Non-metric Continua that support Whitney maps
Except where reference is made to the work of others, the work described in this
dissertation is my own or was done in collaboration with my advisory committee.
This dissertation does not include proprietary or classifled information.
Jennifer Stone
Certiflcate of Approval:
Jo Heath, Co-Chair
Professor
Mathematics and Statistics
Michel Smith, Co-Chair
Professor and Chairman
Mathematics and Statistics
Gary Gruenhage
Professor
Mathematics and Statistics
Piotr Minc
Professor
Mathematics and Statistics
George T. Flowers
Interim Dean
Graduate School
Non-metric Continua that support Whitney maps
Jennifer Stone
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulflllment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
May 10, 2007
Non-metric Continua that support Whitney maps
Jennifer Stone
Permission is granted to Auburn University to make copies of this dissertation at its
discretion, upon the request of individuals or institutions and at
their expense. The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
Jennifer Williamson Stone, daughter of Cecil and Peggy(Duke) Williamson, was
born on October 22, 1973 in Selma, Alabama. She graduated from Morgan Academy
in 1992. In March of 1997, she graduated (with distinction) from Auburn University
with a B.S. in mathematics. In June of 1998, she graduated (Sum Cum Laude) from
Auburn University with a M.S. in Mathematics. She and her husband, Russ Stone,
have a son Nicholas(13) and a daughter Emily(7).
iv
Dissertation Abstract
Non-metric Continua that support Whitney maps
Jennifer Stone
Doctor of Philosophy, May 10, 2007
(M.A., Auburn University, 1998)
(B.S., Auburn University, 1997)
96 Typed Pages
Directed by Michel Smith and Jo Heath
Anexampleofanon-metriccontinuumisconstructed, whereeverynon-degenerate
subcontinuum is non-metric, that supports a Whitney map. Additional non-metric
examples are given and examinations of conditions under which non-metric continua
support Whitney maps are made.
v
Acknowledgments
I would like to thank all the professors here at Auburn that have taught me not
only mathematics but also how to be a good teacher. I?d like to thank Dr. Smith
for allowing me to be his special project. You have been my teacher, my mentor,
and my friend. Without your patience, encouragement, and passion for the material
I would not be where I am today. I would also like to thank Dr. Heath. You were
my flrst topology teacher and showed me how much fun it could be. You encouraged
me to go to graduate school, you directed my thesis, you suggested Dr. Smith as my
advisor, and you tried very hard to improve my grammar skills (sorry for the run on
sentence).
Thank you to all my family for never saying out loud that you thought I was crazy
when at age 30 with two children I said I think I?ll get my PhD. Thank you for praying
for me and with me over these many years. Thank you to God for answering those
prayers and giving me wisdom. Thank you to my parents especially who instilled in
me a desire for education and the ability to believe I could do anything with God?s
help.
Thank you to my husband who has supported me and helped me. I could not
have done it without your help. And lastly thank you to my children. Thank you
for sharing me with my research. Thank you for asking how MY school day went
and when my tests were. Finally, thank you for learning so many new words like
dissertation, defense, outside reader, and of course topology.
vi
Style manual or journal used Journal of Approximation Theory (together with
the style known as \aums"). Bibliography follows van Leunen?s A Handbook for
Scholars.
Computer software used The document preparation package TEX (speciflcally
LATEX) together with the departmental style-flle aums.sty.
vii
Table of Contents
1 Introduction and Background 1
2 A Non-metric continuum that supports a Whitney map 12
3 A hereditarily non-metric continuum that supports a Whitney
map 28
4 a hereditarily indecomposable non-metric continuum that sup-
ports a Whitney map 57
5 Property + Whitney map = metric 63
6 Whitney levels 85
Bibliography 88
viii
Chapter 1
Introduction and Background
Thepurposeofthispaperistoexplorenon-metriccontinuathatsupportWhitney
maps. An indecomposable non-metric continuum, where each proper non-degenerate
subcontinuum is non-metric, that supports a Whitney map is constructed. Several
other examples of non-metric continua that support Whitney maps are given. In
addition, topological properties that willpreventnon-metric continuafrom supporting
a Whitney map are examined.
A non-metric analog of the Cantor set and the Solenoid are used in the con-
struction of an example of a non-metric continuum that supports a Whitney map.
This example will then be used to construct an indecomposable non-metric contin-
uum, where each proper non-degenerate subcontinuum is non-metric, that supports
a Whitney map. Thus a short discussion of the Cantor set and the Solenoid will be
useful.
A metric Cantor set is any uncountable topological space that is compact, each
point of the set is a limit point of the set and the only connected subsets are singleton
points. The most common example of a Cantor set is the \middle third" Cantor set.
This Cantor set is in [0;1] and has the following structure:
M1 = [0;1=3][[2=3;1]
M2 = [0;1=9][[2=9;1=3][[2=3;7=9][[8=9;1]
...etc.
1
At each stage, Mj is formed by removing the middle third open interval from
each segment of Mj?1. The middle third Cantor set is T1i=1 Mi. Every Cantor set
is homeomorphic to the middle third Cantor set. In Chapter 2 an example S is
constructed. S is an non-metric continuum (A continuum is a compact, connected
space) that supports a Whitney map. S will be formed using a subset Z of the
lexicographic arc Lx. The lexicographic arc, Lx, is a non-metric linearly ordered
connected compact Hausdorfi space. ( A space is Hausdorfi if given points x 6= y
there exist open disjoint sets U and V such that x 2 U and y 2 V.) The standard
lexicographic arc is the topological space Lx deflned as follows. Lx = [0;1]?[0;1].
If x1 = (p1;q1) and x2 = (p2;q2), then x1 f(b) so, f0(ak) = (f(a);1?
k) >Z (f(b);1?i) = f0(bi).
Case 2. a = b. Then k = 0 and i = 1. So we have ak = a0 and bi = a1.
Thus, f0(a0) = (f(a);1?0) > (f(a);0) = f0(a1).
(ii) f0 is continuous: Let U be a basic open set with f0(ti) 2 U and U =
[a1;b0]. Then f(a) ? f(t) ? f(b) implies that t 2 [f?1(b);f?1(a)]. Let V =
14
[(f?1(b))1;(f?1(a))0]. So V is open in Z, ti 2 V, and f0(V) ? U since f is order
reversing.
Part 2: Deflnition of f?s. Construction.
Deflne H11 = [01;(12)0] and H12 = [(12)1;10].
Deflne H21 = [01;(14)0]; H22 = [(14)1;(12)0]; H23 = [(12)1;(34)0]; and
H24 = [(34)1;10].
For each positive integer n and 1 ? i ? n deflne Hni = [(i?12n )1;( i2n)0]. Note that
Hni = Hn+12i?1 [Hn+12i .
Claim 2.2.1. For each integer n, fHni g2ni=1 is a partition of Z into disjoint clopen
sets and if U is an open set in Z then there exists integers n and i so that Hni ? U.
Construction: For each n we wish to flnd a homeomorphism fn from Z onto Z
by using order preserving maps to map elements of fHni g2ni=1 onto each other.
Deflne fni;j : Hni ! Hnj , for tk 2 Z with t 2 [0;1] and k 2 f0;1g, by fni;j(tk) =
(j?12n +t? i?12n )k. Note by Background Theorem 2.1.3 that fni;j is an order preserving
homeomorphism.
We deflne fn : Z ! Z inductively.
Let f1(t) = f11;2(t) for t 2 H11; let f1(t) = f12;1(t) for t 2 H12.
For n > 0 let fn+1(t) = fn(t) for t 2 Hni for i > 1 (i.e. t 2 Hn+1i for i > 2).
Let fn+1(t) = fn+11;2n+1(t) for t 2 Hn+11 ; let fn+1(t) = fn+12;2n+1?1(t) for t 2 Hn+12 . What
this does is interchange Hn+11 and Hn+12 before moving them to Hn+12n+1 and Hn+12n+1?1
but keeps the rest of Z in the same order and preserves the previous assignments
15
made at the nth level. We have fn+1(Hn+11 ) = Hn+12n+1, fn+1(Hn+12 ) = Hn+12n+1?1 and
fn+1(rk) = fn(rk) if rk 2 Z ?(Hn+11 [Hn+12 ).
Deflne F : Z ! Z by F(t) = limn!1fn(t).
Note:
1. For each n and each point x 2 Z, the orbit of x under fn intersects every
set in fHni g2ni=1.
2. For every point xk 2 Z except 01, xk 2 Hi2 for some i. This is true since
xk 2 [( 12n)1;( 22n)0] = Hn2 for some n whenever xk 6= 01. Thus Z = [1n=1Hn2 [f01g.
3. Given i is the least integer such that xk 2 Hi2 then F(xk) = fi(xk). This
is true since fi(xk) = fj(xk) for all j > i so F(xk) = limn!1fn(xk) = fi(xk).
4. fj(Hj1) ? fi(Hi1) for i < j.
5. fi(Hi2) ? fi?1(Hi?11 ).
Part 3: F is a homeomorphism.
Claim 1. F(01) = 10:
We have 01 2 Hn1 for every n.
fn(Hn1 ) = Hn2n, thus fn(01) = (2n?12n )1. By the topology on Z,
F(01) = limn!1(2n?12n )1 = 10. So F(01) = 10.
Claim 2. F is well-deflned:
Given a point xk 2 Z, xk 6= 01, we have xk 2 Hi2 for some i; so F(xk) = fi(xk)
which uniquely deflnes F(xk). From above we have 01 mapped only to 10.
Claim 3. F is one-to-one:
Case 1. Suppose ak and bh such that ak 6= bh and ak;bh 6= 01.
16
Case 1.1: a 6= b. Then there exist i;j;m with j and m not equal to 1 such that
ak 2 Hij and bk 2 Him with j 6= m. Then F(ak) = fi(ak) 6= fi(bk) = F(bk) because fi
is one-to-one.
Case 1.2: a = b. Then there exist i;j such that j 6= 1 so that ak;bh 2 Hij. Then
F(ak) = fi(ak) 6= fi(bh) = F(bh) because fi is one-to-one.
Case 2. Suppose ak and bh = 01 such that ak 6= 01.
Then there exists an i such that 01 2 Hi1 and ak 2 Hij with j 6= 1. Thus
fi(ak) =2 fi(Hi1); therefore F(ak) = fi(ak) 6= 10 = F(01).
Claim 4. F is onto:
We know that F(01) = 10; so we need only show onto for ak 6= 10. Since
T1
n=1 H
n
2n = f10g, then given ak 6= 10 there exists i such that ak 2 H
i
j and j 6= 2
i (i.e.
not the last partition element.) Since fi is onto there exists Him and bk 2 Him with
m 6= 1 such that fi(Him) = Hij such that ak = fi(bk) = F(bk) since m 6= 1.
Claim 5. F is continuous:
Suppose that F(xk) 6= 10 and U is an open set such that F(xk) 2 U. There
exists Hij such that F(xk) 2 Hij ? U and j 6= 2i. Let V = (fi)?1(Hij); this set is open
since fi is a homeomorphism and V 6= Hi1 since j 6= 2i. So,
F((fi)?1(Hij)) = fi((fi)?1(Hij)), since j 6= 2i,
fi((fi)?1(Hij)) = Hij ? U.
If F(xk) = 10 and U is the open set [a1;10] then there exists an i such that
Hi2i ? [a1;10]. Since Hi2i ? [a1;10] we have, fi(Hi1) ? [a1;10].
Claim: F(Hi1) ? [a1;10].
17
Suppose for every point wm 6= 01 such that wm 2 Hi1 there exists n such that
wm 2 Hi+n2 for some n. This implies by construction (see notes 3, 4, and 5) that
F(wm) = fi+n(wm).
So F(Hi+n2 ) = fi+n(Hi+n2 ) ? fi+n?1(Hi1) ? fi(Hi1) ? U.
Hence F(wm) 2 U and we know that F(01) 2 U.
Thus F(Hi1) ? U, and F is continuous.
2
Part 4: Deflnition of a non-metric continuum S such that S supports a Whitney
map.
Let X = Z ?[0;1].
Let G = ff(t;0);(F(t);1)gj t 2 Zg[ff(t;r)gj r 6= 0;1g.
Let S = X=G.
Observe that G is an upper semi-continuous collection fllling up X.
Lemma 2.4.1. Let fx0;x1;:::;xng be points in Z so that F(xi) = xi+1 for i =
0;1;::n ? 1; then there exists open sets V0;V1;:::;Vn in Z that are pairwise disjoint
and so that F(Vi) = Vi+1 and xi 2 Vi with i = 0;1;:::;n.
Proof. Step 1. There exists pairwise disjoint sets U0;U1;:::;Un, such that xi 2 Ui
and each Ui = [axi1 ;bxi0 ], with i = 0;1;:::;n.
Step 2. Let W1 = [F(ax01 );F(bx00 )]\[ax11 ;bx10 ]; note [F(ax01 );F(bx00 )] = F(U0).
So we deflne:
W1 = F(U0)\U1;
18
W2 = F(W1)\U2;
...
Wi = F(Wi?1)\Ui;
...
Wn = F(Wn?1)\Un.
