LATTICE JWETRIZED SPACES

LEO LAPIDUS

AN ABSTRACT
Submitted to the School of Graduate Studies
of Michigan State University of Agriculture
and Applied Science in p a r ti a l fulfillm ent
of the requirements for the degree of

DOCTOR OF PHILOSOPHY
Department of Mathematics

1956

Approved

ABSTRACT

LEO LAPIDUS

This th esis is a study in what has come to he known as
the f i e l d of abstract distance spaces, in the s p i r it

of the

l a t e r work of Karl Menger, where the distances of the space
are elements of some abstract algebraic stru ctu re.

In par­

t i c u l a r , the distances of th is study are elements of a
l a t t i c e , more especially of a Brouwerian algebra, a general­
ization of a Boolean algebra.
F irst,

a few l a t t i c e th eo retic properties of Brouwerian

algebras are developed in some d e t a i l , with considerable
attention given the Brouwerian complement, which generalizes
the fam iliar Boolean complement of set algebra.
Next, a Brouwerian algebra Is metrized by symmetric
difference,

a generalization again of the well known

symmetric difference of Boolean algebras. Many properties
of the resu ltin g Brouwerian spaces are then derived and
numerous theorems are obtained which serve to characterize
the Boolean algebras among the Brouwerian algebras. The
congruence order of c e rta in Brouwerian spaces re la tiv e

to

the class of lattice-m etrized spaces is established.
In the fin a l section, properties of lattice-m etrized
spaces in general are obtained and in p articu lar many of the
e a r lie r re su lts are extended. Finally, the notions of metric
and l a t t i c e betweenness are analyzed. By studying the effect
of their coincidence on the algebraic structure

of the under

lying l a t t i c e fthe Boolean algebras are then characterized
among the class of a l l l a t t i c e s with I ,
betweenness concepts.

in terms of the

LATTICE METHI2ED SPACES

By
LEO LAPIDUS

A THESIS
Submitted to the School of Graduate Studies
of Michigan State University of Agriculture
and Applied Science in p a r tia l fulfillm en t
of the requirements for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematic s

1956

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For
Don, David and Jeremy,
and
to my wife

v*.•A

without whose
se lf-sa c rific e ,

i n f i n ite patience, and sympathetic under

standing th is en tire project could not have been brought
to a successful conclusion.

ACKNOWLEDGEMENTS

That the writer i s

sincerely gra teful

to Professor L. M. Kelly for his counsel
and guidance in the preparation of th is
th esis

goes, of course, without saying.

Of much greater significance, however,
is the appreciation of and insight into
the nature of mathematics more generally
which he feels his association with
Professor Kelly has given him.
He also wishes to extend his thanks
to Professor E. A. Nordhaus for his kindly
i n te r e s t In the work.

TABLE OF CONTENTS

Page
Introduction...............................• .................. . . . . . . . .

i

Section 1.

Preliminary r e s u l t s . ......................

Ij.

Section 2.

Brouwerian algebras........................

6

Section

Brouwerian g e o m e t r i e s . . . . . . . .

2^

Section J+. General theorems; Betweenness

Ij-7

Introduction

I f , with each two elements of an abstract set is
associated a number (real or complex), the re su ltin g
structure i s known as an abstract distance space* ( I t is
convenient and suggestive to r e fe r to the elements of the set
as “points1
* and to the number associated with a pair of points
as the “distance 11 between them). This notion plays an import­
ant part in Frechet’s I 9 0 6 th e s is , although the concept was
undoubtedly known to e a r lie r workers in geometry.
The f i r s t systematic study of the geometric properties
of these spaces

was due to Karl Menger [91* who referred to

these structu res as semimetric spaces. In addition, however,
to spaces in which distances were selected from among the
r e a l and complex numbers, Menger [10] and l a t e r Taussky |12]
studied spaces whose distances

were elements of a group.

This has led more recently to the study of spaces ufaosc d is ­
tances

were selected from even more general algebraic

stru ctu res.
E llio tt

In p a r tic u la r , E l li s

15]

9

Blumenthal [ 3 ] and

[ii] have Investigated spaces whose distances are

elements of a l a t t i c e . This notion may be generalized in the
following ways
I f with each two elements (x,y) of an abstract set S,
Is associated an element _a of a l a t t i c e L with le a s t element
0 , the association being denoted by a = d (x ,y ), the r e s u l t ­

ing structure is called a lattice-m etrized ,
an L-metrized) space, provided th a t

(or more b rie f ly

(1) d(x,y) = 0 If and

-

only i f x = y,

2-

(2 ) d(x,y) = d (y ,x ), and ( 5 ) for each three

elements x ,y ,z of S, d(x,y) + d(y,z) > d (x ,z ), where + Is
the addition of the l a t t i c e

and > the order r e la tio n , read

'"over1® (in the wide sense)* That t h i s association may be
reasonably regarded as a "metrization1®of S is suggested by
the formal resemblancer at l e a s t , of the specified conditions
to the usual postulates fo r distance in a metric space* I f ,
in p a r tic u la r , S 5 L, th is association defines a binary
operation on L, termed a metric operation, and the l a t t i c e
Is said to be autometrized.
The studies of E l l i s , Blumenthal and E l l i o t t referred
to above were concerned with a p articu lar autometrization of
a Boolean algebra* E llis observed that in such a l a t t i c e ,
the symmetric difference ab 1 + a*b of two elements a and b
where a 1 Is the complement of a, is a metric operation In the
sense described above* Since, however, a Boolean algebra may
be metrized in other ways,

(for example a l l distances between

d is tin c t pairs of points may be set equal to the same element
a /

0), the term "autometrized Boolean algebra1
* will be used

in th is t h e s i s ,
the postulates

i f the metrization i s any one which s a t i s f i e s
( 1 ) , ( 2 ) ,( 3 ) ,

above, the designation "Boolean

Geometry1
* being reserved for the special

autometrization of

symmetric difference*
I t i s well known that a Boolean algebra is a ring under
the operation of symmetric difference as the addition of the
ring ,

and, indeed, E l l i o t t has shown th a t the only operation

possible in a Boolean algebra which Is simultaneously a

metric operation and a group operation is the symmetric
difference [If]. The group property of th is operation has
in te re stin g consequences fo r Boolean geometries which will
be discussed later*
This thesis is concerned with the extension of certain
properties of Boolean geometries to a somewhat wider class
of spaces called Brouwerian geometries* A further concern is
with properties of L-metrized spaces in general# What i s of
special In te re s t over and above the geometric properties
per se of these spaces i s the Interplay between these
geometric properties o f a space and the algebraic structure
of i t s underlying la ttic e *
Section 1 contains known r e s u lts used throughout th is
thesis* In section 2 are developed in d e ta il properties of
Brouwerian algebras, many of which are stated without proof
in [1],

[2] and [ 8 ]* Brouwerian geometries are introduced

and studied in section 3 *

p a r tic u la r , numerous character­

ization theorems for Boolean algebras are obtained, and the
congruence order of certain Brouwerian geometries is

estab­

lished* In section if, L-metrized spaces in general are
studied* Further, the notion of metric betweenness in these
spaces is

introduced, and consequences fo r the structure of

the underlying l a t t i c e ,

of the coincidence of metric and

l a t t i c e betweenness are derived*

S e c t i o n 1 . P r e li m i n a r y R e s u l t s
The following l a t t i c e theory r e s u l ts are used through­
out th is paper. Details may be found in [1].
A p a r tia l ly ordered set is a co llectio n of elements
together with a binary re la tio n defined on the s e t, which
is re fle x iv e , asymmetric, and t r a n s i t i v e . Denoting the
r e la tio n by the symbol <, read “under11, the three axioms
s a tis f ie d by a p a r ti a lly ordered set are:
(1) For a l l a, a < a
(2 ) a < b

and

b < a

imply

a = b

(3 ) a < b

and

b < c

imply

a < c.

a < b may also be written b > a and read b is "over" a.
I f the order r e la tio n does not ho2.d for two elements a and b,
they are called non-comparable, otherwise , they are called
comparable•
In representing a p a r tia ll y ordered set by a diagram,
a > b

i s indicated thus

, whereas i f a and b are not

comparable, they appear thus, a©

®b.

By an upper bound to a subset X of a p a r ti a ll y ordered
set P is meant an element a£ P such that a > x for every
x€X*

A le a s t upper bound (or l . u . b . )

is an upper bound

which is under every other upper bound. Clearly a l . u . b . i s
unique. Lower bound and greatest lower bound ( g . l . b . )

are

sim ilarly defined. A l a t t i c e is a p a r ti a ll y ordered set in
which every pair of elements has a l . u . b .

and a g . l . b .

In

th is th e s is , these are denoted respectively by a + b and ab

and are called the sum and product respectively of a and b,
(although "join",

indicated by the symbol^ , and "meet",

indicated by the sym bols are also found in the l i t e r a t u r e . )
Each operation i s called the "dual" of the other. They are
readily shown to be idempotent, commutative and associative,
and s a t is f y the absorption laws
Further, a > b i f and only i f

a + ab = a, and a(a + b) = a.

a + b = a and ab = b . A l a t t i c e

is said to be complete i f any collection of elements has a
g .l.b .

and l . u . b .

In p a r tic u la r , a complete l a t t i c e has a

le a s t element 0 and a g reatest element I . A chain is a
l a t t i c e in which each two elements are comparable , and Is
said to be lin e a rly ordered.
A l a t t i c e i n which a < c implies (a ♦ b)c = a + be is
called modular.and t h i s "weak" d is trib u tiv e law is called
the modular law. A d is trib u tiv e l a t t i c e i s one which s a t i s ­
f i e s the d is trib u tiv e laws

a + be = (a + b)(a + c) and

a(b + c) = ab + ac. Each of these implies the other.

S e c t i o n 2 . B r o u w e r ia n a l g e b r a s
In th is section, a Brouwerian algebra i s defined and
some of i t s most important properties are derived. Many of
these will be used in l a te r sections of th is th e s i s .
Definition 2 .1 .

I f with each two elements a and b of a

l a t t i c e L having a greatest element I , there is associated a
le a s t element x such that b + x > a, then x i s denoted by
a - b and the l a t t i c e i s called a Brouwerian algebra [8 ].
Thus, i f b + z > a, then z > x. This association will be
referred to as the subtraction operation.
I t i s of in te r e s t to note th at such an algebra Is
equivalent to a l a t t i c e formulation due to G. Birkhoff [2]
(and called by him a "Brouwerian logic") of Heyting*s post­
u la tes for " i n t u iti o n i s t logic",

(a logic consistent with

the philosophy of the in t u i t i o n i s t school of mathematicians,
whose leading exponent i s L. J. Brouwer). I t i s
dual of a r e la tiv e ly pseudo-complemented l a t t i c e

also the
[ 1 J.

I t i s convenient in practice to make use of the following
Theorem 2 .1 .

A l a t t i c e L with an I is

a Brouwerian algebra,

i f and only I f , for each three elements x ,y ,z of L,
x -

y < z <- —- > x < y +z.
To

show f i r s t

original d e fin itio n ,
poset,
x <

that th is characterization impliesthe
since by the f i r s t postulate for a

x - y < x - y, i t follows that x - y < z

>

y + z Implies x < y+ (x - y ) • Thus (x - y) is an

element

of L which added to y

yields an element over

x.

