ORDERINGS AND BOOLEAN ALGEBRAS
NOT ISOMORPHIC TO RECURSIVE ONES
by
Lawrence Feiner
B.S., Massachusetts Institute of Technology
(1964)
Submitted in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
September, 1967
Signature of Author...........................
........
Department of Mathematics, June 27,
0
a
1967
Certified by .............
esi
Accepted by .......
......
Chairman,
Super'
or
^......
..................
Departmental Committee on
Graduate Students
ACKNOWLEDGEMENTS
Thanks are due to Professor Anil Nerode for
suggesting the topic of this paper, to Dr. Alfred Manaster
who is directly responsible for whatever degree of elegance
the exposition of these results has attained and whose
advice was invaluable in the preparation of this paper,
and to Professom Hartley Rogers and R. Gandy for helpful
advice concerning references.
-
1
-
ORDERINGS AND BOOLEAN ALGEBRAS
NOT ISOMORPHIC TO RECURSIVE ONES
Lawrence Feiner
Submitted to the Department of Mathematics on June 27, 1967
in partial fulfillment of the requirement for the degree of
Doctor of Philosophy.
Abstract
Chapter I:
With each linear ordering,
e,
with first
and last element we can associate a Boolean algebra D . We
discuss a Cantor-Bendixon like classification of order types
due to Erd6s and Hajnal [1], whereby with each order type
and each ordinal v we associate the vth derived linear
ordering, T(V).
?(
)
9
is defined to be the least ordinal,
£(v) = £(vl). We discuss a similar classification
of Boolean algebra isomorphism types due to Mostowski and
Tarski [2], whereby with each Boolean algebra isomorphism
for which
type, B, and each ordinal, v, we associate the vth
derived Boolean algebra isomorphism type, B(v).
8(B) is
defined to be the least ordinal, v, such that B(v) = B(v+1)
We show that, for any linear ordering with first and last
element, £,
and any ordinal, v, (D,)(v) is isomorphic
to D.(v). We use this fact to give non-topological proofs
of some standard properties of Boolean algebras.
Chapter II: We discuss countable Boolean algebras. We define
strict Boolean algebras and we discuss Lindenbaum algebras of
Theories. We define what is meant by a TT1, (F l
1
arithmetic, recursive) Boolean algebra.
Chapter III: We show that, for any subset
numbers, there is a partial ordering, 01 ,
-
ii
-
X
of the natural
such that &
is
r.e. in X but not isomorphic to a X-recursive partial
ordering; there is a linear orderng'l, such that T is
T7 in X, but £ is not isomorphic to any X-recursive
linear ordering; and there is an HX -recursive Boolean algebra,
B, such that B is not isomorphic to any X-arithmetic Boolean
algebra.
strict Boolean algebra with
Chapter IV: We show that any E
a scattered base is isomorphic to a recursive Boolean algebra.
We show that there is a TT' strict Boolean algebra, B,
with a scattered base such that B is not isomorphic to any
E Boolean algebra. We show that if B is the Lindenbaum
TTI
teorythenKleene
en
algebra of a 71 axiomatizable theory, then 6(B)
However, there is a Lindenbaum algebra B of a
-axiomatizable theory such that 6(B) > o leene
'
Thesis supervisor:
Title:
Hartley Rogers, Jr.
Professor of Mathematics
-
iii
-
TABLE OF CONTENTS
Pages
Chapter I
: The Erd6s-Hajnal classification of
denumerable order types, and the
Tarski-Mostowski classification of
Boolean Algebras with ordered bases.
§1:
....
§2:
6. . ..
. .
. . . . . . . . . . . . . . . . . . . .. 2
. . .......
........ .....
..
Chapter II : Countable Boolean Algebras and Boolean
Algebras whose elements are integers . .
§1:
Countable Boolean Algebra$ . . . . . . . . . . . .18
Chapter III:
Coding Functions into the Isomorphism
Type of an Ordering. . . . . . . . . . .
§1.
Preliminaries. . . .
§2.
Partial Orderings .. . .
§3.
Linear Orderings . . . . . . . . .. . .
§4.
Coding
§5.
.17
.27
. . . . . . . . . . . . . . .28
. .
-
.
. . . ..
. .31
. . .
.
.37
H -recursive functions into Recursive
W
Orderings............
.
....
. .. .. .. .
Boolean Algebra
.
.
....
.
..
.
§1.
Preliminaries.
§2.
Analysis of ) by means of the Analytic
Hierarchy. . . . . . . . . . . . . . . . .
.
.
.
.
.
Bibliography . . . . . . . . #.* .
.
.
.
.
.
.
.
.
iv
-
.
.*
.
.
. .
.61
62
.66
.. . .. . . .
Biography. . . . . . . . . . . . . . . . . . . . .
-
. .54
S.
Chapter IV : Analysis of 3 by means of the
Analytic Hierarchy . . . . . . . . .
.
.43
.
.
.78
NOTATION
N is the set of natural numbers.,
is the standard model for arithmetic.
= (N,0,
J(x,y)
+.}
is the
standard pairing function which associates ordered pairs
with natural numbers.
is the
(n+l)-th
by seq
(s),
divides
prime.. s
iff
s)).
J 1(x) = (K(x), L(x)) e N .
(x)(p
is a sequence number, denoted
divides
If seq (s),
y + 1
divides
divides
is the greatest number,
y = s(i),
r
such that p
s,
r,
and
s.
Hi =the complete set
= (ilK(i) e H
}.
H
L(i)
x
H
s ->
(y)<x (p y
1(s) = n, iff n is the
then
greatest number such that p n
iff
pn
= X'.
x
= (ijK(i) e H
). If
CL)
L(i)
(partial orderings), and
Hn+1 = H', for
n+
n-
T
T
and
HX+
n > 1.
,
=(HX)
n
l
9'
for
n > 1.
are linear orderings
is isomorphic to
9', we write
X a 9'.
We say "'e is the G6del number for a recursive linear
or eeLi
ordering",Aiff there is a relation, A1(x,y), whose field is
N,
such that
(A(x,y)
R (x,y) is a linear ordering, and
<-> (e}(x,y) = 0).
(x)(y)
Since every r.e. linear ordering
is isomorphic to a recursive linear ordering (See Crossley [i]),
we consider only recursive linear orderings.
- v -
o
is the least
W, is the least non-recursive
uncountable ordinal, and
ordinal.
If
B
is
a Boolean algebra, we say that
(a)eB (b)eB (a -,
iff
f
say that
a = b).
b ->
If
B
f:B ->
is
strict,
B', we
is a Boolean homomorphism, iff for every a, b, c
e B:
(i)
a U b -
c ->
f(a) U f(b)
f(c);
(ii)
a n b -
c --
> f(a) n f(b)
f(c);
(iii)
a - b
--
> f(a));
a < b ->
(iv)
f(a)<
f(b)
Furthermore, if, in addition to (i)-(iv), we have
(a) eB(b) eB (f (a)
we say that
If
B
is
f
< f (b) ->
a < b),
and
(a) e> tb)
eB(f (b)
is a Boolean isomorphism, and we write
a Boolean algebra and
B
has
IT1
(1
is
T
(1l,
A ,
a(l)
or
B Z B'.
A ,
recursive) field, operations, and relation, we say that
B
recursive).
(2)
written before a logical
w.f.f.
is
not a reference to a footnote, but is a device used to refer
to the formula later on.
- vi -
-a),
INTRODUCTION
In this paper we study countable structures which
are not isomorphic to a structure whose field, operations
and relations are recursive.
(i)
Similar known results include:
the construction of a sentence with no r.e.
models; and
(ii)
the proof that no non-standard model of
arithmetic is r.e.
Most of the constructions of a sentence with no r.e.
models reduce to the fact that Von Neumann Bernays set theory
has no r.e. models.
found in Rabin [ii].
A very nice proof of this fact can be
The proof that there are no r.e. non-
standard models of arithmetic is due to Tenenbaum [iii].
However, in this paper we discuss partial orderings,
linear orderings and Boolean algebras.
The problem which we
discuss is that of finding a partial ordering (linear ordering,
Boolean algebra) at level
9, say, of the Hensel-Putnam
Hierarchy [iv] which is not isomorphic to any partial ordering
(linear ordering,
Boolean algebra) at any level
9'
< 9.
In Chapter III we use a method of "Coding functions
into the isomorphism type of an ordering" to show that for any
subset of the natural numbers, there are:
- vii -
(i)
partial orderings which are r.e. in
X, which
are not isomorphic to any partial ordering recursive in
(ii)
linear orderings which are
1TT
in
X
X.
which
are not isomorphic to any linear ordering which is recursive
in
X; and
(iii)
Boolean algebras whose field operations and
relations are recursive in
which are not isomorphic to
Hx
any Boolean algebra whose field, relation and operations are
arithmetic in
X.
We prove (i) and (ii) by constructing an
partial ordering (X-TT0
X- r.e.
linear ordering) which is not
elementarily equivalent to any
X-recursive partial ordering
(X-recursive linear ordering).
We prove (iii) by constructing a
Boolean algebra whose operations and relations are recursive
in
Hx, but which is not elementarily equivalent in the weak
second order theory (see pp.
54-55
whose operations and relation are
) to any Boolean algebra
X-recursive.
9 < o , if we let
For any ordinal
X = H,
we
observe that, by (i) and (ii), there is a partial ordering
9 + 1
(linear ordering) at level
of the Hensel-Putnam
hierarchy which is not isomorphic to any partial ordering
(linear ordering) at level
(iii) to show that, if
9.
"A<
-
We can modify the proof of
and
1
viii
-
A
is a limit, then there
is a Boolean algebra at level
X
which is not isomorphic
to any Boolean algebra at any level
9 < X.
This fact is
not proved explicitly in this paper.
In Chapter I, we use results of Erd6s and Hajnal [1]
and Tarski and Mostowski [2] to give non-topological proofs
These
of some classical properties of Boolean algebras.
properties when dualized via the Stone representation theorem,
In
[v] become well known theorems of O-dimensional topology.
particular, we assign a rank
6(B)
to each Boolean algebra
In chapter IV, we perform a constructive analysis of
B.
6 by
means of the analytic hierarchy to obtain:
(i)
If
B
is
strict Boolean algebra with a
Fl
scattered base (see Chapter I, p.11 of this paper.)
then
B
is isomorphic to a recursive Boolean algebra; and
(ii)
there is a
7T1
strict Boolean algebra with a
scattered base which is not isomorphic to a
E_
Boolean algebra.