Step 3. Deflne:
Vn = Wn;
Vn?1 = F?1(Vn) = F?1(Wn);
Vn?2 = F?1(Vn?1) = F?1(F?1(Wn));
...
V0 = F?1(V1) = F?1(F?1 ???| {z }(Wn)).
n-times.
We have:
?The Vi ?s are pairwise disjoint since the Ui ?s are pairwise disjoint, Wi ? Ui,
and Vi ? Wi.
? F(Vi) = Vi+1 by construction.
? xi 2 Vi.
Proof: x0 2 U0 so F(x0) 2 F(U0);
x1 2 U1 so x1 = F(x0) 2 U1 and so x1 2 W1 = F(U0)TU1.
Likewise x2 2 W2 and so on: xi 2 Wi for each i.
19
Then we have xn 2 Wn = Vn. So xn?1 = F?1(xn) 2 F?1(Vn) = Vn?1; xn?1 2
Vn?1; and likewise for all i we have xi 2 Vi. Note that Vi ? Ui; so the elements fVigni=1
are disjoint. This establishes the lemma.
2
Part 5: Indecomposablity of S.
Notation:
We have X = Z ?[0;1] and S = X=G. Note that the point (z;0) is identifled
with the point (F(z);1),and that the points z 2 Z are written in the form z = ti for
some t 2 [0;1] and i 2f0;1g.
Let z 2 Z; for each positive integer n we deflne an arc Azn. Let Az0 be the arc
fzg?[0;1] ? S. Az0 is an arc beginning at (z;1) and ending at (z;0). Let Az1 be the
arc fzg? [0;1] [fF(z)g? [0;1] ? S. Let Azn = fzg? [0;1] [[ni=1fFn(z)g? [0;1].
Thus for example Az2 is the arc in S beginning at (z;1) and ending at (F2(z);0).
Deflne Az = S1i=0 Azi.
Theorem 2.5.1. Az is dense and is the union of metric arcs for each z 2 Z.
Proof. First, since each Azi is just flnitely many metric arcs glued together, each
Azi is metric.
Assume that Ax is not dense for some x 2 Z. Let y 2 S be such that y =2 Ax
and y is not a limit point of Ax. Thus there exists a basic open set U in the form
U = [z1;z2]?(r;s) such that y 2 U and no point of Ax is in U. Look at the projection
?1(U) of U onto Z?f1g. Now since y is not a limit point of Ax then ?1(U)TAx = ;.
We also know that there exist i and j such that Hij ?f1g? U which implies that:
20
(Hij ?f1g)TAx = ;.
This implies that (F?1(Hij)?f1g)TAx = ;.
Thus (F?2(Hij))?f1gTAx = ;.
...
(F?2i(Hij))?f1g)TAx = ;.
But then AxT(S2ij=1 Hij ?1) = ; which is a contradiction since Ax is nonempty
and S2ij=1 Hij = Z. Therefore Ax is dense.
2
Similarly deflne Az?n = fzg?[0;1][[ni=1fF?n(z)g?[0;1] and A?z = S1i=0 Az?i.
Then by the same argument we have:
Theorem 2.5.2. A?z is dense and is the union of metric arcs for each z 2 Z.
Theorem 2.5.3. Every proper subcontinuum of S is a metric arc or a singleton
point.
Proof. Let M be a proper subcontinuum of S.
Let (z;t) 2 M. [To use our rough terminology: we wish to flnd points (a;r) and
(b;s) \above" and \below" this point that are not in M. This will show that there
is a metric arc, which will be denoted as L, that is contained in M and then we will
show that L = M.]
Consider the arc fzg ? [t;1]. If this arc is not a subset of M then there is a
number r > t so that (z;r) =2 M. Let (a;r) = (z;r). If fzg?[t;1] ? M then there
is a flrst integer n so that Az?n * M. Otherwise the dense subset A?z would be a
21
subset of M and this would contradict the fact that M is a proper subcontinuum
of S. Then there is an integer n and points (F?n(z);u) and (F?n(z);r) so that
(fzg?[t;1])[Az?(n?1) [(fF?n(z)g?[0;u]) ? M and (F?n(z);r) = (a;r) =2 M; and
furthermore since MT(fF?n(z)g?[0;1]) is closed u has the property that for every
w between u and r there is a w0 so that u < w0 < w and (a;w0) =2 M. We can think
of (a;u) as one endpoint of the arc L that is contained in M.
Similarly, consider the arcfzg?[0;t]. If this arc is not a subset of M then there is
a number s < t so that (z;s) =2 M. Let (b;s) = (t;s). If fzg?[0;t] ? M then there is
a flrst integer j so that Azj *M. Thus, as in the above argument there is a flrst integer
m and points (Fm(z);v) and (Fm(z);s) so thatfzg?[0;t][Az(m?1)[fFm(z)g?[v;1] ?
M and (Fm(z);s) = (b;s) =2 M; and furthermore for every w between s and v there
is a w0 so that w < w0 < v and (b;w0) =2 M. Thus we can think of (b;v) as being the
other endpoint of the metric arc L ? M.
Let L be the arc lying in Az [A?z with end points (a;u) and (b;v) as deflned
above. Note that (roughly) L = Az(m?1) [fFm(z)g?[v;1][fzg?[0;1][Az?(n?1) [
fF?n(z)g?[0;u]. (Roughly in the sense that a slight modiflcation is necessary in the
case that n = 0 or m = 0.)
We now will show that L = M. Assume not so that there exists (q;w) such that
(q;w) 2 M but (q;w) =2 L.
Note that the projection of L onto Z is the set fFi(z)gmi=?n.
Case 1. q =2fxj(x;y) 2 L for some y 2 [0;1]g, i.e. q =2fFi(z)gmi=?n.
Case 2. q 2fxj(x;y) 2 L for some y 2 [0;1]g, i.e q 2fFi(z)gmi=?n.
22
We will now make a tube-like open set in S such that it will contain L but not
(q;w) and whose boundary misses M (thus getting a contradiction). We say an open
set U is tube-like in S if the projection of U onto Z is a collection of disjoint open
sets fU1;U2;:::Ujg such that F(Ui) = Ui+1, for i 2 [1;j].
Case 1: q =2fxj(x;y) 2 L for some y 2 [0;1]g:
We know that (a;r) and (b,s) are not in M; thus there exists Va and Vb, open
sets in S containing (a;r) and (b;s) respectively, that do not intersect M.
By Lemma 2.4.1 (about the V?s), there exists a clopen set V ? Z containing z
so that VL = [fFi(V)gmi=?n does not contain q. Note that VL contains all the points
from fFi(z)gmi=?n. Furthermore, V can be chosen so that F?n(V) ?frg ? Va and
Fm(V)?fsg? Vb.
Thus F?n(V) ? [0;r) [ ([m?1i=?(n?1)(Fi(V) ? [0;1])) [ Fm(V) ? (s;1] is an open
set O in S that contains L. Moreover (q;w) is not in O and, since V is clopen,
Bd(O) ? Va [Vb and hence Bd(O) does not intersect M. But L ? M and L lies in
O and (q;w) is a point in M not in O which contradicts the connectedness of M.
Case 2: q 2fxj(x;y) 2 L for some y 2 [0;1]g.
There are two possibilities: q = Fm(z) or q = F?n(z). Suppose q = F?n(z);
thus (q;w) = (a;w). Since (a;u) is an endpoint of L, there exists a number r0 so that
u < r0 < w so that (a;r0) =2 M.
Then repeat the construction as above but with (a;r0) replacing (a;r) and the
open set Va containing (a;r0) and no point of M. Then the open set O constructed
as above contains L and does not contain (q;w). And again Bd(O) does not intersect
23
M which contradicts the connectedness of M. Therefore (a;w) =2 M for w > u. If
q = Fm(z) then the same argument works by selecting s0 so that w < s0 < v and
(b;s0) =2 M. Thus we have shown that every proper subcontinuum of S is a metric
arc or is a singleton point. Notice that since we showed M = L then the projection
of M onto Z is the set fFi(z)gmi=?n. All proper subcontinua will have this feature.
Theorem 2.5.4. S is an indecomposable continuum such that each composant is
the union of a countable collection of metric arcs.
Proof. Claim 1. S is a continuum:
We know that, since G is an upper semicontinous collection fllling up the compact
space X, S is compact.
For connectedness we know Ax is connected for any x 2 S; thus Ax = S is also
connected.
Claim 2. S is indecomposable:
If S were decomposable then it would be the union of two arcs ( since each proper
subcontinuum is an arc) but S is not the union of two arcs.
Claim 3. Each composant is the union of a countable collection of metric arcs:
Claim: Ax = S1i=?1Axi is the composant of (x;0), x 2 Z. Recall Azn = fzg?
[0;1][([ni=1fFn(z)g?[0;1]) and that Ax is a countable collection of metric arcs.
Assume Ax = S1i=?1Axi is not the composant of (x;0); so there exist a point
(r;w) and a proper subcontinuum B of S such that (x;0);(r;w) 2 B and BnAx 6= ;.
But any proper subcontinuum is an arc and thus we can say B starts at (a;b) and
ends at (Fp(a);d), for some p 2 (?1;1). If (x;0) is on this arc then x = Fn(a)
24
for some n which implies that a = F?n(x). Also Fp(a) = Fm(x) for some m. Thus
B ? Smi=?n Axi ? Ax. This is a contradiction. Thus given any point that is in the
composant of x, that point is in Ax, and we know that given any point p 2Ax there
exist a subcontinuum that contains p and (x;0), namely the arc with p and (x;0) as
the starting and ending points respectively. Therefore Ax is the composant of (x;0)
and actually is the composant of any point in Ax.
Part 6: Deflnition of len and continuity.
From Part 5 we have: If I ? S is a proper subcontinuum of S then there exists
a flnite number of points z1;z2;:::;zn so that:
I ?[ni=1fzig?[0;1]=G:
Let ? denote the usual length of intervals in [0;1] and let ?2 denote the projection
of S onto the second coordinate (loosely deflned by ignoring the decomposition part).
Deflne the \length" of I, len(I), as follows:
len(I) =
nX
i=1
?(?2(I \fzig?[0;1])):
Lemma 2.6.1. len is continuous.
Let M be a proper subcontinuum of S. Let U be an open set in R such that
len(M) 2 U. Now there exists an ? > 0 such that (len(M)??;len(M)+?) ? U.
M ?[ni=0Axi. [We assume that the sets Axi are the \vertical" intervals compris-
ing Z ?[0;1] so that for xi we have Axi = fxig?[0;1].]
25
Thus there exists x0;x1;:::;xn so that M has endpoints (x0;y0) and (xn;yn).
Using the Lemma 2.4.1, we can flnd V0;:::;Vn such that xi 2 Vi, and Vi and Vj
are pairwise disjoint for i 6= j.
Let:
W0 = V0 ?(y0 ? ?4;y0 + ?4), and
Wn = Vn ?(yn ? ?4;yn + ?4).
Let M0 be the arc that begins at (x0;yn ? ?4) and ends at (xn;yn + ?4). Cover
M0 with \balls" of radius ?4, where \ball" around the point (xi;s) would be the open
set Vi ?(s? ?4;s+ ?4). Since M0 is compact, there exists flnitely many of these open
sets, say G1;:::;Gm, that cover M0. (Note: each Gi = Vj ?(yj ? ?4;yj + ?4) for some
j and some yj.) Then G = fG1;G2;:::;Gn;W0;Wng will cover M. R(G) = fK 2
C(S)jK intersects each element of G and is a subset of [Gg is open in C(S).
Thus, by deflnition, if N 2 R(G) and the fact that we know that N is an arc
[previous result], N must start in W0 and end in Wn. Thus len(N) 2 (len(M) ?
?
2;len(M)+
?
2). (Note: this is true since it is at most
?
2 longer than M or no shorter
than ?2 of M.)
So len(N) 2 U and hence len(R(G)) ? U. So len is continuous.
Part 7: Deflnition of ?, where ? is a Whitney map.
Deflne ? : C(S) !R by
?(I) = arctan(len(I))?
2
for I 6= S and ?(S) = 1.
26
Claim 2.7.1. ? is a Whitney map.
Part 1. ? is continuous:
Since len and arctan are continuous it su?ces to show that ? is continuous at
S. Let ?(S) = 12 U where U is an open set in R. There exists an ? > 0 such that
(?(S)??;?(S)) ? U. Furthermore there is a number N such that if len(K) > N +?
then ?(K) > ?(S) ? ?, where K is a proper subcontinuum of S. Now let K be a
proper subcontinuum so that len(K) > N +?. By a previous argument we know that
K is a metric arc beginning at (x0;y0) and ending at (xn;yn).
Using the same method as in proof that len is continuous there exists an open
set R(G) in C(S) such that if J 2 R(G) then len(J) > len(K) ? ? > N; so then
len(J) > N +? and thus ?(J) > ?(S)??.
We will now make a new open set V in C(S). Let V = R(G [fSg). Note V is
open since it is made from a collection of flnitely many open sets from S, and S 2V.
Then, given any proper subcontinuum M 2V, M must intersect each open set from
G which means that len(M) > len(K)?? > N. Thus ?(M) > ?(S)?? which implies
that ?(M) 2 U. Since ?(S) 2 U then ?(V) ? U, and thus ? is continuous.
