THE ALGEBRA OF OPEN BODIES
Let R be the set of real numbers.
Rn is n-dimensional space.
A cut through Rn is a tripartition of Rn <c,EXT(c),INT(c)> such that
for any a  EXT(c) and for any b  INT(c), the continuous line segment from a to b
intersects c.
A bound in Rn is a set b  Rn such that <b,EXT(b),INT(b)> is a minimal cut through
Rn, a cut b through Rn such that no proper subset of b is a cut through Rn.
We choose two objects  and + not in Rn and introduce two improper bounds in
Rn:  = {} and + = {+}, and we set:
EXT(+) = Ø INT(+) = Rn
EXT() = Rn INT() = Ø
We define for bounds b, c in Rn:
b is concentric in c iff INT(b)  INT(c).
It follows that if b is concentric in c then EXT(c)  EXT(b), and it follows that every
bound is concentric in +, and  is concentric in every bound.
Let X  Rn and b a bound in Rn
b is an outer bound for X iff
1. for any bound c in Rn: if b is properly concentric in c then:
INT(c)  EXT(b)  RnX  Ø
2. for any bound c in Rn: if c is properly concentric in b then:
INT(b)  EXT(c)  X  Ø
The intuition is: directly inside an outer bound you find X-elements, and directly
outside an outer bound you find non-X elements.
b is a closed outer bound for X iff b is an outer bound for X and b  X.
b is an open outer bound for X iff b is an outer bound for X and b  X = Ø
b is an inner bound for X iff
1. for any bound c in Rn: if b is properly concentric in c then:
INT(c)  EXT(b)  X  Ø
2. for any bound c in Rn: if c is properly concentric in b then:
INT(b)  EXT(c)  RnX  Ø
The intuition is: directly outside an inner bound you find X-elements, and directly
inside an inner bound you find non-X elements.
b is a closed inner bound for X iff b is an inner bound for X and b  X.
b is an open inner bound for X iff b is an inner bound for X and b  X = Ø
It follows that + is an open outer bound for Rn and  is an open inner bound for Rn,
and that  is an open outer bound for Ø and + is an open inner bound for Ø.
And, it follows that if b is a proper bound in Rn, then EXT(b) has open outer bound +
and open inner bound b, while INT(b) has open outer bound b and open inner bound
.
Let X  Rn.
X is a body iff
for all x  Rn: x  X iff
either a: for some outer or inner bound b for X: x  b
or
b: for every outer bound b for X: x  EXT(b)
or
c: for some inner bound b for X, X  INT(b) = Ø and x  INT(b)
or
d. for some inner bound b for X there is an outer bound c for X that is
concentric in b and for every outer bound c for X that
is concentric in b: x  INT(b)  EXT(c).
X is a regular body iff X is a body and:
1. Every outer bound for X is either open or closed, and the same for every
inner bound.
2. All outer bounds for X and all inner bounds for X are disjoint.
So in a regular body, you don't allow bounds that are half inside, half outside, and no
two bounds touch.
X is a regular closed body iff X is a regular body and:
All outer bounds and inner bounds for X are closed.
X is a regular open body iff X is a regular body and:
All outer bounds and inner bounds for X are open.
Following Tarski 1927, we call regular open bodies solids.
A solid is finite iff it has a finite number of outer and inner bounds.
Sn is the set of all solids in Rn.
Fn is the set of all finite solids in Rn.
Intuitively, a solid s divides Rn into maximal subregions where you can connect any
two points by a continuous scribble without crossing any outer or inner bounds. For
each of these maximal subregions of s you can choose a representing point inside that
region. Since the rational numbers lie dense in the real numbers, you can choose the
representing point to be an n-tuple of rational numbers. This means that you can
make a one-one mapping between a solid and a subset of Qn, where Q is the set of
rational numbers. And this means that you can make a one-one mapping between the
set of all solids and pow(Qn). Since Qn is countable this means:
|Sn| = 20‫א‬
For finite solids, this means that you can make a one-one mapping between Fn and the
set of all finite subsets of Qn. Since the set of all finite subsets of a countable set is
countable, this means:
|Fn| = ‫א‬0
Some examples:
-Rn is a solid, and so is Ø.
-If b is a proper bound, INT(b) is a solid, we can call it a sphere.
-Any number of non-overlapping, non-touching n-place spheres is a solid.
-If we take out from Rn any number of non-overlapping non-touching spheres, plus
their outer bounds, the result is a solid.
-Take out from Rn a sphere, take a concentric subsphere of what you have taken out
and put it back in, so that it doesn't touch, the result is a solid.
Etc.
