C OMPOSITIO M ATHEMATICA
F. R ICHMAN
G. B ERG
H. C HENG
R. M INES
Constructive dimension theory
Compositio Mathematica, tome 33, n
o2 (1976), p. 161-177
<http://www.numdam.org/item?id=CM_1976__33_2_161_0>
© Foundation Compositio Mathematica, 1976, tous droits réservés.
L’accès aux archives de la revue « Compositio Mathematica » (http:
//http://www.compositio.nl/) implique l’accord avec les conditions géné- rales d’utilisation (http://www.numdam.org/conditions). Toute utilisation commerciale ou impression systématique est constitutive d’une infrac- tion pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.
Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
161
CONSTRUCTIVE DIMENSION THEORY
F.
Richman,
G.Berg,
H.Cheng
and R. Mines Noordhoff International PublishingPrinted in the Netherlands
Abstract
The uniform dimension
theory
oftotally
bounded metric spaces isdeveloped
from a constructivepoint
ofview, starting
from amodification of the
Lebesgue
definition ofcovering
dimension. Thesubspace
theorem and the(finite)
sum theorem areproved.
It isshown that dim X ~ n if and
only
if for every map from X into the(n
+1)-cell In+1,
everypoint
of I n+1 is unstable. Anexample
showsthat dim X ~ n is not
constructively equivalent
to the existence of amap from X into In with a stable value. However it is shown that these latter two
properties
areequivalent
if dim X S n. Theproduct
theorem is
proved by
a direct construction of a cover. Alsoproved
are the
(almost)
fixedpoint
theorem forI n,
theP flastersatz,
and theequivalence
of int X ~0
and dim X = n for closed subsets X of I n.1. Introduction
Since the
publication
ofBishop’s
book[3]
on constructiveanalysis,
there has been a resurgence of interest in the constructive
approach
to mathematics. Dimension
theory provides
an historicaltesting ground
for thisapproach.
In 1911Lebesgue [8]
introduced the notion ofcovering
dimension and gave a somewhatfaulty proof
that Euclidean n-space hascovering
dimension n. Two years later Brouwer[4]
gave a correctproof
ofLebesgue’s claim,
introduced the idea of inductivedimension,
andproved
that Euclidean n-space also has inductive dimension n. In 1926 Brouwer[5]
formulated the notion of inductive dimension in intuitionistic terms andproved,
in a con-structive way, that Euclidean n-space has inductive dimension n. The
purpose of this paper is to
develop
thetheory
ofcovering
dimensionfrom a constructive
point
of view.Berg et
al.[2] investigate
therelationships
between Brouwer’s constructivetheory
of inductive dimension and our constructivetheory
ofcovering
dimension.Compact
metric spaces form the naturalcategory
for a constructivetheory (e.g.,
Brouwer’skatalogisiertkompakte Spezies).
Inattempting
to constructivize
Lebesgue’s
definition ofcovering
dimension wewere forced to
adopt
a modification of this definition thatsidesteps
the
problems
ofconstructing
apoint
that is in n + 1 sets. Thisdefinition is
classically equivalent
to theoriginal
forcompact
spaces.However,
thecompleteness
of the space is irrelevant in most ap-plications.
Hence we state our definitions and theorems fortotally
bounded spaces,
although they
agree with the classical onesonly
forcompact
spaces. Our basic definition turns out to beclassically equivalent
to Isbell’s uniform dimension 8d as defined in[7;
p.64].
We assume that the reader is
acquainted
with the constructiveapproach
ofBishop [3].
All spaces will betotally
bounded metric spaces. A set A is located ifd(x, A)
= inf{d(x, a):
a EAI
exists(in
the extended
reals)
for each x. Note that the empty set is located andd(x, cP) == 00
for each x. A subset A is bilocated if both A and its metriccomplement ~ A ={x : d (x, A) > 0}
are located. Aslight
modification of the
proof
of[3;
Thm8,
p.101] ]
showsTHEOREM 0: Let
f
be a continuous real valuedfunction
on atotally
bounded metric space
X,
thenfor
all exceptcountably
many real numbers a, the setsare bilocated and are metric
complements of
each other.We denote
by Se(A),
thee-neighborhood {x : d(x, A) ~}
of A.By
Theorem 0given 0 03B1 03B2,
there exists03B5 ~ (03B1, 03B2)
so thatSE (x )
isbilocated and ~~
S~ (x)
=S~(x).
2.
Covering
dimensionThe central notion in the
development
ofcovering
dimension is thatof the order of a cover.