Part 2: Given A(B then ?(A) < ?(B).
If B 6= S and A(B then len(A) < len(B) which implies that ?(A) < ?(B).
If B = S and A(B then ?(A) < 1 = ?(B).
Corollary 2.1. The same construction can be done using any irreducible continuum
that supports a Whitney map.
27
Chapter 3
A hereditarily non-metric continuum that supports a Whitney map
Note that in this paper, by a hereditarily non-metric continuum we mean a
continuum such that every nondegenerate subcontinuum is non-metric.
Theorem 3.1. If for each positive integer i the space Xi supports a Whitney map ?i
and fi : Xi+1 ! Xi, then X = lim??(Xi;fi) supports a Whitney map.
Proof. We will assume that ?i(Xi) = 1 for all i. Let ?i be the projection map from
X onto Xi. Deflne a map ?i from the hyperspace of X onto the hyperspace of Xi by
?i(H) = ?i(H), where H is any subcontinuum of X. We flrst need to show that ?i
is continuous.
Let U be an open set in C(Xi) such that ?i(H) 2 U, where H is a subcontinuum
of X. Now U = R(fUjgnj=1) where Uj is an open set in Xi. Deflne an open set Vj in X
as Vj = ??Uj. Thus Vj = fx 2 Xj xi 2 Ujg. Deflne eV ? C(X) as eV = R(fVjgnj=1). Let
K be a point in C(X) such that K 2 eV; then KTVj 6= ; for each j = 1 to n. Thus
?i(K)TUj 6= ; for each j = 1 to n and by the deflnition of the V 0js, ?i(K) ?Snj=1 Uj.
Therefore ?i(K) 2 U in C(Xi); thus ?i(K) 2 U, which implies that ?i(eV) ? U. Thus
?i is continuous.
Deflne ? : X ! Xi by ?(H) = P1i=1 ?i(Hi)2i where H is a subcontinuum of X and
Hi = ?i(H). First it is clear that if K ( H then ?(K) < ?(H) since in order for
28
K (H there exist an i such that Ki (Hi; so ?i(Ki) < ?i(Hi) and ?j(Kj) ? ?j(Hj)
for i 6= j.
Second we need to show that ? is continuous. Let U be an open set in R such
that ?(H) 2 U. There exist an n and an ? such that
[Pni=1 ?i(Hi)2i ??] ? U and
P1
i=n
1
2i <
?
2. Note we deflne [A??] = [A??;A+?]
Since each ?i is continuous there exists an open set Vi ? C(Xi) such that ?i(Vi) ?
[?i(Hi)? ?2]. Also since ?i : C(X) ! C(Xi) is continuous there exists an open set V 0i ,
containing Hi, such that ?(V 0i ) ? Vi. Deflne an open set eV ? C(X) by eV = Tni=1 V 0i .
If K 2 eV then ?i(K) 2 Vi for each i = 1 to n. Thus ?i(K) 2 [?i(Hi)? ?2] for each
i = 1 to n. Therefore
Pn
i=1
?i(Ki)
2i < ?(K) =
P1
i=1
?i(Ki)
2i =
Pn
i=1
?i(Ki)
2i +
P1
i=n+1
?i(Ki)
2i <
Pn
i=1
?i(Ki)
2i +
?
2 2
Pn
i=1
?i(Hi)??2
2i +
?
2 ? [
Pn
i=1
?i(Hi)
2i ??] ? U.
Thus ?(eV) ? U so ? is continuous and thus a Whitney map.
? Deflnition of breaking and gluing a copy of S at a point on the arc [e;f].
Let I1 = [a;b] and I2 = [c;d] be two metric arcs. Let S be the space of our
example from Chapter 2, and let p;q be two points from S that are in difierent
composants. Deflne the decomposition space
D = (I1 [I2 [S) ffp;bg;fq;cgg:
29
Now we deflne what is meant by breaking an arc at a point and gluing in S.
(For shorthand we will refer to it as breaking and gluing S at a point). Let [e;f] be
an arc and t 2 [e;f], t 6= e;f. There exist natural homomorphisms h1 : [a;b) ! [e;t)
and h2 : (c;d] ! (t;f]. Deflne a new space
S1 = D[([e;f]nt) fffx;h1(x)gjx 2 [a;b)g
[
ffy;h2(y)gjy 2 (c;d]gg
where x 2 [a;b) and y 2 (c;d]. Note we use this notation loosely since this is not a
decomposition space.
Deflne the topology T by open sets on fx;h1(x)g. If x 2 D , x 2 I1 or I2, and
x =2 fp;qg (thus x is on the arc but not the endpoints where S was glued), then the
basic open sets are the open sets from the normal topology of an arc.
If x 2 D;x 2 S;x =2 fp;qg (thus x is inside the glued copy of S), then use the
relative topology from our example S.
If x = p, or x = q then an open set containing x would be the union of an open
set in S containing p(or q) and a half-open interval on either I1 (or I2). Observe an
important fact that since p and q are in difierent composants of S, D is irreducible
from e to f.
2
Now in the above deflnition we deflned what was meant by breaking and gluing in
a copy of S at a point on an arc. Since our space S is Z?[0;1] with identiflcations we
can think of S as having uncountably many disjoint arcs (except for the endpoints),
30
namely z ?[0;1] for each z 2 Z. Given a flxed z we can use the above procedure to
break and glue at a point in an arc from S. (Note: there are two difierent S0s. First,
we start with S and then we take another S to glue into the flrst one). Thus we can
glue a copy of S into S at a point on the arc z ?[0;1] for some flxed z 2 Z.
For example, let (1=20;1=3) be a point in S. (1=20;1=3) is on the arc 1=20?[0;1],
so we can use the lemma to break and glue a copy of S at the point (1=20;1=3).
Our resulting space would be our example S with one copy of S glued at the point
(1=20;1=3).
In the next theorem we will not just break and glue in one copy of S. We will
want to \break and glue across S", meaning that for some flxed t 2 (0;1), we will
break and glue at all points from the collection f(z;t)jz 2 Zg. Using the previous
example of (1=20;1=3), the term \break and glue across S at (1=20;1=3)" would mean
that you would use the lemma and break and glue at each point from the collection
f(z;1=3)jz 2 Zg. The topology for this space would be locally the product topology
on Z ? [0;1] for points not in a new glued copy of S. For points inside a glued
copy we would use the product topology on Z ? Z ? [0;1]. Note this idea can be
extended if, as we will do in a later step, we glued inside a glued copy; then locally
for the points inside the new glued copy the topology would be the product topology
of Z ?Z ?Z ?[0;1].
Another fact we will use considers the relationship between two spaces made by
using the lemma. Let S1=3 be the space made by breaking and gluing across S at
(1=20;1=3), and let S1=4 be the space made by breaking and gluing across S at the
31
point (1=20;1=4). Notice that each of these spaces look like our original example S
but with a copy of S glued into each arc z?[0;1] for all z 2 Z. Thus it is easy to see
that S1=3 ?= S1=4. This is true because of our construction of S and the fact that it is
again really just uncountably many arcs with identiflcations. Thus for example the
homeomorphism would map the arc z?[0;1] ? S1=3 onto the arc z?[0;1] ? S1=4 by
mapping z ?[0;1=3) onto z ?[0;1=4), the copy of S onto the copy of S, and lastly
z ? (1=3;1] onto z ? (1=4;1]. We will use this fact in the future so that if we have
constructed a Whitney map on S1=3 then we would similarly be able to produce one
on S1=4.
Theorem 3.2. There exist a hereditarily non-metric continuum that supports a Whit-
ney map.
Proof. It has already been shown that the inverse limit of spacesfXfig1fi=1 will support
a Whitney map if each Xfi supports a Whitney map.
Using our lemmas and our space S we will construct a system of spaces fSfig1fi=1
and maps ffflfigfi<
>:P0
0
B@ 1=20
c01
1
CA;P
0
0
B@ 1=20
c02
1
CA;P
0
0
B@ 1=20
c03
1
CA;::::
9
>=
>;be a countable dense sub-
set of R ? S0.
We will break and glue a copy of S at the point P0
0
B@ 1=20
c01
1
CA. Thus the arc
1=20 ?[0;1] in S would now have a copy of S glued in at the point c01 2 [0;1]. We
want to not only break and glue on the arc 1=20 ?[0;1] but for every arc z ?[0;1]
from S0. Thus at every point P0
0
B@ z
c01
1
CA;z 2 Z break and glue a new copy of S. S
1
will be this new space made by breaking and gluing at all the points mentioned.
We need notation for the points in S1. In S1 there are two types of points; points
that are in new glued copies of S and points of the form P0
0
B@ z
t
1
CA for t 6= c0
1. ( Note:
33
the point P0
0
B@ z
c01
1
CA for any z 2 Z does not exist in S
1 since these are the points
that were \replaced" by a copy of S).
Case 1: Points corresponding to P0
0
B@ z
t
1
CA for t 6= c0
1.
These points can still be thought of as a point (z;t) for some z 2 Z;t 2 [0;1];t 6=
c01; thus let P1
0
B@ z
t
1
CA denote the corresponding point in S
1 that is the point that is
on the arc z ?[0;1] at the tth coordinate.
Case 2: Points inside a glued copy of S.
There are uncountably many new glued copies of S in S1. Thus to name a point
inside a glued copy you must indicate which glued copy it is inside and then where
on the glued copy the point is located. To indicate this we will use the same type of
notation but add another pair of coordinates. A point inside the glued copy will be
denoted by
P1
0
B@ z1 z2
c01 ; t
1
CA
for some z1;z2 2 Z. In this notation the flrst column will tell you which glued copy
the point is in ( namely the one glued at P0
0
B@ z1
c01
1
CA ) and the second column tells
the position of the point on the new glued copy.
34
For example:
P1
0
B@ 1=20 1=31
c01 ; 1=4
1
CA
wouldindicatethepointP
0
B@ 1=31
1=4
1
CAfromthegluedcopyofS atthepointP
0
0
B@ 1=20
c01
1
CA.
The topology of S1 is deflned as follows. If p 2 S1 and p is of the form P1
0
B@ z
t
1
CA
fort 6= c01, thenalocalbasicelementatthatpointisaset
8
><
>:P1
0
B@ z
t
1
CAj P
0
0
B@ z
t
1
CA2 O
0
9
>=
>;
for some open set O0 2 S0 not containing P0
0
B@ z
c01
1
CA.
For a point of the form P1
0
B@ z1 z2
c01 ; t
1
CA we will use the open set J from the
deflnition that described an open set in an arc that did not contain 0 or 1 with a
copy of S glued in at the point c01 . Thus an open set in S1 for a point of the form
P1
0
B@ z1 z2
c01 ; t
1
CA would be U ?J where U is an open set in Z. This is topologically
an open set containing P1
0
B@ z1 z2
c01 ; t
1
CA with the product topology.
Note that, like S, S1 is indecomposable. Also recall: given any proper subcontin-
uum M ? S, M was contained in flnitely many arcs joined together ( see explanation
of composants of S). S has a flber-like structure with each flber being an arc [0;1].
35
In S1 if we think of a flber now as an arc with a glued copy of S; the composants
and the description of proper subcontinua will be the same as it was for S. Thus any
proper subcontinuum M will be contained in flnitely many flbers and hence can only
intersect flnitely many of the new glued copies of S.
Now that we have described S1 and the points from S1 we need to describe the
bonding map f10 from S1 to S0. This bonding map will take all the glued copies from
S1 and collapse them down to the point at which they were glued. On all other points
the bonding map will be the identity. In notation this would be represented as
f10
0
B@P
1
0
B@ z
t
1
CA
1
CA = P
0
0
B@ z
t
1
CA;t 6= c0
1
f10
0
B@P
1
0
B@ z1 z2
c01 ; t
1
CA
1
CA = P
0
0
B@ z1
c10
1
CA:
In order to deflne this bonding map we needed notation to describe each point
in S1; in order to describe the Whitney map we will need notation to distinguish
between the uncountably many copies of S that we glued into S0 when we made S1.
The reason this is important goes back again to our example S and what we have
already mentioned about proper subcontinua of S1. A composant in S1 is similar
to the composant from S0 = S except with countably many copies of S glued in at
speciflc points. Observe that if J is a composant of S0 then (f10)?1(J) is a composant
36
of S1. Thus any proper subcontinuum of S1 would be contained in what could be
thought of as flnitely many flbers joined together where each of these is an arc that
has a copy of S glued into it. Because of this fact about the composants, a proper
subcontinuum M will only intersect flnitely many of these new glued copies of S in
S1. ( Note: This fact will hold for any Sfi. If M is a proper subcontinuum then M
will only intersect flnitely many of the glued copies from S1. This will be used heavily
in future levels.)
The easiest way to distinguish between copies of S is to denote the copy by using
the point at which the copy was glued. For instance we glued in a copy of S at the
point P0
0
B@ 1=20
c01
1
CA, denote this copy as
S01
0
B@P
0
0
B@ 1=20
c01
1
CA
1
CA:
Notice that the 0 will indicate that we are denoting a copy of S, the subscript
indicates what space the copy is in ( in our case S1), and P0
0
B@ 1=20
c01
1
CA tells at which
point from the previous space the copy of S was glued.
