Finitely generated closed sets, and the relation between $T_0$-topologies and partially ordered sets

Finitely generated closed sets, and the relation between

$T_0$-topologies and partially ordered sets

Citation for published version (APA):

Bruijn, de, N. G. (1975). Finitely generated closed sets, and the relation between $T_0$-topologies and partially ordered sets. (Eindhoven University of Technology : Dept of Mathematics : memorandum; Vol. 7504).

Technische Hogeschool Eindhoven.

Document status and date: Published: 01/01/1975

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EINDHOVEN UNIVERSITY OF TECHNOLOGY

Department of Mathematics

Memorandum 1975-04

Issued May, 1975

Finitely generated closed sets, and the relation between To-topologies and partially ordered sets

University of Technology Department of Mathematics PO Box 513, Eindhoven The Netherlands by N.G. de Bruijn

N 47 - 1

-Finitely generated closed sets, and the relation between To-topologies and partially ordered sets by N.G. de Bruijn.

I. The purpose of this note is to provide some background to research of

topological models for minimal propositional calculus. Such models were

studied in [IJ; in that report the topologies were a1l finite (due to the finiteness of the alphabet and the absence of disjunction as a logical connective). If we turn to infinite models, however, we may need some of the results listed in this note.

2. Basic definitions.

2.1 ~QEQ1Qgf~~1_~E~£~~' One of the usual forms of the definition of a topological space is the one that uses the closure operator. It runs as follows. If X is a set, then ?(X) denotes the set of all subsets of X.

A mapping cl of p(X) into ~(X) is called a closure operat~r if (i) cl(0) = 0,

(ii) S c cl(S) for all S E J(X)

(iii) cl(Sl U S2)

=

cl(S]) U cl (52) for a1l SI,S2 E 7>(X).

(iv) cl(cl(S»

=

cl(S) for all S E 1'(X).

A set S is called closed if S = cl(5).

We mention a few consequences: (v) monotonicity: if SI c S2 then c1(Sl) c cl(S2); (vi) if S € P(X) then cl(S) is closed; (vii) If SI c S2

and if S2 is closed then cl(S]) C S2; (viii) The intersection of an

arbitrary set of closed sets is closed; (ix) The union of a finite family of closed sets is closed.

If cl is a closure operator on ~(X), then we say that the pair (X,cl) is a topological space.

- 2.2 ~i~if~lX_g~~~E~f~~_!~f!' If (X,cl) is a topological space, and if S E ~(X), then S is called finitely generated if there is a finite set T with S = cl(T).

then S is called an independence ~ if for all XES, YES we have x

f

cl({y}). (The set {y} consists of the element y only, and is often referred to as a singleton).

2.4 §~~iE~!~!!' A semiposet ~s a pair (X,~), where X is a set and ~ is a relation on X which is both reflexive and transitive: we have x ~ x for all

X E X, and for all x,y,z E X with x s Y and y ~ Z we have x s z.

2.5 ~~!~f!. A poset (or partially ordered set) is a semiposet (X,~) that satisfies, for all x E X, Y E X:

if x ~ y and y s x then x = y.

3. Some special conditions for topological spaces. We shall list a number

of conditions, formulated for a topological space (X,cl). Relations between these conditions will be studied in the next sections.

(i) Every independence set is finite.

(ii) (Ascending chain condition for closures of singletons). Ifx

l ,x2, .. • are points of X, and

c1({x]}) c cl({x

2}) c cl({x3}) c

...

,

then there ~s an n such that cl({x }) = cl({x }) for all m > n.

n m

(iii) (Ascending chain condition for finitely generated closed sets):

If T1,T Z"" are finite, and cl(T

l) c cl(T2) c cl(T3) c . . . ,

then there is an n such that cl(T )

=

cl(T ) for all m > n.

n m

(iv) (Ascending chain condition for closed sets). If C

l ,C2, ••• are closed, and C

3

-for all m > n.

(v) Every closed set is finitely generated.

(vi) For every pair (x,y) (x E X, Y E X) the intersection

cl({x}) n cl({y}) is finitely generated.

(vii) The union of any collection of closed sets is closed. (viii) Every subset S of X satisfies

(ix)

c1(S)

=

U

S c1({x}).

XE

For every family {S.}. I of subsets of X, we have

1 1E

(x) (Kolmogorov's separation axiom): If x E X, Y E X, X , Y then

the relations x E cl({y}), y E cl({x}) do not hold simultaneously.