We also prove:
(i)
6(B) < w.
such that
(ii)
If
B
is a
strict Boolean algebra, then
r
However, there is a
T
strict Boolean algebra
6(B) > o1 ,
If
B
is the Lindenbaum algebra of a
6(B) < o1 .
axiomatizable theory, then
Lindenbaum algebra, B, of a
-
However, there is a
7 -axiomatizable theory, such that
8(B) > w1 .
-
TT
ix
-
BIBLIOGRAPHY FOR INTRODUCTION AND NOTATION
i. J. N. Crossley
Constructive Order Types I, Formal
Systems and Recursive Functions, Amsterdam NorthHolland Publishing Company, 1965, c.p. 198.
ii.
M. Rabin
On Recursively Enumerable and Arithmetic Models
of Set Theory, j.
iii.
S. Tenenbaum
S.
L.,
23,
(1958)
4o8-416.
Non-archimedean Models for Arithmetic
Amer. Math. Soc. Notices, 6 (1959) c.p. 270.
iv.
G. Hensel and H. Putnam
On the Notational Independence
of Various Hierarchies of Degrees of Unsolvability,
J. S. L.-, 30, (1965)
v. M. Stone
69-87.
The Theory of Representations for Boolean Algebras
Trans. Amer. Math., 40, (1936) 37-111.
- x -
CHAPTER
I
The Erd6s-Hajnal classification of denumerable
order types, and the Tarski-Mostowski classification of
Boolean Algebras with ordered bases
- 1 -
§1.
3
Let
a subset of
£=
L, we say
is dense in
S
£
no dense subsets, then
S c L
£
&
contains
a-< b
If
L
is said to be scattered.
A
is said to be a segment of
a, b, c e L, (a-< c-< b
is
S
S
a-< c-< b.
such that
implies there is a c e S
iff
a, b e S,
more than two members, and, for every
subset
If
be a linear ordering.
L,-<)
a e S
£
iff, for every
b e S) ->
&
has
c e S.
If two linear orderings are isomorphic, they are said to be
of the same order type.
In Chapter I, we will usually make
no distinction between linear orderings and order types.
However, in Chapters II, III, and IV, the actual presentation
of an order type will become important.
T
is the order
type of any countable dense ordering with no first and no
last element.
(x e L
A point
I a-< x)
a e L
is said to be isolated iff,
has a-<- least member.
and has no isolated points, then
I
If
I
is denumerable
is one of the types
1, 1 + T, 1 + 9 + 1, r + 1, S.
The following definition is due to Erd6s and Hajnal
[1]
I = (L,-< )
Definition 1.1:
Let
each ordinal
we define an equivalence relation,
L
as follows:
T,
If
T = 0, then
-
2 -
be a linear ordering.
a
-
-0£ b
iff
-
a = b.
For
on
Suppose that
every
T > 0, and that
I
has been defined for
v < T. Furthermore, suppose that, for every
v' < v <
and every
(i)
z
a
(ii)
T:
is an equivalence relation on
, b
-
Let
a
->
(a-< c-< b
(iii)
= -V
& a
b)
< T.
b
Define
a-< b
to mean
> a
If
to be the equivalence class of
[b]v
L;
b ;
-
v be an ordinal
[a]v.<
a, b e L,
c.
-V
b e L, define
[b]
under the relation
a
&
-
b
V
£(v) = {{[b]v beL '
Observe that
X(v)
T
is called the
is a limit define
£ b).
a
define
a
" x e([c] v
is
vth
b
L
is a linear odering.
derived linear ordering of
b
ai
If, for some
-
<v
v,
to mean
£.
(3a
v < T
&
[b] -<
x-<
[a] V
(ii)
and
(iii)
T = v + 1,
to mean
[a]
<-< x<
[b]
or
finite"
In either case observe that (i),
hold for every a, b, c e L, every
Let
every
B(9)
v'
be the least ordinal
a e L, b e L, a
-T
17 b <->
a
t
such that, for
1 b.
-T+l
- 3 -
< T+l.
< V
)(9)
is
If
the least ordinal
B(Ts)
T,
such that
has no isolated points.
9(T)
is an invariant of the order type of
Example:
(1=
If
3.
=
the reals, then
X = w + 1,
If
1.
=
Z
9.
3(9) = 0.
F (9) = 2.
then
9 = c, then
If
(W2) (1)
=
H
The following facts can be found in [1].
We shall list
them here, and sketch the proof in some cases.
Lemma 1.2:
)(Z)
If
£
is a denumerable linear ordering, then
If
£
is scattered, then
< 0.
Lemma 1.3:
has a dense subset, then
Proof:
£(v)
()
If
is not difficult to see that
Conversely, if
induction on
S
(
I
£
If
£
has a dense subset, iff
has a dense subset, it
must also have a dense subset.
is a dense subset of
v, that for any
shows that [[a] }aeS
1.
is dense.
Observe that, for any v,
has a dense subset.
(
T, we can prove by
a, b, e S,
a
is a dense subset of
b.
This
(v).
The following facts are not stated explicitly in
[1],
but we shall need them later so we list them here.
Lemma 1. 4:
Let
S be a segment of
restricted to. S.
<->
a-
-v
b.
If
a, b e S
Z.
Let
Z (S
then, for every
Therefore, 3(T P S) <
be
v., a --
Z
b
Lemma 1.5:
If
I
I(v) = 1.
such that
least
v
such that,
Lemma 1.6:
scattered, then
is
If
Z
(9) is the least
In other words,
for every
3 (9)
?(Z) isn't a limit.
Proof:
b
of
£
If
?(9)
and
respectively.
a _-
is
b.
=
?3(T) <
Theorem 1.7:
£
If
v.
Let
£
, and let
a ->
p
p
a.
Suppose we have
limit, then
p
Contradiction.
for
implies
Suppose
[[a] , [pI)
T
c
-
of
Z. Thus,
jj
3((£-o) + 1) >
ith
copy of
. We wish
obvious for
T = 0.
If
2(I).
(~3y)( v< y
a, for some
T <
&
P ~
-
T =
is a
T
a).
v
v + 1.
a / p.
This
contains a finite number of elements,
qn e
Choose
j.,
a-< c-< b.
£,
be the
v < -r<
b.
such that
is a successor.
[q ]v
for some
£.
The fact is
a
T
of
£
-
a e ., and for any ordinal
[qlv ... [q ] , where
j,
c
a
be the greatest element of
proved it
-
Let
Furthermore, suppose that p
means that
< 2(Z),
is scattered, then
to prove that, for any
p /
a v
Contradiction.
t = (T,-o) + 1.
Proof:
-V
Therefore
c, for any element
1, and
in
= 1.
£(())
a limit, there is
a
b.
-_~
be the least and greatest elements
However, for any element
Therefore
(V
a
is the
is scattered and has a greatest and a
least element, then
Let
a
a, b e Z,
v
- 5 -
is the greatest of these elements.
q+1 e
j+2*
We claim that
j+
X, y e
- v y. Thus, by
Contradiction.
,
n+1 ,then
1-1.4, )(1j+1
[qn]v
is greater than
*
qn+ 1
,
p.
but less than
[qnv is the greatest
which contradicts the fact that
element of
for any
By our inductive hypothesis,
[qn+1v
Therefore,
[p]v
v
v
v q1
n
[[a]v, [pv).
We have proved that
least two elements.
scattered, and by
Definition 1.8:
Since
£
We define
ao)
contains at
is scattered,
B(9) <
1-1.5,
+ 1)(
((X,.w)
(1-CO) + 1
((,-co) + 1).
is
II
as follows:
Mo0 = 1 ;
if
T
w T = Aimw v
is a limit,
V<T
if
T = v+,
o' = MO
An ordinal is called a principal number for addition
if it is not a finite sum of lesser ordinals.
Lemma 1. 9:
Proof:
is the
co
principal number for addition.
Straightforward induction on
Theorem 1.10:
(ii)
(a)
T;
=
+
x.
Tr:
For any
(i) 3(CO)
Proof:
ath
1)
>
)(oUT).
The theorem is trivial for
- 6 -
T =
0.
Suppose that for
any
v
< ',
)(W T
to prove that
Case (i):
T
Choose any
and
) = T,
3(oD
1)
+
We wish
(oT).
>
is a limit.
9 < oT
By
1-1.8, there is a
# < T
Therefore, 0
9,
<oc.
<(o)=
WT
<
;.(oy + 1) > B(af).
and
v
=
implies that
0
9.
C
-
This shows that
such that
which
= l.
(1)
Thus,
-T
>
().
Now, suppose that
limit,
=
v + 1
v + 1 >V.
by 1-1.4.
Dv+1
Therefore
case (ii):
T.
Since
?(CD)
+1
T = v +
=
a(W
T
>\ ,
)
must be
+1)
T
B(oT + 1) > B(oD
).
£
of order type
Let
ai e £ be the first element of the
Let
p
be the greatest element of
)(Mv
T
W(oD
CD,
1
Consider a linear ordering
hypothesis,
is a
+
T
1-1.6,
By
.
T
Since
is a segment of
is a limit,
T
aT.
<
M
and, therefore
< T,
Because
a successor.
v <
1) >
+
and
£.
ith
my-D .
copy of
o *?
By our inductive
BQov)
M() =
and
(DV)(v)
1
Using this fact one can easily prove that:
(i) for any
(ii)
c
-
a,
is less than
(iii)
i > 0, a i
for any element
where
i
c
of
a+
1
£,
if
c / p, then,
is the greatest integer such that
c;
for any
'
c / p,
p
-7 -
c.
a
v+1 (v ) =
v+1
)W= and (W + 1)
This shows that
(v+1 (v+l) =
+ 1 ) (V+1) = 2.
1 and (v
= w + 1.
Thus
Therefore
S(o v+1) = j+1
Theorem 1.11:
If
Z
and
(ov+
+ 1) > v + 1.
11
is a denumerable linear ordering,
then:
(i)
where each
is scattered and
re
is one of the types 1,
(ii)
proof:
(i)
(ii)
2(c)
=
1++1,
A-u-b.
n+1, i,
r )Jre
[(X(
1+i;
).
See [1, p.1 19].
Straightforward application of results stated
previously.
- 8 -
§2.
Let
B
be a Boolean algebra.
