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By

T.B.M. McMaster

Pure Mathematics Department, Queen's University, Belfast

and

C.R. Turner

School of Electrical and Mechanical Engineering,

University of Ulster at Jordanstown

[Received 10 November 1999. Read 15 May 2000. Published 16 October 2001.]

Abstract

Reiterative application of Bankston's total negation operator `anti-' upon an

arbitrary topological invariant is known to lead rapidly to repetition in one of

just seven patterns. The authors have recently shown that a great deal of the

total negation procedure can be constrained to take place within a xed class of

topological spaces (the `constraint' for the discussion) without impairing much of

the theory. The present article explores iterative behaviour within a constraint. We

show that, provided the constraint is hereditary, at most eight patterns of repetition

are possible. An example reveals that in non-hereditary constraints the (unending)

sequence of invariants generated may consist entirely of distinct terms, without ever

entering a cycle of repetition.

1. Introduction

In [1] Paul Bankston demonstrated a method for producing a new topological

property anti(

P) that is, in a well-de ned sense, the `opposite' of a given property

P

is used to describe the transition from a property

is known [2] that repeatedly applying this process to a given property

in the generated sequence of properties becoming repetitive in one of only seven

`iteration patterns', and that no more than four distinct properties can appear in

this sequence.

We observe the following notational conventions throughout this article. We shall

use script capitals such as

speci cally, we employ

spaces. We shall, in any given problem, use

the

that the property

replace

represent individual topological spaces.

The following de nition is derived from [1] but presented as in [2]. The three

. This process is known as total negation and, frequently, the term `anti-operator'P to its total negation anti(P). ItP will resultC, P, Q, R and S to represent topological properties and,U to denote the universal property satis ed by all topologicalC to represent the property known asconstraint for the context, which is further discussed below, and we may assumeP is always contained in C (for, were it not so, we could simplyP with C \ P before proceeding). Italic capitals such as X, Y and Z will



Corresponding author, e-mail: T.B.M.McMaster@qub.ac.uk

Mathematical Proceedings of the Royal Irish Academy

succeeding results were established by Matier and McMaster in [2] and, in particular,

Theorem 4 is referred to as the `classical iteration theorem'.

De nition 1.

For a given topological property P we de ne

spec

(P) = f:  is a non-zero cardinal and all spaces on -many points are Pg,

ind

(P) = f:  is a non-zero cardinal and there is a P space and a not-P space on



-many pointsg,

proh

(P) = f:  is a non-zero cardinal and there is no P space on -many pointsg,

anti

cardinality

We may further de ne

(P) = fX: X has no P subspaces, excepting those subspaces Y for which thejY j 2 spec(P)g.anti0(P) = P, anti1(P) = anti(P) and antin+1(P) =

anti

(antin(P)) for n  1.

Lemma 2.

For a given topological property P the sequence

P

, anti(P), anti2(P); : : :

hereafter called the Bankston iteration sequence, contains no new terms after the fourth

term. More precisely, either

antin(P) = U for all n  3, or else both anti2n(P) =

anti

2(P) and anti2n+1(P) = anti3(P) for all n  1.

Lemma 3.

(i)

For a given topological property P :P = anti(P))P= U,

(ii)

P = anti3(P))P= U.

Theorem 4.

topological property

(1) (

The pattern of the Bankston iteration sequence beginning with an arbitraryP is one of the following:U)0 (5) (P, Q)0

(2)

P; (U)0 (6) P; (Q; R)0

(3)

P; Q; (U)0 (7) P; Q; (R; S)0

(4)

P; Q; R; (U)0

where we use

()0 as a short form for an in nite repetition of the sequence segment  inside parentheses. Furthermore, (4) and (7) cannot occur when P is hereditary.

A question frequently posed has been what e
ect the decision to work exclusively

within a given separation axiom might have on the process of total negation. To

investigate this, we consider restricting ourselves to work inside a collection of spaces

C

the following form.

, rather than U. The natural restructuring of the classical de nitions then assumes

De nition 5.

Let C and P be topological properties; then

C

-spec(P) = f:  62 proh(C) and all the C spaces on -many points are Pg,

C

on

-ind(P) = f:  62 proh(C) and there is a C and P space, and a C and not-P space-many pointsg,

C

-proh(P) = f:  62 proh(C) and no C space on -many points is Pg,

23

C

-anti(P) = fX: X is C, and has no C and P subspaces, except for those subspaces

Y

It is easily seen that these de nitions collapse to their classical counterparts (in

De nition 1) when

It is sometimes useful to characterise the

lemma, which follows directly from the de nitions.

for which jY j 2 C-spec(P)g.C is taken to be U.non-C-anti(P) spaces using the following

Lemma 6.

are those which are not

C

Let C and P be topological properties; then the non-C-anti(P) spacesC, or which contain a C and P subspace Y such that jY j 2-ind(P).

We shall further extend the notion of a hereditary property.

De nition 7.

Let C and P be topological properties; then we shall callPC-hereditary

if and only if, when

The following can easily be shown.

X is a P (and C) space, then every C subspace of X is P.

Proposition 8.

