology
for T is metrizable.
(1.14) Theorem. Let X be hemicompact, let (Y,p) be a metric
space, and let F = (RCX x Y I X = RY, xR is closed and bounded for
each x in X, and R is u.s.c. and l.s.c.}. If 51 is the compact open
topology for S(X,Y), I F is metrizable.
Proof: Let Vr(A) = fy Y I p(y,A) < r) for any ACY and real r;
let Z = fACY I A is closed and bounded); and define d: Z x Z*R thus:
d(A,B) = inf (r R I AcVr(B) and BCVr(A)). For a proof that d is
a metric for Z, see p. 167 in [ 6 ]. d is known as the Hausdorff metric.
Let T = ff: X>Z  f is continuous); then if i: T+S(X,Y)
18
maps each f in T onto its graph in X x Y, which is in S(X,Y) by 1.1,
i(T) = F by 1.8. Let & be the compact open topology for (2 ) and
note ZC2 implies TC(2 )X, so I T is the compact open topology
for T. Z is metric, so by 1.13, $ I T is metrizable.
By definition of i: (T, I T) (F, I F) is a homeo
morphism; therefore, since i I T is metrizable, I F is also.
(1.14') Corollary. If *? is the subcompact open topology for
S(X,Y), I F is metrizable.
Proof: By definition of F and by 1.9, \I F = It F.
SECTION 4. A relation space with the compact open topology which is
a semialgebra.
Let (Y,o) be a semigroup and let Y have topology
said to be a topological semigroup iff the function m: Y x Y* Y
defined by m(y,y') = y o y' is continuous, when Y x Y has the usual
product topology.
If A and B are subsets of a semigroup (Y,o), we define A o B
to be fa o b I aeA and beB) Recall that S(X,Y) = fRCX x Y X=RY},
and hence if Q and R are in S(X,Y) and xeX, neither xQ nor xR is null,
so (xQ) o (xR) / d. Let us define Q o R U ffx) x ((xQ) o (xR)) I
xeX), and note Q o RCX x Y and Q o ReS(X,Y). Notice also that for
any KCX, K(Q o R) = (KQ) o (KR). We state these things formally, for
reference purposes.
(1.15) Lemma. If X is a set and (Y,o) is a semigroup, then
(S(X,Y),o) is also a semigroup. If KCX and Q,Re S(X,Y), then
K(Q o R) = (KQ) o (KR).
It is not possible to have S(X,Y) a group even if Y is a group;
for suppose Y is a nondegenerate group with identity 0, and let
P = X x f0). This is the only possible identity for any relation which
is singlevalued at any point, and it is obvious how to define an
inverse for any relation ReS(X,Y) which is the graph of a single
valued function from X to Y. However, if ReS(X,Y) and there is some
x in X such that xR contains more than one point, then there is no
Q E S(X,Y) such that RQ = P. In other words, the only relations which
have inverses are those relations which are graphs of singlevalued
functions.
In 1.16 and 1.18, let the operation on the topological semi
group be o, and let the operation induced in S(X,Y) be o, as defined
above.
If Y is a space and RCX x Y, we define R to be point compact
iff xR is compact for each x in X. Let us define C(X,Y) to be
fRCX x Y I X = RY, R is u.s.c. and l.s.c. on X and point compact).
The following lemma appears in [ 4 ].
(1.16) Lemma. If X is a space, Y is a topological semigroup,
S(X,Y) = RCX x Y I X = RY} and C(X,Y) = (RCX x Y IX = RY, R is
u.s.c. and l.s.c. on X and point compact, then C(X,Y) is a subsemi
group of S(X,Y). That is, if Q,R C(X,Y), Q o RE C(X,Y) also.
Proof: Let m: Y x Y  Y be the continuous operation of Y,
let Q and R be any elements of C(X,Y), and let U be any open set of Y.
(i) Q o R is u.s.c. Let x X and x(Q o R)CU; that is
(xQ) o (xR)CU. m1(U) is open in Y x Y and (xQ) x (xR)Cm1(U);
xQ and xR compact so by a theorem of Wallace (see [ 8], p. 142), there
are open sets V and W in Y such that (xQ) x (xR)CV x WCm1(U).
Q and R are u.s.c., so there are open sets V' and W' in X such that
xe V' n W' and V'QcV, W'RCW. Let 0 = V' n0 ''. Then x 0O, 0 is open,
and O(Q o R) = (OQ) o (OR)CV o W = m(V x W)CU. Therefore QoR is u.s.c.
(ii) Q o R is l.s.c. Let xe (Q o R)U and ye x(Q o R) f U.
y e(xQ) o (xR) implies there are yl xQ and y2 xR such that
yl o y2 = y; y U implies (yl,y2) em1(U), which is open. So there
are open sets V and W in Y such that (y,y2) e V x W m (U). Q and R
are 1.s.c., so QV and RW are open in X. Note x 0 = QV n RW, which
is open; and further, OC(QoR)U: for if x' e 0, there exist
yl' ex'Q n V and Y2' Ex'R n W, so (yl ,y2')CV x W, which implies
Yl' oy21CU. That is, y1' oy2' ex'(QoR) f U. So QoR is l.s.c.
(iii) For each x X, x(QoR) is compact. x(QoR) =
(xQ) o (xR) = m((xQ)x(xR)), which is compact since m is continuous
and xQ and xR are compact.
The following lemma is Theorem 3.2 of [13].
(1.17) Lemma. Let I be compact in a space X, let Y be a
space and let RCX x Y be u.s.c. on X and point compact. Then IR
is compact.
Proof: Let Ue be a collection of sets open in Y which
covers IR. For each x in I, xR is compact and is covered by & so
we can select a finite subcover, c, Then xRCU x open in Y;
X X
R is u.s.c. on X so there is an open set V in X such that xe V
x x
and V R C U x. Find such V for each x in I, and select a finite
subcover of I, Vxi in .Then U [ ki in1 is a finite subcollec
tion of ?j and covers IR.
(1.18) Theorem. Let X be a Hausdorff space, Y a topological
semigroup which is regular, and let A be the compact open topology
for S(X,Y) = fRCX x Y I X = RY}. Then (C(X,Y), 7I C(X,Y)) is a
topological semigroup, where C(X,Y) [RCX x Y X RY, R is u.s.c.
and l.s.c. on X and point compact).
Proof: For brevity we will let C denote C(X,Y) in this proof.
By 1.16, C is a semigroup under the operation o, so we need to prove
that f: C x C  C defined by f(Q,R) = Qo R is continuous, when C x C
has the usual product topology. It will suffice to prove that the
inverse of each member of a subbase for C is open, and by 1.9, since
Y is regular, [A(K,U) n C and Z(K,U) n C I K is compact in X and U
is open in Y} is a subbase for $^ I C.
(i) Let (Q,R) e f(A(K,U) n C). We will find an open set about
(Q,R) which lies inside f (A(K,U) n C). First note that f(Q,R) =
QoRe A(K,U) implies K(QoR)cU, or, by 1.15, (KQ)o(KR)CU. Since K
is compact and Q and R are in C, by 1.17, KQ and FR are compact.
