Derived length for arbitrary topological spaces

(1)

Internat. J. Math. & Math. Sci. VOL. 15 NO. 2 (1992) 273-278

273

DERIVED LENGTH

FOR

ARBITRARY

TOPOLOGICAL

SPACES

A.J. JAYANTHAN

Schoolof Mathematics and

Computer/Information

Sciences Universityof

Hyderabad

Central University

P.O. Hyderabad

500 134 India

(Received February 8, 1988 and in revised form February 15, 1989)

ABSTRACT.

Thenotion ofderivedlengthisasold as that of ordinal numbers itself.

It

is also known as theCantor-Bendixonlength.

It

is definedonlyfordispersed

(that

is

scattered)

spaces.

In

this

paper

this notion hasbeenextendedin a naturalwayforalltopological

spaces

suchthat all its

pleasing

propertiesare retained.

In

this

process

wesolve aproblem posed by

V. Kannan.

([

1] Page

158)

KEY

WORDS

AND PHRASES:

Ordinal invariants,dispersed

spaces,

andderivedlength. 1991

AMS

SUBJECT

CLASSIFICATION CODES.

Primary54(205 Secondary54A25,54 G99.

0.

PRELIMINARIES.

Suppose

(X,x)

is a

topological space.

We

denotethesetof alllimit

points

ofasubset

A

of

X

by

A

1.

Thespace

X

isdenoted

by

X

.

As

inductionhypothesis,weassumethat

X

tisdefinedforall0

I

<t.

Now

wedefine

A

asfollows.

[I,X

ift is a limit ordinal

[(XI)

if t-

[

+ 1.

So

we obtained a chainX

X

D

X

=)...

A

:9

A

m

D

By

_standard_arguments_on_the_cardinality

of the

space

X,

onecan show that there exists an ordinal numbertsuchtharA

-A

m/

PROPOSITION

0.1-

For any

topological

space

X,

the followingareequivalent:

(i)

Thereexists anordinal numbertsuch that

A

m_is_empty.

(ii)

Every

nonemptyclosedsubspaceof

X

contains an isolatedpoint.

(iii)

Every

nonempty subspaceof

X

contains an isolatedpoint.

(iv)

X

does not contain a dense-in-itselfsubspace.

PROOF.

Omitted.

(See

[1]

and

[2])

DEFINITION

0.2:

A

topological

space

X

satisfying anyone of the aboveequivalentconditions is called adispersed

(or

lfdfl!.g.L) space.

From

statement

(i)

of Proposition 0.1,weobserve that for any dispersed

space

X,

there existsan ordinal numbertsuchthat

A

isempty.

We

define the derivedlength ofa

(dispersed) space

X

as

inf{a:

X

O}

anddenote it

by

d(X)

or

d(x)

wherexisthecorresponding

topology

considered on

X. [3,4].

(2)

THEOREM

0.3:

For

adispersedspace

X, d(X)

satisfied thefollowing properties.

(i)

For

asubspace

Y

of

X, d(Y)

d(X).

(ii)

d(X)

issmallerfor finertopologies.

(iii)

d(Y)

d(X)

if

Y

isacontinuousopen imageof

X.

(iv)

If

X

isacompact

space

and

Y

is a

Tt-space

such that

Y

isaclosed continuousimageof

X,

then

d(r)

d(X).

(v)

If

X

isthetopologicalsumofaclass

{X/:

tE

I}

of dispersed

topological spaces,

then

X

is dispersedand

d(X)

sup{d(X/):

EI}.

REMARK:

A

proofof this theorem can befound in

].

This isalso a

consequence

of later results of thispaper.

1.

d(X) FOR

ALL

SPACES.

Suppose

(X,)

is an arbitrary topological

space

and

A

is a subspace of

X.

A

chain,

A

A0

(=

A

C:...(=

A,

C...of subsets of

X

iscalled a K-chain on

X

withbaseset

A

if for allct,

Ao

\Aa

is contained in

A,\A

for all

15

<ct.

By

standardargumentsoncardinalityof

X,

wecan

prove

that there

eists

an ordinalnumberctsuch that

A,-

A,/

1.

We

call

inf{ct:

A,-A,/

x

the

length

ofthe chain.

For

notationalconvenience,hereafter we denote

any

K-chainA0

(=

A1

C:... C:

A

C:

Aa

C:...orsimply

by

withoutmentioningtheindexingsetorbase setexplicitly. The

length

of"the

chain

{A,}

isdenotedby

({A,).

We

definethelengthofatopological

space X

as

sup{l

({A,):

{A,}

isaK-chain in

X}

and denote

itas

(X).

(X)

canbe verifiedtobe atopologicalinvariant.