So, two two-dimensional examples in pictures:
The outer cirle indicates +. Black indicates not in, white indicates in:
Its complement is:
Importantly, the following are not solids:
Let r  Rn and look at Rn{r}. {r} is not a bound in Rn such that EXT(r)=Rn{r},
because then INT(r) = Ø, which is not a partition. This means that {r} does not count
as an inner bound for Rn{r}.
We check:
-r  Rn{r}.
a. r is not part of any inner or outer bound for Rn{r}.
b. The outer bound for Rn{r} is +, and r is not in EXT(+).
c.  is an inner bound for Rn{r}, but r  INT().
d.  does not have outer bounds for Rn{r} concentric in it.
Consequently Rn{r} is not a solid.
This is important, because if it were, the structure would not be distributive, since
both Rn and Rn{r} would have the same complement.
So you cannot eliminate isolated points,. Similarly, you cannot eliminate a set of
points which doesn't form a boundary, like a line segment. The argument runs along
similar lines.
Also not possible is the following:
Let b be a proper bound in Rn and take Rnb.
In that case EXT(b), INT(b)  Rn.
In this case b is neither an outer bound nor an inner bound for Rnb.
b is not an outer bound, since it it not true that for any bound c in Rn: if b is properly
concentric in c then INT(c)  EXT(b)  RnX  Ø. This is obvious, since EXT(b)
 Rnb.
Sinilarly, b is not an inner bound, since it is not true that for any bound c in Rn: if c is
properly concentric in b then: INT(b)  EXT(c)  RnX  Ø. Again this is obvious,
since INT(b)  Rnb.
If b is neither an outer bound not an inner bound, then the argument is as before, you
cannot eliminate b because the points in b are neither in some outer or inner bound for
Rnb, nor in the exterior of every outer band, nor in the interior, disjoint with Rnb of
some inner bound, nor are they in the interior of some inner bound, but outside every
outer bound concentric in that. Consequently, Rnb is not an open body.
The domain of n-dimensional solids is important, since it is the domain on which ndimensional Euclidian Geometry is defined. For three dimension Euclidian
Geometry, we only need to add a relation of 'points x and y equidistant to point z' to
derive the appropriate geometry of three dimensional bodies.
Algebraically, we have the following.
Bn is the set of n-place solids, Fn the set of finite n-place solids.
For set X  Rn let BX be the set of outer and inner bounds for X.
We define Bn = <Bn, v, , u, t, 0, 1> and Fn = <Fn, v, , u, t, 0, 1>
where:
1. a v b iff a  b
2. b = (Rnb)BRnb
(Rnb is a closed regular body. Taking away the bounds makes the result an open
regular body again.)
3. a u b = a  b
uX = X
(The intersection of open regular bodies is an open regular body.)
4. a t b = (a  b)  {r  Rn: for some x,y  Ba  b: x  y and r  x  y}
(The union of two open regular bodies is not necessarily an open regular body,
because the bounds are not necessarily disjoint: adding the points where two bounds
overlap to a  b merges two touching parts into one part with a new open outer
boundary around both. If you do this for any parts whose bounds touch, the final
result will be a regular open body.)
tX = X  {r  Rn: for some x,y  BX: x  y and r  x  y}
5. 0 = Ø
6. 1 = Rn
Theorem: For every n:
Fn and Bn atomless Boolean algebras.
Fn is the countable atomless Boolean algebra.
Fn is not complete, but Bn is, in fact:
Bn is the completion of the countable atomless Boolean algebra.
(the intersection of the complete Boolean algebras of which Fn is a subBoolean
algebra, i.e. the smallest complete Boolean algebra extending Fn.)
Fn and Bn are homogeneous,
(where Boolean algebra B is homogeneous iff for every b  B: {a  B: a v b} is
isomorphic to B).
Summary of important Boolean algebras:
The finite Boolean algebras are:
for each n: the powerset Boolean algebra with 2n elements.
The countable Boolean algebras are:
-The countable atomless Boolean algebra F.
-for each n, the product of F with the finite Boolean algebra with 2n elements,
this gives, for each n, a countable Boolean algebra with n atoms.
The two most important Boolean algebras of cardinality 2o‫ א‬are:
-The powerset Boolean algebra with ‫א‬0 atoms.
-B, the complete atomless Boolean algebra which is the completion of F.
Consequently, also important is:
-The powerset Boolean algebra with 2o‫ א‬atoms: the smallesty free Boolean
algebra of which all the latter are subBoolean algebras.
Note that B is not the only complete atomless Boolean algebra of cardinality 2o‫א‬. But
it is the only complete homogeneous one in which F lies, so to say, extended. It has
in essence the same relation to F as R itself has to Q.