R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
N. D. K ALAMIDAS
Chain conditions and continuous mappings on C
p(X )
Rendiconti del Seminario Matematico della Università di Padova, tome 87 (1992), p. 19-27
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Mappings
onCp(X).
N. D.
KALAMIDAS(*)
ABSTRACT - Let X, Y
Tychonoff
spaces and 8: ~Cp (Y)
a one-to-one, con- tinuous linearmapping.
We prove that if Y satisfies a certain kind of chain conditions (caliber, c.c.c. e.t.c.) so does X. As a consequence of this, weprove
10, 11 1
(Tregular)
cannot be embedded intoCp
(X), if X has r caliber.More
generally,
we prove that if X has T caliber then does not contain compactsubspaces
ofweight
r. It follows,subject
to GCH, that if B is a Ba- nach space and (B, w) has mi and W2 calibers then B isseparable. Finally
weprove that
Cp (X )
with Xdyadic
ofweight r
(of uncountablecofinality)
doesnot admit a
strictly positive
measure.All
topological
spaces are assumed to be infiniteTychonoff
spaces.In the notations and
terminology
leftunexplained below,
we follow[4].
The
symbols X, Y,
Zalways
denote spaces and thesymbols r, A
denoteinfinite cardinals. The
cofinality
of a cardinal r, denotedby cfr,
is theleast ordinal
{3,
such that r is the cardinal sum ofp
many cardinals each smaller than z. A cardinal r isregular
if r = cf z. Thesymbol
N standsfor the set of all
positive integers
and thesymbols k, 1,
m, n are usedonly
to denote members of N. Further d is thedensity, w
is theweight
and
1.1 I
is thecardinality.
Aspace X
satisfies r.c.c. if there is nofamily
Y c
(the
set of allnon-empty,
open subsets ofX)
ofpairwise
dis-joint
elements withlyl I
= r. We set c.c.c. for WI.C.C.. A space X has(r, A)
caliber(pre-caliber)
if for everyfamily
y c5*(X)
withI y I
= rthere is a
subfamily
y1 c Y with|y1| I
= A andn(Yl
iscentered).
Weset r caliber for
(r, r)
caliber. Aspace X
satisfiesproperty Kz
if forevery
family
withI
= r, there is asubfamily
y1 c y with[
= r with the 2-intersectionproperty.
It is well known that if X has (*) Indirizzo dell’A.:Department
of Mathematics,University
of Athens, GR-15781Panepistimiopolis,
Athens, Greece.20
z caliber then X has also cfr
caliber,
and so If F =..., is a
non-empty,
finitesubfamily
of thencal (F)
isthe
largest K
suchthat,
there is ,S c F withS ~ I
= x and IfJ c then
x(J)
=inf ~cal (F)/ ~ F ~ :
F cJ, finite}.
A space X satisfiesproperty (**) if
can be written in theform,
~’*(X)
=U ~,
withK(5g )
>0,
for all n =1, 2, ....
A space X admits astrictly pos it ive
mea-sure, if there is a
non-negative
Borelprobability
measure t4 on X with> 0 for all
non-empty
open U. For acompact
space,Kelley (see [4]), proved
thatproperty (**)
isequivalent
with the existence of astrictly positive
measure on the space. It is well known that if X admitsa
strictly positive
measure then it has(r, c~)
caliber for every cardinal rwith
[4].
If X is a space,
Cp (X )
is the space of all real-valued continuous func- tions on X with thetopology
ofpointwise
convergence. For differentpoints
x1, ..., xK in X andEl ,
...,EK non-empty,
open intervals ofR,
let
It is clear that
V(x1,
..., xK :E1,
9 ..., form a base ofCp (X).
It is wellknown that
Cp (X)
is a densesubspace
of R lxl the set of all real-valued functions on X with thetopology
ofpointwise
convergence. It follows from well knownproperties
ofRlxl,
that haspre-caliber
r, forevery cardinal r with and also satisfies
property (**).
In thecase of a
compact
space X withw(X) = 7
andArhangel’skii
and Tkacuk in
[3], proved
thatCp (X)
does not have cfr caliber and alsoby
a result of Tulcea[9],
it follows thatCw (X) (the
spaceC(X)
with theweak
topology)
does not admit astrictly positive
measure,although
itsatisfies
property (**).
For A c X
and f E Cp (X)
weset f I A
for the restrictionof f on A,
andsupp f = Ix
E X:01
for thesupport
off.
THEOREM 1. Let 3:
Cp (X)
~Cp (Y)
be a1-1,
continuousmapping.
Then we have the
following:
(b)
Let be cardinals with cregular,
c * h andWe suppose that Y has
(r, A)
caliber. Then X has(r, A)
caliber.PROOF.