-

And now x < y + z

> x - y < z

7-

implies th a t x - y is the

le a s t such element*
To show th a t the o rig inal d e f in itio n implies the above
ch aracterization,

suppose x - y is the le a s t element of L

such th a t y + (x - y) > x* Then y + z > x implies z > x - y.
On the other hand, y + (x - y) > x
and i f

implies

z + y + (x - y) > x,

z > x - y, then z + (x - y) = z, hence z + y > x. This

completes the proof*
In what follows, i f
of a l a t t i c e L, U x ^
elements

x^

Theorem 2*2*

X

is a subset of elements

x^

designates the l a t t i c e product of the

of X*
I f L is a Brouwerian algebra, then

(i) L has a le a s t element 0 *
(ii)

L is a d is trib u tiv e l a t t i c e ,

i*e* f o r a l l elements

Y p Z p of L,

(ill)

x

+ yz = (x+ y)(x + z), and dually

x

(y + z) = xy + xz

The d is trib u tiv e law for f i n i t e addition with respect

to i n f i n it e m ultiplication holds, i*e* If X is any subset of
elements
element
(1 )

x* of L such that XIxo< e x is ts ,
a of L,I I (a + x ^
a + Xlx^

then for every

} also ex ists and

= n (a + x<*

)

Proof of ( i ) : This follows immediately from the d e fin itio n
of a Brouwerian algebra, since I - I ex ists and clearly Is
the le a s t element, denoted by 0 .
Proof of

( i i ) : In any l a t t i c e L, for a l l elements

8-

x, y, z, of L)
(x + y) > (x + y)(x + z )
Thus y, z €

u 9 where

such th at x

+u^

I I u^
y >

is the class of elements

u^

> (x + y) (x + z) * By d e fin itio n ,

= (x + y) (x + z) - x e x is ts , and since
(x + y)(x + z) - x, and z >(x + y)(x + z)

yz >(x +
yz + x

X
J

and (x + z) > (x + y)(x + z ).

y)(x + z) - x,

- x, then

hence

>[(x + y)(x + z) - x] * x > (x + y) (x +z) by

d efin itio n i . e . yz + x > (x + y)(x + z).
But in any l a t t i c e , the one sided d istrib u tiv e law
yz + x < (x * y)(x + z)
hence,

holds

yz + x = (x + y)(x + z ),

and the proof is complete.
Proof of ( i l l ) . Clearly, IIx * < x*
a + IIx^

<

for every

a+ x^

for every <

implies

• Suppose, then, p < a + x^

• Then p < (a + IIXo< ) **

Theorem 2*1, p - (a + IIXc< ) < Xc* ,

, hence by
and since th is holds

for every

, i t follows that p - (a + IIx ^ ) < IIx^ , and

Theorem 2.1

again implies p < a * IXx^ •

Thus,

IIx* = I I ( a + x ^ ) by d efin itio n of the l a t t i c e

a +

product of a set of elements.
Remark *

I t should be noted that d is t r i b u t i v i t y alone is

in s u ffic ie n t to insure t h a t a l a t t i c e be a Brouwerian
algebra. The open subsets of the plane, for example,
constitute a d is trib u tiv e l a t t i c e . But If a and b are two
open c irc le s with a non-null in terse ctio n , a - b f a i l s

to

e x i s t , for the le a s t s e t u,

set

such that b + u > a is the

9

■ -

a - ab of elements of a not already in b. This s e t i s not
open, since i t s complement i s not closed* Moreover, any
open set containing those points of a not already in b must
contain^ in p a rtic u la r , a neighborhood of each point o f a - ab
which i s an accumulation element of the c ample ment of a - ab*
Since each of these neighborhoods may be a r b it r a r i l y small,
there i s no le a s t open set with the required property*
However, one does have
Theorem 2*3»

A complete l a t t i c e in which the d istrib u tiv e

law for f i n i t e addition with respect to in f in ite m ultip lica­
tion holds i s a Brouwerian algebra*
Proof•

Let L be a complete l a t t i c e

in which the d is trib u tiv e

law for f i n i t e addition with respect to i n f in it e m ultip lica­
tion holds,. i*e. for every subset X of elements x* £ L ,
a + IIx^ = I I (a + x* )• I f x, y € L, then since y + x > x,
the class U of elements Uo< such that y + u*< > x is not
empty. Since L is complete, II u

and II(y + u* ) e x is t.

Moreover, II(y + u<* ) > x, and since II(y + u^ )-y + II u^
by hypothesis, i t follows th a t y ^ I I u^ > x* Hence
I I u<* - x - y, by d efin itio n of the subtraction operation,
and L is a Brouwerian algebra.
Corollary 2*3*1
Every f i n i t e d is trib u tiv e l a t t i c e i s

a Brouwerian

algebra.
Proof: Such a l a t t i c e i s a complete l a t t i c e
of Theorem 2*2,( i i i ) holds.

in which (1),

-

Remark*

10-

A non-complete l a t t i c e which enjoys the d is trib u tiv e

law in question may or may not be a Brouwerian algebra. The
unit in terv al with an in te r io r point deleted, and the unit
square without the point (1 , 0 ) are l a t t i c e s where
xl*^l > x2 #^2 ^

and °nly ^

X1 —x2* 811 ^

—y2* Each

enjoys the law i n question whenever the products involved
e x is t . Neither is complete.

The former is a

algebra; the l a t t e r i s not,

since [I

for

any x,

Brouwerian

- ( x , l) ] f a i l s

to exist

( 0 < x < 1 ).

Theorem 2.1+.

Every chain with I and 0 is a

Brouwerian

algebra.
Proof.

For any two elements a,b of L with a > b, clearly

a - b = a, and b - a = 0. Hence L is a Brouwerian algebra.
Definition 2.2

An element a of a subset X of a p a r ti a lly

ordered set P is a minimal element of X, i f f o r no element
x of X is a > x.
Definition 2.5

A p a r tia lly ordered set P is said t© s a tisfy

the descending chain condition,

i f and only i f

every non-void

subset X of P contains a minimal element.
Theorem 2 .3 *

Every d is trib u tiv e l a t t i c e L with a g re a te st

element I , which s a t i s f i e s the descending chain condition
Is a Brouwerian algebra.
Proof.

For a rb itrary elements a,b of Lssince b + a > a, the

set X of elements x ^

such th a t b ** x^ > a Is non-void. Then

X contains a minimal element x by hypothesis • Let

y

also

a minimal element of X. Now xy€.X, for xy exists and
b + x » a, b + y > a imply by the d is trib u tiv e law th at

be

-

11-

(b + x)(b + y) = b + xy > a. But then neither x nor y would
be minimal in X since xy < x, and xy < y. Hence there can be
at most one minimal element x in X#
Suppose x is not the le a s t element in X. Thenthere
e x ists an element z in X with z ^

x.

By our previous argument, xz£X, i . e . b + xz > a. Since
xz < x, again x i s not minimal, a contradiction. Thus x is
the l e a s t element of X, and a - b e x is ts .
Definition 2 .3 # A Boolean algebra
tive l a t t i c e ,
la ttic e ,
a

is a complemented d is tr ib u ­

i . e . corresponding to each element a of the

there e x ists an element a* called the complement of

such that a + a* = I and aa’ = 0. I t i s readily shown that

complements are unique and th at complementation i s orthocomplementation, i . e .
Theorem 2 .6.

( a 1)* = a . £ l ] .

Every Boolean algebra B is a Brouwerian

algebra wherein ab® = a - b, fo r a, b 6 B.
Proof.

I t w ill be shown that b + ab® > a, and
i f b + x > a,

Since b
whence

then

+ b®= I and la = a,

x > ah’ .
a > a implies(b + b ') a >

a,

ab 4* ab® > a by d i s t r i b u t i v i t y . Moreover, since

b > ab i t follows that b + ab® > ab + ab® > a, i . e .
b + ab®

> a. Furthermore, i f

and bb®

+ b* x > ab®, i . e . b'x

b + x > a, then b»(b + x) > ab®
> ab® since

bb* = 0 . But

x > b*x > ab® so that x > ab®. Thus ab® =
* a - b, and B is a
Brouwerian algebra.
Thus I t

is evident th at Brouwerian algebras comprise a

ra th e r large class of l a t t i c e s ,

including as they do, all

-

12-

chains with I and 0, f i n i t e d is tr ib u tiv e l a t t i c e s , d is tr i b u ­
tive l a t t i c e s

satisfying the descending chain condition,

complete l a t t i c e s satisfying the d is trib u tiv e law for f i n i t e
addition with respect to in f i n ite m ultip lication,

and the

Boolean algebras* Attention is called f i n a l l y , to the
following
Theorem*

The algebra of closed sets of a topological space,

and every subalgebra of th is algebra i s a Brouwerian algebra*
Conversely, every Brouwerian algebra i s

isomorphic to a

subalgebra of the algebra of closed s e ts cf a topological
space* [ 8 ]*
Theorem 2*7*

In a Brouwerian algebra, the following r e l a ­

(ft) a - b < a

(b) a < b i f and only i f

<c) a - 0 * a

(d) a - a = 0

( e ) a + (b - a ) ® a + b

It)

a < b implies a - c < b - c

(h)

(a ►b) - c = (a - c)+(b - c)

(j)

a < b implies c - b < c - a

( 1)

(a -- b) + a = a

(a) a - b = b - a

a

i
o'
II
o

tions hold;

(g)

( a - b) - b = a - b
(i)a

(k)

- be = (a - b )x (a

(a - b) +■ ab = a

(m) a - b < a + b
i f and only I f

a = b.

(o) a - b < a - be

( p) c + (a - b) = c ^ l(c + a) - (c + b ) ]
Proof of Theorem 2*7*
(a) a - b < a
Proof *

a < a + b

implies

(b) a < b i f and only i f

a - b < aby Theorem

a • b = 0

proof. By Theorem 2*1, a <b + o implies
hence a -

2.1

a - b <0

b = 0* Moreover, i f a - b = 0, then a - b

< 0 and

-

again by Theorem 2.1

a<b

13-

+ 0, i . e . a < b .

(c) a - 0 = a
Proof.

By (a) a - 0 < a,

Hence

and a - 0 > aby d efin itio n

a -0 = a

(d) a - a = 0
Proof. This follows from the defin itio n of subtraction.
(e) a + ( b - a ) = a ^ b
Proof #

By (a) b - a < b. Hence a + ( b - a ) < a ^ b

but a 4 (b
a + (b
(f)

- a) > b by d e fin itio n of (b

- a) > b + a, I . e . a + ( b - a ) = a + b

a < b implies a - c < b - c.

Proof. b < c ^ ( b - c ) b y
implies a < c 4 ( b - c ) ,
(g)

d e fin itio n . Moreover, a < b
hence a - c < b - c by Theorem 2.1

(a - b) - b - a - b.

Proof.
(a-b)

( a - b ) - b < a - b

(a-b) < [(a-b)
a

and

a - b

Hence

< [ (a -

-

(a - b) - b < (a - b) - b,
b] 4-b,

b) - b] + b,

<[(a-b)-b],

by Theorem 2.1

(a-b) - b = a - b

(a + b) - c = (a - c)

Proof. a < c
Hence

by (a). To show th at

- b > a - b, clearly ,

therefore

(h)

- a ) . Hence

* (b- c)

+ (a - c) and b < c + ( b - c ) b y

a + b <c

+ (a-c)

+
■ (b - c) ,

i t follows that (a + b) - c <

(a-c)

d e fin itio n

and by Theorem 2.1,
4- ( b - c ) .

To show the reverse inequality,
by (a),

(a*b)»(a-c)^(b-c)=*[ (a-c)-c]+[ (b-c)-c] by (g).

Hence (a*b) -c> | [ (a-c ) - C '] +[ (b-c ) -c ] ^ -c

-

14-

> j 4 ( a - c ) - c ] - c j + | [ (b-c)-c]-cj by ( f ) .
Therefor©
Hence
(i)

(a+ b) - c > (a - c ) + (b -

(a + b) - c =
a - be = (a

Proof#

- b)

c)

by (g)*

(a - c) + ( b - c )
+ (a - c)

a < b + ( a - b )

and a < c ^ ( a - c ) b y

d e fin itio n ,

hence a < [b + (a-b)][c + (a-c)]
or

a < be + (a-b)c + b(a-c) + (a-b)(a-c)

and a - be < (a-b)c + (a-b)(a-c) + b(a-c) < (a-b) + (a-c)
i*e*

a- bc<

(a-b) +
■ (a-c)

To show that the reverse inequality holds,
since a < a + b + be, then a - b < a + be by Theorem 2*1
But by (e),

a + be = be + (a-bc), hence

i*e*

a

- b < be + (a-bc)

and

a

< b#

by Theorem 2*1

be + (a-bc)

a < b + (a-bc), so again by Theorem 2*1
a - b < a - be

In lik e

manner wecan show th a t
a - c < a - be

Thus (a-b) + (a-c) < a - be, and therefore
a - be = (a-b) + (a-c)•
(j)

a < b implies c - b < c - a

Proof*

I f a < b , then ab = a, hence
c - a b = c - a ? and by (i )

(k)

c

- a = (c-a) + (c-b)

c

-b < c - a

i.e .

(a-b) + ab = a

Proof*

a - b < a

by (a) and ab < a, hence

-

(a-b) 4 ab <

15-

a

To show the reverse

Inequality,

Henoe a(a-b) 4 ab >

a, and since a - b < a, a(a-b) = a - b ,

therefore

(a-b) 4 ab >
(a-b)

(a-b) 4 b >a

by Theorem2.1

a, consequently

4 ab = a.

( 1 ) (a-b) 4 a = a
Proof*

Since a - b

< a, i t follows th at ( a - b )

4 a

= a.

(m) a - b < a 4 b
Proof *

a - b < a < a 4 b .

(n) a - b = b - a
Proof *
a -b

i f and only i f

a = b

The sufficiency is obvious. To show the necessity,

< a , b - a < b by (a). But since a - b = b - a ,

b - a

< a and a - b < b.

Hence by Theorem2.1

b < a

and a < b, i . e . a = b •

(o) a - b < a - be
Proof.

Since be < b, th is follows from ( j ) .

(p) c 4

(a-b) =

Proof.