Now,

wefirst note that satisfiesall those propertiesof derivedlength giveninTheorem 0.3.

THEOREM

1.1:1

(X)

satisfiesall fiveproperties

(i)

to

(v)

ofTheorem 0.3.

PROOF.

Since theproofsof statements

(i), (ii)

and

(v)

areroutine,weomit them.

We prove

statements

(iii)

and

(iv)

here.

(iii)

Suppose

]’is

anopencontinuousmapfrom

X

onto

Y. To prove

(Y)

(X),

weconsider an arbitraryK-chain

{B,}

in

Y.

Define

A,-

]’-(B,).

CLAIM:

{A,}

is also a K-chainand

({A,})

({B,}).

For,

supposexis apointinA,/

\A

and

Ux

isanopen neighborhoodof x. Then,

f(Ux)

isanopen neighborhood

of]’(x)

tE

B,/

\B,.

Since

{B,}

is aK-chain,

]’(Ux) meetsB,,\Bl

for all

I

<a.

Hence

Ux

meets

A,\AI

forall

I

<a.

Thus,

wehave

proved

that

{A,}

is a K-chain.Alsoit is obvious

thatA,

\A,

is

nonempty

for allctsuchthat

B,/

I\B,

isnonempty. This establishes the claim that

({A,})

({B,}).

So

we have

(Y)

sup{/({B,})

{B,}

isa K-chain in

Y}

sup{/({]’-(B)}

{B,}

isa K-chain in

Y}

sup{/({A,})

{A,}

is a K-chain in

X}

(X).

(iv)

Let

X

be acompact

space,

Y

be a

T

space

and

]’be

aclosed continuous

map

from

X

onto

Y.

We

show

(Y)

$1

(X).

As

in

Off),

we start with a K-chain

{B,}

on

Y.

A

K-chain

{A,}

in

X

is definedasfollows:

Ao

f-(Bo)

a,U?a

ifct is a limit ordinal

(3)

DERIVED LENGTH FOR ARBITRARY TOPOLOGICAL SPACES 275

CLAIM:

{A,}

isaK-chain and

({A,})

({B,}).

Obviously A,/

I\A,

iscontainedinA,\A for all < bydefinition.

Hen,

it is

enou

toshow

at

A+ tA

isnonempty forall such

atB+

B=

is

nonemp.

We

prove

ising

(transfinite)

induction.

But

f(A.

A

is confined in

Bo+

tB

bydefinition.

,

it isenoughtoshow

at

B.

tB

is contained in

A.

A). Suppose

-

0 and

y

belongsto

BilBo

Co-

f0)C

) cause

f

is

closed).

en

f-)

Xo

isnonempty.

Hence,

ere

exisma

int

x

in0

f-(BBo)

such

at

f(x)

y.

Now,

aume that

f(Aa)

ntains

Barb

for all

y

< <

Oy induction).

CE

1: + 1

(

isnot a limit

ordinal).

Suppose

y belongs

to

B B.

Since

B

is contained

(BB)

C

AA)

and hence is

contained in

a),

thereexism a

intx

in

f-l(BaBa)

(

a)

such

at

/(x)

y.

]

CASE

2: isa limit ordinal.

Here,

y

is in

B+B

and hence in

BBa

for all

<.

y

belongs to

BoBaC

/

f(AAa)

C

f(AAa)

_’AAa]

(becau

A,Aa

isadecreasing

chain).

Since

X

iscompactand

A,Ap

isnonemptyfor all < thereexistsa

int

x in

A,Ap

suchthat

f(x)

y.

Since is alimit

ordinal,

f(A+A)

contains

B+A

follows

om

induction

hypothesis

and

e

definition

ofA.

us,

(iv)

is

proved.

Our

nextaim istoshow

at

(X)

isanatural extensionof

e

derived

lense.

For

is

puose

we

Oefinea new ordinal invariant

’(X)

ontopologil

spaces.

’(X)

isdefined as

sup{a:

ais

me

derived

lengthof a finerdisperd tologyon

.

We

now

prove

THEOM

1.2:

For any

topological

space

X,

it is

e

at

I’(X)

+1

(X).

PROOF:

Suppose

xis a finerdispersedtopologyonXwith derived

length

d(x).

en,

onecan

prove

that

{:

0<as

d(),

is defined withrespect

m }

is a Khain with

e

tof all isolated

in

of asthe ba t, and’ 1+

e

lengthof this chain’ isequalto

d(x).

so

anyKhain in a finertopology is a K-chain inSe original topologyand hence we conclude that

I’(X)

1+

(X).

For

e

revere

inequality,

we consider aKhain

{A,}

in

X.

Our

aim is

m

define a finerdisperd

topology

on

X

such

at

Sederivedlengthof Sis

topology

coincides with

(A).