Let
be an ideal
I
of
B. There is a natural Boolean homomorphism
If
a e B, we define
jai,
to be
i(a).
i:B _>
The set,
I
B/I.
o B,
of elements bounded by a finite union of atoms is easily seen
to be an ideal of
algebra of
then
f
If
B/I
is
->
B'
f:B
called the first
a
Lemma 2.1:
(i)
-
is a Boolean homomorphism,
f:B ->
B'
(a)B (b)eB
(a
If
Lemma 2.2:
Let
B
element of
B,
then
elements of
is a Boolean monomorphism, then:
< b <->
I .
r
be a Boolean algebra and let
b J I ,
iff
limit, then
I
follows:
I =
I
B
.
If
(B/IB)
= (alaeB
&
lai B
T
= (al aeB
v<T
I
b
b
Xe
bounds
(Tarski-Mortowski [3]).
algebra, then for each ordinal
as
< f(b));
f(a)
be an
disjoint
B.
Definition 2.3:
B
(a)eB(f(a)
0).
(ii) f~(Il )
I Bc
derived
is called a Boolean monomorphism iff
0 <->
-
B.
B.
1
v
- 9 -
T=
If
B
is a Boolean
we define an ideal,
&
a
_
0).
v+ 1, then
If
v
is
a
v, let
For all
B
be
B/I
.
Let
6(B)
If
Examples:
be the least
v
such that
I
=I+1'
is atomless.
8(B) = 1. If
is finite, then
B
B(v )
B , and let
I
B(v)
such that
v
is the least
6(B)
lal
be
jal
B
is the
algebra of all finite-cofinite subsets of the integers, then
B(B) = 2.
If
f:B ->
If
Lemma 2.L4:
B'
is a monomorphism, then, for any
,
)
V
Y
Do induction on y , using
ordinal
f
v,
Proof:
6(B) = 0.
is atomless, then
B
1-2.1.
ff
In the study of the elementary properties of
Boolean
Algebras, one very useful source of examples is the class
of interval algebras.
I = (L, -<)
If
is a linear ordering
with first and last element, then the interval algebra,
of
I
Dg,
is the set
{SIScL
& S = [a1 ,bl) U...U
a -< b 1 -< . ..- < an-< bn' of
for some finite sequence,
elements of
L).
Let
b
It is easily seen that
set of
L - (e).
isomorphic to
example,
Do+1
If
D,,.
be the greatest element of
D,
I
[an, bn)'
is a subalgebra of the power
is isomorphic to
',
then
However, the converse if false.
is isomorphic to
-
£.
10
-
D1+o*+1'
D,
For
is
of
B
S _ B
A subset
Definition 2.5:
is called an ordered basis
iff:
(i) the ordering of
S
s < t,
basis,
a, b e S,
such that for any
c S.
such that
a
last element
If
B
is an ordered basis
b. For any ordered
a
of
B
([a,x)|xeX}
b, respectively, then
a strict ordered basis of
ordered basis, 9,
is
is a linear ordering with first and
I
and
B
a strict ordered basis
B, there is
S, of
t < s;
S.
A strict ordered basis, S, of
B
or
is total,
that is any element of
a Boolean combination of elements of
of
S
restricted to
either
B,
generates
S
(ii)
in
s, t
that is for any
B
B
Conversely, if
T.
of order type
I, then
is
has a strict
is isomorphic to
B
D.. (See [2])
We say that a Boolean algebra has a scattered basis
iff it has a strict ordered basis which is scattered.
Lemma 2.6:
Let
be a subset of
generates
( ai I aeS)
B/I.
I c B
B.
If
If
S
be an ideal of
S
generates
B
and let
B, then
is an ordered basis of
is an ordered basis of
S c B
( Jal, aeS)
B, then
B/I.
We now turn to the main result of this Chapter.
As
far as the author can tell it has not appeared in the literature,
though consequences of it can be found in [2] and Mayer and
-
11
-
Pierce [4 ].
Theorem 2.7:
If
£
is a linear ordering with first and
v, (D )(v )
last element, then, for any ordinal
to
is isomorphic
D (v).
This theorem will be an immediate consequence of the
following lemma.
Let
Lemma 2.8:
where
B = D,
For any a, b e £, any ordinal
If v = 0, then
Proof:
the lemma is true for
a
-
-T
b
&
(3 Y)(v <
(c.}
-0I
[a,b) e IB
v,
b <->
[a,b) =
We prove it for
T.
&
b)
a
[a,b) e Iy) <->
V b.
By
of elements of
(i) for every
(ii)
(iii)
Suppose
V
=
T.
<->
[a,b) e I
I-1.1, there is an infinite sequence
1, such that:
i,
c
ci+1'
either (i)(c 1 -< ci+ 1 ) or
(i)(ci+1 -< ci);
(c } c [a,b).
By our induction hypothesis, for every
U[c i+1, c)
#.
r = v + 1.
a
Suppose
V <
(~v)(v <T
<-
case (ii):
_
has first and last element.
is a limit.
T
case (i):
a
I
I
.i Therefore, I [a, b)
-
12
-
bounds
i,
[c ,i+
disjoint
non-zero elements of
B/I
and, by
,
a
Conversely, suppose
element z
1< i < n.
of
a < z < b
£,
[c 1 ,
c2 )
Let
[ci, ci+ 1 ).
c.
z
Let
i
implies
z
[a
U"o*U[Cn-l-Cn)
.
and
z2
Jab)tJe
We now use the induction
is an atom, for every
c
By our induction hypothesis,
then
[z1 ,z2 ) does not fill
be a member of
[c , i+l
Thus
[c ,ci+
[zi,z 2)
1B
c
C [c, ci+ 1 ). By our
B
1
1 )I
B
or
is an atom.
To conclude the proof of
II-2.6,
11-2.8
{[0, a)
and
11-2.5,
is
I
E
From now on when we write
13
which fills
II-2.7, we note that, by
is isomorphic to
-
, depending
H
an ordered basis for
B
I,
[c ,ci+1), then
such that
D
[z1,z2 )
on whether or not there exists an interval of
).
for some
be an interval such that [z 1 2
We say that [z 1,z2 ) "fills" [cci+1)f
induction hypothesis, [c.,
[c , c
ci
y
[z ,z2 )
if [fz 1 z 2 ) fills [c ,ci+ 1 )
and if
-< b, such that
-<
I[cicc
hypothesis to show that
< i < n.
...
n = b, and for any
l,c )I
U... U[c
I[c1,c2 )1
That is, there is a
b.
-T
'''
C3
[a,b) =
n.2
-
a.< c -< c 2 -< cc<
finite sequence
c2
is
I B/ I Y
not a member of
a=cc
1[a,b)|
I-2.3,
-
B(Y).
Thus
D (i
D£, we shall assume that
9
has first
and last elements.
Lemma 2.9:
Corollary 2.10:
Z=
proof:
1!J
See 1-1.11.
Corollary 2.11:
has no isolated points.
£
is atomless, iff
S(D9).
There are
isomorphism types of countable
X1
Boolean Algebras.
Proof:
+
If
BstaB',
6(B) = 8(B').
then
and,
<
by 1-2.10,
If
T < T',
then
DOT+1 p6 DWTt'+1
See [4, pp. 937-938] for a topological proof of this
Remark:
fact.
The following facts will be useful in Chapter III.
Lemma 2.12:
B
I
is scattered, iff
(D t)
(
Thus1 if
Dl.
has a scattered ordered basis, then every basis of
B
is
scattered.
Proof:
Lemma 2.13:
If
f:Dr
scattered, then so is
Let
be
B
D
-.
a monomorphism from
(B' ) " kal
B(IB'
)(s
-
B
9'
£.
and
B'
be
DL,.
into
B'/I
BI
B/f 1(I
Therefore
Di.
1-1. 3.
is a monomorphism and
Dy,
natural homomorphism from B'
f- f
and
Follows immediately from 11-2.7
i;
B/f 1(I
D
D .
B/I
-
14
-
t
Let
B'
BI/I B1
,
SD
Thus,
1
.
i
be the
i of induces
, .
By
However.,
1-2.1,
B (Y) _< ) (1' ),
and (B)?)(r)
By 1-2.12,
D1 .
Lemma 2.14:
9
If
and
9'
are ordinals and
there is a monomorphism from
Proof:
Let
9 + 1
= 9'+1.
f
into
2 D
Proof:
into
9'.
Let
9' + 1
f
I
If
then
D9'+1'
9 onto an
be an order preserving map
such that
f
induces a monomorphism from
Theorem 2.15:
D
D9+1
9 < @',
be an order preserving map from
initial segment of
from
must be scattered.
£
9 =
and
D9+1
into
f(9+l)
D 9'+1'
is denumerable and scattered then
for some ordinal
9.
See Mazurkievicz and Sierpinski
[8, pp. 2-21.].f1
Remark: In Erd6s and Hajnal [1], we find the following classification of denumerable scattered linear orderings.
O
of
= (0,1}.
Let
L) 0,.
0
=
Then
all
o*-sums of members
w-sums and
0
0=
,
Let
is the collection of
all denumerable scattered linear order types.
express their belief that the classification is
The authors
so natural
that it must have appeared somewhere before in the literature.
This author has not been able to find it
if
it
has.
However
if we look at the Stone spaces of the interval algebras of
the order types, it turns out that
Erdbs and Hajnal's
classification theorem is equivalent to Sierpinski-Mazurkievicz's
theorem that every countable compact space is homomorphic
to an ordinal.
We can do induction on
-
15
-
a
to show that for
every denumerable scattered linear ordering
and last element,
9 < P.
D
is isomorphic to
Z, with first
D8 , for some
Now the Stone Representation theorem tells us that
every countable compact Space (a countable compact Space
must be Boolean) is homeomorphic to an ordinal.
In this non-
topological proof of a classifical topological theorem the
classical fact that every countable ordinal is an
w-sum of
smaller ordinals takes the place of separability and the well
ordering of the ordinals takes the place of compactness.
Conversely for any denumerable scattered linear ordering
with first and last element,
We can do induction on
denumerable scattered
Z
is isomorphic to
D cc
o n+1
(an) to show that, for any
£
with first and last element,
The classification theorem follows easily from this.
also use the Classification theorem to prove that, if
is a countable Boolean algebra and
has a scattered basis, then
QcY
a
B is isomorphic to
<)
-
16
-
£
We can
B
B/(atomless elements)
D
TwQ
where
£
0.
CHAPTER
II
Countable Boolean Algebras and Boolean Algebras
whose elements are integers.
-
17
-
§1.
Countable Boolean Algebras
The following theorem is part of the "folklore" of
the subject of Boolean algebras.
We include it here for
completeness.