(KQ) x (KR)Cm1(U), which is open since m is continuous, so by a
theorem of Wallace (see p. 142 in [ 8]), there are open sets V and W
in Y such that KQCV, ORCW, and m(V x W) = VoWCU. Then Qe A(K,V) n C,
R e(K,W) n C, and both these sets are open in C. Finally, if
(Q',R') (A(K,V) n C) x (A(K,W) n C), f(Q',R') e L(K,U) n C: for
KQ'CV and KR'CW implies E(Q' oR') (KQ')o (KR') Vo WcU, and
f(Q',R') Q' oR'.
(ii) Let (Q,R) e f ((K,U) n C). We will find an open set
about (Q,R) which lies inside f (Z(K,U) n C). First note that
f(Q,R) = QoReZ(K,U) implies KC(QoR)U, so for each x in K, there
is some yex(QoR) U = (xQ o xR) n U. Find yl exQ and y2 exR such
that yloy2 y y, and then use the continuity of m to find V and W
open in Y such that yl e V, y2 W and VoWcU. Find such V and W
x x
for each x in K; note xeQVx 0 RW and this is open in X since Q and R
x x
are l.s.c. on X, for each x in K, so the collection (QV n RW I xe K]
x x
is an open cover of K. K is compact so there is a finite subcover,
fQV. n RW. i n. Since X is Hausdorff, by 1.5 there are closed sets
n
K. such that K = U K. and K.C QV. RW. for each i.
1 i=11 1 1 1
n n
Let V = i Z(KiVi) and W i= Z(K,Wi). It is clear that
(Q,R) e (V n C) x (W n C), and that this set is open in C(X,Y). Finally,
if (Q',R') e (V n C) x (W n C), Q' oR' eC by 1.16; if xeK there is some
i such that xeK.; Q' eZ(K ,V ) and R' eZ(Ki,W ), hence there is
Yl exQ' n V and y2 e xR' n W and then (yloy2) x(Q'oR') n (VioWi).
Therefore (ylo y2) ex(Q' oR') n U, which says xe (Q' oR')U. x was
arbitrary in K, hence Q' oR' eZ(K,U), and this completes the proof.
Let Y be a topological semigroup, let AcY and let pe Y. Then
poA is defined to be U fpoa I aeA). If X is a set and QCX x Y,
poQ is defined to be U ((x) x(o (xQ))I x eX), and poQ is called
a scalar omultiple of Q, or just scalar multiple of Q. Notice that p
is not a relation in X x Y, but poQ is.
(1.19) Lemma. Let X be a space, Y a topological semigroup,
and C(X,Y) fRCX x Y I X = RY, R is u.s.c. and l.s.c. on X and
point compact). Then C(X,Y) is closed under scalar multiplication.
Proof: Let Q C(X,Y) and p e Y. Define PCX x Y thus:
xP p for each x in X. It is easy to see that P C(X,Y), and then
by 1.16, PoQ C(X,Y). Finally, poQ = PoQ: for let xe X; then
x(poQ) = po(xQ) by definition of scalar multiplication, and p = xP,
so x(poQ) = (xP) o(xQ), which is x(P o Q) by definition. Therefore
poQ C(X,Y). This completes the proof.
Let R be the real numbers, let X be a space and let
CCS(X,R) (PCX x R I X = PR). We will call C a semialgebra iff
(i) C is a topological semigroup under both addition and
multiplication; and
(ii) C is closed under scalar multiplication.
(1.20) Theorem. Let X be Hausdorff, R the real numbers, and
let ? be the compact open topology for S(X,R) = (PCX x R I X = PR).
Let C(X,R) fPCX x R X = PR, P is u.s.c. and l.s.c. on X and point
compact]. Then (C(X,R), 92 C(X,R)) is a semialgebra.
Proof: (i) It is well known that R is a topological semigroup
under both addition and multiplication; so by 1.18, C(X,R) is also
a topological semigroup under addition and multiplication.
(ii) Since R is a topological semigroup under multiplication,
by 1.19, C(X,R) is closed under scalar multiplication.
25
(1.20') Corollary. Let X be Hausdorff, R the real numbers,
and let be the subcompact open topology for S(X,R). Then
(C(X,R), l 9I C(X,R)) is a semialgebra.
Proof: C(X,R) contains only relations which are point closed
and u.s.c. and 1.s.c. on X. Certainly R is regular, hence by 1.9,
tI C(X,R) = 94I C(X,R). Then by 1.20, (C(X,R), 7l1 C(X,R))
is a semialgebra.
SECTION 5. An isomorphism theorem for semialgebras of relations.
Let R be the real numbers and, for any space X, let C(X) =
(h: XR I h is continuous]. It is well known that, when we define
(f + h) (x) f(x) + h(x) and (fh)(x) = f(x)h(x), C(X) is a ring,
is closed under scalar multiplication, and, with respect to this
algebra on C(X), the following theorem is true.
(1.21) Theorem (Banach Stone). Let X and Y be compact
Hausdorff spaces. If g: X  Y is given, define g*: C(Y)* C(X) by
g*(h)(x) = h(g(x)). Then g* is an isomorphism onto iff g is a
homeomorphism onto.
The chief result of this section, 1.21', generalizes this
theorem.
Let Z be a space and recall that S(Z,R) = (PCZ x R I = PR.
Let us define addition and multiplication in S(Z,R) as follows: if
A and B are subsets of R, let A + B = (a + b I aA and be B] and
AB = fab I a eA and b B], and then if P and Q are any two elements
of S(Z,R), define P + Q = U ((z) x (zP + zQ) I zeZ} and
PQ = U ((zI x ((zP)(zQ)) I ze Z.
If X and Y are spaces and f: X Y, let us define f* as
follows: for each Q S(Y,R), let f*(Q) U [(x] x ((f(x))Q I x X].
Then f*(Q)CX x R by definition; Q a S(Y,R) implies yQ / 0 for each
y in Y, hence x(f*(Q)) (f(x))Q / J for each x in X. Therefore
f*(Q) ES(X,R): in other words, f* is a function from S(Y,R)
into S(X,R).
(1.22) Lemma. If X and Y are spaces, f: X Y is a function
and f*: S(Y,R)S(X,R) is defined as above, then
(i) f* is a homomorphism;
(ii) if f is continuous, f*~C(Y,R))cC(X,R); that is,
if PCY x R is such that Y = PR, P is u.s.c. and l.s.c. on Y and point
compact, then f*(P) is such that X = (f*(P))R, f*(P) is u.s.c. and
l.s.c. on X and point compact.
Proof: (i) Let P and Q be any elements of S(Y,R). Then
f*(P + Q) = f(P) + f*(Q): for let xeX and note x(f*(P + Q)) =
(f(x))(P + Q) = (f(x))P + (f(x))Q = x(f*(P)) + x(f*(Q)) = x(f*(P + Q)),
all by definition either of f* or of addition in S(Y,R). Similarly,
f*(PQ) = (f*(P))(f(Q)).
(ii) Suppose f is continuous and let Pe C(Y,R). Define
R
p: Y.2 2 by p(y) = yP for each y in Y; since P is u.s.c. and l.s.c.,
by 1.8 p is continuous. The graph of pf in X x R is exactly f*P),
since pf(x) = (f(x))P = x(f(P)) for each x in X. Since pf is contin
uous, by 1.8 again, f*(P) is u.s.c. and l.s.c. on X. Finally,
x(f*(P)) = (f(x))P, which is compact since f(x) eY and since Pe C(Y,R)
implies yP compact for every y in Y.
