On the structure of locally connected topological spaces
by Spencer Edward Minear
A thesis submitted to the Graduate Faculty in partial fulfillment of the requirements for the degree of
DOCTOR OF PHILOSOPHY in Mathematics
Montana State University
© Copyright by Spencer Edward Minear (1971)
Abstract:
In the first part of the paper we prove that every cut point of a connected space is either an open or a
closed point. We also investigate other properties of cut points in a connected, locally connected space,
and the relations amoung the components of the complements of cut points.
We then start a development of cyclic element theory in connected, locally connected spaces "by
extending the work of Albert and Youngs.
We define an A-set and show that it has the classical property, that if H is connected and A is an A-set,
then H A is connected. We also find necessary and sufficient conditions for a non-empty intersection of
A-sets to be an A-set.
In the last part of the paper we impose the condition that every N-set is an A-set. Under this condition
we obtain the result that unicoherence is both a cyclicly extensible and a cyclicly reducible property.
We also show that in a compact, connected, locally connected space, that the fixed point property is
also cyclicly extensible and reducible. ON THE STRUCTURE OF LOCALLY CONNECTED TOPOLOGICAL SPACES
by
SPENCER EDWARD MINEAR
A thesis submitted to the Graduate Faculty in partial
fulfillment of the requirements for the degree
of
DOCTOR OF PHILOSOPHY
in
Mathematics
Approved:
MONTANA STATE UNIVERSITY
Bozeman, Montana
June, 1971
Acknowledgment
I would like to extend my sincere gratitude to my advisor Dr.
Byron L. McAllister for his guidance and encouragement while studying
under him.
I would also like to thank Mr. Orvald Haugshyj, who, hy his
inspiring instruction, showed me how interesting and rewarding the
study of mathematics can he.
iv
Table of Contents
Introduction . . ................
Chapter I
..............
Chapter 2
............ ..
Chapter 3
. . . . . .
Chapter 4
. . . . . . . . . . .
Bibliography ........
........
. . . . . .
. page vi
• page I
. page 10
. page 15
. page 29
page 39
V
Abstract
In the first part of the paper we prove that every cut point
of a connected space is either an open or a closed point. We also
investigate other properties of cut points in a connected, locally
connected space, and the relations amoung the components of the
complements of cut points.
We then start a development of cyclic element theory in connected,
locally connected spaces "by extending the work of Albert and Youngs.
We define an A-set and show that it has the classical property, that
if H is connected and A is an A-set, then H A A is connected. We
also find necessary and sufficient conditions for a non-empty inter­
section of A-sets to be an A-set.
In the last part of the paper we impose the condition that every
N-set is an A-set. Under this condition we obtain the result that
unicoherence is both a cyclicly extensible and a cyclicly reducible
property. We also show that in a compact, connected, locally connected
space, that the fixed point property is also cyclicly extensible and
reducible.
vi
Introduction
The concept of cyclic elements was introduced by G. T. Whyburn
in 1926.
He said that a cyclic element of a Peano space was either:
I) a point p such that X - (p) is disconnected, (a cut point), 2)
a point p with arbitrarily small neighborhoods with singleton boundary
(an end point), or 3) a subcontinuum of the space which is maximal
with respect to the property that each.pair of points in it lie on
a simple closed curve.
This idea proved to be useful when it was
discovered that many properties of the space depend on the analogous
property of all the cyclic elements in the space.
In 1929 R. L. Moore
suggested [6] an alternate definition for a cyclic element.
Using Moore's suggestion in 1930 [3 ] Kuratowski and Whyburn defined
a cyclic element as either: I) a cut point, or 2) a set consisting
of a non-cut point p and all points x such that no point separates
between p and x.
(A subset Y of X is said to separate between two
points a and b if there exist
subsets H and K of X such that X - Y =
H U K, separated, with a e H and b e K.)
This new definition proved
to be very important as it simplified the proofs for cyclic element
theory in Peano spaces and also permitted the extension of cyclic
element theory to more general spaces.
In 1929 C. Kuratowski [2] proved that if each cyclic element of
a Peano continuum is unicoherent, then so is the space and if the
space is unicoherent, then so is each cyclic element.
Such a property
vii
is called cyclicly extensible and reducible.
In the same paper he
suggested that the fixed point property is also a cyclicly extensible
and reducible property.
This was proved by Borsuk in 1932.
In 19^2, Albert and Youngs [l] extended cyclic element theory
to connected, locally connected spaces.
each cut point be a closed point.
However they required that
In this paper we will show that
cyclic element theory can be extended to connected, locally connected
spaces with out this requirement. We will also show that with a
suitable restriction on the.space that unicoherence and the fixed
point property are both cyclicly extensible and reducible.
Chapter I
Let X "be a connected topological space and let Y be a subset of
X.
Then Y is said to be a cutting of X if X - Y is disconnected.
the case when Y is a singleton
(a), then
CL
In
is called a cut point of X.
If Y is a cutting of X and A and B are subsets of X, then Y is said to
separate A and B if there is a separation
of X - Y = H U
and K are separated, such that A <= H and B c K.
where H
If a is a cut point
of X and a and b are distinct elements of X, then CC is said to cut
between a and b if a and b appear in distinct components of X Note that ,if
Ct
separates a and b, then CC also cuts between a and b,
but the converse is not true in general,
is a quasi-component of X component
of X -
(a).
as in the case when there
(cc) which is not a component.
A quasi­
C of X - (cc) is a set such that for each separation H U K
(cc) it is true that C c H or C c K.
Let A and B be subsets of X, then S(A,B)
will be used to denote
the component" of X - B which contains A, if one exists.
B = (b) we use S(a,b) in place of S((a),(b)).
If A = (a) or:
We shall use Ba to
denote the boundary of the set A, A and A° to denote the closure and
the interior of A respectively and 0 to denote the empty set.
The following theorem makes most of the work to follow possible.
It was apparently first observed by my advisor Dr. B. L. McAllister as
a corollary to the main theorem of his "A Note On Irreducible Separa­
tion" [5).
paper.
The proof below is independent of. the results in the above
2
1.1) Theorem:
If a is a cut point of a connected space X, then (a)
is either an open or a closed set.
Proof:
follows that
is open.
A = A or A = A U (a).
Since
is connected,
X - {a) = A U B separated. Since A ("I B = $ it
Let
If
A = A, then B U (a) = X - A
B fl A = 0 it follows that B c B c B U (a) and since X
B ^ B U (a), thus B = B .
which implies that (ctj is open.
open and as above
If
Hence
X - (a) = A U B is closed
A = A U (a), then B = X - A is
B = B U {(%]. Then A H B = (AU (a)) D ( B U (a)) =
(a) is a closed subset of X.
Earlier it was pointed out that it is possible for a cut point
Ct to cut between two points yet not separate them.
If we require the
-
space to be locally connected as well as connected the above situation
may still occur if id) is open: however the following theorem shows
that closed cut points no longer have this property.
1.2) Theorem:
Let X be a locally connected, connected space.
If Oi
is a closed cut point that cuts' between a and b, then a separates
a and b . •
Proof:
Since (o:) is closed, X - (cc) is open, and thus S(x,a)
is an open subset of X for each point x in X.
Thus X - (a) = S(a,a)
(U{ S(x,cx) : x ^ Oi and x ^ S(x,o:))) is a disjoint union of open sets,
one containing a and the other containing h.
BOTE:
For the remainder of this paper the symbol X will denote a
U
3
connected, locally connected topological space, unless stated other­
wise .
The next two theorems are well-known and will he stated without
proof.
1.3) Theorem: If a is a closed cut point of
X, then {a) = ds for each
component S of X - ioc].
I A ) Theorem:
If a is an open cut point of X and x e dS, where S is
a component of X - (a), then x e Ta).
1 .5 ) Corollary:
X -
If a is a cut point of X and S is a component of
(a), then S U {a} is connected.