Classically,
the order of a cover is at least n if there is apoint
common to n sets of the cover. This definition posesproblems
from the constructivepoint
of view. For onething,
it isgenerally
difficult to tell if agiven point
is in agiven
set.Also,
whiletotal boundedness assures us of an
ample supply
ofpoints
that arenear
things,
we are often hard put to constructpoints
that are inthings,
even in acomplete
space. It is thus more natural to demandpoints
that arearbitrarily
close to n sets. In a compact space these notions areclassically equivalent.
Since we will bemeasuring
dis-tances to the sets in the cover we
require
these sets to be located. It will be convenient to have the notion of order relativized to asubspace.
DEFINITION: Let X be a
totally
bounded metric space, n a non-negative integer,
L a subset ofX,
and F a finitefamily
of locatedsubsets of X.
( 1 ) o(3W)
~ n on Lif,
for every 5 >0,
there is an x in L such thatd(x, F)
5 for n distinct elements F in 3W. If L = X we writesimply o (F) ~
n.(2) o(3W) S n
on L withseparation
5if,
for each x inL,
we haved(x, F) 2:: 5
for all but at most n elements F in F. If L = X wewrite
simply o (F) ~ n
withseparation
5.(3) o(F) ~ n (on L)
ifo(F) ~ n (on L)
withseparation 5
for some03B4>0.
The term
finite family
denotes whatBishop
calls asubfinite
set, thatis,
afamily
indexedby
the first kpositive integers
for some k. It isconvenient to allow a finite
family
to beempty.
Two elements Fi andF;
of a finitefamily
are said to be distinct if i4 j
eventhough possibly Fi
=F;.
This allows you to construct covers, and compute theirorders,
withouthaving
to worry whether two elements in the cover areequal,
an undecidablequestion
ingeneral.
Forsimplicity
we shallsuppress reference to the index set and use set
terminology
whendealing
with a finitefamily.
Forexample
we use the notations F G 3W and 3W ={F :
P(F)}
if the latter collection isnaturally
indexed.Classically
eithero(F) ~ n
oro (F) ~ n + 1. Constructively
this isnot so. For
example
let{fm}
be afugitive
sequence and a == L(fm/m),
let F =
{[-1, 0], [a, I]j
and n = 0.However,
thefollowing
weakerstatement holds.
THEOREM 1: Let X be a
totally
bounded metric space and F afinite family of
located subsetsof
X.If o(F) ~
n isimpossible,
theno(F) ~
n + 1.
PROOF: Let 8 > 0. Choose a
03B4-approximation
Y to X.If,
for each y in Y we haved(y, F) 2::
25 for all but at most n elements F ofF,
theno (F) ~ n,
which isimpossible.
Hence there is a y in Y such thatd (y, F) 303B4
for n + 1 distinct element F in F. Thus0 (F) ~ n + 1.
Since the order of a cover F of X is defined in terms of
points being
nearsubsets,
it is natural torequire only
that U F be dense inX,
rather thandemanding
U F = X. This is convenient sincequite
often UF = X in the classical sense but not in the constructive sense, while U F is dense
constructively.
Forexample X = [0,2]
and F ={[0,1], [1,2]}.
Classically,
one needonly
define what dim X :5 n means. Con-structively
we need also to define dim X ~ n. Thesecorrespond
toBrouwer’s oberer
Dimensionsgrad
and untererDimensionsgrad [5].
DEFINITION: Let X be a
totally
bounded metric space.(1)
A cover F of X is a finitefamily
of located subsets of X such that UF is dense in X.(2)
A cover 3W is an E-cover if diam F E for every F in f.(3)
dim X :5 nif,
for every E >0,
there is an ~-cover F of X such that0(F)~ n + 1.
(4)
dim X ~ n if there is an E > 0 such that if F is an E-cover ofX,
then0 (F) ~ n + 1.
(5)
dim X = n if dim X :5 n and dim X ~ n.Our
setting
is the category oftotally
bounded metric spaces anduniformly
continuous functions. Hence ahomeomorphism
comesequipped
with moduli of uniformcontinuity
in both directions. With this in mind it iseasily
seen that these notions of dimension areinvariant under
homeomorphism.
A space X need not have a dimen- sion even when dim X:5 1. To seethis,
let{fn}
be afugitive
sequence and let a == 1fn/n.
Consider the closed interval X =[0, a].
It iseasily
verified that dim X ~
1,
that dim X = 0 if a =0,
and that dim X = 1 if a ~ 0. However we cannot tell whether a is 0 or not, so we cannotassign
a dimension to X.Even when the classical dimension of the space is known to be n
we may be unable to
verify
that dim X = n. Theproblem
is thatdim X ~ n is not
merely
thenegation
of dim X n, but entails the construction of a number E. Anexample illustrating
this distinction is constructed as follows. Let{fn}
be afugitive
sequence and letK(n, N)
={j/2n:
1:5 j :5 2n-N}.