(1.21') Theorem. Let X and Y be compact, Hausdorff and nonnull
spaces. Let R be the real numbers, let C(X,R) = fPCX x R X = PR,
P is u.s.c. and l.s.c. on X and point compact), and let C(Y,R) =
[QCY x R I Y = QR, Q is u.s.c. and l.s.c. on Y and point compact).
Define f*: C(Y,R) C(X,R) by x(f*(Q)) = (f(x))Q, for Q in C(Y,R),
for a given f: X Y. Then
(i) f* is 11 iff f is onto, and
(ii) f* is onto iff f is 11.
Therefore, f is a homeomorphism iff f* is an isomorphism.
Proof: f* is a homomorphism by 1.22(i).
(i) Suppose f is onto and let P / Q in C(Y,R). Then there is
some y in Y such that yP / yQ, and since f is onto, there is some
x in f(y). Since f(x) = y, x(f*(P)) = yP and x(f(Q)) yQ, and
therefore f*(P) / f*(Q).
Conversely, suppose f* is 11. Since f is continuous and X is
compact, f(X) is compact; since Y is Hausdorff, f(X) is closed in Y;
and since Y is compact and Hausdorff, Y is normal and points are closed.
Therefore, if there is any point y~ f(X), by Urysohn's Lemma, there is
a continuous function q: Y R such that q(y) = 0 and q(f(X)'i 1.
Let Q be the graph of q. Let p: Y*R be such that p(Y) = 1, and
let P be the graph of p. It is easily seen that the graph of any
continuous singlevalued function from Y to R is in C(Y,R). Since
yP = 1 and yQ = O, P / Q; however, f*(P) = f*(Q): for let X X;
since f(x) c Y, (f(x))P = 1, and since f(x) e f(X), (f(x))Q = 1.
This contradicts the hypothesis that f* is 11, so we can conclude that
Y\f(X) 12: that is, f is onto.
(ii) Suppose f* is onto and let x / x' in X. Again Urysohn's
Lemma can be applied to find a continuous function t: X R such that
t(x) = 0 and t(x') = 1. Let T be the graph of t and note that
T eC(X,R). By hypothesis, there is some Qe C(Y,R) such that
f*Q) = T. In particular, x(f*(Q)) = xT and (x')(f*(Q)) = (x')T,
which implies (f(x))Q = xT 0 and (f(x'))Q = (x')T = 1. Therefore
f(x) / f(x'), and since x and x' were arbitrary distinct points of X,
we have proved that f is 11.
Conversely, suppose f is 11. Note f : f(X)w X is a con
tinuous function: for f 11 implies f1 singlevalued; and X compact
and f continuous imply f = (f )1 closed, so f1 is continuous.
To prove the assertion, let Qe C(X,R); we will construct Pe C(Y,R)
such that f*(P) = Q. By 1.17, XQ is compact, so there is a closed
I
interval ICR such that XQCI. Define q: X2 by q(x) = xQ for
each x in X, and note that 1.8 implies q is continuous. By Theorem 3
and Corollary 5.1 of [16], the space of closed subsets of a closed
interval of R is a CAR*: i.e., since Y is normal, f(X) closed in Y and
qf : f(X) I is continuous, there is a continuous extension p
of qf to all of Y. p(Y)2 implies p(y) compact in R for each y
in Y. So if P is the graph of p in Y x R, we have Y = PR and yP
compact for each y in Y. Also, p is continuous, so by 1.8 again,
P is u.s.c. and l.s.c. on X. Therefore Pe C(Y,R). Finally,
f*(P) = Q: for let xeX; then x(f*(P)) = (f(x))P = pf(x), and
p = qf1 on f(X) so pf(x) = qflf(x) = q(x) = xQ. That is,
f*(P) = Q, and this completes the proof that f*(C(Y,R) = C(X,R).
PART II
SECTION 1. Background and definitions.
Let X and Y be spaces. A function f: X Y is said to be
monotone iff fl (y) is connected for each y in Y. This is a very
restrictive property, and continuous monotone functions preserve many
topological properties. For example, if f: X Y is a continuous
and monotone function onto a nondegenerate Hausdorff space, it is
known and easily proved that
(1) X an arc implies Y an arc;
(2) X a pseudocircle implies Y a pseudocircle;
(3) X a tree implies Y a tree; and
(4) X unicoherent implies Y unicoherent.
We will call a compact connected Hausdorff space a continuum.
A cutpoint of a space X is a point x in X such that X\x is not con
nected. By an arc we mean a continuum with exactly two noncutpoints.
Notice that an arc need not be metric under this definition; an
example of a nonmetric arc is the "long line" (see [7], for example).
A pseudocircle is defined to be a nondegenerate continuum which is
disconnected by an:, two of its points. A tree is a continuum in i.hich
each two points are separated by a third point. A space is unicoherent
iff it is a continuum and an: two subcontinua whose union is the whole
space have a connected intersection.
See [22] for proofs of (1) (4) in the case where all spaces
are assumed metric. In the following, we will prove generalizations
of (1) (3). So far we have not found a relation to preserve uni
coherence.
Let RCX x Y and define R to be monotone iff Ry is connected
for each y in Y. If R is the graph of a singlevalued function
l
f: X*Y, then Ry = f (y), so R is a monotone relation iff f is a
monotone function. Thus monotonicity for relations is a generaliza
tion of monotonicity for functions. Notice that R is the graph of
a singlevalued function iff X = RY and Ry n Ry' = O for each pair
of distinct points, y and y', in Y.
It is easy to exhibit an u.s.c., l.s.c. and monotone relation
which does not preserve unicoherence or the properties of being an arc,
tree or pseudocircle. The problem seems to be that requiring RC X x Y
to be monotone and u.s.c. and l.s.c. on X does not in any way distin
guish Ry from Ry' for any y and y' in Y. For example, let X be any
connected space, let S be a pseudocircle and define R C X x Y by
xR = S for each x in X. It is easily seen that R is monotone and u.s.c.
and l.s.c. on X, but X could be unicoherent, an arc or a tree and
XR = S is none of these things. Notice that Rs = X for each s in S.
A similar relation can be defined in S x I, where I is an arc, so show
that the property of being a pseudocircle is not preserved by a rela
tion which is monotone and u.s.c. and l.s.c. on S.
Now suppose that I is an arc, Y is a nondegenerate Hausdorff
space, and R C I x Y is such that I = RY. Conditions are known for
R. which iLply that IR is a continuum (see [13] or [15]), and it is well
known that a nondegenerate continuum has at least two noncutpoints.
Thus some additional conditions are needed which will imply that IR. has
at most two noncutpoints. We discovered that it suffices to require
R to be monotone and noninclusive: that is, for each ,y y' in Y,
Ry C Ry' Notice that the graph of a singlevalued, onto function
is noninclusive.
Several lemmas are given before we prove 2.6 and 2.e', the arc
theorem in two forms. Theorems 2.8, 2.15 and 2.15', which exhibit
relations that preserve pseudocircles and trees, are essentially appli
cations of 2.6 and 2.6', using the facts that a pseudocircle is a
certain union of arcs, and certain subcontinua of a tree are arcs.
SECTION 2. Arcs and metric arcs.