1 .6) Theorem:
Let a
component of X - (a)
Proof:
If S is
If S is a
distinct from S(p,a), then S c S(a,f3).
distinct from
S(f3,a),then S <= X - {£) and hy I .5
S U (a) is a connected subset of X - {(3) that contains a.
we know that
Thus
and f3 he distinct cut points of X.
S c (a) U S(a,p), and hence S c S(a,p).
1.7) Theorem:
a e TpT and
If
a and P are distinct cut points of X such that
(a) = ds(a,p), then S(a,p) n S(p,a) = j6.
Proof: Let A =
S(a,p) D s(P,a). Since S(a,P) is closed, A is a
closed subset of
S(p,a). Since a / A, A c S(a,p) - (a) = S(a,p) °,
we have that A =
S(a,p)° H s(p,a). Therefore A is also an open subset
of
S(p,a), but since A ^ S(p,a) and S(Pya) is connected, we have A = 0.
4
I .8) Corollary:
OL
If a and P are distinct cut points of X such that
e TpT and (a) = ds(o:,p), then:
i) S(P^a) =
ii) S(a,P)
Proof:
i)
=
U( S U (p) :S a
component of X - (p) and S / S(a,P) }
U{ S U (a) :S a
component of X - (ct) and S ^ g(p,a) )
Since the right hand side of the equation is connected,
contains P and does not contain
hand side.
x e
CL
it must he contained in the left
If x e X is such that x is not in the right hand side, then
S(a,P) and x 4 S(P,C%) by I .7 . Thus the complementof the right
hand side is contained in the complement of the left hand side and
the left hand side is contained in the right hand side.
The proof of
part ii is identical to that of i.
I .9 ) Corollary:
CX e
TipT
If U and P are distinct cut points of X such that
and (a) = rdS(o:,p) and x e X is distinct from
CL,
then
OL
cuts
between x and P if and only if x e S(a,p).
Proof: Follows immediately from 1.8:
In the classical cyclic element theory in Peano spaces much
of the work relied on the fact that the space was arc-connected.
As pointed out in the introduction the work was soon extended into
spaces that are not arc-connected.
In Whyburn1s work in separable
metric spaces he found a set that possessed the property of tying
the space together much as the arc had done in the Peano space.
shall find later that the same type of set exists in a suitably
We
restricted locally connected space and will be used as a tool much the
same as it Was by Whyburn.
The following definition is essentially
the same as in the separable metric case and is the key to finding
the set mentioned above.
Let a and b be distinct elements of X.
Then E(a,b) denotes
the set of all the cut points of X which cut between a and b.
If
a and b are distinct elements of X and tt and P are distinct elements
of E(a,b), then (X is said to proceed P in E(a,b), CK < p, if a cuts
between a and p.
1.10) Theorem:
The relation "<" on E(a,b) is a linear order for each
pair of distinct points a and b in X.
Proof:
Let Oi and
P
be distinct elements of E(a/b) and assume that
Oi does not cut between a and
P,
thus S(a,o:) = S(P^ct).
a cuts between a and b, so S(a,o:) ^
connected subset of X S(a,p) =
s(b,P)
and
P
(P)
S(b,P)
Since Oi e E(a,b),
and S(b,o:) U (a) is a
that contains both a and b.
Therefore
cuts between Ce and a.
Let Oi and P be elements of E(a,b) such that a < p. . Then S(a,Ci) ^
S(P^ct) and S(a,o:) I) {a) is a connected subset of X - (p) which contains
both a and 0i.
Thus P does not cut between a and Ct hence P ^ Ct.
Let Ct, P and T) be elements of E(a,b) such that Ct < P and P < T).
Since Ct < p, S(a,ct) ^ S(p,Ct)'.
S(a,Ct) = S(^,a).
If Ct
tj,
we know that t] < ct and thus
But then S(a,Ct) is a connected subset of X - (p)
which contains both a and
which implies that P does not cut between
6
a and I), contrary to the hypothesis.
Therefore a < T].
Let a and h he distinct elements of X such that E(a,b) ^ fa.
Then
if B is any nonempty subset of E(a,b), B inherits a linear order
relation from E(a,b).
where
Thus (B,<) is a directed set and hence (ig,<),
is the identity mapping of B onto B, is a net.
For conven­
ience we shall let (B,<) denote the net (i ,<):.
I .11) Theorem:
Let a and P be in E(a,b) with a < p.
Then S(a,a) U (a)
is contained in S(a,|3).
Proof:
Since a < P we know that S(a,Q:) ^ S(p,a), thus S(a,a) U {a}
is a connected subset of X - {(3) which contains a.
Thus we have that
S(a,a) U (a) c s(a,P).
1.12) Corollary:
Let CU and P be in E(a,b) with
i)
S(a,p) =
ii)
S(b,o:) =
iii)
S(b,p) U
iv)
OL
< p.
Then:
S(a,p)
S(p,(%)
(p) c
s(b,a)
S(a,a) nS(b,p) = <j)
Proof:
i)
Follows immediately from 1.11.
ii)
If S(b,(%) ^
S(p,C()} then S(a,a) U (a) U S(p,a) is a connected subset of X Thus S(a,P)
= S(b,P),
know
S(b,P) Cl S(b,o:)
iv)
S(a,a) D
s(b,p)
contrary to the hypothesis. iii)
and by ii,
c
P
e
S(b,o:).
s(a,cx) n s(b,a)
= 0.
By
1.6 we
Thus S(b,P) U(p) c
'
(p).
s(b,oi).
I I H
7
1*13) Theorem:
i)
Let a, b, a, p, and T] be distinct elements of X , then:
If CC and t) are in E(a,b) with o: <
P e E(a,b) and a < p <
ii)
If CC, P and
Proof:
i)
tj
t) and
P e E(a,T)), then
tj.
are in E(a,b) and Cu < P < t), then p e E(a,r]).
Assume that p ^ S(b,a) = S(r],a). Then S(t),«) U (a)
is a connected subset of X - (p) containing a and
that S(a,P) = S(T^p), contrary to the hypothesis.
Similarity P e S(a,rj).
tj
which implies
Therefore p e S( tj,,q :).
Thus we have S(a,(%) U (a) and S(b,Tj) U (tj)
are connected subsets of X - (p), hence S(a,p) = S(a,p) and S(b,P) =
S(Tj,P).
However since P e E(a,Tj), we have that S(o:,p) ^ S(tjjP).
Thus S(a,P) ^ S(bjP) and P € E(a,b).
ii)
By 1.121, CK < P implies
S(ajp) = S(CKjP) and by 1.1211, P < tj implies S(b,p) = .S ( tj, P ) .
But
since p e E(a,b), S(a,P) 4 S(b,p) and thus P e E(a,rj).
I .14) Theorem:
Let a and b be distinct elements of X, and let B be
a nonempty subset of E(a,b).
If S = U{ s(b,Q:) : a e B }, then dS is
contained in the set of limits of (B,<).
If B has no first element,
then S is open and dS is equal to the set of limits of (B,<).
Proof:
If B has a first element P, let x e dp = ds(b,p) and
let V be an open connected neighborhood of x.
Then P e V and thus
x is a limit of (B,<).
Assume that B has no first element and let x e S.
We shall show
that x e S° by considering three cases,which are not necessarily
distinct.
First let x e B and let Ct e B be such that Ct < x.
If Ct is
/ tI
8
a closed cut point, S ( x , a ) = S("b,a) is open and thus x e S°.
is an open cut point, then X - S(a,a) '= (a) U {y e X
S(y,a) ^ S(a,a))
is an open connected neighborhood of x which contains b.
be such that P < a.
x e S°.
Let P e B
Then P e S(a,a) and thus X - S(a,a) c S(b,P) and
Second, if x e S(b,a)° for some a e B, then x e S°.
let x e ds(b,a) for some a e B and assume that x ^ B.
such that P < a.