(xi) If x , y then cl({x}) , cl({y}).

Remark on separation axioms. If (X,cl) satisfies(x) then it is called a

TO-space. We have x E cl({y}) if and only if cl({x}) c cl({y}), and therefore

(x) and (xi) are equivalent.

Definitely stronger than Kolmogorov's axiom is Frechet's axiom T 1:

T₁: If x E X, Y E X, X , Y then x

f

cl({y}).

Again stronger than T) 1S Hausdorff's axiom.

The results of this note will be trivial for Frechet spaces (and a fortiori for Hausdorff spaces).

4. Relations between (i), ••• ,(ix) of section 3.

Theorem 4.). «i) A (ii» ~ (iii) ~ (iv) ~ «v) A (vii»,

(vii) ~ (viii) ~ (ix), (v):::r9- (vi).

We shall now prove «i) A (ii» ~ (iii), (iii) ... (i), (iii)....p (v),

(iii) ~ (iv), (iv) =9 (vii), (vii) ~ (viii), (viii)

==+

(ix), (ix)

"'*

(vii), «v) A (vii» ~ (iii) •

Proof of (i) A (ii) ~ (iii). Assume (iii) false. Then we can find

finite sets T

l,T2, •.• such that all inclusions cI(TI) c cl(T2) c .•• are proper. Now choose any x. in each cl(T. l) \ cl(T.). Obviously x.of cl({x.})

1 1+ 1. 1. J

for all i,j with i > j. Now define a graph on the positive integers: two integers i,j (i ~ j) are to be connected in the graph if one of the relations

X. € cl({x.}), x. € cl({x.}) holds. According to Ramsey's theorem (Ramsey

1. J J 1.

[3J) there is either a sequence i

1,i2, ••• such that i] < i2 < ••• and all pairs ik,i_m, (k ~ m) are disconnected, or a similar sequence such that all pairs

ik,i_m, (k

1

m) are connected. In the first case these xi form an infinite independence set, i.e. (i) does not hold. In the second case we have (note that x.

4

c1({x. }) since i2 > i):

12 11 X. € cl({x. }), ••• 12 1.3 whence cl({x. }) c c1({x. }) c 11 12

These inclusions are proper (note that x. f cl(T. ), 12 12

c cl(T.

»,

i.e. (ii) does not hold. So if (iii) is false, then either (i) 12

or (ii) is false.

Proof of (iii) -+ (i). Assume (iii), and assume that I

=

{x

1,x2' ••• } 1.S an infinite independence set. By (iii), there is an n such that

cl({xl,···,xn+)})·

The right-hand side contains x I ' and the left-hand side equals

u ••• u

5

-cl({x }).

n

Therefore x

n+1 E cl({xi}) with some i, i < n+l. This ~s impossible s~nce I is an independence set.

Proof of (iii) =+ (v). Let C be a closed set. If C is not the closure of a finite set, we can select x

1,x2,x3, ••• , all in C, such that, for all i,

x.

11:

c1(T.), where T. = {xl, ••• ,x). I t follows that all inclusions

~+ ~ ~ •

cl(T.) c cl(T. 1) are proper, and that contradicts (iii).

~ ~+

Proof of (iii)

=+

(iv). If (iii) holds then (v) is true (see above), and then (iii) and (iv) express the same thing.

Proof of (iv) -+ (vii). Let I be an index set, and let C. be closed for

~

every i E I. We put Q

=

LJ

iE1 Ci' Assume that Q is not closed. We select indices i

l,i2, ••• as follows. Take il arbitrarily in I. Now Ci ) is closed and Q is not. Take x E Q, x

i

c . .

Take i2 such that x E C. • Now C. u C.

~l ~2 ~I ~2

is closed and Q not, etc. Putting Dk

=

C. u ••• u C. , we infer that all

~1 ~k

inclusions D) c D2 c ••• are proper. Therefore (iv) is false.

Proof of (vii) ~ (viii). Assume (vii), and let S E ~(X). By monotonicity

cl({x}) c cl(S) for all X E S ; if we put T

LJ

S cl({x}) we infer XE

T c cl(S). By (vii) T is closed; since SeT we infer cl(S) c T. Therefore

cl(S)

=

T.