If R CX x Y, let R(1) (y,x) I (x,y) eR). Then R()
is a relation in Y x X, and R(l)x = xR and yR(l) = Ry.
Notice that saying R( is monotone is just saying that xR
is connected for each x in X. Strother calls such a relation point
connected.
(2.1) Lemma. Let I and Y be spaces, let Y be nondegenerate
and let R C I x Y.
(i) If R is noninclusive, then IR = Y.
(ii) R is noninclusive iff for each y in Y, (I \Ry)R = Y \ y.
Proof: (i) Suppose R is noninclusive and yeY \ IR. Then
Ry = 3 ; since Y is nondegenerate, there is some y' e Y \ y, but then
Ry'D [3 = Ry, which contradicts the noninclusivity of R. Therefore
Y = IR.
(ii) For any relation C I x Y, (I \ Ry)R C Y \ y; so suppose
R is noninclusive and let y' e Y \ y. Since Ry' l Ry, there is some
x in Ry' n (I \ Ry); then y' xR C (I \ Ry)R.
Conversely, suppose for each y in Y, (I \ Ry)R = Y \ y, and let
y / y' in Y. Then y' eY \ y; if Ry' C Ry, then y' (I \ Ry)R, which
is false. Therefore Ry' t Ry.
The equivalence for noninclusivity given in 2.1 (ii) is the
only one found so far.
Recall that R C I : Y is point compact iff for each x in I,
xR is compact.
(2.2) Lemma. Let I and Y be Hausdorff spaces and let R C I x Y
be such that I = RY, R is u.s.c. on I and point compact. Then
(i) if B* is compact in I, (BR.) C (B.iR, and
(ii) for each y in Y, Ry is closed.
Proof: (i) Certairly (BR)* C ((B')R)', and by 1.17. (B)R
is compact. Y is Hausdorff, so (B')R is closed, and therefore
(BRh) C ((B*)R)` = (B')R.
(ii) By Theorem 2.9 (d) of [13], R is closed: R(1) is homeo
morphic to R. hence R(1) is closed. Then by Theorem 2.11 of [1i],
yR() = Ry is closed for each y in Y.
The following facts are well known; it is convenient for us
to use [7] for reference. Let I be a connected space with exactly two
noncutpoints, a and b, and let I have topology The cutpoint order
ing for I is defined as follows: for any p and q in I, define p < q
iff p = a, p = q or p separates a and q. By Theorem 221 of [7], this
is a simple, i.e., linear, order on I. If we define p < q to mean
p < q and p / q, [pa) = x I I p < x < q] and (p,q] = [> I I p < x < q],
then ([a,p) and (q,b] I p U q C I \ (3 U b) is a subbasis for the
cutpoint order topology for I, which we will call L. By Theorem 224
of [7], "' C ,. When I is also compact and Hausdorff, i.e., when I
is an arc, = ' by Theorem 225 of [7].
Henceforth in this paper, whenever we assume I is an arc with
noncutpoints a and b, we will suppose that the cutpoint order and
topology have been defined on I as above, and we will use the fact that
the topology of I is the order topology. In addition to the notation
defined above, we will write (p,q) to denote (x I I p < x < q} and
[p,q] to denote (xe I p < x < q}.
(2.3) Theorem. Let I be an arc with noncutpoints a and b, let
Y be a nondegenerate Hausdorff space, and let R C I x Y be such that
I = RY, R is u.s.c. on I, point compact, monotone and noninclusive.
Then
(i) IR = y,
(ii) Y is compact,
(iii) aR = a / b = bR, where a and b are points of Y, and
y y y y
(iv) each y in Y \ (a Ub y) is a cutpoint of Y.
Proof: (i) Since R is noninclusive, by 2.1 (i), IR = Y.
(ii) Since I = RY, I is compact, and R is u.s.c. on I and
point compact, by 1.17, IR is compact: that is, Y is compact.
(iii) To prove aR is a single point, suppose aR D yl U y2.
Since R is monotone and by 2.2 (ii), Ryl and Ry2 are closed and
connected; so there are points xl and x2 in I such that Ry1 = [a,x1]
and Ry2 = [a,x2]. We may suppose xl < x2, which implies Ryl C Ry2;
since R is noninclusive, this implies yl = Y2. Since I = RY, aR / 0,
which completes the proof that aR is exactly one point. Similarly
bR is a point. Let aR = a and bR = b .
y y
Suppose a = b : that is, aR = bR. Then a U b C Ra Rb ;
Ra is connected by hypothesis, and I is irreducibly connected between
a and b, so Ra = I. But by hypothesis there is at least one y in
Y \ a and of course Ry C I. This contradicts the noninclusivity of R,
and hence a, / b .
(iv) Let Y \ ( \ (a, U b) = Y \ (aR U bR); then Ry C (a,b).
By 2.2 (ii) and since R is monotone, Ry = [p,q]. Let P = [a,p),
Q = (q,b], and note P B/ / Q. By 2.1 (ii), (I \ Ry)R = Y \ y: that is,
PR U QR = Y \ y. Finally, PR and QR are nonnull separated sets. For
suppose y' e (PR)* n QR; by 2.2 (i), y' e (P )R n QR, which says
Ry' n P / 0 and Ry' ,f Q / D Then the structure of I and the fact
that Ry' is connected imply Ry' D [p,q], and hence, since R is non
inclusive, y' = y. But Ry n Q = b by definition of Q, so y' / y.
Therefore no such y' exists. Similarly, PR f (QR)* = C. PR and QR
are each nonnull since P and Q are nonnull and since I = RY. Therefore
Y \ y is the union of two nonnull separated sets, which says y is 3
cutpoint of Y.
(2.4) Lemma. Let I and Y be arcs such that the noncutpoints
of I are a and b and those of Y are a and b For p and q in I,
j1
define p q iff p = a, p = q or p separates a and q; for y and z in
Y, define y < z iff y = a y = z or y separates a, and z. Let
R C I x Y and suppose:
(i) R is noninclusive;
(ii) if A is connected in I, AR is connected;
(iii) aR = a and bR = b ;
y y
(iv) y < Y2 in Y; and
(v) Ryl [pl,ql] and Ry2 [p2,q2].
Then if x e[a,Pi), xR C [a ,yi), and if xe(qib), xR C (yi,b ], for
i 1, 2; also p1 < P2 and ql < q2.
Proof: Let P = [a,pl). Since P C I \ Ryl, PR C Y \ yl =
[a ,yl) U (ylb y. By (ii), PR is connected; also aeP, so PR C [a ,y ).
Similarly, if xe [a,p2), xR C [a ,y2) and if xe (qi~], xR C (Yi,b ] for
i = 1, 2.
Remark: since (iv) implies yl y2, (i) and (v) imply that
either p, < p2 and q < q2 or p2 < p1 and q2 < q1. Suppose the latter
is true; then a 5 p2 < p1. By (iv), yl < y2, and by the previous para
graph, PR C [a ,yl); hence y2/PR, which implies P n Ry2 = B But
p2 eRy2 and by supposition, p2 e [a,pl) = P. This is a contradiction,
and it follows that pl < p2 and q < q2.
The following lemma is due to Strother; he states it for R
either u.s.c. or l.s.c. on I, as Theorem 3.9 in [13], but we need only
the proof for R u.s.c. on I.