Since

(c 4 a )

-

follows

that

(c 4 a)

< (c 4 b) 4 [ (c 4 a) - (c 4 b)]

(c 4 a )

- b < c 4 ((c 4

and

4 [(c+a) - (c4b)]

c

(c + b) < (c4 a) - (c

a) -

+ b), i t

(e 4 b)]

by Theorem 2 .1 .
Now ( a - b )

< a < c 4 a , hence ( a - b )

by (a) and ( f ) ,
Thus

c

4

(a

-

b)

- b<

(c 4 a) - b

and therefore, a - b < (c 4 a) - b
<

c

4

[(c

4

a)

-

(o

4

b)].

The reverse inequality is established thus:
since

c 4 b 4 a > c, i t follows th at c 4b > c - a

by Theorem 2.1, hence

b 4 ( c - b ) > c - a by(e)

by ( g ) .

-

and b + (c - b) > (a +
(b

+ a) + (c

(a - b) + b

Then

and( a - b )

and

so that

- b) > (a + c) by Theorem 2* 1.
(c - b) > (a + c)

^ (b

(a-b)

c) - a by Corollary 2.7 ( h ) . l .

16-

+ c) > a + c by (e)

so t h a t

> (a ^ c) - (b +
■ c) by Theorem 2*1 again,

c + ( a- b) >c

+ c) - (b + c)]

Thus, f i n a l l y , c + ( a - b )

= c + [(a + c)

- (b ^ e)]

and the proof is complete •
Corollaries to Theorem 2.7
Corollary 2.7 ( h ) . l .
Proof.

(a + b) - b = a - b.

Set c = b in (h)

Corollary 2.7 (i)«l#

a - ab - a - b.

.Proof. Set c = a in (1)
d e fin itio n 2.3

The element I - x of a Brouwerian algebra is

called the Brouwerian complement of x,
Similar ly?
Remark.

~|x - I - “ |x .

I t i s clear from Theorem

complement

and is denotedby “j x.

|x coincides

Boolean case. I t

2.6 that

the Brouwerian

with the usual complement

x* in the

should further be noted th a t In the

Brouwerian algebra of the closed subsets of a topological
space K, the Brouwerian complement of a (closed) subset A is
merely the closure of the usual (Boolean) complement of the
set A in the space of a l l subsets of K.

-17-

Theorem 2 .8 .

In a Brouwerian algebra, the following r e la

tions hold;
(a) a < b implies *
”']b <
(c) ~]0 * I ,

a

(b) a + ~ ] a = I

- |l = 0

(d) T ] a
(f)

- 1 a

( a) I l i a

(g) ~] ( a + b) < ~] a • ~| b
(i) a = a

| a

< a

| ( a b ) = ~]a + “ )b

( h ) - ] ( a " | a) = I

+ ] ~) a

( j ) ~ |a = ~j b

= I implieal(a + b) = X

( k ) ~1 (a + b)

= l ”|(~ |a • b)

(1) -] (a - b)

= ~]a + H l b

(m)

(a

(n)

(i)
(ii)

+ b) = I i f and.

only i f

" |» =1& - a
H a

a > l b andb > ~l a

( i i i ) 1 H a = “J H a

= l~ |a - 1 »

<iv) l a

=1&

-~ n «
- H a

(o) 1 x = 0 implies x = I .
Proof of Theorem 2 .8 .
(a) a < b implies 1( b < ~]■ a
Proof.

This follows from the defin itio n o f ~ j x

(b) a + ~“| a
Proof,

= I

a +a > I by d efin itio n

x > I implies
(c) 1 0 = 1 ,
Proof.

x •= I . Hence

of

a

+

a.

But

=I.

11 = 0

These follow d ir e c tly from the d e fin itio n .

( d ) 1 1 a < a.
Proof.
(e) H
Proof.

and Theorem 2.7 (j)«

This follows from (b) and Theorem 2.1
l a

=la

By (d) and (a),

a >

a.

-18-

But again by (d) , H ( 1 a) < ~|a. Hence~n~) a =-] a
(f) 1

(ab) = -|a +“]b

Proof» This follows d ire c tly from Theorem 2 . 7

(i)

(g) " | (a + b) < “|a •~|b.
Proof .

By ( d ) , l " ] a < a#~| ~1^

“I") a + H b < a + b .
hence

(a

1

By

(a)

hence

l ( a + b)

+b) < 1 ( 1 ( ] a * ! b ) )

<1 (H
^ b

a

+ 11b),

by (f) and

(d).

(h>! ( a l a) = I
Proof. By Theorem 2.7 (i)^

1
(i)

( a l a)

= 1 -

a-] a

=

(I

-

a)

+ (X ~1

a)

=l

a

+1 1 * = I

by (b).

a = al a +H a

Proof. a(~ja + ”)”|a) = a I = a, and a( ”|a

= a " |a +~)7a

since “|~|a < a*
(j)~ |x = "7y =I imply"] (x + y) = I
Proof* Let

x+ y # t

Then, t >"")y, i . e .

= I s o t h a t y+ t >“jx, i . e .

y + t -

I.

t - I#

Thus the only element which,

addedto

element over I ,

Hence

is I i t s e l f .

with th is property, i . e . ~ ^ ( x + y )
(k) ~~](a + b) =0~l(~)a

b)

(x + y) yields

an

I is the least element
= 1*

[Dual of Theorem 12.21 (vi) p.i|2

1131Proof. 1 (a"] a) = J<

| b ) = I , by (h) .

Hence by ( j) with x = a~| a,
"] ( a ] a

y = b~] b, i t

follows th a t

+ b~] b) ="] [ (a+b) (a+ ") b) ( *] a+b) (-]a+“7 b) ] = I .

How (a^b)“ja^]b < ( a+b) (a+7b) (7a+b) ("|a+j1 b) implies

(by(a))

"][ (a^b)'|a"]b] >~[l (a*b) (a+”]b) (-]a*b) (^a+’lb) ] = I .
By (f)

then, ~](a*b) + ~)(]a7b)

= I , hence I ~7(7a7b) <7(a+b)

-19-

i .e .77(7a~)b) < 7 (&^b). Further, “7 "]a < a, ”]"jb < b
7 7 a ^ 7 7 b < a^b

imply

so th a t by (a)

”1 (a+b) <"] (-)-j a+T1b) = 77

a *7 b)

( 1

(by f)

or'"'j(a+b) = 77^7 a •‘~]b), which completes the proof*
(l)^ (a -b )

= " ](a ^ b ) = “| a 4 ’7~lb *

Proof * ~~| (a-b) = I - (a-b)

is the le a s t element which

added to (a-b) yields I . What must be shown is th a t
(a-b)

~| a + 7 7 b = I and (2) i f

then

x >

a

(1)

(a-b) + x = I ,

b•

To estab lish

(X), l e t a-b = y, then a < b + y,

7 a > “|(b+y) = T l ( l b

•*] y),"]~[a < ”|( '] b

• “) y) = 17 b +717*

and 7 7 a “l i t ) <7~jy* i •e . 7 7 a “1 7 b < 7 7 ( a*
“b ) < (a“b )
by (a),

(e),

( k) , ( f ) » and (d).

Therefore7 7 a < (a-b) ^ 7 7 b ,
hence I < (a-b) ^ 7 a “^ l l b #
To estab lish

i.e* I - 7 a < (a"b) * 1 1 b,
i*e* X = (a-b) * 7 a ^7~]b.

(2), sine© (a-b) +""|(a~b ) ~

it; follows

from the d efin itio n o f 7 ( a~b) that x> 7 (a*b ). Now a-b) < a
and a-b < 7 b imply

a-b < a 7 b , hence 7 ( a7 b ) < 7 ^ a~b )*

Thus x > 7 ( a*"b) > 7 (a 7 b) " 7 a + "*)7b by (f)# which completes
the proof•
(m) a+b = I i f
Proof.

and only i f

a > 7 b and b > 7 a *

a + b > I implies a >7b and b > 7 a

by ‘Theorem 2.1.

Conversely, a > I - b and b > I - a imply a + b > I ,
a + b = I#

I.e .

-

20-

(n) (i) 7 a = 7 a - a
(ii)77a =77&- ^ a
( i i i ) 7 7 7 a =H7a - 7 7 a
(iv)

7 a =7a -77 a

Proof .
while

(i),

(ii)

and ( i i i )

follow from Theorem 2.7 (g),

(iv) follows from (n) (iii) and (e) of th is theorem.

(o) 7 x = 0 implies

x = I

Proof. I - x = 0 implies I - x < 0, hence I < x by Theorem 2.1,
hence x = I .
Corollaries to Theorem 2.6
Corollary 2.8 ( k ) . l . 77^a4b^ = 7 7 a
Proof. 7 7 ( ft4hb) = 7 7 7 ( 7 a • > )

= 7 ( 7 a * 7 b) =7 7 a *1~lb

by (e) and ( f ) .
Corollary 2.8 (k) .2 . ~~] (x + y) = i implies 7 x =7 7 = 1
(Converse of ( J ))
Proof. i = 7 ( x + y) =77( 7 x
0 =7 1

= 117(7 x

• 7 y) “ 1 ( ] x

*7 y)> and

*7 y) = 1 7 X + 7 1 y b y (e ) and ( f ) .

But7 7 x +77y - 0 implies 1 7X =117 = 0,
hence7l7x =7“)17 = 7 0

= 1 , a n d ^ x = ] y = I by

Corollary 2.8 ( k ).5 » “71 x " 1 1 7

(©).

“ 0

implies (1) 7 ( x + y) = I and ( 2) l 7( xy) - 0.
Proof of ( 1 ) .
hence

By Corollary 2.8 ( k ) . l , 7 7 (x ^ y) = 0,

"^(x + y) = I by (o).

Proof of ( 2 ) . 7 7 (x y ) = 7 ( 7 x + 7 y) =17 ( 1 7 x *71 y) =77 0 = 0,
by (f)
Remark.

and ( d) •
It

should be noted th a t

property (k), since

(2) holds independent of

xy < x im pliesl 7 (xy) <7 l x = 0

-

Corollary 2.8 (k) .U. H l (
Proof*

a *l~]a)

21-

= 0

Follows from (k)j replacing b by ~j a.

Corollary 2 . 8 . ( 1 ) , 1 . ” ) (a - b) < " ] ( T ) a -T? la)
P roof»

Follows from canplementation on-)-) a -'J'jb < ~ ] ~ ] ( a - b ) ,

and (©) •
Corollary 2 .8 .( 1 ) .2 .
Proof*

(a -~| a) = 7 a

Obtained by setting

Corollary 2 . 8 . ( 1 ) . 5 *

b ="] a in (1) •

1 ( 1 a - 1 7 a ) “ “l l a

Proof♦ Obtained by s e ttin g a = “] a in Corollary 2 .8 .( 1 ) .2 .
Corollary 2 .8 . (1) .1+. ~| [(a - b) + (b - a)] = ~[ (a + b) +*7 "] (ab)
Proofs

(a+b) - ab - [(a+b) - a] + [(a+b) - b ]

by Theorem 2*7 (1),

= ( a- b) +( b- a)

(h), and (d).

Hence - 7 [(a-b) + (b -a )] =”][(a+b) - ab] = ”|(a^b) +17 (ab)
Remark.

Due to the prominence of the Boolean algebra of s e ts ,

for example, in many areas of mathematics, i t

is of some

in t e r e s t to see just how the Boolean algebras d iffe r from the
more general Brouwerian algebras. The following properties,
which have been discussed in th is section, show how these
two classes of algebras compare
Boolean algebras

Brouwerian algebras

1.a)

x + x® = I

l#b)

x + ~[ x = I

2 . a)

x • xf = 0

2.b)

x"]x ^ 0 in general

5 *a)

Ij-.a) (xy)
5 . a)

5 .b) ” p x

( x 1)* - x
= x* + y»

(x + y) * = x f • y»
Clearly 1.)

< x

U.b) *7 (xy) = *7 x +7 y
5 .b) “ j (x + y) <”]x • ”"]y

and J+.) hold equally in both cases. In any

chain 5a holds^and a chain of more than two elements Is

-

22-

A Brouwerian algebra which i s not Boolean, so th a t 5a ) does
not have su ffic ie n t strength to make a Brouwerian algebra
Boolean. However, e ith e r of 2a) or Ja) does. Thus one has
Theorem 2.9*
if

A Brouwerian algebra L Is a Boolean algebra

and only i f , for a l l elements x of L,

x^x = 0

Proof. The necessity i s part of the d e fin itio n of a Boolean
algebra. I f ,
also

on the other hand, x 7

x + 7 x —I ,

x = 0 for all x, since

L is a complemented,distributive l a t t i c e ,

hence a Boolean algebra.
Theorem 2.10.
if

A Brouwerian algebra L is a Boolean algebra

and only i f , for a l l elements

Proof .

x of L,7[7X = x.

The necessity i s well known, following from the

uniqueness of complements In the Boolean case [1]. The
sufficiency follows readily from Theorem 2 . 8 . (h), f or ,
~|

(x7 x) = I

im p lies7 7

( x "1 x )

the hypothesis,7'7(x7 x) * x 7

=7 1

= 0. But as follows from

x, hence x “ | x = 0 and L is

a Boolean algebra by Theorem 2 . 9 . A further c rite rio n is given by
Theorem 2.11.
I f and only i f ,

A Brouwerian algebra L Is a Boolean algebra,
each element z of L i s the complemont of

some element x of L.
Proof.