Let C and P be topological properties. Then

(i)

(ii)

C-anti(P) is C-hereditary (but need not be hereditary);if C is hereditary then C-anti(P) is hereditary.

We also use the following lemma, which is an extension of a result found in

classical total negation theory [1].

Lemma 9.

Let C be a hereditary property and let P be a C-hereditary property. Then

C\P)C

-anti2(P).

Proof.

exists a subspace

Therefore there exists a

in turn, there exists a subspace

We can select a

The space

contradicts our choice of

We make a nal observation before embarking on the iteration pathway.

Suppose that X is C and P but not C-anti2(P). Then by Lemma 6 thereY of X which is C-anti(P), and such that jY j 2 C-ind(C-anti(P)).C space Z such that jY j = jZj but Z is not C-anti(P). Hence,W of Z such that W is C and P but jWj 2 C-ind(P).C subspace U of X (and of Y ) such that jUj = jWj.U is C and P and C-anti(P). Therefore jUj 2 C-spec(P). ThisW and hence of X.

Lemma 10.

Then

Let C be a hereditary property and let P be a C-hereditary property.C-proh(P) is an increasing subclass of (spec(C) [ ind(C)).

We now establish a collection of lemmas which form the backbone of the proof

of the

constrained iteration theorem.

24

Mathematical Proceedings of the Royal Irish Academy

Lemma 11.

Let C and P be topological properties. Then

(i)

C-anti(P) = C , C-ind(P) = ;,

(ii)

C-anti(C) = C.

Proof.

Conversely, if we can select

(i) If C-ind(P) = ;, Lemma 6 shows that all C spaces will be C-anti(P). 2 C-ind(P), we may choose a C and P space X on



implies that

(ii) Clearly

-many points. As X is a C and P subspace of itself with jXj 2 C-ind(P), Lemma 6X is not C-anti(P).C-ind(C) = ;, whence the result follows from (i).

Lemma 12.

Let C and P be topological properties such that C\P 6= ;. Then

C

-anti(P) = C\P,C\P = C:

Proof.

[

Lemma 6 we can pick a cardinal

[(] See the preceding result.)] If the rst equality holds but not the second, then C-anti(P) 6= C. From 2 C-ind(P). We can therefore select a C and

P

contradicting our choice of

space X on -many points. Now X is also C-anti(P) and so  2 C-spec(P),.

Lemma 13.

Let C and P be topological properties; then

C

-spec(C-anti(P))\C-ind(P) = ;:

Proof.

can choose a

belong to

Supposing not, we choose  2 C-spec(C-anti(P)) \ C-ind(P). Therefore weC and P space on -many points which is also C-anti(P), forcing  toC-spec(P), a contradiction.

Lemma 14.

Then

Let C be a hereditary property, and let P be a C-hereditary property.C-anti2(P) = C-anti4(P).

Proof.

We suppose that there is a

From Lemma 9 we know that C-anti2(P) ) C-anti4(P) as C is hereditary.C space X which is C-anti4(P) but which is not

C

-anti2(P). It follows that X has a C subspace Y which is C-anti(P) but such that

j

Yj 2 C-ind(C-anti(P)). Therefore we can select a C space Z such that jY j = jZj and Z is not C-anti(P), and Z consequently has a C and P subspace W such that

j

The space

Wj 2 C-ind(P). We also pick a C subspace U of Y such that jUj = jWj.U must be C-anti(P) as it is a subspace of Y , while the space W

cannot be

C-anti(P) as W is a C and P subspace of itself even though jWj 62 Cspec(

P

C

). We must conclude that jWj = jUj 2 C-ind(C-anti(P)), and so jWj = jUj 62-spec(C-anti2(P)) as otherwise Lemma 13 would be contradicted. Additionally, U

is

so by appealing to Lemma 13 again we nd that

C-anti3(P) and C-anti4(P), because U is C-anti(P) (using Lemma 9) and U is a C subspace of the C-anti4(P) space X. Therefore jWj = jUj 2 C-spec(C-anti3(P)) andjUj 62 C-ind(C-anti2(P)). Therefore

j

However,

Uj 2 C-proh(C-anti2(P)).W is a C and P space and, as C is hereditary (and via Lemma 9), a

C

choice of

-anti(P) space, implying that jWj = jUj 62 C-proh(C-anti2P)), contradicting ourW and hence of X.

25

This leads to the following corollary.

Corollary 15.

invariant. Then

Let C be a hereditary topological property, and let P be a topologicalC-anti3(P) = C-anti5(P).

Lemma 16.

Then either

Let C be a hereditary property, and let P be a C-hereditary property.C-anti2(P) = C, or C-anti(P) = C-anti3(P).

Proof.