(2.5) Lemma (Strother). Let RC I x Y and I = RY; let R be
u.s.c. on I and R()be monotone. Then if A is connected in I, AR
is connected in Y.
Proof: Let A be connected in I and suppose AR = P U Q, where
P and Q are nonnull separated sets. Then for each a in A, aR is con
nected so either aR C P or aR C Q. Also, neither A, nor AQ is null,
where Ap = (a A I aR C PF and AQ = fa A aR C Q}; and Ap n AQ = .
Since R is u.s.c., fa cA aR C Y \ Q*} is open in A and
fa eA I aR C Y \ P*} is open in A; but these are just A. and AQ,
respectively, which contradicts the fact that A is connected.
Therefore AR must be connected.
The next two theorems describe monotone relations which preserve
the arc. Actually, the two sets of hypotheses are equivalent, which
is proved by 2.6 (iv) and 2.6' (iv).
(2.6) Theorem. Let I be an arc, let Y be a nondegenerate
Hausdorff space, and let RC I x Y be such that I = RY, R is point
compact, u.s.c. on I, monotone, noninclusive, and R( is monotone.
Then
(i) IR = y,
(ii) aR = a / b = bR, where a and b are points of Y,
Y Y y Y
(iii) Y is an arc whose noncutpoints are a and by, and
y y
(iv) R is l.s.c. on I.
Proof: (i) and (ii) follow from (i) and (iii) of 2.3.
(iii) By 2.3 (ii), Y is compact; by 2.5, Y is connected; and
by hypothesis Y is Hausdorff; so Y is a continuum. By 2.3 (iii),
Y is nondegenerate. It is well known that a nondegenerate continuum
has at least two noncutpoints. By 2.3 (iv), the only possible non
cutpoints of Y are ay and b ; hence we can conclude that Y is an arc
with noncutpoints a and b .
y y
(iv) To prove R l.s.c., it will suffice to let U be any
subbase element of Y and prove RU open. So let yO be any element
of Y \ (ay U by) and let U = (y0,by]. Let x RU and find y in U such
that xeRy. Let Ry = [p,q]. Since yO < y and Y is connected, there
is some y eY such that y < Y1 < y; let Ryl = [pl,ql]. Then by
2.4, pl < p and q1 < q, so x [p,q] C (Pl,b ], which is open in I.
Finally, (Pl,by] C RU; for let x' e (Pl,by]. If x' e (pl,q], then
y e(x')R n U; and if x' e(ql,by ], by 2.4, (x')R C (yl,by] CU.
If we let V = [ay,Y0), it is clear that 2.4 and a dual argument
will imply RV open. Therefore, R is l.s.c. on I.
(2.6') Theorem. Let I be an arc, let Y be a nondegenerate
Hausdorff space, and let R C I x Y be such that I = RY, R is point
compact, u.s.c. on I, monotone, noninclusive, and R is l.s.c. on I.
Then
(i) IR Y,
(ii) aR = ay b = bR, where a and b are points of Y,
y y y y
(iii) Y is an arc whose noncutpoints are a and by and
(iv) R(1) is monotone.
Proof: (i) and (ii) follow from (i) and (iii) of 2.5.
(iii) By 2.3(ii), Y is compact, and by hypothesis, Y is
Hausdorff. Define f: I > 2 by f(x) = yxR for each x in I; since R
is u.s.c. and l.s.c. on I, by 1.8, f is continuous. I is connected,
hence f(I) is connected in 2 and f(a) = a a connected subset of Y.
ThenProposition 2.8 of [11] applies, and we can conclude that
U ff(x) x I) = IR = Y is connected. Therefore Y is a continuum,
and as in 2.6, we can conclude that Y is an arc with noncutpoints
a and b
y y
(iv) Now that we know that Y is an arc with noncutpoints
a and b we can define y < z in Y iff y = a y = z or y separates
a and z, and we know the order topology is the topology of Y. We
y
want to prove xR connected for each x in I. If xe I and xR is a point,
we are done. Otherwise, suppose xe I and xR D yl U y2, where yl < y2.
It will suffice to prove that xR D (yl,y2), so let y (y1,y2). Since R
is monotone and by 2.2 (ii), we may suppose Ryl = [pl,q], Ry = [p,q]
and Ry2 [p2,q2]. By 2.4, since yl < y < y2, we have Pl < p < p2
and ql< q < q2. We know that x e Ryl 0 Ry2, hence p2 < x < ql.
Then x e(p,q) C Ry, which implies y exR.
Now suppose that I is an arc, Y is a nondegenerate Hausdorff
space, and f: I Y is a continuous, monotone, singlevalued function
onto Y. As mentioned on page 30, it is known that this implies that
Y is an arc, and that known theorem is a special case of 2.6. (It is
easy to see that the graph of f is a relation which satisfies all the
hypotheses of 2.6.)
If RC X x Y, Strother calls f: X Y a trace for R iff f is
continuous and f(x) e xR for each x in X. In the next theorem, we
define a trace for the relations of 2.6 and 2.6', and use it to prove
that those relations preserve metricity as well as the arc. In other
words, either of those relations takes a metric arc onto a metric arc.
It is not difficult to see that a metric arc is just a homeomorph of
the unit interval. (A proof can be found under 227 of [7].)
(2.7) Theorem. Let I be a metric arc, Y a nondegenerate
Hausdorff space, and let R C I x Y be such that I = RY, R is point
compact, u.s.c. on I, monotone, noninclusive, and either R(1) is
monotone or R is l.s.c. on I. Then Y is a metric arc.
Proof: If R() is monotone, by 2.6, Y is an arc and R is
l.s.c. on I. If R is 1.s.c. on I, by 2.6', Y is an arc. Therefore
we only need to prove that Y is metric.
Let a and b be the noncutpoints of I; by 2.6 (ii), aR = a
y
and bR = b and by 2.6 (iii), a and b are the noncutpoints of Y.
y y y
Let us define y < z in Y iff y = ay y = z or y separates a and z.
Then the topology of Y is the order topology; the order is linear;
and every closed set has a first element, since g.l.b.(A) exists for
each A C Y (see, for example, Theorem 226 of [7]).
Theorem 1.9 of [11] says that in such case there is a continuous
function g: 2 :Y such that g(A) eA for each A in S. Since R is
point compact and Y is Hausdorff, R is point closed, so we can define
f: I 2' by f(x) = xR for each x in I.
42
Since R is u.s.c. and l.s.c. on I, by 1.8, f is continuous. Therefore
gf: IY is continuous, and hence gf(I) is connected. By definition
of g and f, gf(a) = a and gf(b) = b so a: U b_ C gf(I); Y is irre
ducibly connected between a,, and b hence gf(I) = Y. Finally, since
I is a compact metric space and gf is continuous, gf(I) is metric
(see Theorem 323 in [7] for example). This completes the proof.
SECTION 3. Pseudocircles and simple closed curves.
Recall that S is a pseudocircle iff S is a nondegenerate
continuum which is separated by any two of its points. When a
pseudocircle is metric, it is a homeomorph of the unit circle; that
is, it is a simple closed curve. Theorem 228 of [7] gives a proof
of this.
(2.8) Theorem. Let S be a pseudocircle, Y a Hausdorff space
and a / b in S. Let R C S x Y be such that S = RY, R is point compact,
u.s.c. on S, monotone, noninclusive, and either R( is monotone or
R is l.s.c. on S. Let aR = a / b = bR, where a and b are points
of Y. Then Y is a pseudocircle, and if S is metric, Y is metric.