For

is,wedefinefor each x in

X

a new

neiborhd

systemasfollows, ffxis a

int inA+

A

for me ordinal number

(that

hastobe

ique),

then

{(V

A)

O{x

:

V

is a

neiborho

ofxin

e

Original

topology}

isdeclared asthe newneighborhood systematx,andx is isolated otheise.

One

canverifythat is defines a finer topologyon

X

who

open

tsn be described as follows:

’A

subt

V

of

X

is

open

ifandonlyiffor each ordinal number a and for each

int

x in

V

(A+

A),

it istruethatthereisaneighborhood

V

of x in theoriginal

polo

such

at

V A

is contained

V.’

uivalently,

is

topology

isgeneratedby the

hmily

{V

D

X:

either

V

is

open

in the

original

mpoioor

V

A

U{x }

formepoint xin and formeordinalnumber or

V

is contained in

A0

or

V

isdisjoint

om

OA}.

One

nnow

prove

that

-X

andfor each ordinal number 1,

(in

e

new

pology)-

(Ap)(pAp).

It

follows

at

thisnew

topology

isdisperdandhas derived

length equal

m

’1+thelength of

e

K<hain

{A}’.

us,

I’(X)

+

(X). Hence,

e

proof

of

e

theorem ismplted.

(4)

REMARKS:

1. The abovecorollary 1.3justifiesthe titleof the article. The factthat thedefinition of thelength of aspacedoesnotinvolvedispersednessatallwhereas l’is

completely

dependenton the derived

lengths

of finerdispersed topologies,is

quite

interesting

(.’.

l’ +

).

The above

corollary

that

d(X)

+

(X)

for adispersed

space

X

canbe

proved

directlyalso.

2. Ifweobservecarefullytheorem 1.2,wecaneasilyseethat isuniquelydetermined

among

all ordinal invariants intopology bythefollowingthree

properties.

(i)

issmallerfor finertopologies.

(ii)

coincides with derived

length

fordispersed

spaces.

(iii)

is minimal withrespectto

(i)

and

(ii).

Also,as an extensionof

d(X),

which isour main interestin this article, isuniquelydeterminedby

(i)

and

(iii).

2.

d(X) IN TERMS OF ORDER OF

A

MAP.

In

his memoir

], Kannan,

defined an ordinal invariant6 foralltopological

spaces.

The problemof characterizing6intrinsicallyis leftopenthere

[F.4].

We

solve thisproblemhere in this section.

In

fact, thelengthof a

space

wedefined abovehappenstobe in intrinsic characterization of 6.

We

firstrecall the definitionof6 givenin

].

DEFINITION

2.1:

Let

X

be anarbitrary topological

space

chosen and fixed.

Let

Y

beanyother topological

space

and

f:

Y

X

bea continuousfunction.

For

anysubset

A

and

X,

we let

A-

A

and

A

f(f-(A

)). As

inductionhypothesis,we assume

that/

isdefinedfor all

[

<ct.

Now,

we

letA?

(A))

ift

[

+ 1,anonlimit ordinalnumber,and

A

iaA

iftisa limit ordinal.

Thus,

A

isdefinedfor

every

ordinalt.

By

astandardargument on thecardinalityof

X,

one can verify that there exists an ordinal number t such that

A’-A’

/

Now

we define

o(A,

f)

inf{ct:

A?

A

}.

The order of the

map

f

is defined as

sup{o(,4,

f):

A

is a subset of

X

and is denotedby

o(f).

Finally,

6(X)

is defined as

sup{o(f):

f:

Y

X

iscontinuous,

Y

isany topological

space

}.

The existence of

6(X)

aswell as

o(f)

canagainbe establishedusing cardinality

arguments. One

cancheck that6isalso atopologicalinvariant. Thefollowingtheorem solves theproblemof

Kannan.

THEOREM

2.2:

For any

topological

space

X,

(X)

6(X).

PROOF.

First we

prove

that

(X),. 6(X). For,

suppose

f:

Y

X

is acontinuousmapfrom an arbitrary topological

space

Y

and a is a subset of

X.

CLAIM:

{A’}

is a r-chainand

({A})

o(,jO.

Throughoutthe

proof

of this claimwedenote

A

asA,

for notational convenience.

We

have to

prove

thatfor

any

ordinal numbert,

A,/

\A,

is containedin

A=\Aa

for all

[

<t.

We

use

(transfinite)

induction on to

prove

this. Since

A\Ao

is contained

inA0

canbe verified withoutdifficulty,thecaset 0 follows immediately.

Now,

weassume

thatAl

\Al

iscontained

inAl\A

for all <

[i

<t.

To

show

thatA,/ \A,

is contained in

A,\Aa

for all

[I

<

suppose

x

belongs

to

A,/

\A,.