Proof of (viii) + (ix). Let T

lJ.

lSi- Assuming (viii) we have

~E c1 (T) =

U. IUS

c1 ({x}) , ~E XE i cl(S.) =

LJ

S cl({x}), ~ XE i whence c1(T) _=U·Icl(S.). ~E ~

• Proof of (ix) -+ (vii). I f S. is closed for each 1. E I, we have cl (S.)

=

S.,

1. 1. 1.

whence (ix) shows that

U

iEl Si is closed.

Proof of «v) A (vii» =9 (iii). Assume (v) and (vii). Let T

I, T2, •.• be finite, cl(T]) c cl(T

2) c • • • • Let C be the union of all cl(Ti), then C is closed (by (vii» and therefore finitely generated. So xI, ••• ,x

n exist such that

For every k (1 ~ k ~ n) we have ~ E C, whence ~ E cl(T

i) for some i. Denote this value of i by i(k). Now taking j

=

max(i(I), ••• ,i(n» we find

that x_k E cl(T

j ) (I ~ k ~ n), whence C c cl(Tj ), whence cl(Tj )

=

cl(Tj+l) .•• ,

This proves (iii).

5. Characterizing closed sets by means of m1.n1.ma.

If C is a closed set 1.n (X,cl), and if x E C, then x 1.S called a minimum

of C if for all y E C with x E cl({y}) we have y E cl({x}).

Theorem 5.1. If (X,cl) satisfies condition (ii) (of section 3), and if C 1.S

a closed set, then C

=

cl(M), where M is the set of minima of C.

Proof. For every x E C we have an m E M with x E cl({m}). This can be

shown as follows. If x

f

M we can find Xl with x E cl({x

l}), Xl

f

cl({x}).

That is, the inclusion cl({x}) c cl({x

l}) is proper. If Xl

f

M we can find

x

2' etc. By (ii) this cannot go on for ever, and some ~ satisfies xk E M, X E cl({x

k}). Hence x E cl(M).

Theorem 5.2. If (X,cl) is a topological space, and if C is a finitely generated closed set, then there 1.S a finite independence set I with C

=

cl(I). If (X,cl) 1.S a TO-space, this set I is uniquely determined, and equals the set of minima of C.

7

-Proof. L~t C be finitely generated: C

=

cl(T), where T is finite. Let I be a subset of T such that C

=

cl(I) but C ~ cl(J) for all proper subsets of I. Then it is easy to show that I is an independence set.

Assume furthermore that the space is TO' If m is a minimum of C, then m E _{cl(I). If I} = {x

1' ••• ,xk} we have cl(I) = cl({x1 }) u

...

u cl({~}) , whence m E _{c1({x.}) for some} J. Since m is a minimum, we infer x. E cl({m}),

J J

whence m = _{x. by TO' It follows that Mel. Furthermore, every x. is a}

J J

minimum. For, suppose x. E cl({y}), y

f

cl({x.}), for some y E C. Since

J J

Y E C we have y E cl({x

l}) U ••• u cl({xk}). Putting J

=

I ,,{xj } we

infer that cl(J) contains y, and therefore x., and therefore I. Hence

J

C

=

cl(J), which is impossible.

6. Connections between topological spaces and semiposets.

If (X,cl) is a topological space, we define the relation R by: xRy if and only if y E cl({x}). It is easy to show that R is reflexive and

transitive. Writing R = ~(cl) we thus have constructed a standard mapping

of the set of all closure operators on X into the set of all reflexive transitive relations on X.

If (X,R) is a semiposet, we define the operation cl as follows. If S E :f(X), then cl(S) is the set of all x for which there exists an s E S

with sRx. It 1S easy to show that cl is a closure operator. Writing

cl

=

W(R) we thus have constructed a standard mapping of the set of all reflexive transitive relations on X into the set of all closure operators.

Theorem 6.1. If X is a set, we have ~(W(R»

=

R for every reflexive transitive relation R on X.

*

Theorem 6.2. Let (X,cl) be a topological space, and put cl = W(~(cl».

cl*(S)

=

LJ

cl({x}).

XES

Therefore ~(~(cl»

=

cl if and only if (X,cl) satisfies condition (viii) of section 3 (a condition that was shown to be equivalent to (vii) and to (ix) in Theorem 4.1).

Theorem 6.3. If (X,cl) is a topological space we have: (X,cl) is a TO-space

if and only if (X,~(cl» is a poset.

The proofs of these theorems are straightforward.