Proof: Since S is compact, by lemma 11.19 of [23], S is the
union of two arcs whose intersection is their noncutpoints, a and b.
Let us call these arcs I and J.
By 2.1 (i), SR = Y, so IR U JR = Y. Let P = R f (I x S) and
Q = R n (J x S), and let us consider P as a relation in I x IR and
Q as a relation in J x JR.
Notice that for any x in I, xP = xR, and for any y in IR,
Py = Ry n I. Then it is obvious that P is u.s.c. on I, point compact,
and if R(1) is monotone, P(1) is monotone. To see that P is mono
tone, let y EIR and suppose Py is not connected. Since Ry is connected
and Py = Ry D I, the structure of S implies that both a and b lie in RY.
But then y E aR n bR, which is not true. Therefore Py must be connected.
Finally, F is noninclusive; for let y / z in IR and suppose Py C Pz.
Then Ry i. I, else Ry Py C Fz C Rz, contradicting the noninclusivity
of R. Ry is connected and intersects I, so one of a or b must lie
in Ry, hence in Py. But aP y C Pz and aP = aR imply aR y U z, which
is false; similarly, bjPy. This involves a contradiction, so we can
conclude that Py P Fz, for any y / z in IR.
Dually, Q is point compact, u.s.c. on J, monotone, noninclusive,
if R() is monotone, Q is monotone, and if R is l.s.c. on S,
Q is l.s.c. on J.
Then if R() is monotone, IP and JQ are arcs by 2.6; if R is
l.s.c. on S, IP and JQ are arcs by 2.6'; and in either case, the non
cutpoints of IP and JQ are a and b Further, IP n JQ = a U b :
Y Y Y Y
for if y E IP n JQ, Ry intersects both I and J; Ry is connected since
R is monotone; so the structure of S implies that either a eRy or
b Ry. If a Ry then y = a and if b Ry, y = b by hypothesis.
This completes the proof that Y is a pseudocircle, since Y =
IP U JQ implies Y is a continuum, and surely any two points of Y
disconnect Y.
Finally, suppose S is metric; then I and J are metric so by
2.7, IP and IQ are metric arcs. It is clear that Y is then homeo
morphic to the unit circle.
45
It is quite possible that the hypotheses of 2.8 can be
weakened so that the relation need not be singlevalued at any point
of S. Consider the following example.
Let S be the unit circle in the complex plane and define
R CS x S as follows: zR = (z' I amp z 5 amp z' < amp z + nj. It is
obvious that R is point compact, u.s.c. and l.s.c. on I, and R()
is monotone. Also, R is noninclusive and monotone, since Rz' 
[z amp z' n < amp z < amp z'} for any z' in S. Thus R satisfies
all the hypotheses of 2.8 except that zR is not a point for any z.
SECTION 4. Trees and dendrites.
Recall that a tree is a continuum in which every two points
are separated by a third point.
In [20], L. E. Ward, Jr. has given a characterization of trees
in terms of partial order. We will use this theorem extensively, so
it is quoted below. First several definitions are needed.
A partial order is a binary relation which is reflexive,
transitive and antisymmetric. If (X,<) is a partially ordered set,
then for any x in X we define L(x) = (y I y in X and y 5 x), and
M(x) = I y in X and x < y]. < is said to be semicontinuous
iff L(x) and M(x) are closed for each x in X. < is said to be
order dense iff for each two distinct points of X, x and y, there is
some z in X such that x < z < y. A is a chain in (X,<) iff for each
a and a' in A either a < a' or a' < a.
(2.9) Theorem (Ward). Let X be a compact Hausdorff space.
X is a tree iff X has a partial order 5 such that
(i) < is semicontinuous;
(ii) < is order dense;
(iii) for each x and y in X, L(x) n L(y) is a nonnull chain;
(iv) for each x in X, 1(x) \ x is open.
To prove necessity, Ward lets X be a tree and defines < thus:
choose an arbitrary e in X; then for any x and y in X, define x < y
iff x = e, x = y, or x separates e and y. Ward shows that < is a
partial order on T which satisfied (i) (iv), whatever point
e may be.
If T is a space and a / b in T, we will let C(a,b) denote
(t eT I t separates a and b in T) U a U bo The following lemma is
known; it is an easy result of two of Ward's theorems.
(2.10) Lemma. If T is a tree and a / b in T, C(a,b) is an
arc with noncutpoints a and b.
Proof: Define x < y in T iff x = a, x = y or x separates a
and y. Then L(b) = C(a,b); L(b) is closed by 2.9 (i), and hence
is compact; the order on L(b) is semicontinuous by 2.9 (i); and
L(b) is a chain by 2.9 (iii). L(b) is order dense since, if x < y
in L(b), there is some z in T such that x < z < y by 2.9 (ii), and
z < y implies z eL(b). Then by Theorem 5 of [19], L(b) is connected.
L(b) is Hausdorff since T is, so L(b) is a continuum: i.e., C(a,b)
is a continuum. By definition, C(a,b) has at most two noncutpoints;
it must have at least two, hence C(a,b) is an arc and the noncutpoints
of C(a,b) are exactly a and b.
The following lemma is known. See, for example, p. 74 of [24].
(2.11) Lemma. If a / b in a tree T, if X is a connected
subset of T and if a U b C X, then
(i) C(a,b) C X, and
(ii) x separates a and b in X iff x C(a,b).
Proof: (i) If x / X, then T \ x D X, which is a connected set
containing a and b. Therefore x 'C(a,b).
(ii) Suppose X x = P U Q, separated sets such that a&P and
b Q. By (i), C(a,b) C X, and by 2.10, C(a,b) is connected. Thus
if x/C(a,b), either C(a,b) C P or C(a,b) C Q; but neither of these
is true since a aC(a,b) n P and b&C(a,b) n Q. Therefore x E C(a,b).
The converse is obvious.
A component of a space X is a maximal connected subset of X.
A branch point of a tree T is a point t such that T \ t has three
or more components.
The next lemma is known; in the case of metric spaces, it can
be deduced from 1.1 (iv) of [22].
(2.12) Lemma. If T is a tree, B is the set of all branch
points of T, and J is a nondegenerate connected set in T \ B, then
J* is an arc.
Proof: J* is a nondegenerate continuum, so it has at least
two noncutpoints. To prove there are exactly two, suppose the
contrary and let a, b and c be distinct noncutpoints of J".
Define x 5 y in T iff x = a, x = y or x separates a and y.
Let X = L(b) U L(c). Since J* \ c is connected and a U b C J* \ c,
c / C(a,b) = L(b). Therefore c X \ L(b). Similarly beX \ L(c).
By 2.9 (iii), L(b) n L(c) is a chain; the order is antisymmetric, so
L(b) n L(c) contains at most one maximum element. By 2.9 (i) and the
compactness of T, L(b) 0 L(c) is a compact chain with semicontinuous
order, so it contains a maximum element by a remark by Wallace in [18].
Let x = max (L(b) n L(c)), and notice that X \ M(x) C L(b) n L(c).
Also a X \ M(x) : for a< x, and if a = x, then a separates b
and c; but this is false since J* \ a is a connected set containing
b and c.
