C OMPOSITIO M ATHEMATICA
J. DE G ROOT
R. H. M C D OWELL
Autohomeomorphism groups of 0-dimensional spaces
Compositio Mathematica, tome 15 (1962-1964), p. 203-209
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0-dimensional Spaces
by
J. de Groot and R. H.
McDowell1)
If T is a
topological
space, we denoteby A (T )
the group of allhomeomorphisms
of T onto itself. In[2],
it was shown tlatgiven
anarbitrary
groupG,
one can find atopological
space T such that G andA (T)
areisomorphic;
infact,
such a T can befound among the
compact
connected Hausdorff spaces. Ingeneral,
no such T can be found among the spaces with a base of open - and - closed sets,
i.e.,
thé spaces T such that dim T = 0. Thepresent
paperinvestigates
thefollowing question.
What can besaid,
ingeneral,
about A(T)
if T is acompletely regular
Hausdorffspace and dim T = 0?
If a is any
cardinal >
1, we shall denoteby Sa
the restrictedpermutation
group on aobjects;
thatis,
the group of all thosepermutations
which involveonly finitely
manyobjects.
We willfind it convenient to let
So
denote the group of one element.27
C2
will denote the direct sumof Ml
groups of order two.Throughout
this paper,"space"
will be used to mean"completely regular
Hausdorffspace".
For any o-dimensional spaceT,
we. shall
show thatA (T)
must(1 )
consist of asingle
element(in
which case we say T is"rigid" ), (2)
contain asubgroup Sa
for some (1.;,or
(3)
contain asubgroup
of the formSa
+ ZC2.
This result is bestpossible,
in the sense that for any cardinal a, we can construct spaces whoseautohomeomorphism
group isprecisely Sa
orSa + 1 C2.
Weproduce examples
ofarbitrarily high weight,2)
but we leave open the
problem
ofconstructing compact. rigid
0-dimensional spaces of
arbitrarily high weight.
In
particular,
if T is dense initself, A (T) equals
the unit1) The second author is grateful to the Charles F. Kettering Foundation for its support during the preparation of this paper.
2) The weight of a space is m if there exists an open base of m and not less than m sets.
204
element or contains a
subgroup
ZC2.
On the otherhand,
onecan construct
compact
0-dimensional Hausdorff spacesH,
densein
itself,
for whichA (H)
= 1 orA (H) equals
the direct sum ofcontinuously
many groups of order two(in
the last case one takesthe
Cech-Stone compactification
of[2; §
5,example I]).
Some of the results of this paper were announced in
[3].
I.
A(T)
for 0-dimensionalSpaces
1.1. LEMMA. Let
{aei}
and{Yi}, i
eN(the
naturalnumbers)
besets of distinct isolated
points
in the space T suchthat,
for everyi C N, {x;}
and{Yi)’ i E J,
have identical boundaries inT ;
thenT a,dmits of
uncountably
many distinctautohomeomorphisms
oforder two.
PROOF. It is easy to see that the map
interchanging
xiand yt
for each i in
N,
andleaving
all otherpoints
of Tfixed,
is anautohomeomorphism;
the same isclearly
true for every subsetJ
ofN,
and there areuncountably
many such subsets.In what
follows,
we shall need thefollowing
well known(and easily proved)
result from grouptheory.
1.2. PROPOSITION. If G is a group in which all elements distinct from the
identity
have order two, then G can berepresented
asthe direct sum of
cyclic
groups of order two.1.3. THEOREM. Let T be a 0-dimensional
completely regular
Hausdorff space,
containing
oc isolatedpoints (a
may be0).
Then either
A ( T )
=Sa,
orA ( T )
contains asubgroup
of the formSa. + EC2-
PROOF.
A ( T ) clearly
contains asubgroup isomorphic
toSa,
since every one - one onto map
moving
a finite number of isolatedpoints,
andleaving
all otherpoints fixed,
is a homeomor-phism.
Thus we needonly
show that if T admits any auto-homeomorphism
which does more thanthis,
then T contains asubgroup isomorphic
toSa
+ 27C2.
Note first of all that if a >
Mo,
there is noproblem,
sinceSa
itself contains such a
subgroup.
