By

T.B.M. McMasterPure 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 propertyP

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 willCorresponding 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 nespec

(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) = f*X:**X has no P subspaces, excepting those subspaces**Y for which thej**Y 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 sequenceP

, anti(P), anti2(P)*; : : :*hereafter called the Bankston iteration sequence, contains no new terms after the fourth

term. More precisely, either

anti*n(P) = U for all**n 3, or else both anti2**n(P) =*anti

2(P) and anti2*n+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 assumesDe nition 5.

Let C and P be topological properties; thenC

-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) = f*X:**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 j*Y 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 j**Y 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-hereditaryif 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. ThenC\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 there**Y of**X which is C-anti(P), and such that j**Y j 2 C-ind(C-anti(P)).C space**Z such that j**Y j = j**Zj but**Z is not C-anti(P). Hence,**W of**Z such that**W is C and P but j**Wj 2 C-ind(P).C subspace U of**X (and of**Y ) such that j**Uj = j**Wj.**U is C and P and C-anti(P). Therefore j**Uj 2 C-spec(P). This**W 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 AcademyLemma 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 j**Xj 2 C-ind(P), Lemma 6**X 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= ;. ThenC

-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; thenC

-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 j**Y j = j**Zj 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 j**Uj = j**Wj.**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 j**Wj 62 Cspec(*P

C

). We must conclude that j*Wj = j**Uj 2 C-ind(C-anti(P)), and so j**Wj = j**Uj 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 j**Wj = j**Uj 2 C-spec(C-anti3(P)) andj**Uj 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 j*Wj = j**Uj 62 C-proh(C-anti2P)), contradicting our**W 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**= j**Y 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**< = j**Y j and clearly**62 C-proh(C-anti2(P)).**2 C-ind(C-anti2(P)).C subspace**Z of**Y such that j**Zj =**. 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 j*Zj =**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 topologicalC

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

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

-anti2*n(P) = C-anti2(P) and C-anti2**n+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*

U whose ordering under embeddability conforms to aFurther, 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 generalU 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, we**A**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) = f**X3**; Y3**; Y5g, and that, in general, C-anti**n(C \ P) =*f

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

the loss of hereditariness in

*Yn+3g [ f**Xm; Ym : 3**m*