Finally, X \ x = ( X \ M(x) ) U (X \ L(b)) U (X \ L(c)),
and these sets are easily seen to be disjoint. Also, each is open in X
since points are closed and the order is semicontinuous. Therefore x
separates a, b and c pairwise in X; X is connected so by 2.11 (ii), x
separates them pairwise in T. Thus x is a branch point of T. But
xeJ: for J U a U b is connected, so by 2.11 (i), C(a,b)C J U a U b;
x C(a,b) and xja U b, hence x J. Since J was chosen in the complement
of the set of branch points of T, we have reached a contradiction. It
follows that J* contains exactly two noncutpoints and is therefore an arc.
The following lemma is known; actually, both parts are special
cases of more general theorems (see Theorem 66 of [12]), but we need
only these weaker statements which are easily proved using the fact
that C(a,b) is connected for any points a and b in a tree.
(2.13) Lemma. Let A and B be nonnull subcontinua of a tree T.
(i) If A n B = D, there is some t in T such that t separates
A and B.
(ii) If A n B / D and if A \ B and B \ A are nonnull connected
sets, then A 0 B separates A \ B and B \ A.
Proof: (i) Let aeA and b eB. By 2.10, C(a,b) is connected,
and it intersects both A and B, so there is some te C(a,b) \ (A U B).
T \ t = P U Q, separated sets such that a P and be Q; A is connected,
lies in T \ t and intersects P, so A C P. Similarly, B C Q.
(ii) Let aeA \ B and b B \ A. By 2.11 (i), since A U B is
connected, C(a,b) C A U B. By 2.10, C(a,b) is connected, so there is
some x C(a,b) n (A n B). Therefore A n B separates a and b, and hence
A \ B and B \ A since they are connected.
(2.14) Lemma. Let T be a tree and Y a space. Let RC T x Y
be such that T = RY, R is monotone, noninclusive, and Ry is closed for
each y in Y. Let B be the set of all branch points of T and for each
b in B, let bR be a single point. If y is a point of Y such that
T \ Ry = P U Q, nonnull separated sets, then (P*)R n QR = Q .
Proof: Suppose to the contrary that there is some x in
(P*)R n QR. Then Rx n P* / 3 and Rx n Q / ] The latter fact
assures us that x / y. If Rx n Ry :, then RxC P U Q; Rx n P" / Lthen
implies that FRx n P / 0 ; and Rx n Q / 1 which imply that Rx is
not connected. But R is monotone by hypothesis; hence Rx n Ry must
be nonnull. No point of B can lie in Rx 0 Ry, since for b in B, bR is
a single point and we know x / y. A tree is hereditarily unicoherent
(see Theorem 9 of [21]) and Rx and Ry are subcontinua of T, so Rx n Ry
is connected. Therefore there is some component J of T \ B such that
Rx n Ry C J.
Suppose x / y in (J*)R and Rx J* C Ry N J*. Then Rx J*
since R is noninclusive, and Rx is connected, so there is some
beRx n (J* \ J); this implies that b eB. But b Rx n J* and
Rx P J*C Ry nfJ* imply that be Rx n Ry, which is false. Therefore
Rx n J1* f Ry n J*, and as a corollary, J* is not degenerate. Then
by 2.12, J* is an arc, so we may suppose a simple order has been defined
on J*. Since a tree is hereditarily unicoherent, we may suppose that
Rx n J* = [p,q] and Ry n J* = [r,s]. Since neither of these sets con
tains the other, we may suppose p < r and q < s; and because Rx l Ry
is nonnull and contained in J, we have p < r < q < s.
By 2.13 (ii), T \ [r,q] = P' U Q', where P' and Q' are separated
sets such that [p,r)C P' and (q,s] CQ'; further, since [r,q] C JC T \ B,
P' and Q' are connected sets. Now Rx n Q' = 0 and Ry n P' 0 ; for
suppose Rx n Q' 0. Since s eJ*, J U s is connected, so by 2.11 (i),
C(q,s) = [q,s]C J U s; hence (q,s) C J, and since q / s, (q,s) / [ .
Therefore (q,s) separates T into the sets, P' U [r,q] and Q' \ (q,s);
we know Rx C T \ (q,s), so (q,s) separates Rx 0 P' from Rx n Q'.
Rx is connected, however, and we know that p eRx n P', so Rx N Q' must
be null. Similarly, Ry n P' = .
Therefore, T \ Ry = P' U (Q' \ Ry) and of course these are
separated sets. By hypothesis, T \ Ry = P U Q, nonnull separated sets,
and by supposition, Rx n P* / L and Rx n Q / c. Since PRx n P' /
and this is a component of T \ Ry, it follows that Rx n (Q' \ Ry)* / f3.
But (Q' \ Ry)* CQ', and Rx n Q' = j]. We have reached a contradiction,
so it follows that no such x can exist.
(2.15) Theorem. Let T be a tree and Y a nondegenerate Hausdorff
space. Let R C T x Y be such that T = RY, R is point compact, u.s.c.
(1)
on T, monotone, noninclusive, and R' is monotone. Also, for each
branch point b of T, let bR be a single point. Then TR = Y and Y
is a tree.
Proof: Since R is noninclusive, by 2.1 (i), TR = Y. By 2.5,
TR is connected; that is, Y is connected. By 1.17, Y is compact, and
by hypothesis, Y is Hausdorff, so Y is a continuum. To complete the
proof that Y is a tree, let x / y in Y; we need to find some z in Y
such that z separates x and y.
Case 1. Rx n Ry = L]. By 2.2 (ii) and since R is monotone,
Rx and Ry are subcontinua of T; then by 2.15 (i), there is some t in T
which separates Rx and Ry. Since T = RY, tR is not empty, so we can
choose z in tR. Ileither x nor y lies in tR, so zi(x U y). Now t e Rz,
and since t separates Rx and Ry, Rz separates R: \ Rz and Ry \ Rz.
Since R is noninclusive, neither of these sets is empty. So T \ Rz =
P U Q, nonnull separated sets, and x PR and ye QR. Since
PR U QR = (T \ Rz)R and since R is noninclusive, by 2.1 (ii),
PR U QR = Y \ z. Finally, PR and QR are separated; for by 2.2 (i),
(PR)* C (P*)R, and by 2.14, (P*)R n QR = [. Similarly, PR n (Q*)R = E.
Case 2. Rx 0 Ry / LI. Since T is hereditarily unicoherent
(see Theorem 9 of [21]), since R is monotone and Rx and Ry are
closed by 2.2 (ii), Rx 0 Ry is a continuum. Since bR is a single
point for each branch point b of T, Rx n Ry C J, a component of
T \ B, where B is the set of branch points of T. As in the proof
of 2.14, Rx n J* Ry r J*, henced*is not degenerate. Therefore,
by 2.12, J* is an arc. Let the noncutpoints of J* be a and b, and
define t < t' in J* iff t = a, t = t' or t separates a and t'.
Let S = R n (J* x Y), and consider S as a relation in J* x
(J*)R. It is easily seen that S satisfies all the hypotheses of 2.6,
and hence that (J*)S = (J*)R is an arc whose noncutpoints are aR and
bR. Define w < w' in (J*)S iff w = aR, w = w' or w separates aR
and w'. Now (x U y) C (J*)S and x / y, so we may suppose x < y and
we can find some z such that x < z < y. We will now prove that
T \ Rz = P U Q, separated sets such that x ePR and y e QR.