If a
If p is
Finally
Let p e B be
closed, then S(x,p) is an open connected
neighborhood of x and thus Ct e S(x,P) and S(x,P) = S(o:,p).
S(x,P) = S(b,P)', since S(a,P) = S(b,P), and x e S°.
Then
If P is open,
then x e S(b,P) since S(b,p) is closed and S(b,o:) c s(b,p). Then
as in the first case we have that X - S(a,P) is an open set contained
in S which contains S(b,P) and thus x s S 0.
Therefore S is open.
Again assume that B has no first element.
We shall show that
Bs is equal to the set of limits of (B,<). Let x e ds and let Ct e
If P e B is such that P < a and if S(x,Ct) ^
S(a,Ct),
S(b,P) and x e S which is impossible since S is open.
x e S(a,Ct) for all Ct e B.
Then V H
B.
then S(x,Ct) c
Therefore
Let V be a connected neighborhood of x.
s(b,ct) ^ / for some
Ct e B and since x /
S(b,Ct),
Ct e V.
If P e B such that P < cx, then P cuts between x and Ct since S(x,p) =
S(a,P) ^ S(b,P) =
S(ct,p). Thus P e V. Therefore x is a limit of (Bj<).
Finally, let x be a limit of (B,<).
Recall
shown above there is an Ct e B such that y e
that if
S(b.,o:) °.
y e S, then as
But no element
9
P of
B is such that P < a and is in S(b,a)°.
is a limit of
Therefore no point of S
(B,<). Hence x e S - S = ds, since S is open.
Chapter 2
Let a and b be distinct points in X, then a and b are said to be
conjugate, denoted a 0 b, if there is no point in X which cuts
between them,.
Thus if a O b , then a and b
are in the same component
of X - ( a ) for every a in X distinct from a and b.
2.1) Theorem:
Let a and b be distinct elements of X such that a e T b J .
Then a 0 b.
Proof: Assume there is a point a which cuts between a and b.
If a is a closed cut point then S(b,Ct) is an open neighborhood of a
which does not contain b.
If CK is an open cut point, then S(b,o:)
is a closed set containing b but not a.
contradiction to the hypothesis.
2.2) Theorem:
In either case we have a
Therefore a 0 b.
If a 0 X 1 O i - - O x
O b and z e E(a,b), then z = x.
I
n
'
i
for some i between I and n.
Proof: If z is not equal to any x^, I < i < n, then (a,x^,...x^,b)
must be contained in a single component of X - (z), contrary to the
hypothesis.
2.3) Corollary:
If a 0 X^ 0 ... 0 X^ 0 b and a 0 y^ 0 ... 0 y^ 0 b
and x. ^ y . for all i and j, where I < i < n and I < j < m, then a 0 b.
I J
Proof:
If a and b are not conjugate there is a cut point CK in
X which cuts between a and b.
where i and j are as above.
to the hypothesis.
By 2.1 CK = x^ and CK = y^ for some i and j
But then X^ = y^ for some i and j contrary
Theorem 2.2 and corollary 2.3 along with the following two
definitions are due to Rado and Reichelder [7].
A subset A of X is said to be O-coherent ( or coherent ) if
every two points a and b in
A are conjugate. If a contains every point
which is conjugate to two distinct points in
A, then A is said to
be 0-compIete ( or complete ).
2.4) Theorem:
Let
A be a coherent subset of X, then A - {a) is
contained in a single component of X - (a) for each point a in X.
Proof: let a be in
A
(ct) where a e X.
that b is distinct from a, then a 0 b*
If b e
A - (a) such
Thus S(a,a) = S(b,a).
Therefore
A - (a) c s(a,a).
Let R be a nondegenerate subset of X which is both complete and
coherent, then R is said to be an R-set.
2.5) Theorem:
If a and b are distinct elements of X such that a 0 b,
then there is a unique R-set R which contains both a and b.
Proof: Let R = { x e X : x 0 a and x 0 b } and let x and y b e .
distinct elements of R.
x 0 y, and R is coherent.
Then x 0 a 0 y and x 0 b 0 y.
Thus by 2,3
Row let z e X be such that z 0 x and z 0 y,
where x and y are distinct elements of R . . Then a 0 x 0 z and
a 0 y 0 z, thus a 0 z. Similarly b 0 z.
we have that z e R, and R is complete.
Hence by the definition of R
Therefore R is an R-set.
Let R' be any other R-set containing a and b, and let x e R 1.
Then x O a and x O b since K' is coherent.
Thus x e N and N ’ c u.
If x e N, then x 0 a and x 0 h, hut since N 1 is complete x e W
N c
.
and
Therefore B' = N, and K is unique.
2.6) Corollary: If N and W
are distinct N-sets in X, then M fl N'
is degenerate.
Proof:
Follows immediately from 2.5«
2.f) Corollary:
If N and W
are distinct N-sets in X and {«) = N fl N'
then a is a cut point of X, and if a e N - (a) and b e N 1 - lcc), .
then a cuts between a and b.
Proof:
N*, b 0 a.
Since a and CX are in N, a 0 CK and since b and CC are in
But since N is complete and b is not in N we have that a
and b are not conjugate.
a and b.
Let P be a cut point which cuts between
By 2.2, P = a and thus CX cuts between a and b and is a cut
point.
We should point out that 2.4, 2.5, 2.6 and 2.7 are essentially
the same as in [I] though we did not have to worry about showing
that the cut points are closed points.
Also note that from 2.1 and
2.5 we have the result that every open cut point is in at least one
N-set.
The same is not true of closed cut points.
2.8) Theorem:
Let N be an N-set in X.
If x e N,- N, then there is ■
a unique open cut point a of X in N such that CC cuts between x and
LI IL
I I
13
each point of N - {tij, x 0 a, x ^ N - [ a } , and x e "Taj.
Proof:
Since x is not in K there is a point n in N and a cut
point a of X such that a cuts between x and n.
By 2.4, N - [ a ) is contained in S(n,cx).
Thus S(x,a) /= S(n,a).
If a were a closed cut point
S(x,o:) would be an open neighborhood of x that is contained in
X - ((cd U S(n,ex)).
S(x,a) A H
However N c ({a} U S(n,(%)), thus we would have
^ which is impossible since x e N.
cut point.If a ^ N, then N c S(n,o:)
N c S(n,o:).
x and n.
Therefore a is an open
and since S(n,0!) is closed,
Thus x e S(n,o;) which is impossible since Ct cuts between
Therefore Ct e W.
Since S(n,Ct) is a closed set and W - (a) c
S(n,ct) we have that N - TctJ c s(n,Ct), thus x ^ N - {ct}.
However
H = N - {Ctj U (aj and x e H, hence x e ICtJ and, by 2.1, x 0 Ct.
Let
P be an open cut point of X which cuts between x and each point of
X - (B).
Then since x 0 Ct 0 n where n is any point of N and B cuts
between x and n we have, by 2.2, that Ct = B«
2.9) Corollary: Let W be an W-set in X.
Therefore Ct is unique.
If N contains no open cut
points, then N is closed.
2.10) Lemma: If x e M - W, where H is an H-set, then S(x,W) = S(x,Ct)
where Ct is the open cut point in N whose existence is guaranteed
by theorem 2.8.
Proof: Since Ct e N, S(x,N).is a connected subset of X - (ct)
containing x.
So S(x,N) c s(x,Ct).
Let n e N - (ct).
S(n,ct) ^ S(x,Ct) and by 2.4 S(N - (ct),a) = S(n,a).
Then by 2.8
Thus S(x,a) Cl H
14
which implies S(x,a) is a connected subset of X - N containing x;Therefore S(x,o:) c s(x,N).
2.11) Theorem:
If N is an E-set in X and SI is a component of X - E,
then S is either open or closed.
Proof:
Let S be a component of X - W which is not open.
there is a point x in dS H S:
If V n N = '0, then V c g
V ft N ^
Let V be a connected neighborhood of x.
which is impossible since x e dS.