Thus
8>
<
>:S
0
1
0
B@P
0
0
B@ z
c01
1
CA
1
CA
9
>=
>; for all z 2 Z would be the collection of all glued
copies of S in S1.
37
Next we will deflne the Whitney map. Let ? be the whitney map deflned for our
example S. Since S0 is a copy of S we can deflne a Whitney map ?0 for S0 the same
way we deflned ?.
Let M be a proper subcontinuum of S1.
For each z 2 Z, let Az = MTS01
0
B@P
0
0
B@ z
c01
1
CA
1
CA. Now since M is a proper
subcontinuum, fAzjAz 6= ;g is flnite or empty. (This is true for the previous reasons
stated about the composants and proper subcontinua of S1). Let fAign1i=1 = fAzjAz 6=
;g. Now S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA?= S
0 Therefore each S
0
1
0
B@P
0
0
B@ z
c01
1
CA
1
CA will have a Whit-
ney map deflned the same way as ?0. Call this Whitney map ?10. Note: the superscript
indicates that the Whitney map is on S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA and the subscript indicates
that S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA?= S
0: Now since each Ai is either a proper subcontinuum of or
equal to S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA for some z 2 Z, then each A
i would have a Whitney value
for ?10. Note that at most two of A0is are such that Ai = Az 6= S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA; thus
?10(Ai) 6= 1 for at most two A0is.
Deflne ?1 : C(S1) ?!R by
?1(M) = Arctan(len(f
1
0(M))+
Pn
i=1 ?0(Ai))
?
2
;and
38
?1(S1) = 1:
Since ?1 is the composition of continuous functions, it is continuous. To see that
the Whitney property is satisfled, notice that given M;N 2 C(S1) if M ( N then
len(f10(M))+PnMi=1 ?(A1Mi ) len(f10(N))+PnNi=1 ?(A1Ni ). Thus ?1(M) ?1(N).
Now to make S2 we will again want to break and glue at speciflc points from S1.
The flrst step in this process is again to flnd a countable dense set. This time (and
in inflnitely many future steps) we will choose a countable dense set from one of the
new glued copies from the previous space. In S0 our countable dense set came from
the arc f1=20g?(0;1), we will again look at the arc f1=20g?(0;1) but this time it
be will inside the copy of S glued at the point P0
0
B@ 1=20
c01
1
CA. Recall we denoted this
copy as S01
0
B@P
0
0
B@ 1=20
c01
1
CA
1
CA, and points on this copy would have the form
P1
0
B@ 1=20 z
c01 ; t
1
CA.
Let
8
><
>:P1
0
B@ 1=20 1=20
c01 ; c11
1
CA;P
1
0
B@ 1=20 1=20
c01 ; c12
1
CA;P
1
0
B@ 1=20 1=20
c01 ; c13
1
CA;::::
9>
=
>;
be a countable dense set from the arc f1=20g?(0;1) on S01
0
B@P
0
0
B@ 1=20
c01
1
CA
1
CA . Make
the collection so that none of the c1i?s are 0 or 1.
39
We want to glue inside S01
0
B@P
0
0
B@ 1=20
c01
1
CA
1
CA ( recall this is the copy of S glued at
0
B@P
0
0
B@ z
c01
1
CA
1
CA ) and we will repeat the procedure used in making S
1. We will break
and glue at the point P1
0
B@ 1=20 1=20
c01 ; c11
1
CA. In a similar way that we made S
1, we
also want to break and glue all the way across S01
0
B@P
0
0
B@ 1=20
c01
1
CA
1
CA. Thus for every
z 2 Z we have the point P1
0
B@ 1=20 z
c01 ; c11
1
CA. Break and glue at those points. At
this stage in our construction the arc f1=20g?[0;1] from S0 would now look like an
arc with a copy of S glued in and then within that copy there are uncountably many
copies of S glued on each arc z ? [0;1] . (Hopefully it is becoming clear why you
need notation to indicate all the points and all the copies of S). We have glued in
uncountably many copies of S into the copy of S glued at P0
0
B@ 1=20
c01
1
CA. We want to
do the same gluing on all other copies of S from S1, using the same c11 2 (0;1). Recall
a glued copy was denoted as S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA. And a point inside was named by
P1
0
B@ z1 z2
c01 ; t
1
CA. If we flx z
1 then
8
><
>:P1
0
B@ z1 z2
c01 ; t
1
CAjz
2 2 Z;t 2 [0;1]
9
>=
>; would be
40
all the points from S01
0
B@P
0
0
B@ z1
c01
1
CA
1
CA. We want to break and glue at all points where
t = c11. Thus for a flxed z1 we will break and glue at each point from the collection8
><
>:P1
0
B@ z1 z2
c01 ; c11
1
CAjz
2 2 Z
9
>=
>;. Repeat this procedure for z1 = z; all z 2 Z .
Putting this all together and using the notation, what we have done is break and
glue a copy of S at each point from the collection8
><
>:P1
0
B@ z1 z2
c01 ; c11
1
CAjz
1;z2 2 Z
9
>=
>;.
Denote this new space as S2. Using the same notation as before, the new
glued copies can be represented by the points at which they were glued. Thus8
><
>:S
0
2
0
B@P
1
0
B@ z1 z2
c01 ; c11
1
CAjz
1;z2 2 Z
1
CA
9
>=
>; is the set of all new glued copies of S. Build-
ing on our previous notation we can denote the three difierent types of points from
S2.
Type 1 :
8>
<
>:P2
0
B@ z
t
1
CAjz 2 Z;t 2 [0;1]=c0
1
9
>=
>;: These are the points not in any glued
copy of S.
Type 2 :
8
><
>:P2
0
B@ z1 z2
c01 ; t
1
CAjz
1;z2 2 Z;t 2 [0;1]=c11
9
>=
>;: These points are in the
copies of S that were glued to make S1.
Type3 :
8
><
>:P2
0
B@ z1 z2 z3
c01 ; c11 ; t
1
CAjz
1;z2;z3 2 Z;t 2 [0;1]
9
>=
>;. Thesearethepoints
that lay inside one of the new copies of S from the collection
41
8>
<
>:S
0
2
0
B@P
1
0
B@ z1 z2
c01 ; c11
1
CAjz
1;z2 2 Z
1
CA
9>
=
>;.
For example P2
0
B@ 1=20 1=30 1=40
c01 ; c11 ; 1=5
1
CA would be the point P
0
B@ 1=40
1=5
1
CA on
the copy of S glued at P
0
B@ 1=30
c11
1
CA where P
0
B@ 1=30
c11
1
CA is a point inside the copy of
S glued at P0
0
B@ 1=20
c01
1
CA. In notation P
2
0
B@ 1=20 1=30 1=40
c01 ; c11 ; 1=5
1
CA would be the
point P
0
B@ 1=40
1=5
1
CA on S0
2
0
B@P
1
0
B@ 1=20 1=30
c01 ; c11
1
CA
1
CA.
The bonding map f21 : S2 ! S1 will be deflned in the same matter as before.
New glued copies of S will collapse down to the point at which they were glued and
f21 will be the identify on all other points. Thus f21
0
B@P
2
0
B@ z
t
1
CA
1
CA = P
1
0
B@ z
t
1
CA (these
are the points not in any glued copy of S, they map to themselves), and
f21
0
B@P
2
0
B@ z1 z2
c01 ; t
1
CA
1
CA = P
1
0
B@ z1 z2
c01 ; t
1
CA (these are the points in the glued
copy at the flrst level, they map to themselves), and
f21
0
B@P
2
0
B@ z1 z2 z3
c01 ; c11 ; t
1
CA
1
CA = P
1
0
B@ z1 z2
c01 ; c11
1
CA (these are the points inside
the new copy of S, they collapse down to point at which S was glued).
As before if J is a composant of S2 then f21(J) is a composant of S1, and con-
versely if J is a composant of S1 then (f21)?1(J) is a composant of S2. We need
42
to deflne a Whitney map on S2. In the flrst level, when we deflned our Whit-
ney map, given M a proper subcontinuum we intersected M with the new glued
copies of S and then used the sum of these values to deflne ?1. But that was pos-
sible because M intersected only flnitely many of the new glued copies. Notice that
given M in S2, M can intersect uncountably many copies of S from the collection8
><
>:S
0
2
0
B@P
1
0
B@ z1 z2
c01 ; c11
1
CAjz
1;z2 2 Z
1
CA
9>
=
>; ( this is the collection of all new copies of S
glued into S1 in order to make a new space S2). Therefore we can not use these
Whitney values since we can not sum this uncountable amount. By the deflnition
of f21 if M is a proper subcontinuum of S2 then f21(M) is a proper subcontinuum of
S1. Thus a proper subcontinuum of S2 can only intersect flnitely many copies of S
from the collection
8
><
>:S
0
1
0
B@P
0
0
B@ z
c01
1
CA
1
CAjz 2 Z
9
>=
>;. Thus to deflne ?2 we will use the
intersection of M with the collection
8
><
>:S
0
1
0
B@P
0
0
B@ z
c01
1
CA
1
CA
9
>=
>; for all z 2 Z. To deflne
the Whitney map on S2 let M be a proper subcontinuum. Since at this level the
copies of S were glued inside of S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA, let A
z = M
TS0
1
0
B@P
0
0
B@ z
c01
1
CA
1
CA.
Note that fzjAz 6= ;g is flnite, so let fAign2i=1 = fAzjAz 6= ;g. Also note that each Az
is a subcontinuum.
Now S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA ?= S
1 (because both are the original S with a copy of S
glued onto each arc; thus as previously explained they are homeomorphic.) Therefore
43
each S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA will have a Whitney map deflned the same way as ?
1. Call
this Whitney map ?11. Note: the superscript indicates that the Whitney map is on
S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA and the subscript indicates that S0
1
0
B@P
0
0
B@ z
c01
1
CA
1
CA?= S
1: Now since
each Ai is either a proper subcontinuum of or equal to S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA for some
z 2 Z, then each Ai would have a Whitney value for ?11. Note that ?11(Az) = 1
for all but at most two i = 1;2;::::;n2. (Notice we used the homeomorphism only
to show quickly that Az does support a Whitney map. If Az was not homomorphic
to a previously constructed space then we could prove directly that Az supports a
Whitney map by using the same techniques used to prove S1 supports a Whitney
map. In future stages we may not have spaces homeomorphic to previous stages.)
Let ?2 : C(S2) ?!R be deflned by
?2(M) = Arctan(?1(f
2
1(M))+
Pn2
i=1 ?
1
1(Ai))
?
2
;and
?2(S2) = 1:
Roughly speaking ?2 divides M into difierent pieces where each piece has a
Whitney map associated with it. Note it is not the same Whitney map for every
piece. For the subcontinuum f21(M) we use the Whitney map ?1, for each Ai we use
44
?11. Then we take the sum of the respective Whitney values of each piece and apply
the arctan function and divide by ?2 to bound ?2 by 1. Thus we can deflne ?2(S2) = 1
and satisfy the Whitney property. This idea will be used in all future levels to deflne
the Whitney map; what will change will be the collection that a subcontinuum M
intersects.
To begin the process of deflning S3 we again will flnd a countable dense set from
inside one of the glued copies of S used in making S2. Namely
S02
0
B@P
1
0
B@ 1=20 1=20
c01 ; c11
1
CA
1
CA. RecallpointsonS0
2
0
B@P
1
0
B@ 1=20 1=20
c01 ; c11
1
CA
1
CAhave
the form P2
0
B@ 1=20 1=20 z
c01 ; c11 ; t
1
CA for some z 2 Z;t 2 [0;1]. Let z = 1=2
0 and
thus let
n
P2
0
B@ 1=20 1=20 1=20
c01 ; c11 ; c21
1
CA;P
2
0
B@ 1=20 1=20 1=20
c01 ; c11 ; c22
1
CA;
P2
0
B@ 1=20 1=20 1=20
c01 ; c11 ; c23
1
CA;::::o be a countable dense set from the arc 1=2
0 ?
(0;1) on S02
0
B@P
1
0
B@ 1=20 1=20
c01 ; c11
1
CA
1
CA.
We will again break and glue but not using the flrst element from our count-
able dense set as we have done in previous levels. We need our inverse limit to be
hereditarily non-metric; thus all the points from all the countable dense sets must
eventually end up with a copy of S glued at that point. If we continue gluing at each
level using the flrst element from the countable dense set obtained at that level it
45
will not result in hereditarily nonmetric. We will set up a diagonal system using the
dense sets that are found at each level and a function h from the positive integers
into a diagonal array that will determine which point will be used at which level. For
ease in making our diagonal array, think of the dense sets as fc01;c02;c03:::g instead of8
><
>:P0
0
B@ 1=20
c01
1
CA;P
0
0
B@ 1=20
c02
1
CA;P
0
0
B@ 1=20
c03
1
CA;::::
9
>=
>; , and
fc11;c12;c13;:::g instead of8
><
>:P1
0
B@ 1=20 1=20
c01 ; c11
1
CA;P
1
0
B@ 1=20 1=20
c01 ; c12
1
CA;P
1
0
B@ 1=20 1=20
c01 ; c13
1
CA;::::
9>
=
>;, etc.