So we assume oc:::;: No,
and wedistinguish
two cases.(1)
There is anautohomeomorphism 99
on T which moves anon-isolated
point
p. Then we can find an open - and - closedset U
containing
p such thatU n q;(U) = cp. If U
has no countablebase,
we can find morethan No
distinct open- and -closed subsets K CU,
andinterchanging
K andq;(K) gives
us an autohomeo-morphism
of order two. If U has a countablebase,
let D ={Xi}
be the set of all isolated
points
in U. If D isfinite,
thenM =
( U B D )
ucp(UBD)
isopen-and-closed,
dense initself, separable,
metrizable and0-dimensional,
and is therefore homeo-morphic
to a dense-in-itself subsetqf
the Cantor set. Since M is notrigid, A (M) (and
henceA (T»
contains asubgroup
of theform 27
C2, by [2;
p. 90,(i)].
If D is infinite andclosed,
let{Xi}
be any enumeration of
D;
then{Xi}
and{q:>(Xi)} satisfy
thehypotheses
of Lemma 1.1; if D is notclosed,
it has a limitpoint
q and a
subsequence {Yi} converging
to q. In that case,{y 2i-l}
and
{Y2i} satisfy
thehypotheses
of 1.1.(2)
Noautohomeomorphism
moves a non-isolatedpoint. Let q:>
be a
homeomorphism moving
an infinite set of isolatedpoints {Yi}.
If we can find a set of isolatedpoints {Xi}
such that{Xi}
n{q:>Xi}
=§,
then{Xi}
and{q:>Xi} clearly satisfy
thehypotheses
of 1.1. But such a set
{Xi}
iseasily found,
for if there is a y E{Yi}
with infinite
orbit,
let xi =q:>2iy;
if each yi has finiteorbit,
form
{Xi} by choosing
onepoint
from each of the orbits determinedby Yi.
It follows that
A(T)
contains a groupisomorphic to E C2;
from the
construction,
it iseasily
seen thatby dividing
the iso-lated
points
into twodisjoint
infinite sets if necessary, one can find asubgroup isomorphic
toSa,
+ 27C2.
It should be
pointed
out that inonly
one case in theproof
of1.3 do we fail to find
continuously
many distinct autohomeo-morphisms
of order two. We couldreplace £ C2
in the statementof the theorem
by
the direct sum ofcontinuously
many groups of order two if we could prove thefollowing:
if U and V are 0-dimensional, disjoint homeomorphic
spaceshaving
no countablebase,
and X = U uV,
then A(X )
contains c elements of order two.Il.
Rigid Spaces
In this
section,
we extend the methods of[2]
toproduce rigid
0-dimensional spaces of
arbitrary (infinite) weight.
We shallrequire
some ideas in thetheory
of uniform spaces; the reader is referred to[1]
and[4]
for adevelopment
of these ideas.First,
wc extend a metric space theorem to uniform spaces ina routine manner.
2.1. DEFINITION. An intersection- of m open sets will be called
a
G.8-set;
aG..,a-set
will becalled,
asusual,
aGa-set.
2.2. THEOREM. Let X be a
completely regular
Hausdorff space206
of
weight
m,complete
in auniformitv -9 generated by
a set Dof m
pseudometrics.
Then every continuous mapf
from a subsetH of X into X can be extended
continuously
to amap /
from aGm6-set
G D H into X.PROOF. For each
d c- C,
and each xe 1,
letlOd(ae)
be the oscil-lation of
f
at x withrespect
to d. LetGd
isevidently
aGa-set.
Letthen G D H is a
Gmct-set.
Now
f
can be extendedcontinuously
over G. For let{h0153}
beany net in H
converging
to apoint x
e G.Then,
in theuniformity generated by Ç), {f(h0153)} is
aCauchy
net,by
the definition of G.Hence
{f(x0153)}
converges to somepoint
p eX;
setf(x)
= p.f
isevidently
continuous at x.Now, using
2.2, we extend some of the results in[2].
2.3. DEFINITION. If X is a
topological
space, andf
a map froma. subset of X into
X,
thenf
is called a continuousdisplacement o f
0?’der M iff
iscontinuous,
and is adisplacement
of order M.A continuous
displacement
of order c will becalled,
asusual,
a continuous
displacement [2; § 2].