The necessity follows from ortho-complementation,

since z

= 7 7 £ • For the sufficiency, 5 58 7 x implies

7 7 z =777 x = 7 x = % , i . e . 7^7 2 = 2

svery element z

of L, and L is a Boolean algebra by Theorem 2.10*
I t may be observed f i n a l l y , that a fundamental d i f f e r ­
ence between Boolean algebras and Brouwerian algebras is that
Boolean algebras are dual with respect to the sum and product
operations while Brouwerian algebras may f a i l t o be.

[8].

-23-

S ection
In th is

3* B r o u w e r i a n G e o m e t r i e s

section Brouwerian geometries are introduced,

and many of t h e i r properties are derived. These lead to
numerous characterizations of Boolean algebras.
As indicated in the introduction,

a l a t t i c e may be

(auto)metrized in a variety of ways. In the case of
Brouwerian algebras, the p articu lar metrization which will
be used here is that of symmetric d ifference, namely,
a

b = (a-b) + (b -a ).

The orem 3 .1 .
algebra i s

The symmetric difference in a Brouwerian

a metric operation.

Proof♦ Clearly (1) a * b = (a-b) + (b-a) = b # a
(2) a * a = Oj moreover, i f

(a-b) + (b-a) 38 0, then by

d e fin itio n of the l a t t i c e sum,
Hence

0 >b - a and

0 > a - b.

# b< (b * c)

+ (a

a > b, b > a and a = b.

(3) The triang le

inequality,

a

c ),

is established as follows:
abc + (b # c ) +(a # c) * abc + [ (b-c )+ (c-b) ] * [ (a-c )+(c-a) ]
* abc + [ (a-c)4-(b-c) ]*[ (c-a)+(c-b) ]
- abc + [ (a4-b)-c) +[c-ab] by Theorem 2.7 (h) and (i)
= (ab)c

+[c-ab]

+ [(a^bj-c]

=

c

+ [(a^bj-c]

s=

c + (a+b)

by Theorem

2.7 (k)

by Theorem (2.7) (e) •

Hence
a^b+c <

abc +(b

* c) +
■ (a * c) and

(a^b^c)

- abc < (b # c) ^ (a * c) byTheorem

2.1.

-2 ^ -

Now (a+b+c) -

abc > a - abc > a - b

and (a+b+c) -

abc > b - abc > b - a by

so that

+ (b-a) < (a + b + c) - abc.

(a-b)

Therefore

Theorem 2.7 (f) and ( j ) ,

(a * b) < (b * c) + (a * c)

Theorem

In a Brouwerian algebra,

(a^b) - ab and (a-ab) + (b-ab) are each equivalent to
symmetric d ifferen ce,
Proof.

(a-b) + (b-a).

Immediate from Theorem 2.7 (h),

Definition 3«1«

(I),

and (d).

A Brouwerian algebra (auto)metrized by the

symmetric difference i s called a Brouwerian geometry.
I t Is often convenient to employ geometrical language
and regard a t r i p l e

of elements a,b,c as the vertices of a

triang le with sides a
Theorem 3*1

b,

a * c, and b * c.

asserts that the sides of any triang le in a

Brouwerian geometry s a ti s f y the triangle inequality. The
notation A(a,b,c) will be used to designate the triangle
with vertices a ,b ,c .
One reason why a Boolean geometry has so many novel
properties i s that the symmetric difference in that instance
i s a group operation. This i s not true for Brouwerian
geometries, since Brouwerian symmetric difference is not in
general associative,

(although i t does have the remaining

group p ro p e rtie s). In a Brouwerian chain, for example,
chain ( auto)metrized by symmetric d ifferen ce), for
(a * a) # b = b, whereas a # ( a * b ) = 0 «

(a

a > b,

Thus, because of

the group property, a Boolean geometry can have no isosceles
tr ia n g le s , since the equation a * x = b has one and only one

-

25-

so lution. Brouwerian geometries, on the other hand, may
abound in isosceles tr ia n g le s ,

as happens, for example, in

a Brouwerian chain where one has
Theorem 5«5»

Every trian g le of a Brouwerian chain C is

Iso sceles.
Proof.

For any two elements a > b of C, a

b = a - b = a,

sine© b - a = 0 by Theorem 2*7 (*>) • Hence for any three
elements a > b > c
Theorem

C,

a ' * b —a # c

= a* However,

A Brouwerian geometry is a Boolean geometry,

i f and only i f i t
Proof.

of

Is free of isosceles tria n g le s.

The necessity, as already indicated, is clear.

To establish the sufficiency, le t x be an arbitrary element
of the Brouwerian geometry L, and consider the a ( 0 , I , x 1 x ) .
Now I * x -

(I-x)4 (x -l) =~] x

while I * x~| x = - l ( x l x )

for a l l x, hence 1 * 0

- “[ 0 = 1 ,

= I by Theorem 2.8 (c) and ( h) . Since

there are no isosceles tria n g le s , i t follows th a t
0 * x~]x = x~”]x = 0

and L is a Boolean algebra by Theorem 2.9*

Then a - b = ab®, b - a = a*b and a * b = abf + a fb,

so that

L Is a Boolean geometry.
Corollary 5.J+.1.

A Brouwerian algebra is a Boolean algebra

I f and only i f symmetric difference i s a group operation.
Proof.It has already been indicated, as i s well known, th a t
symmetric difference in a Boolean algebra i s a group opera­
tio n .

If i t I s a group operation in a Brouwerian algebra L,

then the associated geometry contains no isosceles tr ia n g le ,
and L is a Boolean algebra.

-

Remark.

26-

The above corollary holds i f the word, "group" i s

replaced by the word "associative".
Definition 5*2.

Three elements a,b,c of a l a t t i c e , L, are

said to s a tisfy the trian g le inequality i f

each i s under the

sum of the other two, written (a,b,c)T.
Remark.

Although three elements a,b,c of an ( auto)metrized

l a t t i c e may be in this r e l a ti o n , there may not be any
trian g le A(x,y,z)

in L which has sides a,b,c.(^When such a

triangle e x is ts ,

i t will be designated as T(a,b,c) rath er

than A(x,y,z)

the sides, ra th e r than the vertices are to

if

be emphasized).
Theorem 5.3*

In a Brouwerian geometry, the re la tio n

(a,b,c)T

Is equivalent to each of the r e la tio n s
(1 ) a + b = a + c = b + c = a + b + c
(2 ) a - b < c ,

( 3)

b - c < a ,

c - a < b

a # b < c < a 4 > b

(4 ) b - a = c - a, a - b =
* c - b, a - c = b - c .
Proof .
(1 ) a + b
If

= a + c s= b 4 c = s a 4 b

(a,b,c)T,

+ c

then a < b + c implies a + b

but since a + b + c

+ c<b+c^

> b + c , a + b + c = b + c . Similarly

for the remaining eq u a litie s in (1). In words, the sum of
two sides of a triangle equals the sum of any other two sides.
(2 )* a - b < c , b

- c<a,

c - a<b.

These follow from (a,b,c)T and Theorem 2.1.

In words, the

difference of two sides of a trian g le is under the t h i r d side.

-

(j)

a * b < c < a + b.

This follows from (2) by adding

(4)

27-

b - a = c -a,

in e q u a litie s .

a - b = c - b, a - c = b - c.

By Corollary 2*7 (h) , 1, for any two elements a and b,
(a + b)

- a = b - a, and (a + c) —a = c —a

By (1) of th is theorem a + b - a + c ,

likewise*

hence b - a = c - a .

Similarly fo r the remaining equalities*
To prove the converse,
(lj

a +b

+ c= a+ b

(2)

a -b

< c implies a < b + c

(3 )

(a-b)

< (a - b)

implies

( a+b)

> c

by Theorem 2.1

+ (b - a) < c

implies

a <b + c

by Theorem 2.1 again. Moreover (b - a) < ( a - b )

^ (b - a) < c

hence b < a + c «

(4)

a -c

< b - c implies

But

b -c

< b by Theorem 2*7 (a). Hence,

a < (b - c) + c < b + c,

a < ( b - c ) + c

i.e .

byTheorem

2.1.

a < b + c.

The remaining relationships are similarly obtained.
Corollary 5*5»1«

I f a and b are two elements of a Brouwerian

geometry, then ( a, b, a * b)T and ab + (a # b) = a ^ b.
Proof.

In A(0 , a , b) ,

a # 0 = a, b * 0 = b, hence (a,h#a * b)T.

Further, by Theorem 3.5(1), a + ( a # b ) = a + b
Interchanging a and b and multiplying the two equalities gives,
using the d is trib u tiv e law, [a + (a # b ) ] [ b + (a * b ) ]
= (a + b)(a + b ) , or ab + (a * b} = a + b .
Theorem 3.6.

In a Brouwerian geometry the base of an

isosceles triang le is "under" the vertex.

-

28-

P roof.
Xn the isosceles A(a,b,c),

I f b # c = x, a ^ b = a

c = y,

then by Corollary 3 .5*1,
(a »k*y)T,

(a,c,y)T and (b,c,x)T,

Hence, by Theorem 3*5(4)#
b-c = x-c, c-b = x-b, y-c - a-c, y-b = a-b
How in T(x,y,y) clearly x < y, hence
x-c < y-c

and

x-b < y-b

by Theorem 2 . 7 ( f ) .

Hence x = (b-c) + (c-b) = (x-c) + (x-b) < (y-c) + (y-b), there­
f or e, x < (a-b) + (a-e) < a, the l a t t e r by Theorem 2 . 7 (a),
which completes the proof.
I t has been observed th a t whereas a Boolean geometry
has no Isosceles tria n g le s , these may abound in a Brouwerian
geometry. However,
Theorem 5*7*

A Brouwerian geometry contains no eq uilateral

tria n g le •
Proof.

In A(a, b, c),

if

a#b

= b#c

Corollary 3 * 5 (a,x,b)T?hence a + x
by Theorem 3 . 6 , x < a ,

x <b,

= a # c =x ,

then by

= b + x = a + b. But

hence a + x = a, b + x = b,

so th a t a = b^which contradicts our assumption that a,b,c
are pairwise d is t i n c t v ertices of a tria n g le . More generally,
a Brouwerian geometry contains no eq uilateral
n odd.

"n-gon1
* for

[71

Theorem 3.8.

A Brouwerian geometry is a chain i f and only

i f every trian g le Is iso sceles.
Proof* The necessity was proved in Theorem 3*3«

-

29-

The sufficiency is shown as follows;
I f every triang le of a Brouwerian geometry L Is Isosceles,
consider A(0,a,b) where a and b are arb itrary d is tin c t e le ­
ments of L. Then a ^ b = a or a # b = b, since a * 0 —a and
b^O

—b* I f a * b = a, then b < a and L Is a chain.
I t has been indicated e a r li e r that Brouwerian symmetric

difference lacks the

associative property which makes the

Boolean symmetric difference a group operation and leads In
th a t case to the "uniqueness of solution" property. I t is
in stru c tiv e to see just how much of the associativity of the
Boolean symmetric difference remains in the symmetric d i f f e r ­
ence operation in the Brouwerian
E l li s

[ 5 ] has shown that

two pairwise d is t i n c t elements

case.

In a Boolean geometry, for any
a and b. If a * b = c, then

a & c = b and b # c = a. In words, i f a side of a triangle
equals the vertex opposite i t ,

the same i s true of the other

two sides. This is conveniently described by
Definition 5 . 5 .

If each side

opposite vertex, the triangle
(triangular)
It

of a triangle equals the
is said to have the property of

fix ity .

is readily extended by means of

Definition 3.4*

1£ x #y#z are the sides of a triangle in an

(auto)metrized space L, the triang le with vertic es x, y, z is
called the f i r s t distance triang le of the original one.
Similarly, the trian g le whose v ertices are the sides of
the f i r s t distance trian g le i s the second distance t r i a n g l e .

-30-

jlheorem 3*9*

In a Boolean geometry, every f i r s t distance

tria n g le has f i x i t y .
Proof; In A(a,b , c ) l e t a # b = x, b ^ c = y, a

—z. Then

In the f i r s t distance triang le
A(x,y,z), x

* y =(a * b)*(b * c)
= (a * c)#(b # b) associativity)

i.e .

x # y = a * c - z

since

b * b = 0.

Similarly for the remaining distances.
Corollary 3«9»1«

(E llis)

In a Boolean geometry^a # b = c

iraplie s a *■ c = b , b * c = a.
Remark.

I t i s clear from the proof of Theorem 3.9 that the

ass o c ia tiv ity alone of the metric operation implies the r e ­
s u l t . However,

triangular f ix ity

metric are not

quite equivalent. Nevertheless, the following

corollary and Theorem $ . 1 0

and the a ss o c ia tiv ity

of the

show how close this comes to being

the case.
Corollary $ . $ . 2 .