Every denumerable Boolean Algebra
Theorem 1.1:
has an
ordered basis.
Proof:
B.
a1 , a2 ,
Let
be an enumeration of the members of
...
We define inductively an increasing sequence of finite
sets as follows
[ 1 )
A
An = {b 1 , ...
Suppose
(i)
, bn)
0 < b 1< ... _<b n - 1'
(ii)
Each
ai,
of the members of
A .
Let
Now
has the following properties:
c1 = b
1 < i < n, is a Boolean combination
U an+1
c = b
U c
,
for
1 < i < n.
< i < n, and hence:
b < ci, for
0 < b1 n an+ 1 < b1
--
< b
< b
--
b
--
cn-1 n bn < bn < cn *
c
-
18
-
for
1 < i < n; and
r
Thus,
An U (b
an
1
1
)
}
n( c
(b
is totally
l<l<n
ordered by
Let
< .
An+1 =
n an+1
An U (b
U (Cn} U
U
{b
i-
l<i<n
We now claim that
a+1
C .
b 1and
a n+1
computati on.
is
To see this, we perform the following
an+1 =
(b
U an+ 1 ) n (~b
U an+1)
an+
=
(b
U a n+ 1 ) n (b
n an+1
an+1
=
(b
U an+1) U (b
n a+
an+
=
Similarly,
c+'for
c
U (b,
n an+1)
is a Boolean combination of
1 < i < n.
Thus,
(an+1 n b 1 , b 2 n c 1 , b 3
an+1
2,
an+1
. . .,
S =
U
n > 1
S
generates
n bi+1
b
n cn
n-1 ,
Cn},
B.
19
-
and
and, hence,
An+1'
S is totally ordered by
A.
-
c
is a Boolean combination of
is a Boolean combination of members of
Let
and
a Boolean combination of
<
We will now discuss Boolean algebras whose elements
are integers, and whose relations and operations are relations
and function of integers.
points of view.
In doing this, we can take two
We can regard the equivalence relation "-"
as an equality relation, and say that two integers represent
the same elementof the Boolean algebra, iff they are the same
In this case, our Boolean algebra will be a strict
integer.
Boolean algebra, that is, one in which any two equivalent elements
are, in fact, identical.
On the other hand, we can take the
point of view that
is not a strict partial ordering, and
"<"
that two distinct integers may be equivalent in the algebra.
As an example of our first point of view, we will discuss
interval algebras of linear orderings of integers.
seen by our definition of an interval algebra
that an interval algebra is
strict.
It can be
(I-p.11 line 12)
As an example of our
second point of view, we will discuss Ltndenbaum algebras of
theories.
We will regard the elements of the Lindenbaum algebra
to be G6del numbers of sentences.
Certainly, two Gbdel numbers
can represent equivalent sentences or even the same sentence.
We will now begin our discussion of interval algebras
with the following lemmas, which will be useful in Chapter IV.
1
1
Lemma 1.2: If Z is a E
(w1 , recursive) linear ordering
(L,,<}, then
D,
can be presented as a Boolean Algebra whose
-
20
-
operations and relations are
Proof:
Let
a
respectively.
{0)
and
b
1
(V , recursive).
be the first and last elements of
We will present
D
as the union of the set
with the set of all strictly ascending,
s, of elements of
The integer
0
finite sequences,
Z, such that the cardinality of
will represent the
0
element of
s
D .
prove the lemma we shall write down the definitions of
U,
is
(i)
1 (
&
To
<, n,
, recursive) if £
recursive.).
<-> x = 0 V(seq (x)
(i)"O _< i < I(x) ->
x
x
&
(i)<L(x)(K(x(i)-< L(x(i))
L(x(i))-< K(x(i+1)) '
:
C
<-> x = OV (seq (x) &
&~~
(1
(iii)
1
is even.
xe D:
x eD
(ii)
F (
and observe that they are
< i < I (x) ->
(i)<L(x) (K(x(i)-< L(x(i))
L(x(i) )--<
K(x(i+1)) .
x < y:
x <
y <-> x e D
(K(y(j)).-<
K(x(i))
&
y eD
& L(x(i)
-
21
&
-<
-
£,
(1)
L(y(j)))>
(3 j)
/
)(::-o
(i.'
7)
..
y:
x
~
x
y
<->
& y e h9
e
x
(i)<g(X)(K(x(i)
) L(x(i)) -
(x(i)
(x(i)
Q2i)(
y
and
x
says that
(
>
->
(j)(K(y(j)/ L(y(j))
x ~ y
&
y(j)))
=
=
Y(j))
are equal modulo empty
intervals.
(v)
x =
x
e D£
<->
y
=
\{ (y = 0
=
K(z(0)
x =
&
L(z(i)) =
&
(vi)
xU y-z: If
x U y"
term of
x U
&
y <
y
&
(x
y
2 1+-J(ab))
L(z(t(z))) = b
&
= L(y(i-1))
y e D9
&
&
& y =
((x= 0
z
2 1+J(ab)
A(y) + 1
&
&
(i)K 0 <
i < A(z) ->
->
K(z (i) )
(i)(O < i < A(z)
.
x, y
and
z
are sequence numbers let
"z
r
be the recursive predicate which asserts that every
z
is either a term of
z <->
& y e D
x e DI
& ((y = 0
x
or a term of
&
z e D£
x = z) \4 (x = 0
&
-
22 -
&
y.
& x <
z = y)
(z) (~2s)(seq(s)
&
(vii)
x
&
s c x U y
L(z(L(z))) = K(s(A(z))
n y = 2 <
&
&
K(s(O))
K(z(O))
=
(j)<,js) (2 (1+s(j))
<
U y
We observe that, in (i)
a positive fashion.
(vii), "_<"
-
occurs only in
Therefore, we bring the function quan-
tifiers of (i) - (vii)
to the front (as in Kleene
and observe that, if £
is
predicates (i) - (vii).
H
ordering, then
r
If
x c N.
Let
Lemma 1.2?:
[:5 ,p. 315])
recursive), then so are the
x-recursive linear
is an
£
can be presented as a Boolean algebra whose
D
operations and relations are
Lemma 1.3:
X 21+z(i)))
X-recursive.
There exist recursive functions
such that if
e
g1 , g2 ' 93 ' g4 ' g5
is the Gbdel number which defines a recursive
linear ordering, the for every x, y, z
(i) x = 0
x
is the zero element of
(ii)
tg,(e)) (x) iff
(iii)
2e)(xy)
(iv)
x
iff
<1+r+1
iff
(g3 (e)}(x,y,z) = 0
x U1 +
x
++y
iff
x
y
(g5 (efl(xy) = 0,
Similar to the proof of
-
23
D1+r+1'
x e D1+9+1'
(v) (g(e)}(x,y,z) = 0, iff
(vi)
Proof:
iff
-
11-1.2.
H
y
=
=
z;
If
Lemma 1.4:
is a denumerable strict Boolean algebra
B
with recursive field, and
U, n,~
a recursive linear ordering,
£,
are recursive, then there is
such that
recursively enumerable strict ordered basis
has a
B
Use the proof of II-1.1 to show that
Proof:
D
B
Let
S.
9
a recursive linear ordering of the same order type as
Lemma 1.14':
If
B
is a strict Boolean algebra with
fiel d, operations and relations, then
-recursive linear ordering
Corollary 1.5:
B z D
£
S. B
D
X-recursive
for some
£.
If B is a strict Boolean algebra with
B zD
hyperarithmetic field relation and operations, then
where
be
is hyperarithmetic.
We conclude this chapter with a discussion of our
second kind of Boolean algebra.
From this second point of
view)we regard the Boolean algebra as being specified by its
natural ordering,
<.
So when we say that a Boolean algebra
of the second kind is
that
<
is
7T
z , A1 , arithmetic,
recursive), we mean
A}, arithmetic, recursive).
consider the Lindenbaum algebra of a theory, T.
we regard,
where
a
If
For example,
In this case,
"_<"
as being the derivability relation
and
@ are sentences of the language, (,
B is a Lindenbaum algebra, then
following nice properties:
-24
-
B
"a
0"
of
has the
T.
(i) The field of
of
B
is recursive.
(It is just the sentences
).
f , f2 ' f3
(ii) There are recursive functions
every
x, y, z e B, f1 (x,y)
f3 (x)
x U y.
such that for
f2
x
x
(The functions
connectives
f , f2 ' f3
are just the propositional
v, A, ~.)
x n y -
In general, however, the predicates
x U y -
of
z, x
<.
z
-
are arithmetically definable in terms
If
B
is a Boolean algebra whose natural ordering,
is hyperarithmetic (arithmetic) then
a strict Boolean algebra
Proof:
If
a e B, let
B
is isomorphic to
whose operations, relation and
field are hYperarithmetic
f
(arithmetic).
[a]
be
(xjxeB
& x
__-
a).
Let
be a hyperarithmetic (arithmetic) choice function which
chooses a member from each class in the collection
Let
be the range of
x Uy -
of
z,
This gives us the following lemma.
Lemma 1.6:
<,
ly'
z,
<,
Since
x
-
z
f.
{[a]} aeB
Since the relations, x n y -
are arithmetically definable in terms
these relations are hyperarithmetic (arithmetic).
f
relations
is hyperarithmetic (arithmetic) so is $.
The
x U y = z, x n y = z, x = z, when restricted
-
25
-
z,
to
are hyperarithmetic (arithmetic) operations.
relation
"~
",
equality relation
when restricted to
"= ".
B
-
becomes the
is isomorphic to
26
-
The
h
.
|I
CHAPTER
III
Coding Functions into the Isomorphism Type of an Ordering
-
27
-
§1.
Preliminaries
The following lemma and definitions will be useful
in what follows.
Definition 1.1:
predicate.
Let
A(x1 ,...,xn,y)
We say that
y
be an arithmetic
is E.A.N.
in
A(x1,...,x?)
iff,
(x1 )... (xm)(r)(.A(xl,...,xm,r)
Remark:
(
->
E.A.N. stands for "everywhere or almost nowhere."
Definition 1.2:
An arithmetic predicate is said to be in
predicate form iff it is in the form
R(Xz 1 ,..., zn),
Q5,...,
~A(x1 ,...,xm'
Qn
where
Q z1 ,Q z2 '
R(g,z 1 ,...,zn)
'z
is recursive, and
are alternating unbounded quantifiers.
Every
arithmetic predicate is equivalent to a predicate in predicate
form.
(See Rogers [7., p.12 6 ].)