Let Sx = [px^x], Sy = [p y'), and Sz = [pzqz]. Since
(J*)S is an arc, 2.4 applies to tell us that px < z < y and
qx < qz < q. By hypothesis, Rx 0 Ry = Sx 0 Sy / ], and S is
noninclusive, so we may conclude that px < Pz < Py qx < qz < Qr
Then [px,qz] and [pz,qy] are subcontinua of T which satisfy 2.13 (ii);
their intersection is [pz,qz], hence T \ [pz qz] = P U Q, separated
sets such that [p ,pz) C P and (qz ,y] C Q. Therefore x ePR and
y eQR. Finally, PR U QR = Y \ z and PR and QR are separated, just
as in Case 1, since Rz connected implies Rz = [pzqz].
Whether or not 2.15 implies that R is l.s.c. on T is unknown.
While it is true that any union of relations, each in T x Y and l.s.c.
on T, is l.s.c. on T, even if R = U(R n (J* x Y) I J is a component
of T \ B) (which need not be true), we know only that R n (J* x Y)
is 1.s.c. on J*. As mentioned on p. 12, this does not imply that
R D (J* x Y) is l.s.c. on T.
(2.15') Theorem. Let T be a tree and Y a nondegenerate
Hausdorff space. Let R C T x Y be such that T RY, R is point com
pact, u.s.c. on T, monotone, noninclusive, and R is l.s.c. on T.
Also, for each branch point b of T, let bR be a single point. Then
TR = Y, R(F) is monotone and Y is a tree.
Proof: By 2.1 (i), TR = Y. To prove that R() is monotone,
let t eT. If t e B*, let [ba I a D] be a sequence contained in B
such that [ba] converges to t; then (ba)R is a point for each a in D,
and because R is l.s.c. on T, tR is a point also: for suppose
tR 3 (y U y'), where y ./ y', and let U and U' be disjoint neighbor
hoods of y and y', respectively. Then t RU r R(U'), and this set
is open because R is l.s.c. on T; hence there is some a such that
for every a beyond ao, ba E RU 0 R(U'). That is, (ba)R nU / D] and
(ba)R n U' '/ 1. But such ba exists and (b )R is a point, which cannot
lie in both the disjoint sets U and U'. Thus tR contains at most one
point and is therefore connected. If t B*, then let J be the compo
nent of T \ B* such that teJ; T is locally connected (see [20]) so
J is open in T, and hence J is not degenerate. Therefore J* is an arc,
by 2.12. Let S = R D (J* x Y) and consider S as a relation in
J* x (J*)R. It is easy to prove that S satisfies all the hypotheses
of 2.6', and hence S(1) is monotone. That is, for each teJ*,
tS = tR is connected. This completes the proof that R() is monotone.
Now 2.15 can be applied to conclude that Y is a tree.
Let us define a dendrite to be a metric, locally connected
continuum in which each two points can be separated by a third point.
Ward proved in [20] that a tree is locally connected, and hence that
a metric tree is a dendrite.
If (Y,<) is a partially ordered set and A C Y, we will say
a is a zero of A iff a A and A C M(a).
(2.16) Theorem. Let D be a dendrite and Y a nondegenerate
Hausdorff space. Let R C D x Y be such that D = RY, R is monotone,
noninclusive, point compact, and u.s.c. and l.s.c. on D. Also, for
each branch point b of D, let bR be a point. Then X is a dendrite.
Proof: By 2.15', Y is a tree and R(1) is monotone. Define
Y
f: D 2 by f(d) = dR for each d in D; since R is u.s.c. and l.s.c.
on D, f is continuous by 1.8. Since R is point compact and R(1) is
monotone, dR is a continuum for each d in D.
Choose e in Y and define < on Y as follows: let x < y iff
x = e, x = y or x separates e and y; this is a semicontinuous order
by the theorem of Ward's which we quoted as 2.9. Let 2= (A 2Y A
has a zero); Capel and Strother proved in [2] that each nonnull
subcontinuumr of Y has a zero, hence f(D) C 2; they also proved there
that, since Y is compact, the function g: a * Y defined by g(A) =
zero of A, is continuous.
Therefore, gf: D * Y is continuous. D is a compact metric
space, hence gf(D) is also metric (see 323 of [7], for example).
To complete the proof, we need to show that gf(D) Y. We know
gf(D) is connected and Y is irreducibly connected about its set of
noncutpoints, so it will suffice to prove that any noncutpoint of Y
lies in gf(D).
Let y be any noncutpoint of Y. We will prove that there is
some point d in D such that dR = y; then f(d) = y, and therefore
gf(d) = y. It was shown in the proof of 2.15' that for each b in B*,
bR is a point; so if ye (B*)R, we are done. Otherwise, there is some
component J of T \ B* such that ye (J*)R. It was shown in 2.15' that
J* is not degenerate, so by 2.12, J* is an arc; let a and b be the
noncutpoints of J*. Since y does not cut Y, it cannot cut (J*)R; as
was shown in the proof of 2.15, the noncutpoints of (J*)R are aR and
bR, points of Y. Therefore either y = aR or y bR, and we are done.
As was mentioned on p. 31, a relation which preserves uni
coherence has not yet been found. We do know that something more than
the conditions used so far is necessary. In [14] Strother gives an
example of a very wellbehaved relation which takes a 2cell onto its
boundary; we quote this example below.
(2.17) Example(Strother). Let X be the unit disc and S the
unit circle; that is, S is the boundary of X. Define RC X x S as
follows. If x is the origin, let xR = S. If x is not the origin:
(a) Extend the segment from the origin through x until it meets S
in a point A. (b) Draw a perpendicular at x to the radius constructed
in (a) and denote its intersections with S as B and C. (c) Consider
the closed arc BAC on S. Let MBACN be the closed arc of S with
center A, length twice the length of the arc BAC, and having end
points M and N. (d) Let xR = MBACN.
It is geometrically obvious that R is u.s.c., l.s.c., point
compact, monotone, noninclusive, and the inverse relation is monotone;
but X is unicoherent and XR = S is not.
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BIOGRAPHICAL SKETCH
Jane Maxwell Day was born March 12, 1937, at Avon Park,
Florida. She graduated from Avon Park High School in June, 1954.
She received a Bachelor of Arts degree with High Honors in January,
1958, and a Master of Science degree in January, 1961, both from the
University of Florida. She served as a graduate assistant in the
school year 19591960 and as an instructor in the school year 1960
1961, in the Department of Mathematics. She was granted a National
Science Foundation Cooperative Fellowship for study at the University
of Florida from September, 1963, through August, 1964.
Her husband is Walter Ransom Day, Jr., and she has a daughter,
Bonnie Claire. She is a member of Phi Beta Kappa and Phi Kappa Phi.
This dissertation was prepared under the direction of the
candidate's supervisory, committee and has been approved by all
members of that committee. It was submitted to the Dean of the
College of Arts and Sciences and to the Graduate Council, and was
approved as partial fulfillment of the requirements for the degree
of Doctor of Philosophy.
April 18, 1964
Dean, College of Arts and Sciences
Dean, Graduate School
Supervisory Committee:
:'~ 6.