LetHn
=K(n, 0)
iffN
= 0 for all N n, and letHn
=K(n, N)
iffN
= 1 for N n. Let X be the closure of the union of the Hn. Then X is acompact
metric spacewhich, classically,
is either
[0, 1]
or[0, 1/2 N]
Ujj/2’:
1:5 j :5 2NI
for some N. Hence Xhas classical dimension
1,
but to show dim X ~ 1 you would need to find an E thatwould, essentially,
measure the size of the non-degenerate
interval contained in X. This E wouldprovide
unob-tainable information
about {fn}.
The
subspace
theorem takes two forms.THEOREM 2: Let X be a
totally
bounded metric space, L a located subsetof X,
and n apositive integer.
Then(1) If dim X ~ n - 1,
then dim L::5 n - 1.(2) If
dim L ~ n, then there is an E > 0 such thatif
F is an E-coverof X,
then0(F) ~ n
+ 1 on L. Hence dim X 2:: n.PROOF:
Suppose
we havepositive
numbers E and5,
and an E-cover F of X.By
Theorem 0 we canpick 0 ~ 5/2
such that{x
E L :d(x, F) ~}
=S~(F)
n L is located and has diameter less thanE for each F in F. Then
FL
={S~(F)
~ L: F~ F}
is an E-cover of L.If
0 (F) ~ n
withseparation 5,
then0 (FL) ~ n
withseparation 3/2,
so(1)
is verified. Ifdim L ~ n,
choose E so that0 (FL) ~ n + 1
for anyE-cover
$L
of L. Then there is an x in L that comes within8/2
ofn + 1 distinct elements of
0,,
and hence within 3 of n + 1 distinct elements of F. Since this is true for any5, part (2)
follows.If L is a dense subset of
X,
then any ~-cover F of L is an E-coverof
X,
so0(F) ~ n
+ 1 on L if andonly
if0(F) ~ n
+ 1 onX,
and0(F) ~ n + 1
on L if andonly
if0 (F) ~ n + 1
on X.Hence,
if dim L S n, then dim X n, and if dim X n, then dim L n. Thus dense subsets have the same dimensionproperties
as theircontaining
spaces. This is because we are
dealing
with a dimensiontheory
that isinvariant
only
under uniformequivalence, (see [7;
cor.23,
p.66]).
It is a classical theorem that if a space is the union of
countably
many closed
subspaces,
then its dimension is the supremum of the dimensions of thosesubspaces. Thus,
forexample,
the rationalpoints
in the square form a space of classical dimension zero. One
might hope
toget
a constructive version of this theoremby restricting
attention to
compact
spaces for which thetopological
and the uniform theories areclassically equivalent. However,
such a theoremmight
prove to have very little interest since it is
probably impossible
toconstruct a
compact
space that is a union ofcountably
many closedsubspaces
in aninteresting
way. Toappreciate
the difficulties in-volved,
consider the spaceconsisting
of thepoints 0, 1, 1/2, 1/3,....
This space is
certainly
the union ofcountably
many closedsubspaces,
but it is not
constructively compact
because it is notconstructively
complete.
As in the uniform
theory [7;
cor.8,
p.80],
we canget
the sum theorem for finite sums.Constructively
we mustdistinguish
twoclassically equivalent
statements in this theorem.THEOREM 3: Let
{A,B}
be a coverof
thetotally
bounded metric space X. Let dim A :5 n. Then(1) if
dimB :5 n,
then dim X :5 n, and(2) if
dim X 2:: n +1,
then dim B 2:: n + 1.PROOF: Since dim A n we can
find,
for any E >0,
an(E/2)-cover
F of A such that
0(F) ~ n
+ 1 on X withseparation 8 ,E/4. Suppose
G is a
(03B4/ 16)-cover
of B such thato(cg):5 k
on X withseparation
n
03B4/ 16.
We shall construct an ~-cover E of X such that0(E) ~
max
(n
+1, k)
withseparation
17. Partition % intodisjoint
subfamilies11
andHF,
whereF ranges over F,
such that if G EHF,
thend(G, F) 3/2
while if GE 11,
thend(G, F)
>8/4
for all F in F. For each F in F letF
= F UUKF,
and letj7 == {F :
F EF}.