(a)
We can suppose that8(0)
= 0. Let D be a dense subset of Y. Forevery y E D
and n EN,
it follows from thecontinuity
of a at0 E
Cp (X)
that there exist ...,pairwise
different elements ofX,
and~’~...,2~’~
open intervals in Rcontaining
suchthat
’
We claim that the set A =
U
D U is dense in X and
since
|A| |D|, (a)
follows.Indeed,
let then there exists= 0 for every y E D and n E N and so = 0 so
a(f)
= 0.This ~’is
contradiction since a is one-to-one.(b)
i71
Forevery i
r we choosefi
ECp (X) with fi #
0 andsupp f c Ui
and setVi
=lyE
Y:01.
From theregularity
of r, it follows that either exists rVi’s equal
elements or 7pairwise
different. In both cases, since Y has(r, ~) caliber,
it followsthat there exists ACT,
I
=À and yo EA 1.
Since cfA > wand 0 for
every i
E A it follows that either there exists~11
c A,I
= A and r, > 0 such that rl, forevery i E A,,
orA2 c A,
~ ~12 ~
- ~1 andr2 0
such r2 for every i EA2 .
We can sup- pose that we have the first. From thecontinuity
of a at 0 ECp (X)
thereexist xl , ..., xK
pairwise
different elements ofX,
andEi ,
...,EK
open intervals of Rcontaining
0 ER,
such thatThen
f ~ V(xl , ..., xx : E1, ..., EK )
for everyi E Aland
so..., n for every Now
(b)
followsimmediately.
REMARK. We note that the
(a)
of the above theorem follows also from well known results[7].
We also note that in the(b)
of the above theorem theassumption
that Y has(r, À)
caliber cannot be relaxed to havepre-caliber. Indeed,
sinceCp (X)
is containedhomeomorphically
into
Cp (Cp (Cp (X)))
andCp (Cp (X))
has 7precaliber
if ef 7 > (ù, however X ingeneral
does notsatisfy
c.c.c..COROLLARY 2. Let z be an uncountable
regular
cardinal. Then X has r caliber if andonly
ifCp (Cp (X))
has 7 caliber. Inparticular
thespace has not r caliber.
PROOF. From results
in [8],
follows thatCp (Cp (X))
= n e NU Pn,
where
Pn
is a continuousimage
of the space X n x If X hasr
caliber,
it follows that eachPn
has r caliber and so the space22
Cp (Cp (X)).
The «if»part
follows from the fact thatCp (X)
is containedhomeomorphically
intoCp (Cp (Cp (X)))
and Th.l(b).
COROLLARY 3. Let X be a space with r
caliber,
rregular. Let, also, di
E R 1"with di
=(dij)j T, aij
= 0 forand
= 1. Then there is no, one-to-one continuousmapping
from thecompact subspace
i
1"}
U~0~
of R~ intoCp (X)
and so there is no,one-to-one,
contin-uous
mapping from 10, 1}1"
intoCp(X).
PROOF. On the discrete space r, we consider the
family
1"}
and the set of continuous functions i1"}
andrepeat
the
argument
of Th.1 (b).
REMARK. In the case that X is
compact
the abovecorollary
followsalso from well known
arguments.
Indeed in the case that X iscompact
and has r
caliber,
if there exists a: i1"}
Uf 01
-Cp (X),
a one-to-one continuous
mapping
then1"} U (0))
would be acompact subspace
ofCp (X),
ofweight
r, contradiction(see [3]). Also,
if X is com-pact
and10, homeomorphically
then{0, 11’
would be Eber-lein
compact
and since it satisfies c.c.c., would be metrizable[6].
In connection with the above we prove the
following stronger
result.
THEOREM 4. Let X be a space and we suppose that there exists
some F c
Cp (X) compact
withw(F)
= r and Then X has not cf zcaliber.
PROOF. Let
1"}
bea 11 II-dense
subset ofC(F).
We claim that for every i r, there existf , gi
EF, fi ;d gi
and for allj
i. This followseasily
from Stone-Weierstrass Theorem.Since fi ;d gi
there
exist ri
E > 0 such that eitheror
Since without loss of
generality,
we can suppose that there exist T and r E > 0 such thatsuppose, if
possible,
that X has cf r caliber. Then there is a cofinal withB [
= cfr and In case thatthere are no cfr
pairwise
different elements this fol- lowsimmediately.
Otherwise this follows from theassumption
that Xhas cf z caliber. Let x E i E
B}. Then dx E C(F)
and so there existsio
T such thatIf i E B with i >
io
we have and sogi (x) I ~/2
contradiction,
sincef (x)
E( - 00, r)
and gi(x)
E(r
+ 8,+ (0).
COROLLARY 5.
(Arhangel’skii
andTkacuk, [3]).
Let X be a com-pact
space withw(X)
= r and elf 7 > c~. ThenCp (X)
does not have cfrcaliber.