Thus x e N - N,
guaranteed by theorem 2.8.
Then
Therefore
Let (X be the open cut point in N
Then by 2.10 S = S(x,a) which implies that
S is closed.
2.12) Corollary:
Let E be an E-set in X and let S be an open component
of X - N, then 8s
Proof:
E.
Since S is relatively closed in X - E, S contains
dS A (X - E).'
But by the prooof of 2.11
the boundary of S in S, then S is
if
closed in
there is
onepoint of
X.
if S is open
Thus
in X, then dS Tl (X - E) = 0 or dS c E.
If S is a closed component of X -
E, for some E-set E in X we let
b(S) denote the open cut point in E whose existence is guaranteed
by theorem 2.8.
Eote that S = S(S,b(S)).
Chapter 3
A nonempty subset A of X is called an A-set if it satisfies
the following:
i)
ii)
iii)
Every component of X - A is open or closed,
Each open component of X - A has singleton boundary.
For each closed component S of X - A there is a unique open
cut point Ct in A such that S is a component of X - {a) and ds fl A - (ctj
Let A be an A-set in X and S be a component of X - A. Then if
S is open and dS = (cx), b(S) denotes Ct, or if S is closed, then
b(S) denotes the unique open cut point a in A whose existence is
guaranteed by iii of the definition of an A-set.
If S is an open component of.X - A, where A is an A-set, then
since S = (S U (b(S))) fl (X - (b(S))) = S fl (X - (b(S))), S is a
c!open subset of X - (b(S.)).
Therefore S is a component of X - (b(S)).
If S is a closed component of X - A, then by I.4 each point x of dS
is in
TRsJT and
by 2.1, x 0 b(S).
We should point out that the notation b(S) has already been
used in connection with E-sets..
If S is a closed component of X - N,
where N is an E-set in X, b(S) is an open cut point in E such that
S is a component of X - (b) and dS fl E - {bj = 0.
Since the notation
refers to the same type of point in either case no confusion should
arise from its use.
'
If Oi is a closed cut point of X and U is a collection of components
16
of X - (a), then A =
(Uti) U (a) is an A-set since each component
of X - A is a component of X - (o:) and thus is open with singleton
"boundary.
Unfortunately the same is not true in general if (% is an
open cut point.
However, if Oi is an open cut point we do have the
following result.
3.1) Theorem:
If a is an open cut point of X and tiis a collection
of components of X - (a), then A = {a} U (U U ) is an A-set in X if
and only in U U ± s closed.
Proof:
If U ti is closed the only thing to check is that for
each component S of X - (cd which is not in U , dS Tl ( U u )
= (6.
But
this is obvious since U ti and S are disjoint closed sets.
Assume that A = ((%)
of X - A.
U (UU) is an A-set and let S be a component
Then S is a component of X - (ct) and dS fl
dS 0 A - I a J = 0.
Since this is true for each component of X - A
and a is an open point,
3.2) Theorem:
(U U ) =
U u = U U.
Let A be an A-set in X and let x be in X - A.
Then
b = b(s(x,A)) cuts between x and each point of A - (b).
Proof:
Follows immediately from the fact that S(x,A).= S(x,b)
and S(x,A) H A = 0.
3.3) Theorem:
Let N be an N-set in X and let A be an A-set in X.
^
If N Ii A is nondegenerate, then N c A.
'
'
If N is not contained in A
and N fl A ^ 0, then N Tl A = (ct), where a is a cut point of X.
'
I
1!
.1
Proof: Let x e M - A.
of A -
(b).
Then b = h(S(x,A)) cuts x from each point
Since N - (b) c s(x,b) = S(x,A) it follows that
N H (A - (b)) =
and therefore W fl A c (b) . '
A subset C of X is called a true cyclic element of X if it is
both an N-set and an A-set. A point x e X is called an end point of X
if x is a non-cut point which is conjugate to no other point of X.
3.4) Theorem:
Every end point a of X has a basis of neighborhoods
consisting of all sets of the form S(a,a), where a is a cut point of X.
Proof:
(oO =
to a .
If (X is a cut point,
then a e S(a,o:) ° since either
ds(a,ex) or each point of dS(a,o:) is in l<x) and thus is conjugate
Therefore each set of the form S(a,o:) is a neighborhood of a.
By 2.1 we know that (a) is a closed set, thus any neighborhood of a
Let
V be a connected neighborhood of a and let
y ^ a be an element of
V. Since y is not conjugate to a there is
is nondegenerate.
a cut point CU in
V such that
Oi
cuts between a and y .
S(a,ct:) is not contained in V, and let z e S (a,a) - V.
is non-empty and has no first element.
limits of (E(a,z),<) and S =
Suppose that
Then E(a,z)
By I.14, dS is the set of
U{ S(z,CX) : a e Ea,z) ) is open.
Suppose there is a point x e dS such that x ^ a.
Then there is a
cut point a which cuts between x and a and x e S(x,Ct:) °.
Thus
S(x,(X)° n E(a,z) ^ 0, which implies that a cuts between a and
some point of E(a,z), hence Ci e E(a,z).
Let (3 e E(a,z) such that
y /
111
II
18
P < Ct, then S(a,P) = S(z,p).
sible since S is open.
subset of X - (a).
But this implies x e S, which is impos­
Therefore Bs = (a) and S is a clopen proper
Thus a is a cut point, contrary to the hypothesis.
Therefore S(a,Q!) c V. and we have the results.
Let A be an A-set in X and let B be a subset
denote the union of all the components S of
and let B* = B* U B.
of A.
Let B*
X - A such that b(S)e B,
If B is a subset of X and B is not contained
in A, then B 1 = (B H A) * and B* = (B H A)f aslong as there is no
possiblity of confusion as to which A-set A is being used.
3.5) Lemma: Let A be .an A-set in X.
Then if B e A is closed in A,
B e B* .
Proof: Since B = B H A, w e 'need only consider the points in
B - A.
Let x e B - A, then S(x,A) is closed since x e
b = b(S(x,A)).
H S.
Let
If b ^ B, then B c A - (b) and thus B c A - {,bj.
Bufcby the definition of an A-set, x / A - {bj thus x / B, contrary
to the hypothesis.
Therefore b
3.6)
be an A-setin X.
Lemma: Let A
and x e S(x,A) c 3*.
e B
Then if B c A is closed
in A,
B r c B*.
Proof: Let x e B' - B r and let V be a connected neighborhood
of x.
Since x e B 1,
in B' such that
V fl B' /
Thus there is a component S of
X - A
V H & / jZ5. Letb = b(S). Since V ^ V H S ^ 0 and
is connected, b e
V. HoweverS c B' implies that b(s) e B.
19
Therefore V A B ^
3.J)
Theorem:
and thus x e B e B*- "by 3 . k .
Let A be an A-set in X.
Then if B c A is closed in
A, B* is closed in X.
B* = B U B 7 c B* U B* = B*.
Proof:
3.8) Theorem:
Then
Let A be an A-set in X and let B be a subset of A.
X - B* = (X - B)*.
Proof:
Follows immediately from the definition of B*.
3«9) Corollary:
Let A be an A-set in X and let B and C be disjoint
subsets of A, then B* and C* are disjoint subsets of X.
3.10) Corollary:
Let A be an A-set in X and let B c A be open in A,
then B* is open in X.
Proof:
(X - B)* = ((X -B) A A)* is closed in X, thus B* =
X - (X - B)* is open in X.
3.11) Corollary:
Let A be an A-set in X, then the mapping f from X
onto A defined by f(x) = x, if x e A, and f(x) = b(S(x,A)), if x ^ A,
is a retraction of X onto A.
3.12) Lemma: Let A be an A-set in X and let B be a subset of A.
Then B* c B*.
Proof:
Since B c B, it follows that B* c 3 .
—
in X we have B* c B .