Ifd(x, F) 2:: 8
and G EKF,
thend(x, G) 2:: d(x, F) - d(G, F) -
diam(G),
sod(x, F) 2::
8 -
8/2 - 03B4/ 16 ~
~. Hence0 (F) ~ n
+ 1 withseparation
n.Clearly E = F ~ G
is an E-cover of X. We shall show thato(6) s
max
(n
+1, k)
withseparation
~. For x inX,
eitherd(x, G)
> 17 for all G in11,
ord(x, G) 2~
for some G in11.
In the first case the result follows from the fact that0(F)~ n + 1
withseparation
17. In the second cased(x, F) > 03B4/4 - 03B4/16 - 2~ > ~
for each F in Y. Thus ifd (x, E)
17 for E in6,
then either EE 11,
or E =F
andd (E, G)
17 for some G inKF.
For each G in W lett(G)
= G if G~ G,
andt(G) = F
if G EKF.
Since0(G) ~ k
withseparation
n, we may choosea
subfamily J
of W ofcardinality k
such thatd(x, G) ~~
for G inGBJ.
It follows that if E EEBt(J),
thend(x, E)
> 17. Henceo(e)
~ k withseparation
n.Statement
(1)
now followsby choosing
E > 0arbitrarily
andusing
dim B ~ n to
get k
= n + 1. To prove statement(2)
choose E > 0 sothat if ce is an E-cover of
X,
then0(C)
~ n + 2. Then theassumption
that
o(G) ~ n
+ 1 isabsurd,
soo(G)
n + 2by
Theorem 1. But W canbe any
(03B4/16)-cover
of B. Hence dim B 2:: n + 1.Note that we did not prove that if dim X n +
1,
then dim An + 1 or dim B 2:: n + 1.
Indeed,
we cannothope
to show this con-structively.
To seethis,
letfail
be an enumeration of the rational numbers in[0, 1]
and{bj}
an enumeration of the rational numbers in[1, 2].
Let{fn}
be afugitive
sequence and definefx,l
as follows. Letx2;
= bi
iff2n
= 0 for aIl n :5j,
otherwise let x2; = 0. Let X2i-1 = ai iff2n-l
= 0 for all n ~i,
otherwise let X2i-1 = 0. Let X be the closure of the set{xk}.
Then X is acompact
metric space. Let A = X n[0, 1]
andB = X rl
[1, 2].
It iseasily
verified that dim X ~1,
but we have noway of
knowing
whether dim A 2:: 1 or whether dim B 2:: 1.3. Stable values
Let X and Y be
totally
bounded metric spaces andf
auniformly
continuous function from X to Y. Apoint a
in Y is said to be a stable value off
if there is an E > 0 such that if g is auniformly
continuousfunction from X to
Y,
and~ f2014 gll
E, thend(a, g(X))
= 0. Apoint a
in Y is an unstable value of
f
if for every E >0,
there is auniformly
continuous function g from X to Y such that~f 2014 gll
E andd(a, g(X))
> 0. In thesequel
I =[0,1].
THEOREM 4:
If X is a totally
bounded metric space andf :
X ~ I" isa
uniformly
continuousfunction,
then the setof
stable valuesof f
isopen.
PROOF:
Suppose a
is a stable value off,
so that ifIIf - gll
E, thend(a,g (X )) = 0.
Note thatS~ (a) ~ In.
We shall show thatif |b - a | E/2,
then b is a stable value off. Suppose ~f 2014 gli E/2.
We want toshow that
d(b, g(X))
= 0.Let 0
be ahomeomorphism
of In such that~ (b)
= a and~ id 2014~~ ,E/2,
where id is theidentity
map. Then~f - ~o gll
E sod(a, ~(g(X)))
= 0. Thus forany 8
> 0 there is an x in X suchthat la - ~(g(X)) 03C9(03B4),
where lù is a modulus of con-tinuity
for~-1.
Hencelb - g(x)1 8,
and we are done.We need here a few
elementary
facts aboutgeneral position.
Thesefacts will also
play
aprominent
role in theproof
of theimbedding
theorem in section 6. The constructive
point
of viewrequires
aslight
modification of the usual definitions and
proof s.
If 03C51, . . ., v, areelements of R n we say
they
arelinearly independent
if for each E > 0we can find a 8 > 0 such
that Y- 1 ri
Ewhenever ~ |ri03BDi|
5. It is notdifficult to show that v1, ..., vk are
linearly independent
if andonly
ifthere is an invertible n x n matrix T such that
Tvi
= ei for 1 ~ i ::5k, where ei
is the standard i th basis vector of Rn. Elements vo, v1, . . ., vk of R n areaffine independent
if VI - Vo, ..., Vk - vo arelinearly
in-dependent.