Arrange these sets in a diagonal array
0
BB
BB
BB
BB
BB
BB
BB
BB
BB
B@
c01 ; c02 ; c03 ; c04 ; c05 ; ::::
c11 ; c12 ; c13 ; c14 ; : ; ::::
c21 ; c22 ; c23 ; : ; : ; ::::
c31 ; c32 ; : ; : ; : ; ::::
c41 ; : ; : ; : ; : ; ::::
c51 ; : ; : ; : ; : ; ::::
: ; : ; : ; : ; : ; ::::
1
CC
CC
CC
CC
CC
CC
CC
CC
CC
CA
We can construct a function h from N into the diagonal array to indicate which
point to choose at which level. We have already completed the flrst two levels and
thus we know that h(1) = c01 and h(2) = c11. Let h(3) = c02, h(4) = c21, h(5) = c12,
h(6) = c03, etc.
46
We know that in order to make S3 we will use the point in S2 that corresponds
to c02. This point is P2
0
B@ 1=20
c02
1
CA. Just as before we will break and glue at this point
and all points from the collection
8
><
>:P2
0
B@ z
c02
1
CAjz 2 Z
9
>=
>;. These points are not in any
previous glued copy. This new space is S3.
Notice that what we have done is glue a new copy of S not inside a previously
glued copy. Recall in S1 a flber z ? [0;1] had the form of an arc with a copy of
S glued in at the point (z;c01). In S2 a flber had the form of the arc from S1 but
now had uncountably many copies of S glued into the previous glued copy. In S3
a flber would look like the arc from S2 except with a new copy of S glued at the
point (z;c02). In making S3 we notice that the points from the flrst countable set are
special in the sense that when h(n) = c0j for some n;j then we are not gluing inside
any previous glued copy. This is the only time that will happen; if h(n) 6= c0j then we
will always be gluing inside a previous copy . But even in this case the points from
the flrst countable set are unique in that we will always be gluing inside an S that
was glued at the points from this flrst dense set. In other words if h(n) 6= c0j then
the new copies of S will be glued inside S0l
0
B@P
k
0
B@ z
c0m
1
CA
1
CA for some l;k;z;m;. This
is important because it enables us to deflne a Whitney map on any Sn. Before we
deflne ?3 we will deflne the bounding map f31 : S3 ?! S2. This will be similar to
previous levels. Deflne
47
f31
0
B@P
3
0
B@ z
t
1
CA
1
CA = P
2
0
B@ z
t
1
CA, t 6= c0
1;t 6= c
0
2,
f31
0
B@P
3
0
B@ z1 z2
c01 ; t
1
CA
1
CA = P
2
0
B@ z1 z2
c01 ; t
1
CA, t 6= c1
1,
f31
0
B@P
3
0
B@ z1 z2 z3
c01 ; c11 ; t
1
CA
1
CA = P
2
0
B@ z1 z2 z3
c01 ; c11 ; t
1
CA, and
f31
0
B@P
3
0
B@ z1 z2
c02 ; t
1
CA
1
CA = P
2
0
B@ z1
c02
1
CA.
On the flrst three types of points f is the identity and then on the last type f
collapses the glued copy of S onto the point at which it was glued.
Let M be a proper subcontinuum. As before we will think of M in pieces. The
flrst will be f32(M). But notice at this level we did not glue inside S01
0
B@P
0
0
B@ z
c01
1
CA
1
CAas
we did in the previous level; thus the way we deflne Az must change. For each z 2 Z,
let Az = MTS03
0
B@P
2
0
B@ z
c02
1
CA
1
CA. Again each A
z is a subcontinuum and fzjAz 6= ;g
is flnite. Let fAign3i=1 = fAzjAz 6= ;g.
S03
0
B@P
2
0
B@ z
c02
1
CAjz 2 Z
1
CA ?= S
0, thus there exist a Whitney map on the space,
call it ?30. As stated when deflning ?2, if Az is not homeomorphic to a previous space
then we can show directly that Az will support a Whitney map by using the same
techniques.
Deflne ?3 : C(S3) ?!R by
48
?3(M) = Arctan(?2(f
3
2(M))+
Pn3
i=1 ?
3
0(Ai))
?
2
;and
?3(S3) = 1:
To begin on the fourth level we again choose a countable dense set from the arc
1=20 ?(0;1) in S03
0
B@P
2
0
B@ 1=20
c02
1
CA
1
CA. Recall this is a copy of S glued in the previous
level. Let8
><
>:P3
0
B@ 1=20 1=20
c02 ; c31
1
CA;P
3
0
B@ 1=20 1=20
c02 ; c32
1
CA;P
3
0
B@ 1=20 1=20
c02 ; c33
1
CA;:::
9
>=
>;
be a countable dense set from the arc 1=20 ?(0;1) in S03
0
B@P
2
0
B@ 1=20
c02
1
CA
1
CA.
Since h(4) = c21 we will break and glue at the corresponding point which is
P1
0
B@ 1=20 1=20 1=20
c01 ; c11 ; c21
1
CA. Remember this point is inside a copy of S glued
inside another copy of S. We want to break and glue \across" S3, so break and glue
at each point from the collection
8
><
>:P3
0
B@ z1 z2 z3
c01 ; c11 ; c21
1
CAjz
1;z2;z3 2 Z
9
>=
>;. This
new space will be S4.8
><
>:S
0
4
0
B@P
3
0
B@ z1 z2 z3
c01 ; c11 ; c21
1
CAjz
1;z2;z3 2 Z
1
CA
9>
=
>; is the collection of all copies of
S glued into the space S3 in order to make a new space S4.
Deflne f43 : S4 ?! S3 as the identity on all points except those of the form
49
0
B@P
4
0
B@ z1 z2 z3 z4
c01 ; c11 ; c21 ; t
1
CA
1
CA.
For these points deflne f43 by
f43
0
B@P
4
0
B@ z1 z2 z3 z4
c01 ; c11 ; c21 ; t
1
CA
1
CA =
0
B@P
3
0
B@ z1 z2 z3
c01 ; c11 ; c21
1
CA
1
CA.
We want to deflne the Whitney map at this level. Let M be a proper subcontin-
uum. In S4 the new copies were again glued inside of S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA.
Therefore for each z 2 Z, let Az = MTS01
0
B@P
0
0
B@ z
c01
1
CA
1
CA, and fA
ign4i=1 =
fAzjAz 6= ;g. At this level S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA ?= S
2. There exist a Whitney map
?12 on S01
0
B@P
0
0
B@ z
c01
1
CA
1
CA that is deflned in the same manner as ?
2. Again if it was
not homoeomorphic to a copy we could have proved directly that Az will support a
Whitney map.
Then deflne ?4 by
?4(M) = Arctan(?3(f
4
3(M))+
Pn4
i=1 ?
1
2(Ai))
?
2
and
?4(S4) = 1:
50
In each of the previous four cases we used the fact that Az was homeomorphic
to a previous Si in order to show that there was a Whitney map on Az. In some
future levels it is the case that Az will not be homeomorphic to any previously made
space. As previously mentioned this will not be a problem though because we do not
need Az to be homeomorphic to a previous copy we just need to know that Az has
a Whitney map. If Az is not homeomorphic to a previous copy then it is necessary
to determine directly that it has Whitney map. This can be done by using the same
techniques that were used to show S1;:::;S4 have Whitney maps.
For example when n = 8, Az is not homeomorphic to any previous copy. Az
though is homeomorphic to a space call it S3'1, that is S3 with copies of S glued in
at each point from the collection P1
0
B@ 1=20 z
c01 ; c12
1
CA;z 2 Z. We will want to show
that a Whitney map can be constructed on this space. First we need to show that a
simpler space has a Whitney map. Take S1 and break and glue at each point from
the collection P0
0
B@ z
c02
1
CA;z 2 Z. Thus each flber would now have two copies of S
glued into it. Call this new space S1 '1. (Note we have clearly glued uncountably
many copies of S not just one, but the '1 indicates we have we have done one more
step and glued at just one more collection of points.) Now there exists a bonding map
g : S1 '1 ! S1 deflned as the identity except on the new copies of S and those will
collapse down to the point at which they were glued. If M is a proper subcontinuum
of S1 ' 1 then Az will be homeomorphic to S. Thus Az will have a Whitney map,
namely ?0. There are only flnitely many Az say A1 to An. Thus deflne the Whitney
51
map using out technique on S1 '1 as ?(M)S1'1 = ?S1(g(M)) +Pni=1(?0(Ai)). This
can be shown to be a Whitney map by the previous methods. Now the next step is to
prove that S3 '1 has a Whitney map. There exists a bonding map g : S3 '1 ! S3.
g will be the identity on all points except the new glued copies which will collapse
down to the point at which they were glued. If M is a proper subcontinuum of S3'1
then Az will be M intersected with
8
><
>:S
0
1
0
B@P
0
0
B@ z
c01
1
CA
1
CA
9>
=
>; for every z 2 Z. Only
flnitely many of these will be non-empty. Denote those as A1 to Am. Now Az is
homomorphic to S1'1 which was just shown to have a Whitney map . Thus Az will
have a Whitney map. Let ?0S1'1 denote the Whitney map on Az. Deflne a Whitney
map ?S3'1 as ?S3'1(M) = ?3(g(M))+Pmi=1(?0S1'1(Ai)).
Now the same procedure can be used to show that no matter what the structure
of Az, it will have a Whitney map. This fact is what is needed to construct a Whitney
map on Sn, any n. For example, when n = 9, then a typical Az will be homeomorphic
to (S3'1)'1. Since we know that S3'1 has a Whitney map then it can be proven
that (S3 '1)'1 has a Whitney map, regardless of where the copies of S are glued.
When n = 10, Az is homeomorphic to S, thus we know it has a Whitney map.
Deflne Sn inductively assuming that we have deflned Sn?1. At the nth level flrst
deflne a countable dense subset from the 1=20 ?(0;1) arc in the copy of S glued in
the previous level. As before we want to make a new space by breaking and gluing
in copies of S. There are 2 cases that can occur at the nth level.
Case 1. h(n) = c0m for some m. Thus we will beak and glue at
52
8>
<
>:Pn?1
0
B@ z
c0m
1
CAjz 2 Z
9>
=
>;. Note these points are not in any previously glued copy
of S.
The map fnn?1 : Sn ?! Sn?1 will be deflned as in previous levels. It will be the
identify on all points except those points from the new glued copies and those will
collapse down to the point at which the copy was glued.
In this case to deflne the Whitney map let M be a proper subcontinuum of
Sn. Since we have not glued inside any previous copy, for each z 2 Z let Az =
MTS0n
0
B@P
n?1
0
B@ z
c0m
1
CA
1
CA; A
z is a subcontinuum and fzjAz 6= ;g is flnite, so let
fA1igji=1 = fAzjAz 6= ;g.
Note that S0n
0
B@P
n?1
0
B@ z
c0m
1
CA
1
CA ?= S so the corresponding Whitney map on
S0n
0
B@P
n?1
0
B@ z
c0m
1
CA
1
CA will be labeled ?n
0:
Let ?n : C(Sn) ?!R be deflned by
?n(M) = Arctan(?n?1(f
n
n?1(M))+
Pj
i=1 ?
n
0(Ai))
?
2
and
?n(Sn) = 1:
53
Case 2: h(n) = cjm some j 6= 0. Break and glue at the points
8
><
>:(Pn?1
0
B@ z1 zr?1 zr
c0k ; : : : cba ; cjm
1
CA)j(z
1;z2;:::zr 2 Z)
9
>=
>;:
NotethatthesepointsareinapreviouslygluedcopyofS thatwasmadebybreak-
ing and gluing at a point from the flrst countable dense set, namely Ph?1(c0k)
0
B@ z
c0k
1
CA
for some k.
The bonding map will be as in Case 1.
To deflne the Whitney map we previously noted that at this level the new copies
of S are glued inside a previous copy. Call this copy S0h?1(c0
k)
0
B@P
h?1(c0k)
0
B@ z
c0k
1
CA
1
CA for
some k. Let Az = MTS0h?1(c0
k)
0
B@P
h?1(c0k)
0
B@ z
c0k
1
CA
1
CA, and fA
ig
j
i=1 = fAzjAz 6= ;g.
Now we have two cases for Az, either Az is homomorphic to some previous space
(example n = 4) or it is not ( example n = 8). If S0h?1(c0
k)
0
B@P
h?1(c0k)
0
B@ z
c0k
1
CA
1
CA?= S
l for
some l, then the Whitney map on S0h?1(c0
k)
0
B@P
h?1(c0k)
0
B@ z
c0k
1
CA
1
CA will be deflned in the
same manner as ?l and so will be named ?h?1(c0k)l . Let ?n : C(Sn) ?!R be deflned
by
?n(M) = Arctan(?n?1(f
n
n?1(M))+
Pj
i=1 ?
h?1(c0k)
l (Ai))
?
2
54
and
?n(Sn) = 1:
If Az is not homeomorphic to any previous Si, Az can be shown to have a
Whitney map using the same techniques that proved Si has a Whitney map. Denote
that Whitney map by ?h?1(c0k). Let ?n : C(Sn) ?!R be deflned by
?n(M) = Arctan(?n?1(f
n
n?1(M))+
Pj
i=1 ?
h?1(c0k)(Ai))
?