I f the metric operation of an autometrized

l a t t i c e Is associative, every f i r s t distance trian g le with
pairwise d is tin c t vertices has f i x i t y .
Theorem 3.10.

I f every f i r s t distance triangle of an auto­

metrized l a t t i c e L has f i x i t y the metric operation on p a ir ­
wise d i s t i n c t elements Is associative.
Proof •

Let x ,y ,z

be pairwise distinct

elements of an auto­

metrized l a t t i c e L, but otherwise a r b itr a r y . Then, A(x,y,z)
is the f i r s t distance tria n g le of T(x ^ y,
ard since f i x i t y
i t follows that

z) j

implies x * y = z, y *• z = x, and x -> z = y,
(x * y) * z = z * z = 0 = x * x = x * (y *• z ) .

-

31 -

In Brouwerian geometries, although f i x i t y no longer
holds in general, what does hold is
Theorem 5 *11.

In a Brouwerian geometry, each side of a f i r s t

distance triangle
Proof :

is under the opposite vertex.

In A(a,b,c) l e t b * c = x,

then (x,y,z)T and x a- y < z
Since A(x,y,z)

a * c = y, a

*■b = z,

by Theorem 3 . 5 ( 3 ).

i s the f i r s t distance triangle of A(a,b,c),

the theorem is proved. Furthermore,
Theorem $ . 1 2 .

In a Brouwerian geometry, every second

distance trian g le has f i x i t y .
Proof .

Define x, y, z as in Theorem 3 .11. Then A(x,y,z) with

u = y

z, v = x * z, and w = x * y is the f i r s t

tria n g le of A(a,b,c), while
= r

A(u,v,w) with v it w = p, u it w = q,

and u

it v

cient

to prove that p = u. By Theorem 3 .11,

i s the second distance tria n g le .

to show u < p. Since (x,y,z)T,

(4)

distance

It

p < u . It

issu ffi­
remains

(x,w,y)T, and (x,v,z)T, then

of Theorem 3*5# implies

y - z = x - z = v - z ,
inequality

and z - y = x - y

= w- y.

w < z by Theorem 3*11 implies by (j)

th a t v - z < v - w .

The
of Theorem 2.7

Likewise, v < y implies w - y < w - v .

Moreover, since y - z = v - z, i t follows that y - z < v - w,
and z

- y

= w- y

Hence

u =

(y - z ) Mz - y) < (v - w) +(w - v) i . e . u

Corollary 5.12.1.

implies

z - y < w - v.
< p.

In a Brouwerian geometry, every nth d is ­

tance trian g le has f i x i t y f o r n > 1 .

-32-

Theorem $ . 1 $ .

a Brouwerian geometry Is a Boolean geometry

i f and only I f every f i r s t distance triangle has f i x i t y .
Proo£5

necessity was proved in Theorem 3 . 9 .

Conversely,

A Brouwerian geometry which is not a Boolean geometry would
contain an isosceles tr ia n g le , say T(x,x,y).

In the f i r s t

distance tria n g le , A(x,x,y), f i x i t y would imply y = x

x = 0,

a contradiction. Hence the theorem.
Theorem 3»l4*

A Brouwerian algebra Is a Boolean algebra if

and only i f i t

admits a metric group operation. Furthermore,

the operation must be symmetric difference.
Proof.

If a Brouwerian algebra is a Boolean algebra i t

admits

a metric group operation, namely symmetric difference. As
E l l i o t t has shown [41# th is is the only metric group opera­
tion possible in a Boolean algebra. Conversely, i f

0

Is a

metric group operation in a Brouwerian algebra L, since
0

0

0 - 0, the zero element of L is the group identity and

a

0

0 = a. The A(G,a,b)

b - a < a

0

b

I

jZf

a0 b

and

by Theorem 2.1 so that a* b < a $ b.

Hence, in p a rtic u la r,
But since

implies a - b <

0

=

I,

I j Z f a ' ^ a > I * a"]a =I ,

i.e.

1 0

i t follows that a“Ja = 0 since I

must have a unique solution,

0

a~| a = I .

0 x=X

being a group operation• There­

fore by Theorem 2*9# B is a Boolean algebra, and

0

is conse­

quently the symmetric difference.
Remark.

Since any f i n i t e d is trib u tiv e l a t t i c e for example is

a Brouwerian algebra,

the above re s u lt implies that any f i n i t e

d is trib u tiv e l a t t i c e which admits a metric group operation is
a Boolean algebra. I t i s natural to inquire as to whether

-

th is is true for a rb itra ry d istrib u tiv e

33-

l a t t i c e s . The

theorems which follow have a d irect bearing on th i s question*
Definition 5»^*

I f , to each member c< of an index set A,

there corresponds a l a t t i c e
{

°T the l a t t i c e s ,

p

, by the d ire c t product*

Lo< » is meant the set of a l l

functions x on A such that x(<* ) & L «<;
Thus, an element

{

of |

for each

is a collection of

elements ^ x ^ selected one from each l a t t i c e L
\

i y »d

>

®eans

x o< * y *

ۥ

L*

every c< of A* The l a t t i c e L ©
<
. is the
and the element
^ x

x ^

is the c< th

of the d ire c t product

in £ L ^
Proof *
x^

•
of a collection of

is a l a t t i c e , with addition and m ultiplication

If

x ^ , yo<

6

+
■ y^J

> { x <\

z^>

7»c

L *< , then x ^

+

>Xo<

and { z -d

and

> x

>

D u ally,

y^

Definition 5*6*

♦ y^

. Hence

{ z^j.

+ yx^ = £X«<J

+

?

if

» then, 1x1 L =
>
<» z

= { x <* y-<J » 8X1(1

> x <* ,

> -jx ,* + y^J•

is a la ttic e .

Let S be a su blattice of a d irect product

of l a t t i c e s
d ir e c t product

»and

+ yx} >

by d e fin itio n . Furthermore,

for every «< of A, and ^ x ^

L ^

for

componentwise•

^x*<

^ z

x ^ > yu

coordinate of the element

+ ye( > yc< for every ©
< of A« He n c e j x^

and

of

and

• Furthermore,

<< th coordinate l a t t i c e ,

Theorem 5*15* The d ire c t product
la ttic e s L

of A.

L *< . Then $ is a su blattice of the

^S

of the sublattices

appearing as

of elements of S.

S o< of elements

-components (or <
=
< -coordinates)

S is called a subdirect pro due t of the

-

S oC and denoted by

54-

I t should be noted that every

element of S »< is an «>
( * component or some element o f S.
To c la r if y the concepts involved in the above de f i ni ­
tio n , l e t

L

be the di r ect product

| xy} of the chains

X and Y consisting of the points 1,2, and 3 of thex-axis
and the points 1 and 2 of the y-axis respectively.

^XY^

consistsjthen^ of the six points ( 1 , 1 ) , (l ,2 ) , ( 2 , 1 ) , ( 2 , 2 ) , ( 3 , 1 ),
( 5 , 2 ) of the xy-plane.
The subset S of points

(1,1)

and (2,2) constitutes a

su b -la ttic e of ^XY \ and at the same time a sublattice of the
d ire c t product of the sublattices

X of X (consisting of
s
the points 1 and 2 of the x-axis) and Y of Y (consisting of
s'
the points 1 and 2 of the y-axis,

i.e.

Y = Y in th is case.)
s

Then S is a subdirect product of the sublattice a X and Y ,
s
s
but not a subdirect product of X and Y, since the point 3
of X for example, does net appear as a coordinate of any
point of the su b lattice S. The sub l a t t i c e S* of ^XY^
consisting of the points

,

(1 , 1 ), (2 , 2 ), and ( 3 ,2 ), on the other

hand, jLs a subdirect product of X and Y, since each point
of X and each point of Y appears as a coordinate of some
point of s \
Attention is called to the following representation
theorem for distributive la ttic e s
Theorem.

Any d is trib u tiv e l a t t i c e

[ l ] p . l 40 :
(of more than one element^

is isomorphic with a subdirect product of chains of length
two ( i . e .

chains consisting of ju st two elements each.)

-

35 -

One shows readily that a d irect product of d istrib u tiv e
l a t t i c e s is a d is trib u tiv e l a t t i c e ,

and in p a rtic u la r,

a

d ire c t product of Boolean algebras is a Boolean algebra.
Theorem 5*16.

Any d ire c t product of Brouwerian algebras

is a Brouwerian algebra and subtraction is component-wise.
be a d ire c t product of Brouwerian
algebras

L^

•

Then, for every
and for
by d e fin itio n , y^
and i f

z

is

) > x^

+ (x ^

,

any element such that
*

then
Now, since

y*< +

(x*

- y*

) > x^

K “ y* V > {x -<[

, i t follows
fey definition

of a d ire c t product of l a t t i c e s .
Moreover, since y^ +

> x^

implies
>

then
implies
Thus

x^

-{y 4

Hence

L

is a Brouwerian algebra and subtraction

is component-wise •

exists and equals

36 -

-

Corolhary ^ . 1 6 . 1 .
is

Any d irect product of Boolean algebras

a Boolean algebra.

Theorem 5•17 ♦

If a d is tr ib u tiv e l a t t i c e representable as

a d ir e c t product of chains (each having a greatest element I)
admits a metric group operation, j6

i t must be a Boolean

9

algebra.
Proof ♦

I t will be shown th at in these circumstances, the

chains must each be of length two • Let
product of a co llec tio n of chains C ^

be the d ire c t

. Let

0^

and 1 ^

be

the le a s t and g reate st elements of C <
*. respectively. Then
^ 0 j,| and -[i/^ are the le a s t and greatest elements of
Let

, 0^

x oc ^ Q u

be an element of

and a l l

•

having one component

other components

0^3 , (3 f&ol . Let

6

with 0 oc < x aL < I o< • Since jii is a metric group operation,
\ 0 ,4

0

{

0

^

= -fo_( }
■Implies

Hence

\o+\ 0 { x u

,

THua

A( {O*} ,

I

Letting

0^}

= {x u.

, {x*

,

0^ 0

implies (1 )

{a*];

»

Op

=

and

.

, Op J ) implies

I 1 *-If}* the triang le inequality
*

is coordin atewise, for a given

£ {Cv} , the re la tio n

(2J

^ a4 + ^ Xj- = ( 1^

implies ( 3 )

for every =< .

Moreover, since
( 3 ) holds i f

]

+ i z'<} "

How, since addition in
element

is the group Id e n tity .

C.( ia a chain, then for a„c 0

and only i f

x* = lot.

. (For

a^

1^
= I^

,
, xx

37 -

-

may he a r b itr a r y ) . Thus the solution of (2) i s
if

and only i f

0 ^ = 1 ^

,

for no °< . Consequentlyj in (1),

{ zu] = {i^l .
0

Thus

{ i^

=

(x^

, 0^}

0 {

=

l4

{ l4 .

But th is contradicts th© unique solution property of the
group operation 0 ,

Hence an element

x ^ with

0^ < x u

<

cannot ex ist f o r any &L , and each chain contains precisely
two elements. Since such a chain is a Boolean algebra, ^ C >
is a Boolean algebra by
Definitlon 3 » 7 *

lt«wc^3.l6.1.

A (d istrib u tiv e)

subdirect product

l a t t i c e representable as a

of V chains

( V any cardinal), will

be cailed V-dimensional.
Theorem 5«l8»

A f i n i t e dimensional l a t t i c e | C/}

which

admits a metric group operation is a Boolean algebra.
Proof.

I t will be shown th a t each chain

of

has

\

only two elements.
Assume as in Theorem 5 .17 the existence of an element
x ei 6

such th a t

0 u

<x*<

< I

• I f -[x

; 0^6-(c^ ^ ,

Theorem 5*17 applies. If th is is not the case, then
contains an element £y«<V

* 0^} $ but with y^

= x^

for some ^

• This means that at le a st one coordinate other

than the

th coorinate does not equal

Case 1.

y^
>{7*]

/

?or
>

^ 7 ^

0^

for

any

•

• A consideration of

^ leads b y the argument of Theorem 3 .17

to the same contradiction.
Case 2.
equal to

At le a s t one coordinate of {y^} other than
I^

x^

is

for some oC of the index set A* Consider again

^X^ij) • To sa tisfy the trian g le Inequality

in

th is case, vi z.
& { * 4 \ = { ZA
{ 7*1 $

j; = ^ X ^ ,

l z *\ “

& ^ I,4 i

{ 2 *} ^ G<*}

as was observed in Theorem 3.1?. Now i f

9 "t l i e n

and{c/ L
3

^

y aL < z cl

fo r some
and i f

or

\

s

s i n c e -{y=<I e

,

wo< = y«aC °r

,

2^ <

In any case, therefore,

•

w^; =

according

since G<< is a chain. Hence, i f
but

z^

^ X^

z ^ ~ X^

, then

w^

and again,

^ X^
.