A finite subset V of R n is ingeneral position
if every finite subset of V ofcardinality
notexceeding n
+ 1 is affineindependent.
If V is in
general position
inRn
then there is a 5 > 0 such that forany two
disjoint
finite subsets S and T ofV,
if card S + card T ~ n +1,
then the distance between the convex hulls of S and T is at least 5.Finally,
if xl, ..., Xm are anypoints
inRn,
and E >0,
then there existpoints
y1, . . ., ym ingeneral position
in Rn such that xl = y, andlxi - yil
e for2:5 j
:5 m. This can beproved by
induction on musing
the characterization of linearindependence
stated in thepreceding paragraph.
THEOREM 5: Let X be a
totally
bounded metric space andf
auniformly
continuousfunction
X to In+1. Then(1) if dim X ~ n,
then eachpoint
in In+1 is an unstable valueof f,
and
(2) if f
has a stablevalue,
then dim X ~ n + 1.PROOF: Let be a modulus of
continuity
forf.
Fix apoint a
inIn+1,
a number e >0,
and anw(e/3)-cover f¥
of X. Choose 8 such that 0 03B4w(e/3).
For each F inf¥,
choose apoint b,
in In+1 so thatd(bF, f(F)) e/3
and thebF, together
with a, are ingeneral position.
Let
AF(X)
= sup(0,
1 -d(x, F)IS). Define g by
Since F is a cover of
X,
the denominator 1Àp(x)
is at least 1. Note that ifd (x, F) ~ 03B4,
then03BBF (x ) = 0,
whereas ifd (x, F) úJ(E/3),
thenlb, - f(x)1
E. HenceIif
-gll
E.To prove statement
(1),
supposeo (F) ~ n
+ 1. Then we can choose8 such that for every x in
X,
there is asubset
of F ofcardinality
n + 1 so that
d(x, F) ±
8 if F is inFBJ.
Since a and thebp
are ingeneral position,
and since no more than n + 1 of the numbersÀp(x)
are different from
0,
we haved(a, g(X))
> 0. To prove statement(2),
suppose
d(a, g(X))
= 0 for aIl 8(note
that gdepends
on8).
Theno(F) ~ n
+ 1 isabsurd,
and henceo(F) ~ n
+ 2by
Theorem 1.The converse of the first
part
of Theorem 5 isproved
with the aid of an intermediate notion that isclassically equivalent
to[7;
Th.24,
p.87], and,
forcompact
spaces, to[6; Prop. C,
p.35].
We say that B isan
E-enlargement
of A if A C B andd(b, A)
E for each b in B. If K and L are any subsets of a metric space, thend(K, L)
> 0 means that there is an n > 0 such that if x E K and y E L thend(x, y)~
TI.DEFINITION: Let X be a
totally
bounded metric space.(1)
Xsatisfies
conditionDn
if for any located subsetsAo, ..., An
ofX,
andE > 0,
there exist bilocatedE-enlargements Bo,..., Bn,
and 8 >0,
such thatd(~ S03B4 (Bi ), ~S03B4(~ BJ)
> 0.(2)
Xsatisfies
conditionDn
if there exist located subsetsAo, ..., An
ofX,
and E >0,
such that for any bilocatedE-enlargements Bo,
..., Bn we haved(~S03B4(Bi) ~ S03B4(~ Bi))
= 0 for all 8 > 0.We shall show that if each
point
in I"+1 is an unstable value of each mapf : X ~ In+1,
then X satisfiesDn (Theorem 6),
and if X satisfiesDn
then dim X:5 n(Theorem 7). Together
with Theorem 5 this will show that these three conditions areequivalent.
ConditionDn provides
acharacterization of dim X * n that is new even from the classical
point
of view. The situation is notquite
sosimple
for dim X ~ n + 1.We have shown that if there is a map
f : X ~ In+1
with a stablevalue,
then dim X ~ n + 1(Theorem 5).
We shall show(Theorem 6)
that if X satisfiesDn
then such a mapf
exists. We shall also show(Theorem 8)
that if such a mapf exists,
then X satisfiesD’n.
However anexample
shows that we cannot show
(constructively)
that if dim X > n then X satisfiesDn.
In the next section we willdevelop
moremachinery (Theorem 9)
that will enable us to show that if dim X = n + 1 then X satisfiesDn.
THEOREM 6:
If
eachpoint
a in In+1 is an unstable valueof
eachuniformly
continuousfunction f :
X -In+1,
then Xsatisfies
condition Dn.If
Xsatisfies
conditionD’n
then there is auniformly
continuousfunction f:
X ~ In+l with a stable value a.PROOF: Let
Ao,..., An
be located subsets ofX,
and 1 > ~ > 0.Define f: X ~ In+1by fi(x) = d(x, Ai ) 039B
1.Let g : X ~ In+1
be such that~f - gll ~/4.