Definition 1.3:
If
Q z1 ,...,Q zn R(g,z 1 ,...,zn)
predicate form where
QSk+1 zk+1 ''
Lemma 1.4:
R
is recursive, then Q1 z 1 ,...Qnzn
is said to be in E.A.N.
form iff, for every
R(g,z 1 ,..., zn)
1 < k < n,
is in
Qk =
V
implies
Qnzn R(E
z
is
E.A.N.
z ,5...,lzp...,zn)'
Every arithmetic predicate,
a predicate in E.A.N.
in
form.
-
28
-
A,
is
equivalent to
Proof:
By 111-1.2, we may assume
A
is in predicate form
We do an induction on the number, N, of universal quantifiers
preceding the recursive predicate to show that
equivalent to a predicate in
A
E.A.N.
form.
is in one of the following forms where
A
N = 1, then
If
R
is
is recursive:
(i) (y) R (1,y); or
(y)(z)R(,yz); or
(ii)
In the first case
A
is equivalent to
which is in E.A.N. form.
A <->
In case (ii),
(y)(u)<y (3z)R( ,u,z)
<-5%
where
(y)(u)<y R(9,u),
) (-g Z) (u) <yR( ,u, (z)uI'
(1) is in E.A.N. form.
The third case reduces to
the second by observing that, in general, if
E.A.N. form, then so is (3z)A(1,z).
A( ,z) is in
Before proceeding to
the induction step, we prove the following claim.
Claim:
If
y
is E.A.N.
in A( ,u,y), then it
E.A.N.
in
(u)<v A( ,u,y).
Proof:
Suppose
implies
(3u)<v
Because
y
~(u)<v A(9,u,r)
A(9,u,r).
is E.A.N.,
Thus, for some
-. L
29
-
2,r.
This
t _<v,
~A(,V1,y) holds.
(y).
-
holds for some
This implies
r (au)<
(Y)>r ~(u)<
or equivalently
A(-,u,y)
Using the claim, the induction on
N = n.
Suppose the theorem is true for
(y 1)(3z) (Y2 )
n + 1
) ... R(2,y 1 ,zly
2 ,z 2 ,...),
(2).
in E.A.N.
is
3z2)..R(9,y1 ,z1 ,...)
E.A.N. in
R( ,u,zly
We now drive
1
(
2
u'
yk
' '
therefore, by our claim every
be
where
A has
yk
form.
A <>
where
y
(2)
is
inwards (as in Kleene
is E.A.N. in
z''
-
y ,... )
and,
is E.A.N. in
1u' '.
' ' 'k-
() zk ) ... R(2 ,
Therefore after
2 ,...)
"(u) y1
]) and observe that, every
(zk )...R(,u,(zl)u'
(u)
A
By our induction hypothesis, we can assume that
(yl)(u)<y()z)(y2 ) ...
[6,p.
Let
may be completed.
universal quantifiers preceding the recursive
predicate.
(y2
2
N
A(Z,u,y).
l(u)
is driven all the way
inward, the resulting predicate is in E.A.N. form.
-
30
-
Partial Orderings
§2.
The proof that every recursively enumerable linear
ordering whose field is total is a recursive linear ordering
depends on the fact that a linear ordering
that is,
is an
for any
a, be £, either
r.e. partial ordering and
as we will prove below,
a-< b
9
or
is connected,
b-< a.
If 9
is not connected, then,
need not be isomorphic to a
recursive partial ordering.
Theorem 2.1:
There is a recursively enumerable strict partial
ordering which is not isomorphic to a recursive strict partial
ordering.
Proof:
Let
be the language of the elementary theory,
of partial orderings,
Let
=
P,.<}
and let
w<
be a model of
be its
T.
relation symbol.
is a strict partial
,
ordering iff, for any two elements
a, b e
a-b.
is called an antichain
iff
A set
S
of elements of
any two elements of
S
are
-<-
a = b implies
incomparable.
It is
clear that any anti-chain can be extended to a maximal antichain.
It is also clear that, for every integer n, there is
n
n
a sentence
n of T such that
n asserts the existence
of a maximal anti-chain of cardinality
Lemma 2.2:
(i) If
n.
is a recursive partial ordering, then
-
31
-
ago---- --,--
(ii)
If
ordering then
Proof:
(i)
Ay
(1)
(i)<
is a recursively enumerable partial
1=
hi(t
"f
I=
J") e
n-
(3s) (seq(s)
A(s) = n
&
(i) / j ->
(j)
~ (s(i) -< sj))
< z
(z) (3 j)<n (s(j) --
V z -.<
where
a-< b
a-< b
relation.
means
We apply the
[7', pp. 131-133])
is
E3'
or
a = b.
(i)
in
r2* (ii)
I|
r.e. partial ordering
R(m,y,y
is a recursive
t
is
to see that (1)
Thus, to prove the theorem, it
Let
k"
Tarski-Kuratowski algorithm (see
to (1)
proved similarly.
s(j)),
2 ,'y
3 ,n)
n > 2
6 such that
suffices to find a
A
n)
C 73
be a recursive predicate such that:
&
(3m)(y) (3y
3 2
2)
and
(ii)
is
in
E.A.N.
(:]y 2 )R(m,y 1 ,y 2 ,n)
We seek a r.e. partial ordering
(n) >2 ($ ,n
<-> Y
2'
P
such that
(3m)(y1 )(y 22)R(m,y 1 ,y 2 n)).
-
32
-
For notational convenience, we will deal with a
symbol * in addition to the integers.
That is, the field
of the relation we will construct will be a subset of the
fourth cartesian power of
Let
P
(*,0,1,2,3,4,...}
be the following set of
4 -tuples.
(n,m,x,i)n > 2 & O < x < n & (x < n <-> i*)
xi*
P
n / *
&
&
m
*
can be visualized as follows:
I
a
&
A*
%
&
9
.
4
£
~
4k
p
a
p
If
n '-
a
e
a
*
S
a
We define the following relation on
P:
(n,m,x,i).< (nt,m',x',i') <->
(n,m) <
i < i
&
x
=
(2Y 2) R(m,i,y
x
< n
&
x
1
=
Let
((n,m)
x' = n
2
(n',m')
=
&
(gy2) R(m,i,y 2,n) &
&
,n)) V ((n,m) = (n',m')
n
=
&
&
(Sy 2 ) R(m,i',y 2 ,n))
{(P, -<).
is
Claim (i):
Proof:
/
(n',m')
'l<"
is
an r. e.
relation.
a strict partial ordering.
Follows from an examination of the definition ofS.
The reader will get some idea of what
looks like by checking
that this claim is true.
Claim (ii):
If
S
is an antichain of
and (n',m',x',i') are members of
Thus if
S2
A =
S
then
(n,m)
is an anti-chain we define
m, where (n,m,x,i)
is
an element of
(n',m',x,i) (n',m',xi) e P
&
Bn=
S,
and both (n,m,x,i)
,
(n,m) =
34
n
=
(n,m)
and
Let
& 0 < x < n
&
R(m,i,y 2,n)
-
=
(n',m')
(n',m',x,1) (n',m',x,i) e P
(y2
S.
S1
(n',m').
=
-
x
=
&
n
&
(n',m')
(n',m',x,1) e P
Cnni
& ~(3 Y2)
Claim (ii):
x = n
R(m, i,y2 ,n)
(n',m') =
&
(n,m)
P
a maximal anti-chain of
is
S
S2 = m,9 and the cardinality of
S 1 = n,
and that
S
Suppose that
&
(card S)
> 2, then:
is
(y 1 ) (2
(i)
2
S = n,
card
,n) ->
~(yl )(:A2) R(M,Y, , y 2 ,m) ->
(ii)
Proof:
)R(m,y,y
and
S =
card
.
An examination of the definition of 6 will be the
justification for the assertions made in this proof.
(i)
If
(yl) (
y2)R(M, y , y2, n),
S
then
n
Am*
must be
An = n.
Card
(ii)
Since
there is an
T
C=
m
card
y1
is
E.A.N.
such that
Since
in
2,n),
2 ) R(m, yVy2 , n).
-;
(yl )>r
S
(.5y2 ) R (m,yy
is maximal,
Cn c
S,
so
Thus,
card
s='YC 0 .H
We now conclude the proof of Theorem 2.1.
(2M) (y )3y2)R(m,yl
An
By Claim (ii),
(ff,y, ,y,,n).
cardinality
y2,n), then for some
n, and,
n.
thus,
if
S
=
'
/
n,
S
(y 1) (
Conversely, suppose
= n, then card
then either card
-
35 -
y2) R
a maximal anti-chain of
In this case, if S
(m)~(y )( y2 ) R(m,yl ,y2 ,n).
anti-chain, and
is
R,
If
S
S
=)
=
is
a maximal
On the other hand,
or card
S = hn
n.
MANUS-__
Therefore,
Theorem 2.1:
)
n
n.
Let
There is a
X c N.
ordering which is not isomorphic to an
ordering.
-
36
-
X-r.e.
partial
)( -recursive
partial
3. Linear Orderings
The proof that every r.e. linear ordering is isomorphic
to a recursive linear ordering depends upon the fact that
every r.e. relation is isomorphic to an r.e. relation whose
field is total.
(This latter fact will be proved in chapter
III, §1).
In this section we will construct a R
ordering,
£,
ordering.
In particular,
linear
0
which is not isomorphic to a recursive linear
relation whose field is
is
£
.(y.-< x),
and, thus,
For if
total.
I
is
T
X,
TT
y e T,
r.e. which means that
is
TT
is not isomorphic to any
then for every
total,
field of
I
0
and the
0
x-< y <->
£
is
recursive.
Theorem 3.1:
There is a J7
scattered linear ordering not
isomorphic to a recursive linear ordering.
Proof:
Let -<
be the relation symbol in the language
of the elementary theory,
be a model of
chain
T.
iff for any
T,
A set
of linear order.
S c L
Let
£
=
{L,z}1
is said to be a successor-
a e S, b e S, [a,b)
is finite.
It is
clear that any successor-chain can be extended to a maximal
successor chain.
It is also clear that for every integer n,
n ofn
there is a sentence
of
such that i
asserts the
existence of a maximal successor-chain of cardinality
-
37
-
n.
Lemma 3. 2:
n()=
(1)
3
(ii)
fn)
Proof:
If
T
is a recursive linear ordering then
If
I
is
(i) £|= #n
(~3s)(seq(s)
(z)(s(0)
a
I=A where
is equivalent to
,(s) = n
&
linear ordering then
I 1
< z < s(n) ->
(s (i)
& (1)<
(~ ci (z
=
(z)(z < s(0) ->
(-3y)(z < y < s(0))
(z) (s(n) < z ->
(2y)(s(n)
y
<
<
< s(1+1))
A is
&
s(1 )
z).