Then, xis in

A,

f(f-(A=)).

So

(5)

DERIVED LENGTH FOR ARBITRARY TOPOLOGICAL SPACES 277

CASE

1: a

[

+ for someordinalnumber

In

thiscase,it isenoughtoshow thatxbelongsto

A,\A.

Suppose

weassume that thereexistsa

neighborhood

Ux

of x such that

U,

f3(A,\Ao)

isempty. Then,

J’-(Ux)

is aneighborhoodof

y

such that

J’-(U,)

f3

]’-(A,,)is

contained in

f-(A).

But

xis notin

A,

]’(f-(Al)

andhence

]’-(x)fl

J’-(A)

isempty. Therefore there exists aneighborhood

Vy

of

y

such that

Vy

f3/’-(A)

isempty.

Hence,

Vy CI/’-(U,)

does

not meet

f-(A,\A)

U.f-(A)

f-(A).

This is a contradictiontothefact that

y

isin

f-(Aa).

CASE

2: c is a limit ordinal number.

Here,A,

U

AI.

In

thiscase,xis notinA, U

A,

[’-J

f(f’-(Ao))

_impliesthatxis notin

f(f-t(Ao))

forany <

.

Hence,

for

any

givenordinalnumber < there exists aneighborhood

V

₀of

y

such that V

f3f-(A0)

isempty.

Now,

ifx doesnotbelong

toA\A

₀for some < then thereexists aneighborhood

U

of x such that

U

N(AQ\A)

isemptyand hence

f-(U0)

does notmeet

f-(A\A0).

So

theneighborhood

/’-(U)

V

ofyloes not meet

f-l(A)

(as earlier).

Thisagainis a contradiction.

Hence,

the claim is proved. Thus,we have

o(A,f)

({AQ})=

({A’})

and hence the

inequality

(X)

X).

For

thereverseinequality

(X)

(X),

weconsider anarbitraryK-chain

{A,}

in

X

withbasesetA. Weconstruct atopological

space

andacontinuous function

f:

X

such that

o(f)

({A,}).

For

each ordinalnumbero.,let

I.

A

\A

with relativetopologyfrom

ith

thetopologygenerated by

{

V: V

is asubset

ofA,

,

eitherVis an

open

subset of

A

or

V

(AQkAo)

U{x

for some xinA,/

lkA

and some <

a}.

Finally, isthedisjoint topologicalsum of all these

,’s.

A

map

(canonical)

in

,

for some

.

So y

is in

A

\A

for aunique a. Define this

point

in

X

astheimageof

y

under/(. It

is

easy

toverifythat

f

is continuous and

o(A0,

({A}).

This

proves

thereverseinequality

o()

().

Thus,theproofof the theorem iscomplete.

REMARK. One

can

prove

more results onthe

length

of a

space

such as

perfect maps

decrease the length,etc. The readershould consult

1]

for moredetails.

One

naturalquestionisthat "whichproperties of the derived lengthofdispersed topological spacesremaintrue for the lengthofgeneral topological

spaces?" A

mapis called finite-to-one ifthe inverseimageofevery pointis a finite subsets of the domain. Itcanbeprovedthat if

X

isa compact Hausdorffdispersed spaceand

f

is a finite-to.one continuousmap from

X

onto where is aHausdorff space,then

d(X)

d().

But

wedonotknow whether this istrue

for the

length

ofarbitrary compact Hausdorff

spaces. Let

X

and bearbitrary compact Hausdorff

spaces

such that is a continuousimage ofX. Thenusingtheorem 1.1

(iv),

one can

prove

that

()

(X).

It

is

interestingtoknowwhether

(X)

()

ornotwhen isafinite-to.one coatinuousimageof

X.

ACKNOWLEDGEMENT.

The author

expresses

hisdeep gratitudetoProfessor

V. Kannan

for his

helpful

suggestions duringthepreparationof this

paper.

The author likestothank the referee for suggestingthepresenttitleandforpointingoutanerror in the earlier version of the article.

Financially

supported

byCouncil of Scientific and IndustrialResearch,India.

Current

Address:

Department

of Mathematics,

Goa

University, Bambolim,

Santa Cruz

P.O.,

Goa,

India 403 005.

REFERENCES

KANNAN, V.,

Ordinalinvariants in

topology,

Memoirs,

Amer.

Maths.

Soc.,

Number245

(1981).

[2]

SEMADENI,

Z.,

Sur

lesensembles clairsemes,

Rozprrrawy Mat.

19

(1959).

[3]

KANNAN,

V.

and

RAJAGOPALAN, M.,

Scattered spaces

II,

I!1.

’.

Math.,

21

(1977)

735-751.