Let a be apoint
in In+1 such thatE/2
ai3E/4
for1 ~ i ~ n + 1. To prove the first statement,
choose g
so thatd(a, g(X))
> 0. Nowadjust a slightly
so that Bi =gal ([0, ai])
is bilo- cated for 1 ~ i :5 n + 1. The{Bi}
show that X satisfies condition Dn with 8 =03C9(d(a, g(X))/2),
where l1) is a modulus ofcontinuity
for g.To prove the second statement, note that
d(a, g(X))
> 0 isabsurd,
for itimplies
that X satisfies condition Dn. Henced(a, g(X))
=0,
so a is astable value of
f.
We say that the bilocated sets
Bo,...,
Bn are wellplaced
ifd(~ S03B4(Bi), ~S03B4(~ Bi))
> 0 for some 8 > 0.LEMMA 1:
If Bo,...,
Bn are wellplaced,
then there is an a > 0 such that any bilocateda-enlargements Eo,...,
En are also wellplaced.
PROOF:
Suppose d(~S03B4 (Bi), ~S03B4)(~ B,»
> 0 for 03B4 > 0. Let a =S/2,
and let
Eo,..., En
be bilocateda-enlargements
ofBo,...,
Bn. ThenS03B1(Ei)
CS03B4 (Bi )
and ---Ei
C -Bi,
whence the lemma follows.LEMMA 2:
If
Xsatisfies
condition Dn and L is a located subsetof X,
then Lsatisfies
conditionDn.
PROOF:
Suppose
E > 0 andAo,..., An
are located subsets of L.Then there exist well
placed,
bilocated(E /2)-enlargements Bo,..., Bn
ofAo,..., An
in X.By
Lemma 1 we can choose a in the interval(0, E/2)
so thatS03B1 (B0),
...,S03B1 (Bn)
are wellplaced,
andCi
=S03B1(Bi)
~ Lis bilocated in L for 0 ~ i ~ n. Then
Co,..., Cn
are wellplaced,
bilocated
e-enlargements
ofAo, ..., An
in L.THEOREM 7:
If
Xsatisfies
conditionDn,
then dim X ~ n.PROOF: We shall
show,
for fixed E > 0 and allpositive integers
m, that if X satisfiesDn,
and X has an e-cover ofcardinality
m, then X has an ~-cover F such that0 (F) ~ n + 1.
Ifm ~ n + 1,
there is noproblem. Suppose
the conclusion is true for spaceshaving E-covers
ofcardinality
notexceeding
m, and letCo,..., Cm
be an E-cover of X.Then, by
Lemma2,
L =CI
U ... UCrn
has an e-cover 6 such thato(E) ~ n + 1.
If card(E)~
n we are done. If not,modif y Z by successively enlarging n
+ 1tuples
of elements of 6 so that E is bilocated for each E in 6 and such that every n + 1tuple
of sets of 6is well
placed, keeping 0(E) ~ n
+ 1. Choose 6 > 0 such thato(6) S
n + 1 with
separation 5,
and such that ifEo,..., En
arein 6,
thend(~S03B4(Ei), ~S03B4(~Ei)) > 0.
Let F =~E~ES03B4/2(~E)~S03B4/2(Co).
We candecrease 6 so that F is located and diam F ,E. Let F = E U
{F}.
Tosee that F is a cover of
X,
note that if x EX,
then either(i)
xE F,
or(ii) d(x, - E)
> 0 for some E in6,
and henced(x, E)
=0,
or(iii) d (x, Co) > 0
and sod (x, L) = 0.
It isreadily
seen that0(F) ~ n + 1
with
separation 3/2.
THEOREM 8:
If f is a uniformly
continuousfunction from
X to I"+’with stable value a, then X
satisfies Dn.
PROOF: Since the set of stable values of
f
is open, we may choosea =
(a1, . . ., an+1)
so thatAi
=fj-1
1([0, ad)
is located for 1n
+ 1.Since a is a stable value of
f,
there is an Eo > 0 such thatif g :
X ~ In+1is
uniformly
continuous and~if - g~ ~o,
thend(a, g(X))
= 0. We canchoose Eo so that
Seo(a)
C In+l(in fact,
we cannot avoidit).