Since B
is closed
20
3*13) Theorem:
Let A "be an A-set in X'and let B and C be subsets
of A which are separated in A.
Proof:
B
H C* =
B*
n C*
3.1^)
Then B* and C* are separated in X.
B and C separated in' A implies that B D C = j#, thus
Since
c B , we have that B* D C* = $•.
Similarly
=
Theorem:
subset of X.
Let A be an A-set in X and let H be a connected
Then H H A is connected.
Proof: I f H c z A o r H r i A = ^ the result is immediate.
assume that 0 ^ H
fl A ^ H.Let x
H is connected and
^ H H S ^ H, b = b(S)
(H n A)*.
Therefore H c: (H Pl A)*.
So
e H - A and let S = S(%,A).
Since
e H and thus x e S(x,A) c
If H Pl A = B U C, where B and C
are non-empty and
separated in A,then H c B* U C*, where B* and C*
are non-empty and
separated in X. Thus H = (B* Pl H) U "(c * Pl H)
separated and B c B* Pl H and C c: c* Pl H.
C* Pl H is empty.
Thus neither B* Pl H nor
Thus H is not connected, contrary to the hypothesis.
Therefore H Pl A is connected.
3.15) Corollary:, If A is an A-set in X, then A is connected.
3 .16) Theorem:
A =3 Pl U ^ 0.
Let U be a collection of A-sets- in X such that
If S is a component of X - A, then S is either open
or closed and for each closed component S of X - A there is a unique
open cut point P in A such that S is a component of X - {(3) and
S Pl A - IpJ / 0.
I I Il
W
21
Proof: Let S Le a component of X - A such that there is a point
x in Ss H S
j
(i.e. assume that S is not open) thus x e A - A.
x d A , there is auA-set A^ in U such that x / A^.
Since
Let S^ = S(x,A^)
and let b = b(S^), since (b) is open, (b3* is an open neighborhood
of x, thus {b)* H A ^
thus (b)* A A =
(b).
Since A c A , (b)* n A c {b}* A A =
(b)
Since A^ is an A-set in X, S^ = S(x,A^) = S(x,b)
and since b e A it follows that S c s(x,b).
Also since A c A^ .it
follows that S = S(x,A) 3 S(x,A^) = S(x,b).
Therefore S = S(x,b),
which implies that S is closed.
3*17) Corollary:
A = Al/ /
an A^ in U such that
A = AU
- (aJ = 0.
Let U be a collection of A-sets in X such
and let
3-18) Theorem:
Also S A A - {o:J c S A
that
S be a closed component of X - A. Then there is - •
S is also a component of X - A ^ •
Let U be a collection of A-sets in X such that
^ 0 and let x e X - A. Then S(x,A) = lLS(x,A^), where
e U
such that x / A^.
Proof: Let S =
S(x,A_) c S(x,A) for
S(x,A).
Since A c A^ for all A^ eU we have
all A^ e U •
Thus ILS(x,A^) C S . If S is closed
then S = S(x,A^) for some A^ e I/by 3-17 and thus S
So in the following We shall assume that S is open.
c U S(x,A^') •
Let y e dS and
consider the case where there is a point z e S such that y 0 z.
Since
z e S, there is an A^ e U such that z / A^ . Then, by 3-2, b = b(S’(z,A^)) •
cuts between z and each point of A -- (b).
I
Thus b = y, since y e A ■
I
22
and y O z.
But y e A , so b e A.
S(z,b) = S(z,A^).
Since b e A we have that S(z,A). <=
But z e S implies that S(z,A) = S(x,A^).
S(x,A) c: s(z,A^), which implies that S(z,A^) = S(x,A^).
S(x,A) c 8(x,A1) and S c Uj3(x,A^).
Hence
Therefore
Next, still letting y e ds,
consider the case
where y is not conjugate to any point in S.
A^ be any element
of U such that x / A^.
Let
If b^ = b(S(x,A^)) = y,
then as in the case when S is closed S(x,A^) = S and S cr U^S(x,A^).
So assume that for all A^
e U , such that x / A^, b^ / y, where
b^ is as above.
cuts between x and each point of A^ - (k )
Since b^
and since S U (y)
is a connected set containing both y and x it
follows that bu e
S.
such that K
^ A^.
But b^ e S implies that there is an A. e U
Let b = b(s(b^,Aj)).
so by I .13, b cuts between y and x.
Then b cuts between y and b^,
Thus b < bu in E(y,x) and S(b^,b) =
S(x,b) = S(x,A.) so b = b . = b(S(x,A.)).
0
J
J
Therefore B = { b. : b, =
i
I.
b(S(x,Aj,)), A^ e U , and x, ^ A^ ) is a subset of E(y,x) with no first
element.
By 1.14, tU3(x,A^) = U^S(x,b^), b^ e B, is an open subset
of X and 3(lLS(x,A^)) is equal to the set of limits of (B,<). Let
z e B(U^S(x ,A^)).
We shall show that z ^ S.
Then we know that E(y,z) ^
Let Ct e E(y,z).
connected and contains both y and z. Ct e S.
conjugate to y.
P < Ct.
V(
But if f3 cuts between y
Since S U {y) is
Therefore Ct is not
and Ct, then P e E(y,z) with
Hence E(y,z) has no first element.
S(z,a)
Hence suppose z e S.
By 1.14, we know that
: Ct e E(y,z) } is an open neighborhood of z.
Therefore
(U{ s ( z f a )
a
: OC e E(y,z)
and an A
e S(y,z)
e
)) H (U^S(x,A^)) ^
U such that S(z,a) fl S(x,A ) ^ j6.
then b = b(s((X,A^)) cuts between y and O'.
S ( ^ A 1)'.
But s(z,a)
S(ZjA 1) = S(XjA 1).
n
I f OC ^ A ^
Thus S(z,a) c g(a,b) =
S(XjA 1) yS J& and S(z,a) c S(G)IjA ) implies that
Therefore z e S(XjA 1) which is impossible.
Oi e A 1, then SfxjA 1) c S(ZjOl).
that Ql ^ Ag.
Thus there is an
If
Since Cd e S there is an A^ in U such
Then SfzjCd) c SfzjAg) since b = B(SfzjAg)) cuts between
y and Cd, and S(ZjAg) = SfzjB).
SfzjAg), and z e U^S(xJA^).
Thus SfxjA 1) c SfzjAg).
But this is impossible.
So SfxjAg) =
Therefore z / S.
3.19) Corollary: If U is a finite family of A-sets in X j and if
A =
fl U
is non-empty, then A is an A-set in X.
Proof: Let A and B be A-sets in X such that C = A D B in non­
empty.
if S
By the definition of an A-set in X we need only prove that
is a component ofX - A and T is a component of X - B such that
S A T ^ 0, then either
S c T or T c 8.
For then each component of the
complement of X - (A H B) will have the necessary properties to make
A H B and A-set.
Let x e S H T and assume that S is not contained in T.
b(T)
€ S.
If b(S) e A flB j then
bfs)
^ T so T is a connected subset
of X - (b(S)) containing x and thus T c SfxjB(S)) = S.
bfs)
A H B and let y e A fl B.
b(S).
Assume that
Then b(S) cuts between b(T) and y
and b(T) cuts between x and y, thus
B(T) <
Then
bfs)
and b(T) are in E(x,y) with
By 1.11, SfxjB(T)) c Sfb(T)jB(S)).
By 1.12i,
S(b(T),b(S)) = S(x,"b(s));
Therefore T = S(x,"b(T)) c S(x,"b(S)) = S.
In 3-22 and 3-23 we shall find a necessary and sufficient
condition to extend the above result to the infinte case.
3.20) Theorem:
Let X be compact.
If U is a family of A-sets in X
with the property that each finite subfamily of sets has non-empty
intersection then n{
: A_^ e U
} ^
Proof: Let U = ( A^ : A^ € U
}.