Lemma 9 we may then select a

Thus we can select a subspace

C

Suppose that both C-anti2(P) 6= C and that C-anti(P) 6= C-anti3(P). FromC space X which is C-anti3(P) but not C-anti(P).Y of X such that Y is both C and P but  = jY j 2-ind(P). Therefore Y is C-anti2(P) by Lemma 9, and is C-anti3(P) because

C

-anti3(P) is C-hereditary. It follows that  2 C-spec(C-anti2(P)). By Lemma 13,



As

62 C-spec(C-anti(P)), and  62 C-ind(C-anti(P)). Therefore  2 C-proh(C-anti(P)).C-anti2(P) 6= C we can select  2 C-ind(C-anti(P)). By Lemma 10,  < 

as

 2 C-proh(C-anti(P)) which is an increasing subclass of spec(C)[ind(C). As



2 C-ind(C-anti(P)),  62 C-spec(C-anti2(P)) by Lemma 13. Also  62 C-proh(Canti2(

P

Therefore

Choose a

as

)) as Y is C-anti2(P) and  <  = jY j and clearly  62 C-proh(C-anti2(P)). 2 C-ind(C-anti2(P)).C subspace Z of Y such that jZj = . Then Z is a C and P spaceP is C-hereditary, and so Z is C-anti2(P) by Lemma 9. Therefore Z cannot be

C

Now

-anti3(P) as it is a C-anti2(P) subspace of itself and jZj =  62 C-spec(C-anti2(P)).Z is a C subspace of X and thus is C-anti3(P), contradicting our choice of Z

and thus of

X.

Theorem 17.

property. Then in the constrained Bankston iteration sequence, no new terms appear

after the fourth term. Additionally, either

Let C be a hereditary topological property, and let P be a topological

C

-antin(P) = C for all n  3; or

C

Further, the iteration pattern will be one of the following:

-anti2n(P) = C-anti2(P) and C-anti2n+1(P) = C-anti3(P); for all n  1.

(1) (

C)0 (5) (P;Q)0

(2)

P; (C)0 (6) P; (Q;R)0

(3)

P;Q; (C)0 (7) P;Q; (R;S)0

(4)

P;Q;R; (C)0

Further, sequences

properties

(4) and (7) cannot occur when P is C-hereditary. Note that theQ, R and S are all subfamilies of the constraint family of spaces C.

Proof.

It is apparent that this result is based upon the hereditary character of the

constraint

also losing the result.

We shall now exhibit a certain non-hereditary constraint

family of topological spaces

qoset proposed by Matthews and McMaster in [3]. This counterexample shows that

not only does the theorem above fail to hold without the hypothesis of hereditariness

of the constraint, but that we can in fact form a never-repeating chain of `negated'

properties.

This follows easily from the lemmas proved previously.C, but it is (as yet) unclear whether we can relax this requirement withoutC from the general
U whose ordering under embeddability conforms to a

Example 18.

discrete and trivial topological space respectively on

shall use the notation

For each positive integer n let us use D(n) and T(n) to denote the@n-many points. Further, weA  B to denote the topological direct sum of the spaces A

and

B. We now de ne a family of topological spaces Xn and Yn for n  3 as follows:

X

n = D(n)  T(n);

Y

n = D(n)  T(n - 2) if n is odd;

Y

Let

shown that

n = D(n - 2)  T(n) if n is even.P be the topological invariant comprising simply Y3. Then it can easily beC-anti2(C \ P) = fX3; Y3; Y5g, and that, in general, C-antin(C \ P) =

f

will never repeat itself, and the collapse of the iteration theorem is attributable to

the loss of hereditariness in

Yn+3g [ fXm; Ym : 3  m  n + 1g. Clearly, the Bankston iteration sequence hereC.

Example 19.

sequence

anti(compactness) = totally disconnected, and thus the Bankston iteration sequence

for

in the constraint

ind(

totally disconnected space which is embedded in every in nite

in nite

nite

that

iteration pattern in

It is noted in [2] that connectedness exhibits the Bankston iterationP;Q; (R;S)0 in the context of all topological spaces. In fact, Q =P = totally disconnected is P; (Q;R)0. Now consider P = totally disconnectedC = T2. Observe rst that T2-spec(P) = (1, @0) and that T2-P) = [@0;1). Note that the natural numbers with the discrete topology form aT2 space. Thus everyT2 space contains an `indecisively' totally disconnected space, whereas everyT2 space only contains totally disconnected spaces of `spectral' size. It followsT2-anti(totally disconnected) = f nite T2 spacesg. From Lemma 11 above, theT2 is therefore

C\P

Clearly, the imposition of this constraint, even though it is a hereditary one, has

completely altered the Bankston iteration sequence of the property under scrutiny.

It may be noted that other properties are `pre-antis' for totally disconnected. For

example, anti(not totally disconnected) = totally disconnected. It can be shown that

if we take

iteration sequence is

;Q;C; : : :.P to be `not totally disconnected' and C = T2 then the absolute Bankston

P

;Q; (R;S)0

and that the constrained sequence is

C\P

;Q;R; (C)0.

References

[1] P. Bankston, The total negation of a topological property,

241{52.

[2] J. Matier and T.B.M. McMaster, Iteration of Bankston's `anti'-operation,

Mathematics and Computer Sciences (Mathematics Series)

[3] P.T. Matthews, Quasi-order in relation to the total negation operator, unpublished Ph.D. thesis,

Queen's University, Belfast, 1994.

Illinois Journal of Mathematics 23 (1979),Journal of the Institute of3 (1990), 31{5. 


, 101A (1), 21{26 (2001) c Royal Irish Academy