We apply the Tarski-Kuratowski algorithm to (A) to
show that (A) e 73.
The proof of (ii) is similar.
Thus to prove the theorem it will suffice to construct
a
TT 0 linear ordering
(
n )
', F
'3
£
such that
*
111-1.4 will be useful in constructing such an
R(m,y,y
2 'Y3 ,n)
Let
be a recursive predicate such that;
(i) n > 2 & (21 m)(y)( y2)
(ii) "(2m)(y)(3y
2
r
3
)R(m,yly
form.
-
38
-
3
)R(m,y ,y2 ' 3 ,n)
2
'y3 ,n)"
is in
C F: -
E.A.N.
5:3;
As in the proof of Theorem 2.1, we deal with a symbol
*
Let
in addition to the integers.
L
be the following set
of 5-tuples
&
j(n,,MX,,i-,i)ln>2
&
(i / *
(x = n ->
& (x < n ->
O < x < n
&
i / *)) &
x / *
(i,j)
& m
*J.
We define the following linear ordering on
(n,m,x,i,j) -<
((n,m) = (n',m'
x = x'
i = i'
&
(n',m',x' ,i',j') <->
)&
x <x')Vg'
i > i') I
((n,m)
((n,m) < ex.
&
(n',m')
=
(ni,m')
((n,m)
L.
&
x
=
x'
=
n
&
j < ji).
&
It is not too difficult to see that the order type of
(L,-<)
is
(n +
W-.
*)-o.
n>2
subset
We now define a
(n,m.x. i. j)
e t
i = 0) V (x = n
<->
(n,m,x, i,j)
& i > 0
&
e L
t
of
&
(x < n \{ (x = n
(3y2<j N
- 39 -
3
L.
)R(m,1-1,y2 y 3,n)).
If
A. Let
A c L, let
A
be the restriction of
£
to
be the set
A
( (n',m', x, i,j)|(n',m')
=
(nm)
&
x = n).
We wish to show that,
(2)
(n) ,( t t = n < _
(-*a (l 1
_3 2) (3)
R(m, y , y2'y 3n)).
The following observations, which will be useful in
establishing (2), are direct consequences of the definiticns
of
T', L,
and
(i)
(y )
(ii)
2
()
where
(iii)
~y
2
If
S
For every
q
n
and
m,
rfAS
3)R(m.y ,y2 'Y3 ,n) ->
y2)
For every
.
n
and
n s w.o*.
m,
(y332)
)R(m,y,y
2
'y3,n)
Oz w. q
£
-- >
is the least integer such that
)(Y3 )R(m,r,y2 y 3,n).
is a maximal successor-chain of
have a least element.
Furthermore,, if
x r
S
,
is
then
S
must
to be finite,
it is necessary (though not sufficient) that the least element
of
S
be of the form
where
(n,mx,i,j)
We now conclude our proof of
of the definitions of L, t
and £
for the assertions made below.
-
40
-
x = 0.
111-3.1.
An examination
will be the justification
Suppose that
(3m)(y )(y
y 3 )R(m,y1 ,y2
2
Therefore, for some
holds.
M,
'
3 ,n)
(y1 )(Gy 2)
By (i),
a lower limit point.
n.
'y
3 ,n)
(n,)
x< n
is a maximal successor chain of
=
3 )R(r,y1 ,y2
(n',m') =
\&
n
holds for some
f.
£
((nj,n-l,*,*)
is
(n,i,*,*) is an initial point if
2, and an upper limit point if
n > 2.)
Conversely, suppose that, for some
(3 m) (y 1)( y2 )(y 3 )R(m,y
1 ,y2y
3 , n).
n,
Furthermore, suppose
that S isamaximal successor-chain of £ and that S
has least element
(n.,m,o,*,*).
Let
B
be the set
m
((n',m',xi,j)|(n',m',x;i,j) e L
If
£IAZ
(
Ai w.w*, then
S = B
m
of
S
integer
is
,
(n',m') = (nff) &
x < n').
and, therefore the cardinality
m
K.
On the other hand,
if
I
A
o
L
.
$,for
some
q, then one easily checks that the cardi.nality of
S is infinite.
follows from
In fact, in this case,
Lemma 2.6, that either card
£ S s o. Thus, it
s = '
or card S =N
0
depending upon whether or not
holds.
&
(yl) (y
2 ) (Y3 )R(m',y ,y2 'y3'
However, it follows from our assumption that if n =
-
41
-
",
then, for any
hold.
Thus,
Theorem 3.1':
%, (yl)( Y 2)(y
either card
Let
to any
(m, yly
S = n/ n
X c N.
ordering which is ~lT
3
in
, .')
2' 3
or card S
does not
=N
.
There is a scattered linear
X
and which is not isomorphic
X-recursive linear ordering.
- 42 -
§ 4 Coding
H -recursive functions into Recursive Orderings
03
The following theorem will give us a method for
constructing a recursive structure whose isomorphism type has
a non-arithmetic collection of elementary properties.
In
particular, at the end of this section we will construct a
recursive linear ordering,
elementary statements true of
Definition 4.1:
lel =
We say
J-'(We)
(see
e
I
is not arithmetic.
be a number, and
iff for every
n,
T be an order type.
(e}(n)
is a linear ordering of order type
Theorem 4.2:
that if
£,
Let
such that the set of all
£,
There is a recursive function
is defined and
£.
e(e,a,m,n)
n + 1
e defines a recursive function of
such
arguments
Kleene [8, pp. 288-289]), then for every integer
a:
(i) (2z 1 )(z2 )( z3)(z 4 )...({e)(zl,z 2,z 3 ,...,a)= 0)
->
(11)
~(4z
->
le(e,a,m,n)
1 )(z2 )(:2z 3
le(e,a,m,n)
=
om+n+l
)(z4 )...({e)(z,z 2 ,z3 ,...,a) = )
= om+n
The proof of this theorem will require several lemmas.
Lemma 4.3:
that, if
There is a recursive function
e
7r(ea,m,n)
defines a recursive function of
-
43 -
n + 1
such
arguments,
z1 ,.. . ,
then, for every
a:
0)
--
>
1(e, a,m,n)
=m+1
(ii) ({e)(z 1 ,...,zna) / 0)
--
>
?(e,a,m,n)
=m
(i) ({e)}(z,
Proof:
...,zn,a)
=
Straightforward.
The function,
|
r, will correspond to the recursive
matrix of an arithmetic predicate in predicate form.
will now define two recursive functions,
V
will correspond to
14:
Lemma 4.
e
3,
and
There is a recursive function
I{e)(n)l = In,
1E(e)l
then
and
1, which
respectively.
is a number such that, for every
and
T
We
F
n, { e}(n)
such that, if
is defined
=
Proof: Suppose we are given a number
e.
Let
r(e)
be the
Gadel number of the following partial recursive function:
Given
n, to find
(F(e))(n), we:
Qn c (0,... ,n x {0,...,n) be the set of all pairs (i,j)
2
e(O, ...,n)
such that the computation of ( e}(i) terminates
(i) let
(ii)
in at least
n
((e)(i)}(j)
terminates in at least
let
=
lij=L
steps, and, furthermore the computation of
K({(e}(i)J(j)), and let
{
(
)
;
n
steps;
--
(iii)
-- -
-
-
I
let
(jOP kij,i )' J( IP
An=
&
&
(i?,jI)e,
& k' < 1
k < 1
& (i<i'V(i,j)=(i'
(iv)
Let
r
be the least
m
such that
and
Am
let
) &k<k)
[E(e)}(n)
be the least element of
Cn = Ar+n-
if
C /
,I and let
(Y.,(e ) }(n) = (E(e)}(n-1), otherwise.
Z
is recursive and has the desired properties.
Lemma 4.5:
e
is a number such that, for every
and
(e}(n)J
Proof:
define
(ii)
Let
f(z, y)
A(z, y) ->
~A(-Zy)
= Zn,
A(z,y)
( If (Z,Y)I =
value depends on
is
(e} (n)
defined
IE(e)|
a
i j<i
II
be an arithmetic predicate.
recursively.
(if( -,y)I
->
n,
such that, if
then
Similar to that of 4.4.
Lemma 4.6:
(i)
There is a recursive function,
If for every
m+1
;
where
z,y; then, for every
45
-4
Z-1Y;
and
om. q),9
-
Let
-
q
is
an integer whose
(1)
(
(2)
~.y)A(Zy)
Proof:
< 91
Ag
<
If
M+ 1.If
Lemma 4.7:
(1)
(ii)
Let
and,
for every
~A(z,y) ->
is
w-sum of ordinals.
wherefor every
holds, then
(:y)A(Z,y)
i,
Mom+l
occurs
w-sum, and, therefore, the sum will be
A(z,y)
f(2,y)
G
m+1
<
om+1
be an arithmetic predicate.
recursively.
If
y
is
E.A.N.
in
Let
A(Z,y),
z,y;
(If(*, y)1 =
and
m+1)
(If(Z, y)j = wm)
then, for every
z;
()(y)A(Zly) ->
(2)
Ir~.(S'(e,-z)))
2; and
d
m+l
(S (e, Z))
(
o-sum will be
define
A(-Z, y) ->
(m
(y) ~ A(zy), then, for every i,
Therefore the
if
(e,))
1E(Sn (e,Z))I = E9
times in the
m+2.
(a(Sn
>
->
By 111-4,5,
That is,
Co
--
Ay)A(Z,y)
~(y)A(Z,y)
->
m+2
( IE(Sn (e,-Z))lI
') I
(17, (Sn ( e., 'z
integer whose values depends on
=
q)
; and
where
q
is
-4
z.
Proof :By 111-4. 4,
where, for every
ieN
-
46 -
either
9
m
=
m
m)
(i)(0
or
and
9 i=
29
Therefore, if
A(Z,y).
m+1.
If
(y)A(Z,y),
= CDm+1 -o =
m+2..
(y)
~ A(Z,y).
In this case,
m+T
G. = W
.q where q < r.