PROOF. It follows from the fact that X is contained
homeomorphi- cally
intoCp (Cp (X))
and Th. 4.COROLLARY 6.
Suppose
that 2Wl = W2. Then we have the follow-ing :
(a)
If X has c~l and c~2calibers,
everycompact F c Cp (X)
ismetrizable.
(b)
If X iscompact
andCp (X)
has oil and W2 calibers then X is metrizable.PROOF.
(a)
We claim that F isseparable.
Ifnot,
there exists{fi: i w1} c F such that fj E {fi: i j}. Then w({fi: i w1}) 2w1 =
= W2 and so = c~1 or c~2 contradiction
by
Corol. 5. There- fore F isseparable
and sow(IJ £
2W :::; 2Wl = W2 - It follows as before from Corol. 5 that F is metrizable.(b)
It follows from the fact X qCp (Cp (X)) homeomorphically
and(a).
NOTE 1. I have recent information that Th. 4 follows also Corol. 5 and results in
[8].
NOTE 2. The
(b)
of the abovecorollary
hasalready
beenproved
in
[3].
NOTE 3. The above theorem is not valid if the
assumption
r caliberis relaxed to
pre-caliber. Indeed,
ingeneral X
qCp (Cp (X))
andCp (X)
has 7
pre-caliber
if cfr > or,although X
may haveweight
z.24
THEOREM 7
(GCH).
If B is a Banach space, such that the space(B, w)
has oil and (V2calibers,
then B isseparable.
PROOF. It is well known that
(,SB * , w * )
the unit ball of B* with thew*-topology
is containedhomeomorphically
intoCp (B, w),
and also that B is containedisometrically
intoC(,SB * , w * ).
The result follows from Corol. 6.For a set 1’ we set
~(II~r) _ It e R : ly: t(y) ~ 01
iscountable}
withthe relative
topology
inPROPOSITION 7. We assume that
Cp (X)
has oil caliber and there exists a:Cp (X)
~£(R~),
a1-1,
continuousmapping.
Then X is separa-ble.
PROOF. We claim that there exists a countable A c 1’ such that
8( f)(y)
= 0 for every y Er~A. Indeed,
ifnot,
there exists cui in r and fE, w1 inCp (X)
with8( fj )(yj )
> 0 forevery E
We may suppose that8~fj ) % r
for some r E R. SinceCp (X)
has oil caliber there existI B
oiland f E Cp (X)
such that8( f)
E B(r, + oo ))
E E B
contradiction,
because8(f)
It follows that the continuous remains
1-1,
and the result follows from Th.l(a).
In Corol. 2 we have that
Cp
has not rcaliber,
if r is an uncount-able
regular
cardinal. In connection with this we have thefollowing stronger
result.PROPOSITION 8. Let T be a cardinal with cfr > (v, then the space
Cp(R~),
does not have(z, c~)
caliber(and
so it does not admit astrictly positive measure).
PROOF. For
every i
T,let 8i E
Rrwith 8i
=(8ij)j
T’ and8ij
= 0 fori
~ j and 8ii
= 1. Then thefamily
does not contain an infinite
subfamily
withnon-empty
intersection. In-deed,
let A c r andf e fl V(8i, 8i + 1: (o,1 ), (2, 3)).
We set 0 =(0)j ~ .
i EA
We may assume that
(o,1).
We consider r > 0 with+
r)
n(0, 1)
= 0. From thecontinuity of f at
0 there existil ,
9 ...,ix
in r and
I1, ..., IK
open intervals ofR, containing
0 such thatIf
then 8i
Enil 1 (I1 )
rl ... and so(0, 1),
contradiction.COROLLARY 9. Let X be a
dyadic
space withw(X) =,r
and cfr > w. ThenCp (X)
does not have(r, to) caliber,
so it does not admit astrictly positive
measure.PROOF.
By
a result ofEfimov, [5],
it follows that10,
If werepeat
theproof
ofProp.
8 we can prove thatCp ~0,
does not have(T,W)
caliber. Now themapping
a: witha( f ) _
=
f
iscontinuous,
linear and onto. So ifCp (X)
had(T, w) caliber,
then
Cp (f 0, 1)~)
would have(T, c~) caliber,
contradiction.The above theorem
gives
apartial
answer to thefollowing.
PROBLEM. Is there a non-metrizable
compact
Hausdorff space such thatCp (X)
has astrictly positive
measure?D. Fremlin in note of 10 Oct. 1989
proved
that thisproblem
is con-nected with the
following
PROBLEM.
(A. Bellow).
Is there aprobability
space(Z,.~, v)
with a Y c20 (~) ( =
the space of real-valued measurable functions onZ)
such that Y iscompact
and non-metrizable in thetopology
ofpointwise
con-vergence and any
pair
of distinct members of Y differ on anon-negligi-
ble set?