Then Tf also has the property
that each finite subcollection has non-empty intersection.
X is compact, DU
/ j6.
Since
Let b e DU , then b e A^ for all A^ e If.
If
b e A^ for all A^, e U we are through, so assume that b ^ A^ for some
A^ e Lf.
Let L 1 = b(S(b,A1)).
If L 1 is a closed cut point then
S(b,A1) is a neighborhood of b and S(b,A1) A A 1 = 0 which is impossible
Thus b 1 is an open cut point, and hence
b.
Therefore Cb1 }* H
)* is a neighborhood of
/ 0 for all A^ e U.
is an Ag e Cl such that b 1 /I Ag.
But since
If b 1 / AU , then there
)* A Ag ^ 0 and b1 / Ag
we have that (bj)' A Ag ^ 0.. Thus by the definition of (b1 } ’ there
is a closed component S of X - A 1 such that Ag A S ^ 0.
S is also a component of X - (b1).
B 1 / A ^ , we have that Ag cr S.
hypothesis.
Recall that
But since Ag is connected and
Therefore A 1 A Ag ^ 0, contrary to the
Thus b e A^ for all A_ e U and AU / 0.
3.21) Theorem:
Every E-set in X is an intersection of A-sets in X.
Proof: Let S be a component of X - N.
We shall show that there
Ill [I W
25
is a collection of A-sets, A(S), such that S is a component of X - A(S)
and N is contained in each member of A(S).
If S
A(S)
is closed, then b = b(S) is an open cut point
= S(N - (b),b)
and by 3*1 >
U (b) is an A-set containing N and S is a closed
component of X - A(S).
If S
that
S
is open we consider the following two cases.
is an open component of X - N such that there isa -point s in S
which is conjugate to some point x
e dS.
If y e E
not conjugate to y . . So there is acut point
and y.
First assume
But since s 0 x 0 y we have x = a.
- (x), then s is
OC which cuts between s
Since x e
BS,
(x} is not
open so x is a closed cut point and hence S(E - (x),x) is an open
set containing E in
Bs.
But
BS c
tx) not meeting S, so that no point of E - (x) is
E, so that (x) =
BS.
Consequently A(S) = S(E - (x),x)
(x) is an A-set in X containing E. and S is an open component of
X - A(S).
Second let S be an open component of X - E such that no point of
S is conjugate to any point x e
E(x,s) ^ 0.
BS.
If a e E(x,s), then a
contains both x and oc.
e S since (a)
BS:
.
Then
U S is connected and
But Ct e S implies that there is a cut point
P which cuts between x and ct.
E(x,s) has no first element.
an A-set containing N.
Let s e S and x e
Then P e E(x,s) and p < ct.
For each Ct e E(x,s), S(x,ct)
By 1.14, T =
Therefore
U (a) is
U{ s(s,Ct) : Ct e E(x,s) ) is an
open set, and since S(s,Ct) c S for each Ct e E(x,s), T C S.
Let z € BT
U
IU L
26
and assume that % e S.
thus R =
Then as above E(x,z) has no first element and
U{ S(z,a) : Ct e.E(x>z) ) is an open neighborhood of z.
R D T /
Thus
So there is an a e E(x,s) and a P e E(x,z) such that
S(s,a) n S(z,p) £ (.
If p ^ S(s,a), then S(s,a)c
cuts between x and s so P e E(x,s) with P<. Ct.
S(z,p).
But then S(z,p) =
S(s,p) and s e T which is impossible since T is open.
then CL cuts between x and p.
Then p
If P e S(s,a) ,
Thus CX e E(x,z) with a < p.
Therefore
S(s,a) = S(p,a) = S(z,a) and again we have that z e T which is impos­
sible.
Thus if z e ST^ then z ^ S.
c!open subset of S.
Therefore T is a non-empty
But since S is connected T = S .
So A(S) =•
{ s(x,a) U (a) : a e E(x,s) 3 is a collection of A-sets, each of
which contains N and, by $. 18, S is a component of X - (HA(S)).
Then E c
H( A : A e A(S), S a component of X - N 3 and since
each component of X - N is contained in X X - E c X N =
(HA(S)), we have that
H( A : A e A(s), S a component of X - N 3.
Therefore
H{ A : A e A(S), S a component of X - N 3.
3*22) Theorem:
A necessary and sufficient condition that every E-set
in X is a true cyclic element in X is that every non-empty intersection
t
of A-sets in X is an A-set in X.
Proof: First assume that every N-set in X is a true cyclic
element, and let U be a family of A-sets such that A = HU
empty.
is non­
In view of theorem 3•16 , it is sufficient to show that each
open component of X - A has singleton boundary.
Let S be an open
_ LI IL
27
component of X - A and let x and y be distinct elements of dS.
If
x 0 y, then there is a unique cyclic element C containing both x and y
which, by
is contained in A.
■is not closed.
Hence S(S,ti) is an open component of X - C with at
least two points in its boundary.
an A-set.
But it is impossible since C is
If x is not conjugate to y, let a be a cut point which cuts
between x and y.
CK e S.
Since x and y are in dS, S(S,C)
Since S U (x,y) is connected and contains x and y,
Similarly since x and y are in
A i is connected we have CK e A i for all
for all
e U and each
e U and thus CK e A.
But
CK e A and CK e S implies that CK e A 0 S which, since S c X - A, is
impossible.
Therefore dS is a singleton (note that BS ^ ^ is
guaranteed since X is connected.)
If every non-empty intersection of A-se.ts is an A-set, then
by 3-21, we have that every N7Set in X is an A-set in X and hence
every N-set is a true cyclic element.
3.23) Theorem:
A necessary and sufficient condition that every non­
empty intersection of A-sets in X is an A-set in X is that for each
two points a and b of X, if B is any subset of E(a,b), such that B
has no first element, then (B,<) has a unique limit, which is either
a closed cut point or an end point.
Proof:
First assume that every intersection of A-sets in X
is an A-set in X.
Then for each CK e B we know that S(a,CK) U (a) is .
an A-set in X. containing a.
Thus A = D( S(a,CK) U (a) : CK e B } is
28
an A-set in X..
By 3 .18 , S = S(b,A) = U{ S(b,a) : a € B }.
S is open and dS is equal to the set of limits of (B,<).
By I .lU,
Since S
is an open component of X - A t -then dS = (b(S)) and thus (B,<) has
a unique limit which is an end point if it is equal to A and it is a
closed cut point if it
is not all of A.
Let U be a family
of A-sets and assumethat A = DU
is non-empty.
As in the proof of 3*22# in order to show that A is an A-set in X
we need only show that if S is an open component of X - A such that
no point of S is conjugate to any point of dS, then BS is a singleton.
Let s e S, and let x e dS.
S =
U{ S(s,A_) i
where K
e Li
= b(S(s,A_)).
}.
Then x is not conjugate to s and by 3*18,
But for each A^ e Li , S(s,A_)
If there is an
result, so assume that
b^ ^ x for each A^
= S(s,b^)
such that b^ = x we have the
e
Li.
Thus
b^e E(x,s) and
b. e S for each i. Since b. e S there is
an
A-set A .
e Li such that
1
1
J
b. 4 A.. Then b(s(b.>A.)) cuts between x and b., thus b(s(b.,A.)) =
1
j *
1
J
x
1
J
b^ with b^ < b^ in E(x,s). Therefore B = ( b^ : b^ = b(s(s,A^)) ) is
a subset of E(x,s) with no first element.
But d(Uj3(s,b^)) is the set
of limits of (B,<) and therefore dS is the singleton set consisting
of the unique limit of (B,<) which is necessarily a closed cut point
or an end point.
Chapter 4
A property is said to he eyelieIy extensible if whenever each
cyclic element has the property then the entire space has the property.
A property is said to be cyclicly reducible if whenever the space has
the property then each cyclic element has the property.
A space X is said to be unicoherent if whenever X is expressed
as a union of two closed connected sets A and B, then A H B is
connected.