Lemma 4.8:
There is a recursive function
e defines a recursive function of
(i)
(ii)
A(e)
(
(3z
(z 2 ) (.
z)
r
(1)>r(Oi = om)
A
such that, if
arguments, then:
n + 1
arguments;
)(z2)(: z3)(z4)... ({e)(zl,...,zn a) = 0
(3zl)(z2)('3z3)... ((X(e))(z 1
<->
(iii)
n + 1
defines a recursive function of
for every a,
E.A.N. in
~(y)A(z,y), then there is an
such that
and
is
y
then
z3 )... ((e))(z,
.1,zna) =
... ,zm, a) = 0)
0);
is in
E.A.N.
form.
Proof:
Look at the proof of 1-1.4.
This proof gives an
effective procedure which given the G6del number defining the
recursive matrix of an arithmetic predicate, A, yields a
G8del number of the recursive matrix of an arithmetic predicate
A, where
A
is in E. A.N. form and
We now complete the proof of
e(e,a,m,n) for the case where
n
A
is equivalent to
IV-4.2.
is odd.
We will define
(The case where
is even is similar and will not be discussed further).
(e}(z
1
,..., zn,a)
be a recursive function.
e1 (e,a,m,n)..., en(e,a,m,n)
We define
inductively as follows:
- 47 -
A. 1
Let
n
(i) ~e(e,a,m,n)
as a
(ii)
For
defines
~,,(n-1
r(S
(r(A(e),a,m,n),z
place function;
n-1
e 2 i+1(e,a,m,n)
0 < 2i+l <n,
f(Sn-(2i+1)(e2 1,z1 ,...,zn(21+1))
function.
For
7s(Sn-2(1+1
0 < 2(i+l) < n,
2
1
defines
n
as a
e2 (i+l)
, z,...,zn-2(i+1)))
place
(2i+l)
-
defines
n - 2(i+l)
as a
place
function.
Let
e(e,a,m,n)
inductive proof, using
111-4.6
and
It is a straightforward
I11-4.7, to show that
e(e,a,m,n)
has the desired properties.
e(e,a,m,n)
is recursive.
Definition 4.1':
type.
and
J~
(We)
that, if
(2z
e
be a number, and
iff
for every
Z
n,
be an order
(e}(n)
is a linear ordering of order type
There is a recursive function
X c N
recursively in
every
Let
Observe also that
H
!ex! = £,
We say
Theorem 4.2':
()
en(eea,m,n).
=
and if
X
e
defines an
n + 1
is defined
£.
w(e,a,m,n), such
place function
(see Kleene [8, pp. 266-281]), then for
z ,.. .zn,a:
)(z
2 (3z 3 )... ({e}x(z,..., zn,a) = 0)
-
48
-
->
(
e(e,a,m,n)X =om+n+1
(ii)
~3z)(z2 )(3z 3 )...({e)x(z 1,...,z,a)
x
(Tr(e,a,m,n)Xl
Proof:
=
0)
m+n
mn).
Alter the proof of 111-4.2
by making the following
an expression of the form "(t)."
replacements:
replaced by
=
is
"(tX)?.
an expression of the form
" It1 = I"
is replaced by
an expression of the form "t defines a recursive function..."
is replaced by
function
...
"t defines, relative to
x, an x-recursive
"
Observe that this altered proof is a proof of 4-2'.
The assertions in the altered proof are shown.to be correct
by an argument
very similar to the proof of the relativization
Sn theorem (Kleene [10, pp. 150-155].)
The following corollaries will give us 111-4.2
of the Kleene
1
in
the form we need it.
Lemma 4.9:
every
There is a recursive function
n,
G(n)
defines an
n + 1
9 such that for
place recursive function,
and
(a)
(n)
(aeH
n
(
1->
z2) (2z3)z4)...
-49 -
({-(n))(z9,-.
..
,zna)=0))
Proof:
Look at the proof of the Post representation theorem.
(See Davis [ 9, pp. 158-161])
Observe that this proof gives
us an effective procedure which given a number,
n, yields
the G6del number which defines the recursive matrix of
Corollary 4.10:
Let
(a,m,n) =
r(9(n),a,m,n).
Hn'
For every
a, n:
(i)
(ii)
a e Hn ->
a
Hn ->
( 1 P(a,m,n)
= om+n+1
(IP(a,m,n)?
= W
If in
111-4.9
listed in the proof of
and
).
111-4.10, we make the replacements
111-4.2', we obtain the relativized
versions, 111-4.9' and 111-4.10'.
Corollary 4.10':
such that if
(i)
(ii)
a e H
a
H
>
n
--
There is a recursive function
X _- N, then for every
(1P(a,m,n)x,
a, n:
= om+n+l)
(xP(a,m,n)XI =
>
p(a,m,n)
m+n
We will now use IV-4.10 to construct a recursive linear
ordering,
in
T
£,
such that the set of elementary statements true
is Turing equivalent to
H
.
First, we will need
several preliminaries.
Definition 4.11:
Let
be a structure and let
language of the same similarity type as
-
50 -
Z.
II
be a
The truth set of
is the set
Lemma 4.12:
(11§
is a sentence of
&
is a denumerable structure of finite
If
similarity type , whose field is an arithmetic subset of
N,
and whose relations and operations are arithmetic, then the
is Turing reducible to
truth set of
Proof:
H.
and keep in mind that every member of
reducible to
Lemma 4.13:
Hn
n
Let
of linear order.
(
TT
n0 U E0n
T"
is Turing
be the language of the elementary theo ry
The predicate "There exists an
upper limit which is also a lower limit point."
insi
in
"
Look at the model-theoretic definition of
nth
induc tive
is expressa ble
.
Proof:
Let
v,
v 2 , v 3 ...
be the variablesand
-<
be the
relation symbol of G;. We define the w.f.f.'s L~(v
1
)
inductively:
L0 (v)
Ln+1
(3v2)(v2<vl)
<->
v1)
L 0 (v l)
<->
<->
(v
(3V2) (v-<
1
)(L~(vl)
(V2)(V2<Vl
relativized to
w.f.f. 's
Now we define the
L (v1 ) <->
&
v2
&
&
L (v))
-
51 -
--
3v3)(v
>
v2-<vl).
L(v 1 ).
L (v,)
(v2)(vl-< v 2
and
->
n
(v3)(vv
3
<v2
n
asserts the existence of an
nth
inductive
I
upperlimit point which is also a lower limit point.
Lemma 4.14:
The exists a 1-1
recursive function
cp
such
that:
(i)
(ii)
0 /, range
w;
for every integer
n,
+ 1,
((n)
disjoint from the range of
...
p(n) + L(n) + 1)
,
is
m.
We are now in a position to construct a recursive
linear ordering whose truth set is Turing equivalent to
Consider an ordering of the form,
I =
7 (om+1+o*), where
meQ
(ii)
m
9 |=
4
Q r N.
It is not difficult to see that
m e Q.
We are looking for a recursive linear ordering of the
form
(i)
H .
E (wm+l+o*), such that, for every
meQ
p(J(a,n)) + n + 1 eQ <->
e(J(a,n))+ n e Q <->
,
iff
a,n:
a eH
H
a /
If we can find such a recursive linear ordering
(a)(n)(a e Hn <-> r
then
H
is 1-1
in 111-4.10.
e'
J(an)+n+1)
reducible to the truth set of
^A(K(i)), o(i), L(i)I
number
=
where
p
£.
£,
and we see that
Let
is the function defined
It is not difficult to prove the existence of a
such that
(i)(1(eJ)(i)| = Ai+1+w*).
-
52
-
Let
£ = IF(e)I.
By III-4.14, for every
or
A
1
= o((i)+L(i))
K(i) e HL(i).
K(i) e HL(i)
see that
i, either
A
=
o~ep(i)+L(i)+1)
depending on whether or not
Therefore, by our construction of
i
=
(i)(K(i) e HL(i)
w(co(i)L(i)+1))
__
-
53
-
p,
Hence, we
(to(i),L(i)+1)
Boolean Algebra
§5.
In this section we apply the methods of
III-§3
and
III-%2
to the problem of constructing a Boolean
algebra which is not isomorphic to a recursive one.
In attempting to code a non-recursive function into
the isomorphism type of a Boolean algebra, we immediately run
up against that fact that every
Boolean algebra has a recursive
truth set in the elementary theory of Boolean algebra [See
Tarski [11, pp. 62-64].
Therefore we look at a variant of
the weak second order theory described in Ehrenfeucht [ 12
Let
A
be a set of relation and operation symbols.
"Indiv"
be a one variable relation symbol such that
We will denote by
e A.
].
Let
"Indiv"
the set of all formulas of
k(A)
the lower predicate calculus with identity,"=" which contains
the predicates
of
e
and predicates from
A
only.
As models
F(A) we will admit those models for the set formulas
in which:
(i)
(ii)
(Indiv ('xJ) is a set of individuals;
IM
X
(iii)
(the set of elements of the model M) is the smallest set
such that
i(Indiv (x) c X and,
and, for every
1 < i < k,
the members of
A
on
A
x (Indiv
if
Indiv (xi),
x
then
e
,..,Xk e X
{xl,...,xk) e X;
are interpreted as relations ard operations
(x)) ;
-
54
-
(iv)
e-relation in
is the set-theoretical
e
"equality. "
means
(v)
M;
Let 2 = (A 1, R ,2
be a structure (where
...
). We designate by h", the
is the universe of(
structure
Lemma 5. 1:
/
If
)
card S <
&
(S
e
, Indiv, R ,R , .. }
is an arithmetic structure of finite
R
.
.o ,Rn )
Proof:
_<T HW .
is
is an arithmetic structure, there is a
Since
sequence of integers,
R 1 :T
2
Hm
Using
m 1 ,..*,mn+1, such that
I4(
Hm
< H
.
Choose m > max(m 1 ,.. .mn+1'
n T mn+l
as an oracle,we can G6del number the members of
,
Hm
(in
Indiv,
n
similarity type, then the truth set of
.., R
in such a way that:
(i)
* (ii)
(iii)
(iv)
xe
1 Indiv
(x)"
is
is
Hm-recursive;
is
H*-recursive;
Hm-recursive;
R,5...,R n
(For every
are
Hm-recursive.
1 < i < n,
its arguments is in
Now that A
Let
will fail to hold unless each of
R
A
x(Indiv (x).)).
*
is
A = (Indiv.,
presented in this way, we use 111-4. 12. II
<,
-
n, U, -}.
55
-
If we can find an
H -recursive Boolean algebra
B
reducible to the truth set of
then
B
1-1
is
H
such that
-(Indiv., <, n, U,
in
),
can't be isomorphic to a Boolean algebra whose
B
will be
III-4.10'
relations and operations are arithmetic.
useful towards this end.