LetE =
03C9(~o/2),
where w is a modulus ofcontinuity
forf,
and letBj
beany bilocated
E-enlargement
ofAj
for 1:5 j :5 n
+ 1. Define h : X ~Inllby
hj(x)
=[(fj(x) - E./2)
v(a;
~t(x)
+d(x, Bj))]
A(t(x)
+Eo/2)
A 1.Note that
~ f - h~ ~ ~o/2.
Ifx E B;,
thenhj(x) ~ (f1(x) - ~o/2) v ai = a;
since
d (x,Aj ) 03C9(~o/2)
andf(y) ~ aj
for y inAj.
Ifx ~~Bj,
thend (x, Bj)
> 0 andfj(x)
’2: a;, sohj (x)
> a;. Define g : X ~ In+1by
gj(x) ==
0 v(hj (x)
+(co/3)(- 1
v(d(x, Bj)2014 d(x, ’- Bj)) 039B 1))
A 1.Clearly Ilg - hll:5 Eo/3,
soIif - gll
Eo. Given 0 03B41,
choose x in X such thatd(a, g (x)) ,EoS/6.
Ifd(x, - Bj)~ 5/2
for somej,
thend(x, Bj)
= 0 sogj(x) ~ aj - Eo8/6
which isimpossible.
On the otherhand,
ifd (x, Bj) ~ 8/2
for somej,
then x ~ ~Bj,
sogj(x)
> a; +Eos/6,
also
impossible.
Hence x ES’s(B;)
and x ES03B4(~ Bj)
for allj,
sod(~ S03B4(Bj), ~S03B4(~~ Bj))
= 0.We shall now exhibit a
space X
such that dim X ’2:1,
but we cannot assert that X satisfies conditionD’.
Let si, s2, ... be a sequence ofpoints
in12
such that for eachpositive integer k,
if N ±4B
then{s1, . . ., s,l
is a21-k-approximation
toI2.
Let pi, p2, . . . be anenumeration of the
polygonal paths
inI2,
with rationalendpoints,
thatjoin opposite
sides ofI2.
Let ani, an2, ... be a sequence ofpoints
in Pn such that for eachpositive integer k,
if N ~4B
then{an1,
...,anNI
is a2’-’-approximation
toPn.
Letlfn)
be afugitive
sequence.Define xi
to be si iffn
= 0 foraIl n :5 i,
and xa = ani iffn
= 1 for some(unique)
n ~ i.Let X be the set of all such xi. Then X is a located subset of
I2,
sincefor each
positive integer k,
ifN ~ 4k,
thenlx,,..., xN}
is a 21-k-approximation
to X.We shall show that dim X ’2: 1. In
fact,
we shall show thato(F) ~
2for any
1/2-cover F
of X.Suppose F
is a1/2-cover
of X. If0(F) ~
1 withseparation 8 1/2,
let A =i{x1, . . ., xN}
be a(5/3)-approximation
to X. Then A contains a
(03B4/3)-approximation Ao
to apolygonal path joining opposite
sides ofJ2. Hence,
for anypair
ofpoints a, b
inAo,
we can find a sequence of
points a
= ao, ai,..., an = b in Ao such thatd(ai-,, ai)
,b for 1 ~ i :5 n. Sinceo(fF) :5
1 withseparation 5,
we must haveAo C F
for some F in F. But diamAo > 1/2
and F is a1/2-cover.
This is
absurd,
so0(F) ~ 2 by
Theorem 1.On the other
hand,
there is nohope
ofshowing
that X satisfiesD’0.
To do so, we would have to exhibit a located subset A of X and a
number E >
0,
so that if B is any bilocatedE-enlargement
ofA,
thend(B, - B) = 0.
Inparticular,
we would have toproduce
an(E/2)-
approximation
T to A. Then we could find a 5 betweenE/2
and E such that 203B4 ~lx - y |for
all x and y inT,
and 8:;ilx - Si for
all x in T andall i. Let K = UxET
Si) (x)
in 12. It isreadily
seen that there arepaths Pn,
with narbitrarily large,
that are either bounded away fromK,
or bounded away from - K.Clearly
B = K ~ X is a bilocated E-enlargement
of A.However,
iffn
=1,
andPn
is bounded away from Kor ~
K,
thend(B, - B)
> 0.Hence, knowing
thatd(B, - B)
= 0gives
us information about the sequence
fn
that we cannotnecessarily
obtain
by examining
a finite number of its terms.Alexandrov gave a classical characterization of dimension for se-
parable
metric spaces in terms of extensions of continuous functions intospheres [6;
Theorem VI4].