^ I ^

w* j = -j[y-tj{z^ has a t le a st one

X
<*/ than does ^ y * | • I f now w<x = X^ for

no <
?< , Case 1 again applies. I f th is is not the case, i te r a t e
the above process with «[ w*<| in the role of '[y*<j' •
la ttic e

Since the

i s f i n i t e dimensional, in a f i n i t e number of steps

there w ill emerge an element
no «?< •

such that

u^

= X^

for.

The argument of Theorem 3 . 1 7 will then apply. Thus

each chain must contain precisely two elements. Hence the
l a t t i c e i s f i n i t e and is therefore a Boolean algebra in
accordance with the remark following Theorem 3*1^-*
Definition 5«8«
space S ! i f

A space S is said to be congruent to a

there e x is ts a one-to-one distance preserving

map of S onto 3 *•
A space 3 is said to have congruence order k re la tiv e
to a class of spaces M containing S, provided th at any space
of M is congruent (isometric) to a subset of S whenever each
k of i t s

^

>

^ Xe* , then

fewer coordinates

t

is a sublattice of J

, y^ = I^

yu

= i 7 +1 i z °

3

Moreover, for every d
as

, doe s not require that

points are, and k Is the smallest number with th is

,

-

39 -

property# This concept is due to Menger L9]* (p#ll 6 ), who
proved, fo r example,

that the congruence order of

n-dimensional Euclidean space, En , re la tiv e to the cl ass of
metric spaces is n + 3 * X
n 151 i t was shown that the
congruence order of a Boolean geometry re la tiv e to the class
of

L - metrized spaces i s three. I t i s natural to seek the

congruence order of Brouwerian geometries. The theorems
which follow hear d ire c tly on th is question.
Theorem 5«19« If the distance function of an autometrized
space i s

a group operation, the congruence ©rder of the space

r e la tiv e to the class of L-metrized spaces i s three.
Proof; Let L denote the autometrized space whose distance
function ^ is a group operation, and suppose S is any
L-metrized space with the property th at every three of its
points can he congruently embedded in L# Consider any fixed
element a of S, x an a r b itr a r y element, and l e t d(a,x) = u,
where u 6

L. I f a i s any point of L, then there exists

uniquely a point x £ L such that a # x = u,

since the

distance function i s a group operation. This implies the
mapping x ■
—
-» x of

S

onto a subset of L Is clearly

single valued. Moreover if y 6

S, y / x, and d(a,y) = u,

then the isosceles triang le with vertices
congruent, by hypothesis

(a,x,y)

is

to an isosceles triang le in L,

a contradiction. Thus the inverse mapping is single valued
and the described mapping i s one-to-one. In p a rtic u la r,
if

x = a, then u = 0

and

a*

> a.

-Uo-

W
© prove next that t h i s mapping is dist&noe preserving*
Let y e-

s

and d(a,y) = v. Then a

#

y = v. Suppose d(x, y)

=

w,

w 6 L* Then a triangle exists in L with sides u,v,w. However,
if 1 two sides of a triangle in L are respectively equal to two
sides of another triangle in L, then the third sides are equal*
This follows from the associative law of the group operation,
since i f A(a,b,c)

in L has a * h = u, b * c = v, then

(a * b)^(b * c) —a * (b # b) * c = a * c, i*e.

the third

side of the triangle i s uniquely determined by the other two
sides* Thus x * y = w*

A three point

space with a l l distances

equal to the same non-zero element of L shows that the con­
gruence order i s not two, since L contains no equilateral
triangl e •
Corollary 5 * 1 9 (El l is)* The congruence order of a Boolean
geometry r e l a t i v e to the class of L-metrized spaces is three.
Theorem 5*20*
and only i f

A Brouwerian geometry i s

a Boolean geometry i f

i t has congruence order three rel a t i v e to the

class of L-metrized spaces*
Proof *

In view of Corollary 3 . I 9 . I ,

it

is necessary only to

show that a Brouwerian geometry with congruence order three
i s a Boolean geometry* Suppose x " ] x /

0, and consider the

L-metrized space consisting of the four d i s t i n c t elements
a , b ,c , d with a * c = b

d = x”| x and the remaining distances

equal to X. Each three points of S are congruently embeddable
on the points 0, I , x “| x ,
hypothesis, the entire

of the Brouwerian geometry, and by

space i s so embeddable. This config­

u ration, however, i s impossible In an arbitrary Brouwerian

-1 + 1 -

geometry, for i f

a,b ,c,d map respectively into a

of L, then x ”|x < a-^> x~~|x <
isosceles triangle
and

since the vertex of an

is M
over” the base. Then x~[x < a-^c^»

+ x"]x =But by Corollary 3.5.1,

“ al +

cl* 3 0 that ai ° i + x ~lx = a 1 + c •

al ° l ~

al + C 1 * therefore a.^ = c ^ i . e .

a contradiction,
Theorem 5»£I«

^

and L is a Boolean

*2 °^

*ai

* ci

Hence
^

* c 1 =x ”| x

=0 ,

geometry.

A Brouwerian chain has congruence order four

r e l a t i v e to the class of L-metrized spaces.
Proof. Since any chain with I ,

0 is a Brouwerian algebra,

then metrized by symmetric difference,
for a > b) i t

(wnerein a - b = a

i s a Brouwerian geometry. Let C be such a

chain, i . e . Brouwerian. Then (1) every triangle i s isosceles,
(2 ) i f

a > b > c > d, then a # b —a * c = a * d

b**c=b^d

= b, c * d = c,

quadruple cannot be equal,

= a,

(3) Opposite sides of a

since a Brouwerian geometry has no

equilateral triangle.
Let S be any L-metrized space with the property t h a t
every four points of S are congruently embeddable in C.
I f S contains a point 0 f which is not the vertex of an
Isosceles t r i a n g l e ,

l e t 0! be mapped into the 0 of C and

every point x* of S into i t s distance x from O’ . This
establishes

a one-to-one distance preserving map x»

> x

of S onto a subset of C. The one-to-oneness Is obvious. That
the mapping Is a congruence i s seen as follows:

If x 1 and y 1

are d i s t i n c t elements of S whose distances from 0* are x
and y respectively, then d(x,y) equals x or y, according as

- Jl? *“
x

>y

or y > x, since by hypothesis, the three points

O'fXSy 1 are congruently embeddable in C.
If S contains no point 0 * as described above, then
every point

is the vertex of an isosceles t r i a n g l e . However,

two isosceles triangles with the same vertex must have t h e i r
legs equal, otherwise a quadruple including the vertex is
determined, which maps into a quadruple in C not satisfying
( 3 ) above. Let each point x* of S therefore be mapped into x,
the leg of an isosceles triangle with vertex x 1. This i s a
one-to-one mapping of S onto a subset of C. The single
valuedness of the mapping i s obvious. That the inverse mapp­
ing is

single valued i s

seen as follows: I f x 1 and y* are

d i s t i n c t points of S, each the vertex of an isosceles triangle
with leg x, say, then a quadruple including x f and y* is
determined having opposite sides equal to x,

a contradiction,

since t h i s implies a similar configuration In C, violating (3)*
To see t h a t the mapping i s a congruence, consider two
d i s t i n c t points x*,y* of S whose images under the mapping
are x and y.

(Suppose x > y ,

sox*y=x.)

Then x ! is the

vertex of an isosceles tr i a n g l e with leg x. Let x = x* # u*,
and l e t y*

u» - z. Then z / x by ( 3 ) so that d(x»,y#) = x

or z. I f dCx^y1) = z, then z = y by definition of the mapp­
ing and x < y, a contradiction. Thus d(x», y 1) = x = x * y .
Thus the congruence order i s ,

at most, four. That i t i s not

less than four i s shown by a four point L-metrized space S
with two opposite distances equal to a, and the remaining
distances equal to b where a < b and a,b are d i s t i n c t

elements of C. Each three points are embeddable in C, but
the entire space i s not,

since ( 3 ) is violated. This completes

the proof* Examples show that a Brouwerian geometry may have
congruence order four without being a chain*
Remark:

Theorem 3*2.1 i s the analogue of a classical metric

theorem, namely; The congruence order of the Euclidean line
(Ex ) r e l a t i v e t o the class of semi-metric spaces is four.
Moreover i t s

congruence order rel a t i v e to the class of semi­

metric spaces containing more than four points is t h r e e . One
says in t h i s case, that the quasi- congruence order of E^ is
three. This is not so for a Brouwerian chain, however, as
shown by the following example. Let S be an L-metrized space
of five points a ,b , c , d , e
a # b —b # e

with

= c -«*d = d # a

a * e = b * e = c
= y, and a

e

= d # e

c = b * d = z.

Let the Brouwerian chain C contain the elements x > y > z > 0 .
Then each three points of S are congruently embeddable in C.
but S i s not, for the quadruple

(a,b,c,d)

embeddable in C, since i t violates
Definition 3 * 9 *

The d ir e c t product

autometrized l a t t i c e s

is not congruently

( 3 ) of Theorem 3 .21.
of a collection of

A©
< i s defined as the l a t t i c e direct

product with the metric operation coordinate-wise•
Theorem 3.22.
product

If each autometrized l a t t i c e
has congruence order k

Ac< of a d irect

rel a t i v e to the class

of L-mietrized spaces, then the

congruence order of ^AoJ*

r e l a t i v e to that class i s also

k.

Proof.

space, each k of whose points

Let S be an L-metrized

are congruently embeddable in |Ax^ • L e t p ± ,

be arbitrary

-in­
d i s t i n c t points of S with d(p±,p } =

.

For each ‘K , l e t A^°^ be a space whose points
correspond one-to-one to the points of S, with pJ ^
i

J

the respective correspondents of p . , p . , with d(p. ^

,p ^

J

J

}

58 z&
< 6 Ao< • Thus the distance between any two points of
( ^

A

)

is the &
<th coordinate of the distance between the

corresponding two points of S.
For each triangle A(pi#Pj,Pk) of S with sides
{ Z » 1 » {u -<} » {v»<} * < { z < \

» {u =t} » {v<*} )T Implies

(by d e f in itio n of the l a t t i c e d ir ect product)
* u
then, i f ,

* v* )T in Ao< > for every o( . In p articular,
say,

z^ 38 0 o< for

(Zo<, Uo<, v o< )T implies
v^= u

. Thus, i f

ua< > v 0< , v c^ > u 0< ,

in A^

we shall identify p^

some c< , i t follows that

^ for

hence

some ©
< , d(p^*^

) - 0c< ,

( <K )
(
§ with pj

, and only in these
(c* )
circumstances. Then, c l e a r l y , for every^ , A
is an

L-metrized space.
Consider now an arb itrary k-tuple
k points, P1 ,P2#»** Pk ) £

(P),

(a set of

containing P±*Pj*

with
^

d(p. ,p ) = {z ^

by hypothesis, l e t

(Q) be a k-tuple of^A^to which (P) Is

congruent. Let q±s

,q,s { y ^ € (Q)

be respective

correspondents of Pj^P. under this congruence, so that
d(qi#qj) = { z^

"i

♦ Since (P) is congruently embeddable in-^A^Jj

= {x * * y*] * <since distance in ^

is

coordinate-wise) • Then, by d efinition of the autometrized
d ir e c t product, there e x is t s a k-tuple
every << , with q ±

i x*

5 7 ^

(^^)

in

for

Ac* corresponding

-Ir­

respectively to q ^ q ,

of -^Ajwherein d(q
/

^

\

/

Clearly, the k-tuple Q
k-tuple

/

,q .

v

in A

) = x^

# y^ •

oC

*

corresponding to the

(F) in S is congruent to the k-tuple Q
,^

in

,

for every ^ . Thus each k points of
^
are congruently
( oC ^
embeddable in A e* » hence A
is congruently embeddable
in Ac< for every cK , under the hypotheses of the theorem.
Now, l e t P^#Pj
d tp^P j)

l>e arbitrary d i s t i n c t points of S with

=

, l e t P1( °< ) , P j (0<) <
£ A(o< }

as previously constructed,

and l e t q,

s x^ , q.

foO ^

be respective correspondents of p^'
congruence of A
^ 0<.

\ y , e A^

under the

into A©
< already established. Then

) =

* y*

= 2 ^

, and there exist points

' qj s {y"^ 111 ( A-<> with

Then the mapping

> q^, P j

-

qi :

J o<

be

qi * qj = I * *

> q^

of S into

* y

^

[ A^j is a

congruence, established as follows:
According to the construction of the spaces A ^ °^ , the
mapping p^-—> p^

( u )

, Pj

> p^

( u )

of S onto

( <=< )
A
for

every ©
< i s one-to-one. Moreover, under the established
congruence of A^
p
i

^ q
1o<

into Ac*,

q.
is also one-to-one. Finally
«
Jc<

J

the mapping q± -----> ^i*1
^
every

-----* qj

of A<* into

{A^j-for

is one-to-one by the definition of the direct

product. Therefore the mapping
into

for every <
=
>
< , the mapping

^ A^

p^-------> Q-j^Pj*--> Qj

of

Is one-to-one as well. Since

d(p #P ) = {
i
j
^
as claimed.

y

=

* <*'•
J

the ^aPPing is a congruence,

S

*

-1*6-

Corollary 3 . 2 2 . 1 .