Theorems 5.1 and 5.2 express that ~ and ~ provide a one-to-one

correspondence between semiposets and topological spaces satisfying (vii).

If a topological space satisfies (vii) then we can build a new

topological space in which the open sets are the closed sets of the old one,

and vice versa. This relation between topological sets corresponds to a

relation between semiposets, viz. the one that replaces ~ by ~.

7. Examples of posets. We shall present a number of examples in order to show that Theorem 4.1 is best-possible as long as we restrict ourselves to

topological spaces that satisfy (vii) (of section 3). According to Theorem 6.2 these are just the spaces which can be represented as semiposets. Theorem 4.1 says that

(i) A (ii) ~ (vi);

we shall show by examples a, b, c, d, e, f, g that this ~s all we can prove

about (i), (ii) and (vi). The role of these examples is indicated in a Venn diagram.

• 9

-(vi)

(i)

(ii)

d _e

It has to be read like this: b is an example of a topological space

satis-fying (i) and (vi) but not (ii). All these examples are given in the form

of posets. That is, the topological spaces are of the form (X,$(~»,

where (X,~) is a poset.

Example a. The space consists of one point only.

Example b. X consists of the negative integers, with the usual meaning of ~.

If k E X, ~ E X and k ~ ~, then cl({k}) A cl({~})

=

cl({k}), whence (vi)

holds. The only non-empty independence sets are singletons, so (i) is clear.

But (ii) does not hold: cl({-I}) c cl({-2}) c ••• , where all inclusions are

proper.

Example c. X LS a countable set, and x ~ y only if x

=

y. Now (i) is false

(X itself is independence set). Both (ii) and (vi) follow from the fact that

for each x E X the set cl({x}) has just one element.

Example d. We take X

=

NuQ, where N consists of the negative integers, and Q has just two elements, QnN

=

0.

In. N we define ~ as usual (just as in example b), furthermore we put q ~ n for all q E Q, x E N, but no relation ~ is taken

between the two elements of Q. Now (ii) is false (cf. example b), and (i) LS

true (the independence sets have at most two elements). And (vi) is false

since cl({qI}) n cl({q2})

=

N (if QI,q2 are the elements of Q) and N is not finitely generated.

Example e. Let H be some countable set, and Q a set with two elements QI,Q2.

among each other, and so are the elements of Q. Now (vi) is false since

cl({q)}) n cl({q2})

=

H. And (i) is false S1nce H is an independence set. But (ii) is true: a chain cl({x

l}) C cl({x2}) c ••• can have at most two different elements.

Example f. Let X be the set of all pairs of integers (m,n) and let

(ml,n) s (m

2,n2) mean that both m1 s m2 and nJ

=

n2• We get infinite independence sets by keeping m constant, and infinite chains by keeping n

constant. Hence both (i) and (ii) are false. But (vi) is still true: cl({x})

n

cl({y}) is either cl({x}), or cl({y}), or empty.

Example g. We add to the space of example f two new points Qj,q2' which are incomparable, but s all other points. Now (vi) is also violated.

8. An example that is no semiposet. We shall present an example where

(vii) is not satisfied, where (v) is true but (iii) is not.

Example h. Take X

=

{0,1,~,1,!, .•• }. For any subset S (S

1

~) we define cl(S) as the set of all x E X with x ~ inf(S). (the infimum is taken in

the normal sense: X is a set of reals).It is easy to check that cl is a closure operator. The union of cl({)}), cl({!}), .•• is the set {I,i, ••• },

and that is not of the form cl(S). Every cl(S) is of the form cl({x}); therefore (v) is true. But (ii) is false: cl({l}) c cl({i}) c ••• is

properly ascending. Therefore (iii) is false.

9. Concluding remarks. The author does not think that much of the material of this note has never been written before, but it is hard to trace such elementary material. He will be happy to receive references if there are any.

For the case of totally ordered spaces a quite extensive study was recently published by H. Kok [2J, who compared a large number of conditions,

..

11

-References.

[IJ N.G. de Bruijn Exact finite models for minimal propositional calculus over a finite alphabet, Technological University Eindhoven, Department of Mathematics, T.H.-Report 75-WSK-02, February 1975.

[2J H. Kok

[3J F. P. Ramsey

Connected orderable spaces, Mathematical Centre Tracts MCT 49, Amsterdam 1973.

On a problem of formal logic, Proe. London Math. Soc. (2) 30 (1930) pp. 264-286.