2
and
?n(Sn) = 1:
Let X = lim??fSn;fg To show the inverse limit space X is hereditarily non-metric,
let M be a nondegenerate proper subcontinuum of X. There exist a;b 2 M such that
a 6= b. By the nature of our inverse limit space and the fact that M is a continuum,
we can flnd an n and an r and points ar;br 2 ?Sr(M), such that ar 6= br and
ar;br 2 S0n
0
B@P
n?1
0
B@ z1 zr
c0k ; : : : ; cjk
1
CA
1
CA. Furthermore a
r;br are such that
ar = Pr
0
B@ z1 zr zar
c0m ; : : : cjk ; tar
1
CA and
br = Pr
0
B@ z1 zr zbr
c0m ; : : : cjk ; tbr
1
CA , where z
ar = zbr, and tar tbr.
55
By the way we chose our countable dense sets at each level there exists a cvw such
that tar < cvw < tbr. Since M is connected there exists an x 2 M such that
x = Pr
0
B@ z1 zr zar
c0m ; : : : cjk ; cvw
1
CA.
Now at some level, call it p, h(p) = cvw; a copy of S was glued at x, namely
S0p
0
B@P
r
0
B@ z1 zr zar
c0m ; : : : cjk ; cvw
1
CA
1
CA. Thus in S
p, ?p(M) must contain a point
from this copy of S. But it also contains ap; and bp which, because of the irreducibility
of the continua at each level, lies on either side of this copy of S; thus the whole copy
of S must be in ?p(M) . Therefore ?p(M) is non-metric and so M must also be
non-metric.
56
Chapter 4
a hereditarily indecomposable non-metric continuum that supports a
Whitney map
Theorem 4.1. There exist a hereditarily indecomposable non-metric continuum that
supports a Whitney map.
Proof. It has already been shown that the inverse limit of spacesfXfig1i=1 will support
a Whitney map if each Xfi supports a Whitney map.
Using the space S from the example we will construct a system of spaces fSfig1i=1
and maps ffflfigfi 0 such that [f(x;y) ??] ? U, where
[f(x;y)??] = [f(x;y)??;f(x;y)+?]. Let ? = minf?4; ?(y;t)??(x;t)2 g. By the continuity
of ? there exist a
Vx, open in C(ConeX), such that x;t 2 Vx ? C(Cone(X)) and ?(Vx) ? [?(x;t)?
?]
and
Vy such that y;t 2 Vy ? C(X) and ?(Vy) ? [?(y;t)??].
64
Since x;t 2 Vx and Vx = R(fVigni=1) we can do the same construction as before
and get open sets such that x;t ?Smi=1(Wxai ?(cai;dai)) ?Sni=1 Vi ? Cone(X).
Let eV ? X ?X be deflned as
eV = (Tmi=1(Wxa
i))?(
Tj
i=1(W
y
bi)). Now if (r;s) 2 eV then ?(r;t) 2 [?(x;t)??] and
?(s;t) 2 [?(y;t)??].
Therefore minf?((r;t);?(s;t)g = ?(r;t) 2 [?(x;t)??] ? [f(x;y)??] ? U. Thus
f(eV) ? U.
Case 2.?(x;t) = ?(y;t).
Same proof except let ? = ?4 so that if r;t 2 Vx then ?(r;t) 2 [f(x;y)? ?4], and if
s;t 2 Vx then?(s;t) 2 [f(x;y)??4]. Thereforeminf?(r;t);?(s;t)g2 [f(x;y)??4] ? U,
regardless of which arc is the minimum. So that f(eV) ? U and thus f is continuous.
So we have shown that F(x;y) is a difierence of two continuous functions and
thus is continuous. We will now show that F(x;y) = 0 , x = y.
Assume that F(x;y) = 0. Then ?(x;t [ y;t) = minf?(x;t);?(y;tg, but x;t ?
(x;t [ y;t) so by deflnition of a Whitney map ?(x;t) < ?(x;t [ y;t). Thus x;t =
(x;t[y;t) which implies that x = y.
Assume that x = y. Then x;t[y;t = x;t = y;t so ?(x;t) = ?(x;t[y;t) which
implies that F(x;y) = 0.
Therefore X is metric.
The next proof is a generalize of Theorem 5.1
65
Theorem 5.2. Given X is contractible let ` : X ? [0;1] ?! R be the contraction
map, where `(x;1) = p for any x. Let xp = f`(x;t)jt 2 [0;1]g. If xp = yp if and
only if x = y and X supports a Whitney map, then X is metric.
Proof. Let ` : X ?[0;1] ?! X be the contraction map, where `(x;1) = p for any x.
Let xp = f`(x;t)jt 2 [0;1]g. This will be a subcontinuum of X.
Deflne G : X ?X ?!R by G(x;y) = ?(xpSyp)?minf?(xp);?(yp)g.
We will show that each part of G is continuous and thus G is continuous. We
will then use the fact that if you have a continuous function G from X ?X into R
such that G(x;y) = 0 if and only if x = y then, X is metric.
Deflne F : X ?X ?!R by F(x;y) = ?(xpSyp).
Let (x;y) 2 X ? X and let U be an open set in R with F(x;y) 2 U. By the
continuity of ? there exists a basic open set V 0 in C(X) such that (xpSyp) 2 V 0 and
?(V 0) ? U. Now V 0 = R(fVigni=1) where each Vi is an open set in X. First make
this collection such that for each i there exists a z 2 (xpSyp) such that z 2 Vi and
z =2 Vj for every j 6= i. There exist subcollections fV xj gmj=1 and fV yk glk=1 of fVigni=1
such that ?fV xj gmj=1SfV yk glk=1? = fVigni=1, xp ?fV xj gmj=1, and yp ?fV yk glk=1.
For each z = `(x;t) 2 xp there exists a j such that z 2 V xj and by the continuity
of ` there exists a basic open set Wt = Ut ?(ct;dt) ? X ?[0;1] such that (x;t) 2 Wt
and `(Wt) ? V xj .
S
t2[0;1] Wt covers x?[0;1] thus there exists a flnite subcover fWtrg
s
r=1 that covers
x?[0;1]. Furthermore the subcover can be chosen to satisfy the condition that for
each V xj there exists Wtr such that `(Wtr) ? V xj . ( If this condition is not satisfled
66
then just add to the collection a flnite number of the Wt?s so that the needed property
is satisfled.)
Let A = Tsr=1 ?x(Wtr), where ?x(Wtr) is the projection of Wtr onto the space X.
A is open in X and if z 2 A then
1. Given anyt there exists r such that (z;t) 2 Wtr which implies that `(z;t) 2 V xj
for some j. Thus zp ?Smj=1 V xj .
2. Since we added in the extra condition on the Wtr?s, given any V xj there exists
Wtr such that `(Wtr) ? V xj , which implies that zp intersects each V xj for j = 1 to m.
Therefore, using the above two facts, zp 2 R(fV xj gmj=1).
Repeating the same procedure using yp we obtain an open set B = Tdq=1 ?x(Oq)
such that if z 2 B then zp is contained in Slk=1 V yk and intersects each V yk ;k = 1 to l.
Now A?B is an open set in X ?X. If (a;b) 2 A?B then apSbp intersects
each Vi since ap intersects each member of the collection fV xj gmj=1 and bp intersects
each fV yk glk=1 and ?fV xj gmj=1SfV yk glk=1? = fVigni=1.
Also apSbp ? Sni=1 Vi. Therefore apSbp 2 V 0 thus ?(apSbp) 2 U which
implies that F is continuous.
Now it can be shown that if F(x;y) = minf?(xp);?(yp)g, then F is continuous.
Thus G is continuous.
We just need to show that G(x;y) = 0 if and only if x = y. If x = y then xp =
xpSyp which implies that minf?(xp;?(yp)g = ?(xp) = ?(xpSyp), so G(x;y) = 0.
If G(x;y) = 0 then ?(xpSyp) = minf?(xp);?(yp)g. We know that xp = yp if
and only if x = y thus ?(xpSyp) = ?(xp) only if x = y.
67
Therefore X is metric.
Theorem 5.3. Given X is contractible, let ` : X ?[0;1] ?! X be the contraction
map, where `(x;1) = p for any x. Let xp = f`(x;t)jt 2 [0;1]g. If X supports a
Whitney map and if Q is a compact subset of X such that xp = yp if and only if
x = y for every x and y in Q, then Q is metric.
Theorem 5.4. Let S be the example from Chapter 2. C(S) is contractible, and
furthermore that contraction satisfles that conditions of Theorem 5.2.
Proof. Before we begin, a lemma will be useful.
Lemma 5.1. Let V be an open set in C(S). Let M be a proper subcontinuum of S
that starts at the endpoint (a;t) and ends at the point (b;s), such that M 2 V. There
exists an open set eV lying in V and associated ? > 0 so that
1.
eV = R(fV 0a;V 0b;V 0cg):
The union of V 0a;V 0b;V 0c (the open sets that make up eV) is a tube-like set in S (recall
deflnition of tube-like in S from Chapter 2)
2. M 2 eV
3. Suppose (aN;tN) and (bN;sN) are endpoints of a subcontinuum N. Let
(aN;tN) be the starting point and (bN;sN) be the ending point. If N 2 eV then
jt?tNj < ? and js?sNj < ?, for some ? > 0.
68
Proof. Let M be a proper subcontinuum. M is a metric arc with endpoints (a;t) and
(b;s). Let (x;y) be the midpoint of the proper subcontinuum M. Let F?n(x) = a,
Fj(x) = b, and F0(x) = x. Then
M = (F?n(x)?[0;t]) S (F?n+1(x)?[0;1]) S . . . . . S (F?1(x)?[0;1]) S
(fxg?[0;1]) S (F(x)?[0;1]) S. . . . . S (Fj(x)?[s;1]).
V = R(fVkgmk=1). Let f?Z(Vk)gmk=1 be the projection onto Z of fVkgmi=1. For each
i 2 f?n;:::?1;0;1;:::jg, let fVkgk2Ji, Ji 2 f1;2;:::mg, be the collection of V 0ks such
that Fi(x) 2 ?(Vk): Let Wi = Tk2Ji Vk for each i 2f?n;:::?1;0;1;:::jg. Note Wi is
an open set in Z, and Fi(x) 2 Wi. Using the methods from the proof of lemma 4.1
in Chapter 2, the W0is can be reflned into U0is so that
1. Ui is open,
2. Fi(x) 2 Ui,
3. F(Ui) = Ui+1 for each i 2f?n;:::?1;0;1;:::jg, and
4. fUigji=?n are pairwise disjoint.
We will use these open sets in Z to create open sets in S. We will also need open
sets in [0;1]. The endpoint (a;t) is in at most m open sets from fVkgmk=1. Thus there
exists an ?a such that if (a;t) 2 Vk then (a;t+?a) 2 Vk and (a;t??a) 2 Vk. Likewise
there exists an ?b such that if (b;s) 2 Vk then (b;s + ?b) 2 Vk and (b;s ? ?b) 2 Vk.
Let ? = minf?a;?bg. Deflne V 0a = U?n ?(t??;t + ?), and V 0b = Uj ?(s??;s + ?).
Note V 0a and V 0b are open sets in S and are contained in any Vk that contained (a;t)
and (b;s) respectively. Let V 0c = (U?n ?[0;t? ?2)) S (U?n+1 ?[0;1]) S . . . . . S
(Uj?1?[0;1]) S (Uj?[s;s+ ?2)). Note that V 0c is tube-like in S. Thus by construction
69
SfV 0
a;V
0
b;V
0
cg is tube-like in S. Deflne
eV = R(fV 0a;V 0b;V 0cg):
Given N is a proper subcontinuum of S if N 2 eV then N must begin in V 0a and
end in V 0b. Let (aN;tN) and (bN;sN) be the endpoints of N; then jt ? tNj < ? and
js?sNj < ?. Also, by the construction of eV, N 2 V. Therefore eV lies in V.
Before we begin,we need some notation.
1. If N is a proper subcontinuum of S then N + fi is the arc in S made by
extending N by fi2 in each direction. Thus len(N +fi) = len(N)+fi.
2. If aN is an endpoint of N then aN + ?, ? < 1, is the endpoint on the new
arc N + 2?. Given aN = (x;t) then aN + ? would be the point (x;t + ?) or (x;t??)
depending on which direction we want to go. Note, if t is within ? of 0 or 1 then a
slight modiflcation must be made. If t is within ? of 1 then aN + ? , the new point,
would be (F?1(x);t+??1). If t is within ? of 0 then aN ?? would be (F(x);t??+1).
3. Let (a;b) and (c;d) be two points in S. \(a;b) is within ? of (c;d)" will mean
jb?dj < ?.
70
Now we need to deflne the contraction map f : [0;1] ? C(S) ! C(S). Let
f(?;M) =
8
>>>
>>>>
>>>
>><
>>>
>>>
>>>
>>>:
M if ? ? ?(M);
M0 if ? > ?(M); where M0 has the same midpoint as M
and ?(M0) = ?;
S if ? = 1;
S if M = S:
Notice that M0 is uniquely deflned by the nature of S since if M0 = M +? where
?(M0) = ? then ? = arctan(lenM0)?
2
.