THEOREM 10. Let 3:
Cp (X) -~ Cp (Y)
be a1-1, continuous,
linearmapping
and 7 be an uncountableregular
cardinal. Then we have thefollowing implications.
(a)
If Y has(r, to) caliber,
so does X.(b)
If Y admits astrictly positive
measure then X has(r, c~)
cal- iber and satisfiesproperty KT .
PROOF.
(a)
iT}
be afamily
ofnon-empty
open sets in X. For every i r wefind f
ECp (X)
withsupp f
cU, fi #
0 and setVi
==
~y
E Y:01.
From theregularity
of T, it follows that either there exist rVi’s equal elements,
or Tpairwise
different. In both casesthere exists A c r, infinite and Then there exist xl , ..., x,,
pairwise
different elements inX,
andE1,
...,E,,
open inter- vals of R eachcontaining
0 such thatWe claim that if
f ~ fxl, ..., xK} = 0
thenIndeed,
if0,
there is a ~ E R such that9(Af)(yo) = ( -1, 1), by
thelinearity
ofa,
butA/’eV(~i,...,~:Oi,...,0~),
contradic-tion.
26
(b)
If Y admits astrictly positive
measure u, then Y has(z, to)
cal- iber and so X has(~, to)
caliberby (a).
In the
following
we shall prove that X satisfiesproperty K-r; .
Wesuppose, if
possible,
that there exists afamily {Ui: i rl
of non-emp-ty,
open subsets of X which does not containsubfamily
of the same car-dinality
with the 2-intersectionproperty. Let f , Vi ,
i r as in(a).
Wecan suppose that
,u(T~2 ) ; ~
for all i r, for some 8 > 0.For every A cr, we set
~’A
= the has the2-int-property}.
The set
~A
isnon-empty by (a), partially
orderedby
inclusion and satis- fies theassumptions
of Zorn’s Lemma. LetBA
be maximal. ThenI
r. For every i EA B BA
thereexists ji
EBA with Ui n Uji
= 0.Then
"
From the
regularity
of T, thereexists ji
EBA ,
such that the setAl
=li
EA: ji
= hascardinality
T. Werepeat
the sameargument withAl
in
place
of A.Inductively
we findjI, j2 , - - -, 9 jn 9 ... pairwise
different elements ofz, such that
U~1
fl0,
for every 1 ;,-, m,1, m
=1, 2, ....
Now since=
1, 2, ...
it follows that there exists a infinite with 0. Now similararguments
as in(a)
lead to contradiction.1,E B Ji
COROLLARY 11. Let r be an uncountable
regular
cardinal. If X ad- mits astrictly positive
measure, there is no,1-1,
linear continuousmapping
from R" intoCp (X).
REMARK. Let
X,
Y becompact
spaces and a:Cp (X)
zCp (Y)
be aone-to-one,
continuous linearmapping,
then themapping
a:
(C(X), II il)-(C(Y), ~’ ~~)
is also continuous(see Arhangel’skii[2]).
However the existence of a
1-1,
continuous linearmapping
a:
(C(X)g 1111>
-~(C(Y), ’~ ~~)
does notimply
the existence of a1-1,
contin-uous
~: Cp (X)
-Cp (Y). By
Dixmier’s Theorem L 00[0, 1]
=C(Q)
whereS2 is a
compact extremelly
disconnected space. On the other hand L °°[0, 1]
isisomorphic
to Thespace 13
is notseparable,
so fromTh.
l(a)
there exists no aone-to-one,
continuous~:
-3
Cp @N).
REFERENCES
[1] A. V. ARHANGEL’SKII, Function spaces in the
topology of pointwise
conver-gence and compact sets,
Uspekhi
Mat. Nank., 39:5 (1984), pp. 11-50.[2] A. V. ARHANGEL’SKII, On linear
homeomorphisms of function
spaces, Soviet Math. Dokl., 25, No. 3 (1982).[3] A. V. ARHANGEL’SKII - V. V.
TKA010CUK,
Calibers andpoint-finite cellularity of the
spaceCp(X)
andquestions of S.
Gulko and M. Husek,Topology
and itsApplications,
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Cam-bridge
Tracts in Mathematics, Vol. 79,Cambridge University
Press, Cam-bridge
(1982).[5] B. A. EFIMOV,
Mappings
andembeddings of dyadic
spaces, Mat. Sb., 103 (1977), pp. 52-68.[6] H. ROSENTHAL, On
injective
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measures 03BC, Acta Mathematica, 124 (1970), pp. 205-248.
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TKA010CUK,
The spacesCp(X): Decomposition
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andcompleteness properties, Topology
and itsApplica-
tions, 22 (1986), pp. 241-253.[8] V. V.
TKA010CUK,
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~U iseverywhere
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Manoscritto pervenuto in Redazione il 4