The cyclic extensiblity and reducablity of unicoherence
for Peano spaces was first proved by Kuratowski.
We shall prove that
unicoherence is a cyclicly reducible property in a general connected,:
locally connected space.
Then we shall restrict the space in order
to utilize some of the results of chapter three and obtain the result,
in 4.j), that in these restricted cases unicoherence is cyclicly exten­
sible.
4.1) Theorem:
Unicoherence is a cyclicly reducible property.
Proof: Let C be a cyclic element in X and let C = A H B ,
A and B are connected and closed in C.
where
Then A* and B* are closed
connected sets in X and since X = A* U B* we have that A* H B* is
connected.
Then, by $.14, C H (A* H B*) is connected.
c n a* n
=
C H
a
NOTE:
b*
cn
H B = A H B .
(a
u a ') n
(b
u b
') = (c
na n
b)
However
u (c n a 1 n
b
1) =
Therefore C is unicoherent.
For the remaining part of chapter 4 we will assume that X has the
property that each N-set in X is a cyclic element.
30
\
Note that when a space has the property that every N-set is a
cyclic element then any point of the space is either a cut point,
an end point, or a member of a true cyclic element.
If a and b are two distinct points of X and C is a cyclic element
then C 0 ((a,b) U E(a,b)) = D has at most two points.
b
C and S(a,C) = S(b,C), then D = 0.
If a ^ C and
If a ^ C and b ^ C and S(a,C) ^
S(b,C), then D = ( b(S(a,C)),b(s(b,C)) }, which could be a singleton.
If a e C and b ^ C, then D = (
\
in C, then D = (a,b).
a,b(s(b,C))
3 and if both a and b are
■
If a and b are distinct elements of X let G(a,b) = (a,b3 U
E(a,b) U { C : C is a cyclic element in X such that C fl ((a,b) U E(a,b))
is exactly two points].
4.2) Theorem:
If a and b are distinct elements of X then G = G(a>b)
is an A-set in X.
Proof:
If a 0 b, then a and b are in a true cyclic element and
the result is immediate.
a and CC are in
If E(a,b) = (ct), then G =
U Cg, where
and Ct and b are in Cg and C^ D Cg = (a).
Then
any component of X - G is a component of X - C^ distinct from S(b,C^)
or a component of X - Cg distinct from S(a,Cg); thus G is an A-set
in X.
For the remaining part of the proof we shall assume that E(a,b)
is nondegenerate.
If CC and P are distinct elements of E(a,b), then
P e S(a,cc) or P e S(b,(%); thus no component of X - (a) distinct from
H
31
both S(a,a) and S(b,o:) meets E(a,b) U (a,b).
If C is any true cyclic
element of X in G and a e E(a,b), then (C - (a)) D (E(a,b) U (a,b)) ^ 0
and thus C - (a) is contained in S(a,(%) or in S(b>C().
S(a,cc) U s(b,a).
Hence G - (a) c
Therefore if S is a component of X - {a} distinct
from S(a,cc) and from s(b/(X), then S fl G = 0 and S is a component of
X - G.
Let C be a true cyclic element contained in G.
If a e C let
{aj = C H E(a,b); then E(a,b) - {a} is contained in S(b,C) = S(b,C%)
and thus G - C c
If C
n E(a,b)
S(b,C).
= (a,p) with a < p, then E(a,b) -'{a,p} c
and as above G - C c
X - C
S(a,C) U S(b,C).
S(a,C).
s(a,a)
U
s(b,P)
Thus if S is a component of
distinct from S(a,C) and S(b,C), S 0 G = 0 and S is a component
of X - G.
a
Similarly if b e C, then G - C e
If a is
component of X -
X - G.
a cut point, then G e S(b,a) U {a}.
(a)distinct
from S(b,a),. then S is
Thus if S is
a
component of
A similar result occurs if b is a cut point.
At this point we have that for each cut point Ct in E(a,b) U (a,b)
every component of X -(ct), distinct
a component of X - G. Also for
from both S(a,ct) and S(b,ct), is
each cyclic element C of X in G every
component S of X - C,distinct from both S(a,C) and S(b,C), is a component
of X - G.
In the remaining part of the proof we shall show that every
point of X - G is in one of the above types of components.
We will then
have that G is an A-set of X since each component of X--- G will have
the necessary properties.
Let x e X - G.
0
If there is a cut point OC in E(a,b) U {a,b} such
32
that S(x,cc) is distinct from "both S(a,(%) and S(b,cx), then S(x,G) is as
required.
If there is a true cyclic element C of X in G such that
S(x,C) is distinct from S(a,C) and S("b,C), we have the same result.
So assume that for each CC e E(a,t>) we have x e S(afcc) or x e S(h,a).
Let H = { a e E(a,b)
: x e S(b,a) } and K = { a e E(a,b)
Then E(a,b) = H U K ,
and if a e H and P e K ,
: x e S(a,cc) ).
then a < P; for if p < a,
then S(a,p) H S(b,0!) = j6 but x e S(a,p) fl S(b,a).
element CC and K has a first element P, then a 0 p.
If H has a largest
So in this case let
C be the cyclic element containing Ce and P which is contained in G.
Since x ^ S(a,a) = S(a,C) and x 4 S(b,P) = S(b,C), S(x,C) is distinct
from both S(a,C) and S(b,C) and we have the result.
If K has no first
element let y be the unique limit of (K,<). ' If Ct e K, then { P e K
P < Ct } c S(a,Ct).
If a ^ S(y,Ct), then either S(y,Ct) or X - S(a,Ct)
is a neighborhood of y which cannot contain any P e K
such that P < a.
But this contradicts the fact that y is a limit of (K,<).
a e S(y,ct) for all Ct e K.
For each Cte
of X which contains both a and y.
By 3*18, S(b,A) =
Ct e K }, which by I .14, is open and y =
is a cut point. But x e S(b,a) =
b(S(b,A)).
U {a} : Ct e K )
U[ S(b,Ct) :
If y = a, then a
U( s(b,ct) : Ct e'/K ), thus x e S(b,ct)
for some ct 6 K, which is impossible.
y ^ K since K has no first element.
that x e S(b,y) =
K, S(y,ct) U {ct} is an A-set
Thus A = fl( S(y,Ct)
is an A-set of X.containing a and y.
Therefore
If a ^ y, then y e E(a,b) and
But then y e H which implies
U( s(b,Ct) : a e K }, which leads to the same
L
I ?
!
33
contradiction.
A similar argument will lead to the same contradiction
if H has no largest element.
Thus in each case, s(x,G) must he a
component of the complement of a cut point in E(a,b) U (a,b) or of
a true cyclic element of X in G, and we have that G is an A-set of X.
b .3) Theorem: Unicoherence is a eyelieIy extensible property.
Proof: Assume that each cyclic element in X is unicoherent
and let X = A U B ,
where A and B are closed, connected subsets of X.
Then for each true cyclic element C of X, C =
(C fl A) U (C fl B) where
C H A and C fl B are both closed, connected subsets of C. Thus
(c n A))n (C fl B) = C n A n B is connected.
elements in A A B.
Let a and b be distinct
If a 0 b, then a and b are elements of a true
cyclic element C of X and C A A A B is a connected subset of A A B
containing a and b.
If Qi is a cut point of X which cuts between a
and b, then CC is necessarily in A and in B, so that E(a,b) c A A B.
Assume that E(a,b) ^ 0 and let G = G(a,b) and let G^ = G A A A B =
{a,b} U E(a,b) U (U{ C A A A B = C
Assume that G^ =
U
is a true cyclic element in G )).
separated.
If C is a true cyclic
element in G, then C^ = C A A A B is connected and hence C^ c
or C1 c K 1.