We wish to prove that f is continuous. Let V 2 C(S) be an open set such that
f(?;M) 2 V. Let M be a proper subcontinuum and ? 2 [0;1); then we have three
cases:
1. ?(M) > ?,
2. ?(M) = ?, and
3. ?(M) < ?.
Case 1. ?(M) > ?.
If ?(M) > ? then f(?;M) = M. Let V 2 C(S) be an open set such that
f(?;M) = M 2 V. Using the lemma we can flnd a tube-like reflnement eV with
associated ? containing M so that eV is tube-like in S, eV = R(V 0a;V 0b;V 0c), and the
associated ? is small enough so that if N 2 eV then
?(N) > ? + j? ??(M)j2 > ?:
71
Let W = (? ? j???(M)j4 ;? + j???(M)j4 )? eV. If (j;N) 2 W then j is within j???(M)j4 of
?. If N 2 eV then ?(N) > ? + j???(M)j2 > ? + j???(M)j4 > j. So, for any (j;N) 2 W,
f(j;N) = N. We already know N 2 eV which is contained in V. Therefore f(j;N) 2
V. Thus f is continuous if ?(M) > ?.
Case 2. ?(M) = ?
Let f(?;M) 2 V where V is an open set in C(S). We can again reflne V into a
tube-like eV with associated ?, such that M 2 eV. Also we can choose ? small enough
so that M + 2? 2 V and if N 2 eV then the endpoints of N are within ?2 of the
endpoints of M. (Recall: \within ?" and M + 2? are deflned in the notation section
at the beginning of this proof).
Let N 2 eV. If N0 is the subcontinuum made by increasing N equidistance at
each end then by the deflnition of ? and eV if len(N0) < len(M +2?) then N0 2 V:
There exist an fi 2R such that if N 2 eV then
? ?fi < ?(N) < ? +fi < ?(M +2?).
Let W = (? ?fi;? +fi)? eV.
Claim: f(W) ? V. Let (j;N) 2 W. We have two cases
Case 2.1. ?(N) ? j. If ?(N) ? j then f(j;N) = N 2 eV ? V.
Case 2.2. ?(N) < j. If ?(N) < j then f(j;N) = N0 where N0 is made by
increasing N on each end until ?(N0) = j. We know if N 2 eV and len(N0) <
len(M + 2?) then N0 2 V. Thus we just need to prove that len(N0) < len(M + 2?).
Now j 2 (??fi;?+fi) so we know that j = ?(N0) < ?(M +2?). Therefore len(N0) <
len(M +2?). Thus f is continuous if ?(M) = ?.
72
Case 3. ?(M) < ?. Now since ?(M) < ?, f(?;M) = M0 where ?(M0) = ?. Let V
be an open set in C(S) and f(?;M) = M0 2 V. Now assume that M =2 V. (If it were
we could reflne V using the lemma and ? ? len(M0)?len(M)4 ). Assume that V is tube-like
in S. Now there exists a such that for any continuum N 2 V then the endpoints of
N are within of the endpoints of M0. Note since M =2 V that the endpoints of M
are more than away from the endpoints of M0. Let O = (?(M0 ? 4);?(M0 + 4)).
Recall that M0+ 4 was made by adding 8 to each end of M0 and likewise M0? 4 was
made by subtracting 8 from each end. Now len is continuous so there exist an open
set U containing M such that len(U) ? (len(M)? 4;len(M)+ 4).
Assume that U is a tube. We know that V is also a tube. Let eU = UTV. This
makes the tube corresponding to eU have radius less than the tube corresponding to
V. (We do this because we will extend continua in eU and we want to make sure that
when we extend a continuum we stay inside the tube corresponding to V.) Notice
that if N 2 eU then len(N) is within 4 of len(M). Thus the endpoints of N are within
4 of the endpoints of M. Also note that the endpoints of M are at least away from
the endpoints of M0. Thus if N 2 eU then N =2 V which implies that ?(N) 2 O.
Let W = O ? eU. If (j;N) 2 W, then N 2 eU and ?(N) =2 O. Therefore for all
(j;N) 2 W, f(j;N) = N0 where len(N0) > len(N).
Let fi? 2 [0;1) be the unique number such that ?(M + fi?) = ?(M0) = ?. For
any j 2 O there exist a number fij such that ?(M +fij) = j.
73
For j 2 O, j = ?(M + fij) < ?(M0 + 4) = ?(M + fi? + 4). Thus fij < fi? + 4.
Similarly j = ?(M+fij) > ?(M0? 4) = ?(M+fi?? 4). Thus fij > fi?? 4). Therefore
for any j 2 O, fij is within 4 of fi?.
Given (j;N) 2 W, f(j;N) = N0. N0 is made by adding a certain length onto
each end of N. We have three cases.
Case 3.1. If the len(N) = len(M), then we add fij2 to each end of N.
Case 3.2. If len(N) < len(M), then add fij+jlen(M)?len(N)j2 to each end of N.
Case 3.3. If len(N) > len(M), then add fij?jlen(M)?len(N)j2 to each end of N.
Let aN and bN be the endpoints of N, aN0 and bN0 be the endpoints of N0, aM
and bM be the endpoints of M, and aM0 and bM0 be the endpoints of M0. We will
show in all three cases that aN0 is within of aM0, and bN0 is within of bM0. Thus
N0 2 V.
Case 3.1. len(N) = len(M).
Now aN is within 4 of aM so aN + fij2 is within 4 of aM + fij2 . aM + fij2 is within
4 of aM0. Thus aN +
fij
2 = aN0 is within
4 of aM0: Likewise bN +
fij
2 = bN0 is within
4
of bM0. Therefore N0 2 V.
Case 3.2. len(N) < len(M).
In this case in order to make N0 we will add fij+jlen(M)?len(N)j2 to each end of N.
Thus aN0 = aN + fij+jlen(M)?len(N)j2 . The endpoints of N are within 4 of the endpoints
of M so if we add fij+jlen(M)?len(N)j2 to each endpoint then aN + fij+jlen(M)?len(N)j2 is
within 4 of aM + fij2 + jlen(M)?len(N)j2 . If we show that aM + fij2 + jlen(M)?len(N)j2 is within
3
4 of aM0 then we will know that aN0 is within of aM0. Thus N
0 2 V.
74
Recall that
1. jlen(M)?len(N)j < 4.
2. fi? ? 4 < fij < fi? + 4.
3. aM + fi?2 = aM0.
Using these facts we have
aM + fij2 + jlen(M)?len(N)j2 < aM + fij2 + 8 < aM + fi?2 + 8 + 8 = aM0 + 4.
Similarly we have
aM + fij2 + jlen(M)?len(N)j2 > aM + fij2 +0 > aM + fi?2 ? 8 > aM0 ? 8.
ThereforeaM+fij2 +jlen(M)?len(N)j2 iswithin 4 ofaM0. ThusaN+fij2 +jlen(M)?len(N)j2 =
aN0 is within 2 of aM0. Thus N 2 V.
Case 3.3. len(N) > len(M).
In this case in order to make N0 we will add fij?jlen(M)?len(N)j2 to each end of N.
Thus aN0 = aN + fij?jlen(M)?len(N)j2 . The endpoint of N are within 4 of the endpoint
of M so if we add fij?jlen(M)?len(N)j2 to each endpoint then aN + fij?jlen(M)?len(N)j2 is
within 4 of aM + fij2 ?jlen(M)?len(N)j2 . If we show that aM + fij2 ?jlen(M)?len(N)j2 is within
3
4 of aM0 then we will know that aN0 is within of aM0. Thus N
0 2 V.
aM + fij2 ? jlen(M)?len(N)j2 > aM + fij2 ? 8 > aM + fi?2 ? 8 ? 8 = aM0 ? 4.
Similarly we have
aM + fij2 ? jlen(M)?len(N)j2 < aM + fij2 +0 < aM + fi?2 + 8 = aM0 ? 8.
ThereforeaM+fij2 ?jlen(M)?len(N)j2 iswithin 4 ofaM0. ThusaN+fij2 ?jlen(M)?len(N)j2 =
aN0 is within 2 of aM0. Thus N0 2 V. Thus if ?(M) < ?, then f is continuous.
75
We still need to show continuity if M is a proper subcontinuum and ? = 1.
If ? = 1 then f(?;M) = S. Let S 2 V where V is open in C(S). By deflnition
V = R(fVigni=1) for some open sets Vi in S. There exist an i such that Vi = S.
Also there exist a number r 2 [0;1) such that if ?(N) ? r then N 2 V. Since ? is
continuous there exists an open set U in C(S) so that
M 2 U, and ?(U) ? (?(M)? jr??(M)j2 ;?(M)+ jr??(M)j2 ). Let W = (r;1)?U. If
(j;N) 2 W then
Case 1. f(j;N) = N. Thus ?(N) = j > r which implies that N 2 V.
Case 2. f(j;N) = N0. Thus ?(N0) = j > r which implies that N0 2 V.
Therefore if M is a proper subcontinuum then f is continuous for all values in
[0;1].
Lastly we need to prove continuity if M = S. If M = S then f(?;S) = S for any
?.
Case 1. ? < 1.
Let f(?;S) = S 2 V, V is open in C(S). Since we know that one of the open
sets that make up V must be S, if N 2 V then N +? 2 V for any ?.
Let W = (?2;1) ? V. Given (j;N) 2 W, then N 2 V. If f(j;N) = N then
N 2 V. If f(j;N) = N0 then since N is in V so is N0.
Case 2. ? = 1.
Let f(?;S) = S 2 V, V is open in C(S). Let W = (r;1] ? V. If (j;N) 2 W;
then N 2 V. If f(j;N) = N then N 2 V. If f(j;N) = N0 then since N 2 V then
76
N0 2 V. Given f(j;N) = S, then S 2 V. Therefore we have proved continuity if
M = S. Since f is continuous in all cases then C(S) is contractible.
Let Q = fH 2 C(S)jH is a singleton point in Sg; then the previous contraction
for C(S) has the property that xp = yp if and only if x = y for every x and y in Q. If
we assume C(S) supports a Whitney map then Q is metric by Corollary 5.1, which
is a contradiction; thus C(S) can not support a Whitney map. This is an example
of a space supporting a Whitney map but whose hyperspace does not. This is also
an example of a space that is arcwise connected by metric arcs that does not support
a Whitney map thus proving that being arcwise connected by metric arcs is not a
su?cient condition for supporting a Whitney map in the non-metric case.
Theorem 5.5. Suppose X and Y are a continua and Y is nondegenerate. If X ?Y
supports a Whitney map then X is metric.
Proof. Let p be an arbitrary point in Y. Deflne F : X ? X ! R as F(x;y) =
?((x?Y)[(y?Y)[(X ?p))? min f?((x?Y)[(X ?p));(?((y?Y)[(X ?p))g.
We will show that each part of F is continuous and thus F is continuous. We will
then use the fact that if you have a continuous function F from X ?X into R such
that F(x;y) = 0 if and only if x = y, then X is metric.
Let f(x;y) = ?((x?Y)[(y?Y)[(X?p)). We want to show f is continuous. Let
(x;y) 2 X?X and let U 2Rbe an open set such that f(x;y) 2 U. We need an open
set eV 2 X?X such that f(eV) ? U. Since X?Y supports a Whitney map we know
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there exist a basic open set V 2 C(X?Y) such that (x?Y)[(y?Y)[(X?p) 2 V and
?(V) ? U. Now there exist open sets V1;V2;:::Vn ? X?Y such that V = R(fVjgnj=1).
For each (x;a) 2 x?Y there exist an open set Wxa ?Oa such that (x;a) 2 Wxa ?Oa
and Wxa ?Oa ? Vj for some j. Since x?Y is compact and x?Y ?Sa2X Wxa ?Oa then
there exist a flnite subcover of x?Y, namely x?Y ?Smi=1 Wxai ?Oai ?Snj=1 Vj. We
also want to ensure that for every Vj intersecting x?Y there exist a Wxai ?Oai lying
in Vj. If this does not happen we can just add in flnitely many open sets from the
collection fWxa ?Oaga2X so that above condition will be met; thus we can just assume
that the collection fWxai?Oaigmi=1 satisfles the fact that for every Vj intersecting x?Y
there exists Wxai ?Oai that lies in it.
Let fWx = Tmi=1 Wxai. If z 2 fWx, then z ? Y ? Smi=1 Wxai ? Oai ? Snj=1 Vj and
z ?Y intersects each Vj, j = 1 to n.
By a similar construction using y ? Y instead of x ? Y you can flnd a fWy so
that if w 2 fWy then w?Y ?Ski=1 Wybi ?Obi ?Snj=1 Vj and w?Y intersects each Vj
intersecting y ?Y, j = 1 to n.
Let eV = fWx?fWy. If (z;w) 2 eV then f(z;w) = ((z?Y)[(w?Y)[(X?p)) ?
Sn
j=1 Vj and f(z;w) intersects each Vj. Thus if f(z;w) 2 eV then (z?Y)[(w?Y)[
(X?p) lies in V. Therefore ?((z?Y)[(w?Y)[(X?p)) 2 U. Thus f is continuous.
Let g(x;y) = min f?((x?Y)[(X?p));(?((y?Y)[(X?p))g. Using a similar
procedure we can show that g is continuous. Thus F is continuous.