As usual the constructive version of this theorem must take two forms. Theproofs
are modifications of theproof
in[6]
and are omitted.THEOREM A: Let X be a
totally
bounded space. Then dim X:5 nif
and
only if for
every located subset Cof X,
each mapf :
C - S" can be extended to X.THEOREM A’ : Let X be a
totally
bounded metric space. Then thefollowing
areequivalent:
(1)
There exists a located subset Cof X,
a mapf : C ~ Sn-1 ~ In,
and a
positive
number 8 such thatif
g : X ~ I n withIlf - gllc 8,
then
g(x)
is dense in I n.(2)
There exists a located subset Cof X,
a mapf :
C ~Sn-’,
and apositive
number 8 such thatif
g :X ~ Sn-1,
thenIlf - gllc
> 8.(3)
There is a mapof
X onto I" with a stable value. The con-structive Tietze extension theorem
[3;
page107]
isrequired
inthe
proofs of
these theorems.4.
Complementary
coversIt is a classical theorem that dim X ~ n - 1 if and
only
if X =¥1
U ... UYn
where dimYi
= 0 for1 ~ j ~
n. This is false in the uniformtheory (see
Theorem3).
At the level of E-covers,however,
there is a(constructive)
uniformanalog.
Given an E-cover of order n we can find an E-cover that is the union of n families of order 1.Moreover,
we can do this in such a way that the union of the two E-covers has order n + 1(see
Theorem9).
LEMMA 3:
If
F is an E-coverof X,
and Ek is a located subsetof
Xsuch that
0(F) ~
k onEk,
then there is a located subset Ek-1of
Ek such that0(F) ~ k 2014
1 onEk_1,
and afinite family Ck of
located subsetsof Ek, of
diameters less than E, such that0(Ck) ~
1 andCk
U{Ek_1}
is aco ver
of
Ek.PROOF: Choose 8 such that
k
onEk
withseparation S,
and F is an(E - 8)-cover
of X.Choose q
between6/4
and8/2
so thatC(1)
=f x
EEk : d(x, F)
TI for all F inJI
is bilocated in Ek for each finitesubset J
of F.Let ’Ck
=IC(j):
card(1)
=kl.
Note that ifC(J,)
and
C(j2)
are inCk,
andJ1 ~J2,
thend(C(J1), C(J2))
>8/4,
forotherwise there would exist an x in
C(Ji)
and a y inC(J2)
such thatd(x, y) ,6/2,
so x would be within 5 of every F inIl
UJ2,
whichcontains at least k + 1 sets. Thus
0(Ck) ~
1.Also,
if C E’6k,
then diam(C)
E.Now choose a in the interval
(0, ô18)
so that Ek_1 ={x
EEk : d(x, - C)
a for all C in(Ckl
is located.Then ’Ck
U{Ek-1}
is acover of Ek
since,
if x EEk,
then either x EEk-¡,
ord(x, --- C)
> 0 forsome C in
Ck,
in which cased(x, C)
= 0.Moreover, 0(J) ~ k 2014
1 on Ek-1 withseparation ôl 16.
Otherwise there would be an x in Ek-1 such thatd(x, F) 5/8
for all F in asubset 1
of F ofcardinality
k. Sincex ~ Ek-1,
there is a yin ~ C(J)
such thatd(x, y) 8/8.
But thend(y, F) 3/4
for all F inJ,
so y EC(j),
which is absurd.THEOREM 9:
If F is
an E-coverof
X such thato (F) ~
n, then thereis an e-cover C =
C1
U ...~ Cn of
X such thato(Ck) ~
1for
1 ~ k ~ nand
o(F ~C) ~ n+1.
PROOF:
By repeated application
of Lemma3,
we construct located subsets X = En ~En-1, ~· · ·~
Eo =0
and finite familiesck
of located subsets of diameters less than E, such thatk
onEk, lek
U{Ek-1}
is a cover for
Ek,
and0(Ck) ~ 1,
for 1 ~ k ~ n. Hence 16 =C1 U · · · U Cn
is an E-cover of X with0(Ck) ~ 1
for 1~ k ~ n. It remains to show thato(F
UC) ~ n
+ 1. Choose 6 such thato(F) ~ k
on Ek withseparation 36,
and0(Ck) ~
1 withseparation 5,
for 1 S k Sn. For any x E
X,
thereexists î ~ F
such thatd(x, F)
26 for all F inJ,
andd(x, F) 2: 8
for all F not in1. Let J
havecardinality j
+ 1.Then
d(x, Ej) ~ 8,
sinceo(F)~ j
onEj
withseparation
38. ButEj
contains every set C in C1 U ...
U ’6j, and,
sinceo(Ck) ~
1 withseparation 5,
we haved(x, C)
2: 8 for all but atmost n - j
sets C in C.Hence