2.4. THEOREM. Let X be a
completely regular
Hausdorff spaceof
weight
m,complete
in auniformity e generated by
mpseudo- metrics,
and letIXI
= 2"’ = M.Further,
let{KI1}
be anyfamily
of M subsets of
X,
each of cardinal M. Then there is afamily
{F,y}
of 2M subsets of X such that(1)
Fory =1=- y’, IF yBF y’ 1 = M.
(2)
NoFy
admits of any continuousdisplacement
of order Monto itself or any other
F y’.
(3)
For everyp,
y,1 F^i
nKg = M,
and1 (XB FY)
nKpl
= M.PROOF. There exist
only
MGmct-sets
inX,
and a fixed subsetof X admits at most M continuous maps into
X,
and thereforeat most M continuous
displacements
of order M. Letf03B2
be acontinuous
displacement
of order M whose domain is aGg,a-set.
The
family {fp}
of all suchmappings
has cardinal at most M.This
family
isnon-empty (otherwise
the theorem istrivial),
soby counting
agiven displacement
M times if necessary, we mayassume that
1 {Ill})
= M.Now we
apply [2;
Lemma1],
with X =N,
M = m, andffgl.
We obtain a
family {FY}
of 2M subsets of Xsatisfying (1 )
and(3).
Suppose (2)
isfalse,
and there is a continuousdisplacement
oforder
M,
({J, fromFy
ontoFy.. This 99
can be extended(Theorem 2.4)
to a continuousmap e
of aG.,a-set Gly D
F intoX, so §5
=fp
for some
p. Hence, by [2;
Lemma 1,(2.3)],
for everypair
y,yi,
f pF "1 B F "1’ =1= cp,
and so,since p = tq
onFy, ({JF ’1 B FY, §, i.e.,
99 maps
F ’1
onto no member of{Fy}.
2.5. LEMMA. Let P be a space in which every open set has cardinal at least M. If 99 : P --> P is
non-trivial,
and is eitherlocally topologically
into P or continuous ontoP,
then ç is adisplacement
of order M.PROOF. The
proof
is word for word theproof
of[2;
Lemma2],
with
"K" replaced by "M",
and"continuous displacement"
replaced by
"continuousdisplacement
of order M".2.6. THEOREM. Let X be a
locally compact
Hausdorff space ofweight
m,complete
in auniformity generated by
mpseudo- metrics,
such that every open set in X has 2’npoints.
Let K bethe set of all
compact
subsets of X whose cardinal is 2m. Then the sets{Fy}
constructed in Theorem 2.4 are such that no{Fy}
can be
mapped topologically
into orcontinuously
onto itself orany other
Fy,.
PROOF. Each open set in each
Fy
will have 2mpoints. By
Lemma 2.5 and
(1),
Theorem 2.4, anynon -trivial 99 satisfying
either condition of the theorem is a continuous
displacement
of order M. But this contradicts
(2),
Theorem 2.4.2.7. EXAMPLE. Theorem 2.6 enables us to construct many
examples
ofrigid
0-dimensional spaces ofarbitrary weight.
Forinstance,
letwhere 1 AI == m, and,
for each a,Xa
is a discrète space of cardinaltwo. Then X has
weight m, X
iscompact,
and hencecomplete
in any
uniformity,
so X iscomplete
in auniformity generated by
m
pseudometrics. Further,
every open set in X contains 2mpoints. Now, applying
Theorem 2.6, weget
a collection of 221sets
{F a.}’
each ofweight
m and dimension 0, such thatFy
isrigid
foreach y,
and theF y
aretopologically
distinct.2.8. PROBLEM. The
rigid
spaces constructed in thepreceding
example
are proper dense subsets of acompact
space, hencethey
208
are not themselves
compact.
We have not been able to constructexamples
ofcompact, rigid
0-dimensional spaces ofarbitrarily high weight;
such spaces would be of interest in thestudy
ofBoolean
rings (see,
forexample, [2; § 8.1]).
III.