If F(x;y) = 0 then ?((x?Y)[(y ?Y)[(X ?p)) = min f?((x?Y)[(X ?
p));(?((y?Y)[(X?p))g. Withoutlossofgeneralityassumethat?((x?Y)[(X?p) =
78
min f?((x ?Y) [ (X ?p));(?((y ?Y) [ (X ?p))g. Now we know that since Y is
nondegenerate that if x 6= y then (x?Y)[(X ?p)((x?Y)[(y ?Y)[(X ?p);
thus by the Whitney property ?((x?Y)[(X?p)) < ?((x?Y)[(y?Y)[(X?p)),
which is a contradiction with ?((x?Y) [(y ?Y) [(X ?p)) = min f?((x?Y) [
(X ?p));(?((y ?Y)[(X ?p))g; thus x?Y = y ?Y, which implies that x = y.
If x = y then (x ? Y) [ (y ? Y) [ (X ? p) = (x ? Y) [ (X ? p) so then
?((x?Y)[(X ?p)) = ?((x?Y)[(y ?Y)[(X ?p)) = min f?((x?Y)[(X ?
p));(?((y ?Y)[(X ?p))g = ?((x?Y)[(X ?p)). Thus F(x;y) = 0.
Therefore X is metric.
Theorem 5.6. Let f be a function f : X ?! C(C(X)) such that f(x)= xX, which
is the point in C(C(X)) where xX represents an order arc in C(X) from fxg to X.
If f is continuous and C(X) supports a Whitney map then, X is metric.
Proof. DeflneG : X?X ?!RbyG(x;y) = ?((f(x)Sf(y))?minf?(f(x));?(f(y))g.
We will show that each part of G is continuous thus G is continuous. We will
then use the fact that if you have a continuous function G from X ?X into R such
that G(x;y) = 0 if and only if x = y, then X is metric.
Let F : X ?X ?!R be deflned as F(x;y) = ?(f(x)Sf(y)). We need to show
F is continuous. Let U be an open set inR, (x;y) 2 X?X and F(x;y) 2 U. By the
continuityof?thereexistsanopensetV inC(C(X))suchthat(f(x)Sf(y)) 2 V and
?(V) ? U. Now V = R(fVigni=1) where each Vi is an open set in C(X). There exists
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subcollections fV xj gmj=1 and fV yk glk=1 of fVigni=1 such that ?fV xj gmj=1SfV yk glk=1? =
fVigni=1,
xX ?Smj=1 V xj and intersects each Vj and
yX ?Slk=1 V yk and intersects each Vk.
Now let R(fV xj gmj=1) = V x this is an open set in C(C(X)) such that xX 2 V x,
and likewise V y is an open set that contains yX. By the continuity of f there exist
open sets A;B in X such that
x 2 A and f(A) ? V x and
y 2 B and f(B) ? V y.
A?B is an open set in X ?X.
Claim: F(A?B) ? U.
We need to show that if (a;b) 2 A?B then f(a)Sf(b) 2 V. Thus we need to
show that f(a)Sf(b) ? Sni=1 Vi and that f(a)Sf(b) intersects each Vi. We know
that f(a) 2 V x and f(b) 2 V y, thus f(a)Sf(b) ? Sni=1 Vi. Since f(a) 2 V x we
have that f(a) intersects each member of the collection fV xj gmj=1 and likewise f(b)
intersects each member of fV yk glk=1. Thus f(a)Sf(b) must intersect each member
of fVigni=1 since ?fV xj gmj=1SfV yk glk=1? = fVigni=1. Therefore f(a)Sf(b) 2 V which
implies that ?(f(a)Sf(b)) 2 U. Thus F is continuous.
By a similar argument if F(x;y) = minf?(f(x));?(f(y))g then F is continuous.
Therefore G is continuous.
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If G(x;y) = 0 then ?((f(x)Sf(y)) = minf?(f(x));?(f(y))g. But because of
the Whitney property and the fact that xX = yX if and only if x = y the only time
that will happen is when f(x) = f(y) which implies that x = y.
If x = y then ?((f(x)Sf(y)) = minf?(f(x));?(f(y))g, which implies that
G(x;y) = 0.
Therefore X is metric.
Theorem 5.7. If each X and Y support Whitney maps and XTY is a singleton
point, then XSY supports a Whitney map.
Proof. Suppose that K is a subcontinuum of XSY and X \Y = fzg. Let KX =
KTX and KY = KTY. Let ?X be the Whitney map on X and ?Y be the Whitney
map on Y. We will deflne the Whitney map on XSY as
?(K) = ?X(KX)+?Y (KY ):
First it is obvious that if H (K then ?(H) < ?(K). Since, in order for H to be
a proper subset of K, then either HX (KX, which implies that ?X(HX) < ?X(KX),
or HY (KY , which implies that ?Y (HY ) < ?Y (KY ).
In order to show ? is continuous, let U be an open set inRand let ?(K) 2 U; then
there exists an ? > 0 such that [?(K)??] 2 U. Now since both ?X and ?Y are Whitney
maps, for KX there exist an open set V X in C(X) such that ?X(V X) ? [?X(KX)? ?2]
and likewise for KY .
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Next we will reflne each V X and V Y . Given V X there exist an open reflnement
eV X = R(fWXl gml=1) such that
1. KX 2 eV X.
2. If z is the intersection point of XTY and z 2 KX then z 2 WXl for only one
l, assume that l = 1. Thus z 2 WX1 .
3. In the open set WX1 there exist another a point x 2 KX such that x is in the
interior of WX1 and x =2 WXl for l 6= 1.
4. z is not a limit point of any WXl for l > 1.
Notice that WXl for l > 1 is an open set not only in X but in XSY, but WX1 is
not an open set in XSY. This is important because that means that we can not use
WX1 to make an open set in C(XSY). To flx this problem we will make two open
sets in XSY using WX1 .
Make a new open set CX contained in X such that
1. CX ? WX1 ,
2. CX TKX 6= ;,
3. CX TWXl = ;, for any l > 1,
4. z =2 CX.
Notice that since CX does not contain the intersection point that means that CX
is an open set in XSY. Also since CX ? WX1 that if we add CX into the collection
of open sets that made up eV X we will just reflne eV X thus reflning V X even more.
Doing the same procedure will yield similar open sets in Y.
Now we will make our second open set using WX1 and WY1 .
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For z 2 (XTY)TK, let Dz = WX1 SWY1 . Note that since z 2 WX1 and z 2 WY1
then z 2 Dz. Also most important Dz is an open set in XSY.
Let
O = R?fWXl gml=2;fWYj grj=2;CX;CY ;Dz?:
O is open in C(XSY). Given H 2 O, let HX = HTX. H must intersect each
WXl and WYj for every l > 2 and j > 2 and CX;CY ;Dz, and be contained in their
union. Thus HX 2 V X. Note this was the reason CX was necessary so that HX
would be forced to intersect WX1 . Similarly HY 2 V Y .
Thus ?X(HX) 2 [?X(KX)??2] and?Y (HY ) 2 [?Y (KY )??2], so ?(H) 2 [?(K)??].
Therefore ? is continuous and thus is a Whitney map on XSY.
Corollary 5.1. There are continuum many nonhomeomorphic decomposable non-
metric continua that support Whitney maps.
Theorem 5.8. There exist X and Y, that both support Whitney maps, such that
XSY does not support a Whitney map.
Proof. Our example S was made by taking the cross product of Z with [0;1] and then
making the proper identiflcations. If instead of [0;1] we use [?1;0] then a similar non
metric space can be made that also supports a Whitney map. Denote this space as
S[?1;0].
The intersection of S and S[?1;0] has uncountably many points. Observe that
STS[?1;0] = Z ?f0g. SSS[?1;0] is a continuum.
Assume SSS[?1;0] supports a Whitney map.
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Let Z1=2 = f(z;t) 2 Sjt = 1=2g. Note that Z1=2 ?= Z and thus is non-metric.
Given a point x 2 S, x = (zx;tx) for some zx 2 Z;tx 2 [0;1]. Let xS = f(zx;t)jt ?
txg. Let F : Z1=2 ?Z1=2 !R be deflned as
F(x;y) = ?(xS [yS [S[?1;0])?minf?(xS [S[?1;0]);?(yS [S[?1;0])g:
Using a similar construction to the one in the proof of \If the cone over X supports
a Whitney map then X is metric" (where S[?1;0] behaves in a similar manner as the
top point of the cone), F can be shown to be continuous and it can be shown that
F(x;y) = 0 if and only if x = y. Thus Z1=2 is metric which is a contradiction.
Therefore the continuum SSS[?1;0] can not support a Whitney map.
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Chapter 6
Whitney levels
Let X be a continuum and ? a Whitney map onC(X). For t 2 [0;?(X)] and for
x 2 X let Wtx = fH 2 ??1(t)j x 2 Hg.
Theorem 6.1. Wtx is a subcontinuum of C(X).[10]
Theorem 6.2. If t > 0 then ??1(t) is a subcontinuum of C(X).
Proof. CLOSED: Let L be a continuum such that ?(L) 6= t. Then ?(L) = r where
r < t or t < r. Assume r < t
Let U = (r? t?r2 ;r+ t?r2 ). ?(L) 2 U. Since ? is continuous there exists an open
set V containing L such that ?(V) ? U. That implies that all continua that are in
V must have Whitney value in U. Thus UT??1(t) = ;. Therefore V is an open set
that contains L and misses ??1(t). Thus L is not a limit point of ??1(t).
CONNECTED: Assume ??1(t) in not connected and thus is the union of two
disjoint compact sets A and B. Then ??1(t) = ASB. Now ??1(t) = Sx2X Wtx.
Let A0 = fxjWtx ? Ag and B0 = fxjWtx ? Bg. Since Wtx is connected, A0 [B0 =
??1(t).
First: A0TB0 =?.
Assume not. Then A0TB0 6= ? so there exist a z such that Wtz ? A and
Wtz ? B. But Wtz is connected and thus can not be contained in and intersect two
disjoint separated sets, so A0TB0 =?.
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Second: A0 and B0 are separated. Let y 2 B0 . We want to show that y can not
be limit point of A0. Notice for each continuum Kfi in A, y =2 K ( because if so then
Wy would intercept A and B). Therefore for each Kfi in X such that Kfi 2 A there
exist open sets Ufi and Vfi in X such that Kfi ? Ufi;y 2 Vfi and UfiTVfi =?.
In C(X) deflne open sets U0fi = R(Ufi). Then A ? Sfi2C U0fi. A is compact
in C(X) so there exist flnitely many of these open sets that cover A, namely A ?
Sn
i=1 U
0
fii in C(X).
A0 ?Sni=1 Ufii. Look at the corresponding Vfii?s. Now y 2Tni=1 Vfii; this is open
and contains no points of A0. Thus y is not a limit point of A0.
Therefore A0 and B0 are separated sets but X = A0SB0 which is a contradiction
since X is connected. Thus ??1(t) is a subcontinuum of C(X).
Theorem 6.3. If X is hereditarily indecomposable then ??1(t) is hereditarily inde-
composable.
Proof. First we want to show that ??1(t) is indecomposable. Note that if H and K are
two subcontinua of X such that H;K 2 ??1(t), then since X is hereditarily indecom-
posable, HTK =?. Nowassumethat??1(t)isnotindecomposable; ??1(t) = ASB
where A and B are two proper subcontinua. Let A0 = fxjx 2 H;H 2 Ag and
B0 = fxjx 2 K;K 2 Bg. Now A0SB0 = X. Also A0 and B0 are proper subcon-
tinua since there exists H 2 AnB and K 2 B nA, and we know that HTK = ?.
But we have now shown that X is the union of two proper subcontinua, which is a
contradiction since X is indecomposable.
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Next let M be a proper subcontinuum of ??1(t), then deflne
M0 = fxjx 2 H;H 2 Mg. M0 is a subcontinuum of X and thus is indecompos-
able so the previous argument will show that M must be indecomposable.
Theorem 6.4. If X supports a Whitney map ? then the order arcs are metric.
Proof. We will use the same fact as in Theorem 5.1. Let O denote the order arc.
We want to flnd a continuous function f from O ?O into R such that f(H;K) = 0
if and only if H = K. Deflne f(H;K) = j?(H) ? ?(K)j. First, since O is an
order arc it is obvious that f(H;K) = 0 if and only if H = K, so we need to
show f is continuous. Let U be an open set such that f(H;K) 2 U; there exist an
? > 0 such that f(H;K) 2 [f(H;K) ? ?] ? U. Assume that ?(H) < ?(K). Let
? = minf?;?(K) ? ?(H)g. Since ? is continuous there exists an open set VH such
that
?(VH) ? [?(H)? ?4] and likewise there exists a VK such that
?(VK) ? [?(K)? ?4].
Deflne eV = VH?VK. If (R;S) 2 eV then ?(R) 2 [?(H)??4] and ?(S) 2 [?(K)??4].
Thus f(R;S) = j?(R) ? ?(S)j 2 [j?(K) ? ?(H)j ? ?2] ? [j?(K) ? ?(H)j ? ?] =
[f(H;K)??] ? U. Therefore f is continuous and thus O is metric.
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