H = R
Let R = ( (a,b)
U E(a,b)) A E^, S = ((a,b) U E(a,b)) A K1,
U (U{ C : C1 c H1 )) and K = S U (U{ C : C1 c K1 }).
definition of H and K we have immediately that G = H
1V .■
‘
By the
U K, where both
■
H and K are non-empty since B1 c H and K 1 c K.
If x e S, then there is a connected neighborhood W of X that does
not meet
.
Thus W fl R ^ 0.
with C1 c H1; then x ^ C.
Let C he a true cyclic element of X
IfW A C / ^
then W A C A (E(a,h)) ^ $
which implies that W A R ^ 0, which is impossible.
If
x € C f where C is a
C A ((a,bj LI E(a,b)).
x ^ H.
Thus x / H.
true cyclic element of X in K, let {a,p} =
If x = a or x = P, then x e S, so by the above,
Hence we may suppose that a ^ x / p.
If a = a, then either
S(a,3) or X - S(b,p) is a neighborhood of x which does not meet
G - C and thus does not meet H.
Similarly if (3 = b, there is a
neighborhood of x which does not meet H.
If CX, (3 e E(a,b) with a < S,
then let U = S(b,a) if a is a closed cut point of X or U = X - S(a,o;)
if CX is an open cut point.
Similarly define V to be S(a,f3) or
X - S(b,P) depending on whether P is a closed or open cut point
respectively.
Then U A V is a neighborhood of x and U A V A G c
Therefore again x / H.
Thus, in any case, H A K = 0.
H A K = 0, arid thus G = H U K
c.
Similarly
separated, which is impossible.
Therefore G1 is connected.
Hence given a e A A B, we have that A A B = U{ G^(a,b) : b e A A B J
and thus A A B is connected.
Another of the classical cyclicly extensible and reducible
properties is the fixed point property as was first proved by Borsuk.
The question of redueibiIity is easily answered in view of the fact
that the fixed point property is invariant under retraction in general
spaces and that each cyclic element in X is a retract of X as proved
LI Jl L
35
in 5.11 .
Let f be a continuous map from X into X and let A be an A-set of
X.
A is said to have property-A relative to f if for each X in A
such that x = b(S) for some component S of X - A,:.f(x) £ S.
Generally
we shall just use the term property-A unless there are several
mappings in question.
The idea for the definition of property-A
is supplied by a similar unnamed property used by Whyburn, see page
241 of [8] .
4.4) Theorem: If X is compact and if f is a continuous map from X
into X, then there is a cyclic element in X with property-A.
Proof: Let U be the family of all A-sets in X with property-A.
Since X e U we know that U
U , then, by 3.20, A = HF
A-set of X.
is not empty.
Let E be a nested family in
is non-empty and thus,
by3*23, A is an
If S is a closed component of X - A and a = b(S), then,
by 3 - 1 J ) S is a component of X -
A^ for some A^ e F and thus f(a) ^
If S is an open component of X -
A^ such that a 0 s for some s e S,
then as in 3.18 (first paragraph
of the proof), S is again a component
of X - A 1 for some A1 e F and again f(a) ^ S.
Now assume that S is
open and a is not conjugate to any point s e S.
S = lis(s,A^) where s is a
element of F.
If B = ( OL
Then by 3*18,
fixed element of S and each A^ is an
: OL = b(s(s,A_)), A ■ € F ), then (B,<)
is a subnet of (E(a,s),<) with no first element and a is the only
limit of (B,<).
If f(a) e S, then again as in the proof of 3*18,
S.
f(a) G S(s,A^)° for some
G F.
a such that f[w] cr s(s,A^)°.
an
€
B
fl
Thus there is a neighborhood W of
Since W is a neighborhood of a there is
W with Ct^ < CX^, thus f(o:^)
g
S(s,A^).
However Ct^ <
implies that S(s,A^) c s(s,A^) and thus f(o:^) e S(s,A^) which is impos­
sible , since A^ has property-A.
Therefore f (a) £ S and A e Li.
Thus
by Zorn's Lemma the family U o f A-sets of X has a minimal element C.
If C is a singleton, then C is a cut point or an end point, and
we have the result.
So let a and b be two distinct elements in C and
assume that there is a cut point a which cuts between a and b.
C is connected CC must be in C .
If f(cc) = cc, then {cc} is an A-set of X
which is properly contained in C and which has property-A.
is impossible since C is minimal.
Let D = C H (S U {a}).
Since
Hence we may let S =
But this
S(f(cc) ,a).
Then D is an A-set and any component S^ of
X - D is either a component of X - C or of X - (cc) distinct from S
and thus f(b(S^)) / S^.
Therefore D is an A-set with property-A
which is properly contained in C.
point
cc cuts between a and b.
But this is impossible.
Thus no
But then C is a 0-coherent A-set of X,
hence it is a cyclic element.
4.5) Corollary:
If the cyclic element C is a singleton, f has a fixed
point.
Proof: Follows immediately from the fact that if CC is a cut point
then
cc = b(S) for each component S of X - [cc).
I I'
IIlH
c.
\
37
Our terminology has "been so arranged that the proofs of the next
four theorems are exactly the same as in the classical theory and
thus need not be included here.
4.6) Theorem:
See [8] page 242.
If f is a continuous map from X into X with no fixed
point and A is an A-set of X with property-A relative to f, and g
is the retraction of X onto A then g o f is a continuous map from
A into A with no fixed points.
4.7) Theorem: Let X be compact, and such that every K-set o f X is
a cyclic element.
Then the fixed point property is cyclicly extensible.
4.8) Theorem: Let X be compact.
If f is a continuous map from X into.
X and C is a cyclic element of X such that
f[c] is
nondegenerate
and C has property-A relative to f and no end point or)cut point is
fixed under f, then C H
4.9) Theorem:
f[c] is
Let X be compact.
nondegenerate.
If h is a homeomorphism of X onto
X, then there is a cyclic element C in:X such that h[C] = C.
For the following we remove all conditions from the space X.
We do not even need to require that the space is connected or
locally connected.
4.10) Theorem:
is closed, x e
If x and y are distinct points of X such that (x)
TyT
and f(y) = x, where f is a continuous map from
z
r
Ill
I V
-I/
II
III
38
X into X , then f(x) = x.
Proof:
f \ x ) is a closed set containing y and thus x e T y J c
f~1(x).
4.11) Theorem:
If a and y are distinct points of X such that (x)
is open, y e TxT and f(y) = x, where f is a continuous map from
X into X, then f(x) = x.
Proof:
f
4.12) Theorem:
*I
—I
(x) is a neighborhood of y and thus x € f (x) •
Let x and y he distinct points of X such that (x)
is closed and let f he a continuous map from X into X.
If f(y) ^ T x T ,
then f(x) ^ T x T .
Proof:
f"1[X - (x)].
f ^[X - TxJ] is a neighborhood of y and thus x e
LI
39
Bibliography
[I ]
G. E. Albert and J. W. T. Youngs, The Structure of Locally
Connected Topological Spaces, Trans.. Amer. Maty. Soc., $1 (1942),
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[2]
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654.
C. Kuratowski, Quelques applications d'elements cycliques de M.
Whyburn, Fund. Math., 14 (1929), 138 - 144.
[3 ] _____ and G.
T. Whyburn, Sur Ies elements cycliques et leurs
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[4]
B. L. McAllister, Cyclic Elements in Topology, A History, Amer.
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[5 ] ______> A Hote On Irreducible Separation, Fund. Math., 63 (1968),
143 - 144.
[6]
R. L. Moore, Concerning Upper Semi-continuous Collections,
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[7 ]
T. Rado and P. ReicheIder, Cyclic Transitivity, Duke Math. J.,
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G. T. Whyburn, Analytic Topology, Amer. Math. Soc. Colloq. Pub.,
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MONTANA
-
, .«(mimiMtmiH mimini
3 1762 10011003 8
#
cop. 2
Minear, Spencer E.
On the structure of
locally connected
topological spaces
NAM*
AbowiEgy
WTERUBRAmr