Spaces
whoseAutohomeomorphism Groups
areIf a is
finite,
the discrete space of cardinal cc hasSa
as itsautohomeomorphism
group. This is not the case for ainfinite,
of course. In
Example
3.1,however,
weproduce
for each infinitea a space
having
ce isolatedpoints
whoseautohomeomorphism
group is
precisely Sa.
InExample
3.2, we find spaces whose auto-homeomorphism
group is the direct sum ofS.
and the sum ofcontinuously
many groups of order two; this group is then iso-morphic
to S + ZC2
if we assume the continuumhypothesis.
In this connection one should recall the remark
following
theproof
of Theorem 1.3; it is conceivablethat Ml
can bereplaced by
cthroughout
this paper.In both 3.1 and 3.2, the spaces
SI’
whichplay
apart
in the construction canevidently
be chosen to havearbitrarily high weight,
hence the same is true for ourexamples.
3.1. EXAMPLE. Let P be a discrete space of cardinal a, and let
PP
be its(o-dimensional ) Cech-Stone compactification.
Witheach p E
P,
we associate a 0-dimensional spaceSI’
such that(1 )
foreach p
EP, Sp
isrigid
anddense-in-itself,
and
(2) if p and q
are distinct elements ofP,
then nonon-empty
open subset of
S1J
ishomeomorphic
to an open subset ofSa.
Such a collection
{S1J}
can be constructedby using Example
2.7,as follows: with each p e
P,
we associate a cardinal a1J such that ifp =1=
q, 211-- e 2eo.Taking
aD = m in 2.1, we obtain arigid
space which we can denote
by S1J
such that each open subset ofS1J
contains 2eppoints.
The collection{S1J}’
p E Pevidently
satisfies
(1)
and(2).
Now let
We
topologize X by prescribing
a base for the open sets,consisting
of
(i)
the sets{p}, p e P,
(ii)
theopen-and-closed
sets inSp
for each p eP,
(iii)
the setswhere U is
open-and-closed
in P.The space X so defined is
evidently
a 0-dimensionalcompletely regular
Hausdorff space. Thetopology
on eachSp
as asubspace
of X is the same as its
original topology.
Every mapping
of X onto X whichpermutes
a finite number of the(isolated) points
of P and leaves all otherpoints
of Xfixed,
isclearly
ahomeomorphism.
These are theonly
auto-homeomorphisms
of X. For if anautohomeomorphism p
leaveseach p
e Ppointwise fixed,
then thepoints
ofPP
arefixed,
somust be
mapped topologically
on itself. But from(1)
and(2),
this space is
rigid,
so 99 is theidentity
map. On the otherhand,
ifq
displaces
an infinite subset D ofP, then p
must move somepoint
ofPPBP (since
the closures of D and99(D)
inPP
arenon-empty
anddisjoint),
hence there is a p e P such thatSp
nq;S’J)
=cp.
But99S,
nPP
=§,
since no open set in5p
containsan isolated
point.
It follows thatq;S’J)
nS,, # §
for somep =1= p’, contradicting (2).
3.2. EXAMPLE. For each oc, we construct a space
Ta
such thatA
(T.)
isprecisely Sa,
+ ZC2 (assuming
the continuumhypothesis).
Let M be a 0-dimensional subset of the
real
numbers such thatA (M)
is the direct sum ofcontinuously
many groups of order two[2; §
5,Example I],
and let X be the space constructed inExample
3.1, so that A
(X)
=Sa.
LetTd
= X u M.If 99
is any autohomeo-morphism
ofT,
then x e M if andonly
ifrp(ae)
eM,
since x e Mif and
only
if the least cardinal of a base at x isKo.
It followsthat
A(Ta)
=A(X)
+A(M) = S.
+E C2.
REFERENCES L. GILUIAN and M. JERISON
[1] "Rings of Continuous Functions", New York 1960.
J. DE GROOT
[2] Groups represented by homeomorphism groups I, Math. Annalen 138, pp.
802014102 (1959).
J. DE GROOT
[3] On the homeomorphism group of a compact Hausdorff space, Amer. Math.
Soc. Notices 7 (1960), p. 70, Nr 5642014264.
J. L. KELLEY
[4] "General Topology", New York 1955.
(Oblatum 10-9-62).