Any autometrized l a t t i c e L which i s

a d ir e c t product j C^j of Brouwerisn chains

has

congruence order four*
£r££f.

The congruence order of each Brouwerian chain

r e l a t i v e to the class of L—
metrized spaces is four*

(Theorem

3#21).
Corollary 5»22*2*
d ir e c t product

If each autometrized l a t t i c e A c< of &
has a f i n i t e congruence order relative

to the class of L-metrized spaces, the congruence order of
i s the maximum of the congruence orders of the A ^
if

,

I t exists*

iftroof.

Let the maximum congruence order be the k of the

theorem* I f the congruence order of A*< for some
is
( oC}
n < k, then, since each k points of A
are congruently
embeddable in A

, each n points are certainly likewise
(o< )

embeddable, hence so Is A

. All other d e ta il s of the

proof are Identical with those of the theorem.
Remark*

The congruence order of an a rb itra ry Brouwerian

geometry r e l a t i v e t o the class of L-metrized spaces is s t i l l
an open question*
(This r e c t i f i e s an e a r l i e r statement of L.M.Kelly and
the writer

[Bulletin of the American Mathematical Society,

Vol. 62 Number 2, March 1956, PP.172“33 that the congruence
order in the general case was shown to be four. Further
Investigation to date of subdirect products of Brouwerian
algebras continues to suggest strongly that this is indeed
the case.)

-k7-

S e c t i o n J+. G e n e r a l T h e o r e m s j B e t w e e n n e s s
In t h i s section properties of L-metrized spaces in
general are established. Further, metric

and l a t t i c e between­

ness are introduced, and the consequences of their coincidence
i s studied.
Theorem 1^.1.

In any Lmetrized space, the sum of the

distances of the elements of a subset S from any element of
the subset i s constant and equal to the sum of a l l the d is ­
tances of the subset, provided the sums e x i s t .
Proof♦

Let p be any element of S, x and y arbitrary d i s ti n c t

elements of S, with d{x,y) = d
, d(p,x) = p and d(p,y) = p .
xy
x
y
Then the triangle
hence

inequality asserts that p + p > d
x
y
xy 9

p > d
and y
P > /
d
.
px + Py = ^—
z
x
xy
^—
x
^—
xy
x #y€ s
xe s
x,y e s
x65
/

d
> y
p ,
z
xy
^— x 9
X € S
X,ye s
Corollary 14-. 1 .1 .

>
p = >
d .
^—
x
^—
xy
x€ S
x,y € S

In any L-metrized space, the sum of any

two sides of a triangle
Corollary if.1. 2 .

henc e

equals the perimeter.

In any L-metrized space, the sum of two

sides of a triangle equals the sum of any other two sides.
Definition 4.1.

Three points u,v,w, of an L-metrized space

are called l i n e a r i f d(u,v)
Definition

+ d(v,w) = d(u,w).

In any L-metrized space, a triangle with

sides a,b,c will be designated T(a,b,c).
( a ,b ,c ) T ) .

(Note: t h i s implies

-usCorollary

in any L-metrized apace, the vertices of

an isosceles triangle are li n e a r .
P roof.

In T(a,a,b),

Theorem I4-.2 .

a+b

= a + a = a

In any L-metrized space, a triangle T(a,b,c)

Is isosceles i f

and only i f a,b,c form a chain (or a,b,c are

pair-wise comparable).
Froof.

I f a = b, say, then a + a

Conversely,

since a + b = b + c =

= a>c.
a+c,

a > b > c

Implies

a = b.
Theorem 1+.3*

In any L-metrized space, i f ,

then a > c and b + c = a ( i . e .
Moreover, I f T(a,b,c)

in T(a,b,c),

a > b,

the vertices are l i n e a r ) .

i s not isosceles,

then b and c are

non-comparable•
Proof.

Since a + b = b

a = a ^ c = b + c

+ c-

i.e.

a + c, a > b

implies

a > c. I f b and c were comparable,

the elements would form a chain and the triangle would be
Iso s celes.
Corollary Ii.5.1.

In anyL-metrized space, a non-Isosceles

triangle has either precisely one pair of non-comparable
sides, or a l l three sides
Proof.

are pair-wise non-comparable.

Exactly two pairs of non-camparabi e sides leads

immediately to a contradiction of the theorem.
Corollary

In anyL-metrized space, I f precisely

one p air of sides

b,c of a triangle are non-comparable,

th ir d side

uniquely determined as t h e i r sum, a = b + c,

a

Is

and again the vertices are l i n e a r .

the

Corollary Ij,.3»5«

In any L-metrized space, a tr i a n g l e with

precisely one pair of non-comparable sides is uniquely
determined by them.
Corollary

In any L-metrized space, the vertices of

a triangle are l i n e a r , i f and only i f

the triangle has a

pair of comparable sides.
Proof.

The sufficiency of the condition i s a conclusion of

the theorem. On the other hand i f a = b + c, a > b
by d e f i n i t i o n of the l a t t i c e
Corollary U . 3 . 5 *

and a > c

sum*

In any L-metrized space, the only triangles

whose vertices are non-linear are those whose sides are p a i r ­
wise non-comparable.
Proof.

Any triangle without this property satisfie s the

conditions of the theorem.
Definition lj-.3.

An L-metrized space whose distance l a t t i c e

Is a chain C, Is called a C-metrized space.
Theoremii.ij.
Proof.

Every triangle of a C-metrized space

is

isosceles

Tliis theorem follows from Theorem J+.2*

Corollary l+.l+.l.

Every triangle of an autometrized chain

i s Isosceles.
Remark*

Every triangle of an L-metrized spacemay be

iso sceles, even though i t s distance la ttic e is
An example i s
a,b ,I,0

the autometrized four element Boolean algebra

with d(a,b) = a, d( 0 ,I )

equal to I .

not a chain.

= b and a l l

other distances

50-

Definitlon lul|.

An autometrized l a t t i c e L as well as i t s

metric operation i s called regular i f a # 0 = a for every
element a of

L. (Thus Boolean and Brouwerian geometries are

readily seen

to he regular.)

It

should he noted that every l a t t i c e admits the

regular

metric

operation a #

h = a+ h for a ^ b, and

a * a — 0, for a # O s s a + Q= a, and since fo r any A(a,b,c),
(a ^ h)
a+h

+ (h **

+ c>

a +

c) = (a +h) + (h +c ) = a + b + c ,
c = a * c, i . e .

the triangle inequality holds.

Theorem it-.5*In a regular autometrized l a t t i c e ,
Proof ♦

(a,b,a

# h)T.

Evident from a consideration of A(a,0,b).

Although, as indicated above, an autometrized l a t t i c e
in which every triangle Is isosceles need not be a chain,
one does have
Theorem It-.6 .

A regular autometrized l a t t i c e L In which every

triangle i s isosceles i s a chain.
Proof.

Let a and b be arbitrary d is t i n c t elements of L and

consider A(a,b,Q). Since a * 0 = a and b * 0 =b, i t follows
that a # b

= a o r a # b = b s o

Theorem if. 7.
a > b,
Proof.

a

that either a

> b or b >

In a regular autometrized chain, for

b = a,
In A(a,b,0) , a # 0 - a, b * 0 = b. Since every triangle

Is isosceles,

a # b = a for a > b*

Corollary Lt-.7.1.

Every regular autometrized

chain is a

Brouwerian geometry •
Proof.
a

a.

A chain i s a Brouwerian algebra and for a > b,

b = a = (a-b) + ( b - a ) .

-5 1 -

Corollary i^.7.2.

A regular autometrized. chain has no equi­

l a t e r a l tr ia n g le s.
Proof.

A Brouwerian geometry has no equilateral tri a n g l e s .

Theorem 1+.8.

An autometrized chain C of more than three

elements i s regular, i f

and only i f ,

elements a > b > c > d

has the distance pattern,

a * - b 5!:a jif-c = a * d = : a , b # c = b
Proof ♦

The regularity

every quadruple of

d = b, c

d = c.

and the fact that every triangle is

Isosceles yields the above pattern.
Conversely, if x Is an arbitrary element of C, then in
any quadruple z > y > x > 0, the prescribed pattern yields
x

0 = x.

Remark.

A non-regular autometrized chain may have a

quadruple whose distances have the indicated pattern but
the vertices of the quadruple need not be those indicated.
Theorem i+.9«

An autometrized l a t t i c e L is regular i f

only I f a + (a *-b) = a ^ b
Proof.

and

for a l l elements a, b of L.

I f L i s regular, then (a,b,a * b)T by Theorem J4..5

and the condition follows from Corollary I4.. 1.2* On the other
hand the condition implies,
0,x,

in p a r ti c u l a r , for the elements

that 0 + (0 * x) = 0 + x = x, i . e .

Corollary ii.9*!*
a * b < a + b
P r o o f.

Remark.

0 * x = x.

In a regular autometrized l a t t i c e L,

for a l l elements a,b,

of

L.

Implied by (&,b,a * b)T in &(a,b,0).
The condition a * b < a + b

is not sufficient to

ensure reg ularity as shown by the chain I > a > b > 0
I

a = I # b = I * 0 = I,

a # b = a # 0 = b -> 0 = b .

with

-5 2 Definition 1+.^.

An L-metrized space is called distributive

i f i t s distance l a t t i c e
Corollary ii.9.2.
lattice,
Proof •

is d is tr i b u t i v e .

In a regular distrib utive autometrized

ab + (a * b) = a + b .
Identical with that of Corollary 5 . 5 *1 *

Remark.

The above condition even with regularity i s not

su f f ic ie n t to yield d i s t r l b u t i v i t y • The condition holds for
every pair of elements of the non-modular five element
lattice

[l](page 6 figure Id.) with elements 1,0, a > b,

and c, i f

the autometrization i s regular,

a * b = a, and a l l

other distances are equal to I .
Definition i+.6.

An autometrized l a t t i c e

is called symmetric

i f the distance between every two of i t s elements is equal
to the distance between t heir sum and product. Thus
Brouwerian, and Boolean geometries are read ily seen to be
symmetric.
Theorem 4.10.

An autometrized l a t t i c e which i s symmetric

and contains no isosceles triangle i s d i s t r i b u t i v e .
Proof.

A non-distributive l a t t i c e must contain one of the

two special five element l a t t i c e s shown below.

[l](page 1 5 ^)*

The symmetric property implies in each case the existence of
an isosceles t r i a n g l e ,
In figure 1, i f x ^ y

contrary to assumption.
= u ^ v = y * z = x ^ 2 , then x , y, z

are vertices of an isosceles triang le.
In figure 2, i f y '> z = x * z ~ u * v, then x,y,z are
v ertices of an isosceles triangle.

-5 3 (A

A
X or

©

*

©

V

ir

f ig.l

D e f i n i t i o n 1+.7*

f i g .2

I n an L - m e t r i z e d s p a c e , t h e e l e m e n t b i s

m e t r i c a l l y betw een

a and c ,

if

d (a ,b )

+ d(b,c)

The points

(a,b,c)

indicated,

and the relation i s written (a,b,c)M.

It

= d(a,c).

are said to be linear as already

should be noted at the outset that this i s not

a

betweenness r e l a t io n in the usual sense, since i t f a i l s in
many instances to have the special inner point property, i . e .
(a,b,c)M and (a,c,b)M may both p e r s i s t even though b and c
do not coincide. None the l e s s j i t i s convenient to use the
terminology which has been indicated. However, the r e l a t i o n
does have the other basic betweenness property, viz.,symme­
tr y in the outer points,

i.e.

(a,b,c)M i f and only i f

(c,b,a)M,

since the metric operation is commutative.
Theorem ij-.ll.

Three linear points of an L-metrized space

f a i l to have the special inner point property i f and only i f
they are the vertices of an Isosceles t r i a n g l e .
Proof.

I f in A(a,b,c), d(a,b) = d (a , c ) , then (a,b,c)M and

(a,c,b)M by Corollary l+.l.2. Conversely, i f

these relations

hold, then d(a,b) + d(b,c) = d(a,c) and d(a,c) + d(b,c) = d (a , b ) .
Hence d(a,c) > d(a,b) and d(a,b) > d (a ,c ) , i . e . d(a,b) = d( a, c) .

-5 k-

D e f i n i t i o n ii+S,
property that

r ela tio n

If a b e t w e e n n e s s r e l a t i o n

R h a s the

( a ,b ,c ) a snd {&,x,b)R imply (x,b,c)R, the

is said to have t r a n s i t i v i t y t^

• If the same two

re l a t i o n s imply (a,x,c )R, the relation is said to have
transitivity
Theorem
Proof ♦

[1 1 3 *
Metric betweenness has t r a n s i t i v i t y

I f * denotes the metric operation, then

(a,b,c)M

and (a,x,b)M imply (a * b)4*('b * c)=(a* c) and (a *-x)+-(x
Hence,

b)=(a * b ) #

(a * x)+*(x * b)^(b * c ) = (a * cJ.Now, since

x * c < (x # b)^(b * c) by the triangle inequality, i t
follows that

(a # x)^(x

Inequality implies that
(a

c) < a * c, but the triangle
(a

x)^(x * c) = (a *-c),

Remark,

x) +(x * c ) > a *■ c, hence ,
i.e*

(a,x,c)M as claimed.

In general, metric betweenness f a i l s to have t^ ,

notably in the case of an isosceles t r i a n g l e , for i f &(a,b,c)
has d(a,b) = d( a, c) ,
c,b,c)M i s not valid .

then (a,b,c)M and (a,c,b)M hold, but
Indeed, this may happen in an

L-metrized space without isosceles t r ia n g le s, as the follow­
ing example shows:

fig.5

-5 5 In S,(figure 3 ),

(1,3,)+)M and ( 1 , 2 , 3 )M but (2 , 3 ,lj.)M does

not hold.
Definition lt-,9*

An element b of a l a t t i c e i s l a t t i c e

between1
* a and c, written (a,b,c)L, i f and only i f
ab + be = b = (a + c)(b + c) ,

[11] p,105* This relation has

t r a n s i t i v i t y t ^ [ l l ] , but not in general t^ ,
modularity (and conversely)

[11], I t

l a t i o n in the usual sense. Further,
implies a c < b < a + c ,

but not,

since

implies

is a betweenness r e ­
in any l a t t i c e ,

(a,b,c)L

in general conversely.

However, in a d is trib utive l a t t i c e ,

(a,b,c)L i f and only i f

ac < b < a + c [11] , Metric and l a t t i c e betweenness will be
said to coincide in an autometrized l a t t i c e provided (a, b,c)M
i f and only i f

(a,b,c)L,

Theorem i+,13*

In an autometrized l a t t i c e L, metric and

l a t t i c e betweenness coincide, i f

and only i f

betweenness has t^,(2) L i s symmetric,

( 1 ) metric

( 3 ) a < b < c implies

( a,b,c)M,
Proof m

Use i s made here of a theorem due to Pitcher and

Smiley [11, Theorem 10,1]# In verifying the hypotheses of
that theorem, one observes that

( 1 ) implies t h a t L is free

of isosceles triangles and so by Theorem 4.11 that metric
betweenness has the special inner point property,
( 3 ) imply (a, a + b,b)M and (a,ab,b)M, for
(a + b,

a, ab)M and (a + b, b,

(2 ) and

( 3 ) implies

ab)M,hence, letting

x = (a 4’ b) * a, u = a # ab, y = (a + b) * b, and v = b * ab,
we have x + u = y + v = (a + b) * ab, Moreover, by (2)
(a 4- b) # ab = a * b,

so that x 4 >y +- u + v = a * b. Hence

-56-

x + y < a # b , and u + v < a /$«'b.

u + v > a 4b fa
58 a 4k fap i . e .

But x +

y > a 4b b and

t)y the triangle inequality, hence, x + y = u + v
(a,

a + b,b).M and (a,ab,b)M.

Conversely, i f metric and l a t t i c e betweenness coincide,

(1 )

and ( 5 ) are immediate. To establish the symmetric property,
consider the quadruple (a,b,ab,

a + b ).

(a +
* b, a, ab)L and

(a h*3 b,b,ab)L imply, by the assumed coincidence,
( i )[(a ^b )* a] +(a * ab) = [(a+b)* b] +[b # ab]=(a+b)* ab.
Similarly,
(i i )

(a,a + b,b)L

and

(a,ab,b)L imply

[ ( a+b)# a] +[ (a^b)* b]-(a * ab) +(b * ab )= a 4b fa ,

Adding f i r s t and second members of(i)

gives the same re su lt

as adding f i r s t and second members of ( i i ) .

IdLempotency

yields
(a + b ) ^ a b = a

b•

Thus the theorem is established.
Corollary it-. 15.1.

An autometrized l a t t i c e in which metric

and l a t t i c e betweenness coincide is d i s t r i b u t i v e .
Proof.

By Theorem 1^.10

Theorem 4.14*

(Ellis)

In a Boolean geometry, metric and

l a t t i c e betweenness coincide.
Proof.
follows:

A somewhat briefer proof than E l l i s has given i s as
(a,b,c)L implies ac < b < a + c, in any l a t t i c e .

Taking complements of ac < b yields b* < a ’ + c». Moreover^
(a 4b b)^(b 4b o) = ab * + a*b + be * + b fc
= (a + c ) bf +b( a*

+ c f ). But b < a + c implies

b(a* + c *) < (a +c) ( a*

+c»)

and b* < a f * c*

implies

-57-

b , (a+c)<(a+c)(a,+c ») so that b(a»*c»)+b*(a+c)<(a+c) (a»*c*),
i.e .
(a

(a * b) + (b * c) < a * c.

But in any case,

b)+(b 4b c)> a * c, hence (a * b)+(b * c) = a * c. i . e .

(a,b,c)M.
Conversely (a,b,c)M, i . e .
implies

(ab*+a*b)+(bc*^b*c)=ac*^a!c

(a + c) bf + b(a* + c 1) = ac* + a*c. Multiplying the

l a t t e r f i r s t by ac, then by a fc*, we obtain acb * = 0 and
a*c*b = 0. But xy * = 0 implies xy + xy* = xy or x = xy so
that x < y. Hence ac < b. Similarly b(a*c*) = 0 implies
b ( a^c) * = 0 , hence b < a ^ c. i . e .

ac < b < a + c or since

a Boolean algebra i s d is t r i b u t i v e ,

(a,b,c)L. This completes

the proof.
Remark.

Since t^ f a i l s for an isosceles triang le, i t

fails

in any brouwerian geometry which is non-Boolean, hence
metric and l a t t i c e betweenness do not in general coincide in
Brouwerian geometries. Indeed this

coincidence to -gether

with regularity i s suffic ient to characterize Boolean
geometries among the Brouwerian geometries as the following
theorems show.
Theorem 4.15.

An autometrized l a t t i c e with an I Is a Boolean

geometry i f and only i f

it

is regular,

and metric

and

lattice

betweenness coincide.
Proof.

Theorem Ij-.l4 establishes the necess ity^ since symmetric

difference is a regular metric operation. To prove the
sufficiency, l e t
points

a * I = x, I # x = y

and consider the

( 0 , I , a , x , a x ) . Regularity implies (a,I,x)T and

(x,I,y)T or a * x = I and x + y = I . Moreover (a, I, x)L

Implies x + y = a * x since M
B ^ LB. Hence, a * x = I .
Symmetry (by Theorem

1 3 ) implies further that

a * x = (a + x) * ax = I

ax = I . But since

1 * 0 —1, A(0,I,ax) is isosceles. This is impossible since
t-^ must hold under our assumptions. Therefore ax * 0 = ax = 0.
Thus the l a t t i c e

Is complemented, and, being distributive by

Corollary It-.lj^.l is a Boolean algebra.
To show that a * b is symmetric difference, consider
the points

(0 , a , b ,1)^where a * I = a t , b * I = b » ,

in a Boolean algebra,
(due to regularity)

complements are unique, and ( a, I , a * I)T

Implies,

Clearly a * b < a + b ,

since,

in particu lar, that a & (a * I) =

and a * b < a f + b* (since A(a,b,I)

Implies (a*,b*,a * b ) T) , whence a * b < (a^b)(a* + b*)
- ab* + a*b. But,
implies th at
a fortiori,

(a * b) + ab - a + b (Corollary 1+.9.2),

(a+b) - ab < a * b, since a Boolean algebra ia ,
a Brouwerian algebra. Thus a * b = (a-b)^(b-a),

a * b is therefore Boolean symmetric difference, and the
lattice

is a Boolean geometry.

Corollary J+.15 +1.
i f and only i f

it

A la ttic e with an I is a Boolean algebra
admits a metric group operation under which

metric and la ttic e betweenness coincide.
Proof 1 The necessity is c le ar . The sufficiency is assured
since a metric group operation is regular.
Definition 4.10.
constant width,
m, i . e .

An L-metrized space S is said to be of
if

(1) There exists in S, a maximal distance,

such that m > x for every distance x in S, and

(2) Corresponding to each element a e S there exists

-59-

an element b 6 - S such that d(a,b)

= m.

Theorem lj-.16.

A l a t t i c e with an I is a Boolean algebra

i f and only i f

i t admits a metrization such that the space

is of constant width, and metric and l a t t i c e betweenness
coincide.
Proof: The necessity is clear,

since (1) Boolean symmetric

difference i s a metrization under which metric and l a t t i c e
betweenness coincide and (2 ) being a group operation,the
equation

a* x = X always has a solution, namely y =

a*.

(Clearly

X
i s the maximal distance in the space

It

actually occurs,

and

since X * 0 - X)

To establish the sufficiency, l e t a and b be arbitrary
d i s t i n c t elements of the lattice * Then, since metric and
l a t t i c e betweenness coincide,
(a * 0 )

+ (a * X) = (b * 0 )

A( I , a, b) , ( a * b) < (a * X )

(X,a,0)M and (X,b,0)M, I . e .

+ (b * X) = X * 0. Since, in
+ (b * X), then a * b < X * 0

since (a * X) + (b * X) < (X * 0)^and X * 0 i s the maximal
distance which occurs. Now, since the space i s of constant
width, corresponding to an element

x, there e x is t s an

element y, such that x * y - X * 0. More overly is unique,
since there can be no isosceles tria n g les. Furthermore,
since the l a t t i c e must be symmetric by Theorem J4..I 5 ,
(x+y) 4k xy = X * 0, and [ (x+y), (xy) ,0]M implies
[ (x+y) *(xy) ]+[ (xy) * 0] =(x*y) *Q, i . e. ( X *0) *( (xy)

* 0) =(x+y) *-Q

But X * 0 > (xy) * 0, hence (x^y) * 0 = X * 0* Since there
can be no isosceles t r i a n g l e s , x + y = X. Hence
X * xy =

X
* 0, and again, the absence of isosceles triangles

-6o-

implies xy —0. Thus the l a t t i c e
Corollary I4-.I 6 . I .

is a Boolean algebra.

A f i n i t e autometrized l a t t i c e in which

metric and l a t t i c e betweenness coincide i s a Boolean
algebra.
Pr°££«

By Theorem l+.l^ the l a t t i c e can have no isosceles

triangles.

If the l a t t i c e has n elements, then each element

is a “vertex** of (n-X) distances. Since none of these can be
repeated, each element of the l a t t i c e
at each vertex as a distance,

(except 0 ) must occur

hence being f i n i t e , the l a t t i c e

has an I which occurs at each vertex. I being clearly maximal,
the l a t t i c e i s of constant width, and i s therefore a Boolean
algebra.
Remark.

In the I n f in ite case even though no distance may be

repeated at a vertex,

there Is no assurance that every distance

must occur. Thus whether in an autometrized l a t t i c e with an I ,
the coincidence of metric and l a t t i c e betweenness in and of
Itself Is
still

sufficient to characterize a Boolean algebra is

an open question.
E l l i s has observed [ 5 ] that the group of motions of a

Boolean geometry Is simply t r a n s i t i v e ,

i.e.

for any two points

v

a,b of the geometry, there i s

a motion (a one-to-one distance

preserving map of the space onto i t s e l f )

which carries b into a.

An examination of the Brouwerian chain of three elements shows
that

the group of motions of a Brouwerian geometry i s not

simply t r a n s i t i v e , and, indeed
Theorem i-|..17.
and only i f

A Brouwerian geometry is a Boolean geometry i f

i t s group of motions is simply t r a n s i t i v e .

-61-

Pir©£jfcThe necessity as indicated has been observed by Ellis*
Suppose then that a Brouwerian geometry L has a simply
t r a n s i t i v e group of motions, and consider that motion which
c arries I into 0* Then i f x
i*e*

x

> y, I * x = 0 ** y =~| x = y

»~]x* Hence 0 —►I , x

so that 0 = x l x

x

(x ~| x) = I ,

and L i s a Boolean geometry*

- 62BIBLIOGRAPHY
1* Garrett Birkhoff, Lattice Theory, Amer. Math. Soc.
Colloquium Publications 2 5 ,

(1 9 J4.8 ), revised edition.

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4* J. G. E l l i o t t , Autometrization and the Symmetric Difference.
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6 .______, Autometrized Boolean Algebras I I , Canadian Journal

of Math.,

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7« £• A. Nordhaus and Leo Lapidus, Brouwerian Geometry,
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On closed elements in

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53(1951), 596-418.

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