First we will list
some properties of Boolean algebras
which can be expressed in the weak second order theory.
v 1 , v 2, .
stand for individuals and
V,
V2 ,
Let
stand
...
for finite sets of individuals.
(i)
"v
is bounded by the union of the members of
we shall write,
"
(vl' V 1 ) <->
(ii)
"lVln
A(v 1 )
Suppose
((v2 )((v 3 )(v3 e 1 ->
((v 1
0
be the
N3...V\/An(v
3
))
vl:v2))
v3 < v2) ->
An(vl) (See 1-2.3):
&
v2 -- 0 y' v<
(v2)(v
A0 (v1 ,...,An(v 1 ) have been defined.
(See 1-2.3)
or as
V(v 1 , V1 ):
is an atom", or
<->
V "
w. f. f. ,
Let
(3 Vl) ( (v 3 ) (v 3 e V 1
"vle In"
->(yV
v1 , v 1 ). We now define
&
v2
A+1 (v1 )
as follows:
An+(v)<>-~(vleIn)
"lv-n
&
(v2 )(vl0v
5v26n'
-56 -
2
C In
V
2 -v
Q
e In))
(iv)
(v)
V
n
" Vt
"i1 v1 ! n
n
V
(vi)
21n <
Iv3In
Ivln
" V
is
" lvlIn
Let
£n
V21 n <
be
_)
I!
(2v3)(
-
&
(V2)((v)(v 2e 1->(An
O(v4e
V1 ->
r =
(m) (meQ iff
Ivn
->
(2v
) (An(v 3 )
3
&
31n <
Iv2
n)'
atoms, but no completely atomic element
IVlIn
bounds
atoms",
Cn(V
(3vV)Cn(v9 ).
( E
meQ
(Wm+ 7)) +
(Wm'-k + q1)) +
F e B.
If for every
an interval of order type
is atomic and
Q r-- N.
( m,
mCQ
Let
and
B = D,
We wish to prove that
We shall do this in several steps.
B i= m).
Claim (1):
Claim (ii):
n
completely atomic"
We now consider Boolean algebras of the form
where
Iv3
Iv4n ;
2
& An (V3)
)
bounds
bounded by
(viii)
&
1vi 1n1)
(v2) ( Iv2 in
(vii)
atoms"
bounds
(3v2 )(An (v2 )
I
I
F2
om
then
+
is atomless.
-
57
-
(See 1-1. 1)
m,
F =
does not bound
tl U t2,
where
m
F =
Proof:
[a ,bl) U...U[an,bn),
2
', then
below
[a ,bi)
om
for
is
[a.,b.)
and
T
also lies in
lies
a
bounds a segment of type
a
Therefore
m.
some
where
If
T.
lies in some summand
b.
Suppose
U
=
is completely atomic.
is atomless and
case I:
<an
-< b 1-< a2-< ...-
[a.,b.)
1 < i < n,
We claim that for every
a
where
atomless.
in
T
must either lie in
a
[a.,b.)
then
oP
is
lies in
.
then [ai,b ) bounds a segment of type
w
Therefore
op
a
If
.
o+
q
preceded by a summand of the form
or below
o .
lies in some summand
b
Suppose
case II:
least element of
wo.
= [ai,*) U [*,bi)
a.
or in
Let
is completely atomic.
If
a
lies in
*
lies
be the
[a.,b )
then
is atomless and
[ai,*)
where
T
a
If
[*,bi)
is
atomic.
To complete the proof, we let
F
-
l
a
in the first
n
l<i<n
Claim (iii):
Proof:
If
B
1 e Q,
rational interval.
then choose a point
Let
F
atoms, but no completely
[O,a) bounds
atomic subinterval of
1 , Q.
1 e Q.
iff
=
[O,a)
B.
be a member of
-
58
atoms.
bounds W
-
If
F
Suppose
bounds a segment of
bn'
om + I
type
for some
m > 1, and, therefore, a
m, then
completely atomic subinterval of some interval of
bound W
atoms.
If not, then
completely atomic and
E2
F
F
will
(1 U (2 where
=
is atomless.
g
If
l
is
bounds {
atoms, then
(1,
which is completely atomic, must bound
atoms, since
(2
is atomless.
Claim
(iv):
Proof:
B# |=
Observe
#n,
n + 1 e Q.
iff
B(n) = D [
n
1]
- meQ
Since
B
ff
n.,
1 e x(x = m'-n
Corollary
&
meQ).
A
o,
B,
Hence
B4
we see
=
iff
i
20
B
Q
n + 1
Q
5.
There is a strict Boolean algebra
whose field, relation
that
iff
completions.
We now turn to the main theoren of
Theorem 5.3:
fn
The weak second order theory of Boolean
5.2:
Algebras has
(B (n)
and operations are
B
(See II)
H -recursive such
isn't isomorphic to any Boolean algebra whose field,
relation and operations are arithmetic.
Recall 4.10',
Proof:
for the case where
X = H .
Let
H
Ai =
p(k(i), ep(i),
there is
a number
1.
L(i))
e
It is not difficult to see that
H
(n))wI
|(('
such that
=
A. + T.
Let
H
£
=
I (r~e))
"), and
let
I
+ 1.
=
HW -recursive linear ordering.
-
59
Let
-
1
is isomorphic to an
B = D
91
By an argument
very similar to that given in the example at the end of
§4, we can show that
((+i)L(A)
K(i) e Hi.
<->
Therefore by claim (iv),
(,
(i)KK(i) e H L(i
<->
B
=
(i)+L(i))
H
Thus
by
H
is
111-5.1,
1-1
B
reducible to the truth set of
the field, relation and operations of
5.3':
Let
X r N.
B
are
B
Thus,
By
II-1.2',
H -recursive.
There is a strict Boolean algebra,
B, whose relation, field and operations are
that
.
is not isomorphic to a Boolean algebra whose
field, relation and operations are arithmetic.
Theorem
B
H -recursive such
is not isomorphic to a strict Boolean algebra whose
operations, field, and relation are arithmetic in
- 6o -
X.
CHAPTER
Analysis of
3
IV
by Means of the Analytic Hierarchy
-
61 -
-
§1.
Preliminaries
The following facts will be needed in
§2
wherein
the main results of this Chapter are discussed. We give the
preliminaries all at once in order to facilitate the
exposition in
Lemma 1.1:
§2.
Any infinite recursively enumerable relation
r.e.
is isomorphic to an
fj (x,y)
relation whose field is
total.
Proof:
Let
f be the recursive function such that
(x)(y)(6(x,y) <->
We define
J(x,y) e Range f).
(O) = J(0,O), if
K(f(O)) = L(f(O)).
have been defined.
then let
some
let
Tr(O) = J(0,l)
as follows.
If
0 < i < n, then let
0 < i < n,
then let
If
be the least number
x / K(T(i)),
or
K(f(i))
L(T(i))
respectively.
x / k(r(i)),
x /
x
f(O),. .. , f(n)
If
L(f(i)), then let
x
-
/ k(?(n+1)).
62 -
Otherwise,
L(f(n+l)) = K(f(n+l)),
i, K(f(n+l))
L( (i))
Otherwise, let
such that, for any
L( (i)),
0 < i < n,
such that, for any
If, for some
K(f(i))
be the least number
x
x / L(ff(i)).
is equal to
L(f(O)).
K(f(n+l)) = L(f(i)), for
K(I'(n+l)) = L(?(i)).
L(f(n+l)) = K(f(n+1)).
or
Suppose that
K(f(n+l)) = K(f(i)), for some
K('T(n+l)) = K(?(i)).
K(f(n+l))
K(f(O)) /
if
be
L(f(n+l))
1 < i < n,
"J(x,y) e Range I"
Lemma 1.1':
is
isomorphic to
X c N.
Let
if= N, and the relation
Range
is recursive.
Any
R .
X-r. e.
relation isomarphic to an
X-r. e. relation whose field is total.
Corollary 1.2:
Any hyperarithmetic relation is isomorphic
to a hyperarithmetic relation whose field is total.
Lemma 1.3:
Let
(P ieN
be a sequence of ordinals.
If
B
a,
then:
(i) oco;
(ii)
E
on' =
o
.
ieN
Proof:
(i) Follows from the definition of
(ii)
By I-1. 9,
>
co
5
o
> o n
o7.
,
for every n.
I
i<n
Lemma 1. 4:
Let
0
be the standard
7Il'1
the recursive ordinals (See Kleene [11,
set of notation for
pp. 51-52].
a partial recursive function, f, such that if
[e]
is the ordinal named by
e, then
|f(e)I
e e0
= o[.
There is
and
(See
111-4.1).
Proof:
We use effective transfinite induction.
partial recursive function such that:
-
63
-
We seek a
4
f(1) = dl,
f(2 e)
Idll = o,
where
7(d2 (b)),
=
where
d2 (b)
is the G8del
number of the constant function which assigns
f(b)
to every integer.
(See 111-4.4);
f( 3 .5Y) = r(d 3 (b)), where
d3 (b)
number of the recursive function
is the Gbdel
f((y}(n)).
We see by the Rogers' recursion lemma that such a
partial recursive function exists (See Rogers [1$, p.849].
Using
IV-l.3
we prove by induction on
the desired properties.
<0
that
f
has
f
Corollary L 5:
(i)
If
(ii)
M1
(iii)
Proof:
(i)
(ii)
x
is a recursive ordinal, then so is
1 .
9 < o<->
)(9)
<o.l
Immediate from 1.4
Any finite sum of recursive ordinals must clearly
be recursive.
Thus, by 1-2.9,
w
cc < o ,then
a
is recursive.
(iii)
If
9 < w, then by
such that
9
>
o1,
then
(A.
9 < co
< o 1.
>_
= w , for some
Contradiction 4 Thus
1-1.8, there is a
Thus, ? (9) < p < ol.
o.
- 64
c.
-
If
c = o
P < w
Conversely, if
Lemma 1.6:
If 9 is isomorphic to a r(
ordering, then so is
Proof:
Let
Define
x << y
T, )
linear
1 + I + 1.
£ = (L,, .<}
where
as follows:
L
and -<
are
1(
T
x << y <-> x = 0 V y = 1V{
x + 2 -< y + 2.
=
(x(x=O \/ x=1
is
F ( TT)
V x=z
and
+ 2
where
zeL),
<<}
is isomorphic to
- 65 -
1 + I + 1.
)
Analysis of
§2.
3
Definition 2.1:
by means of the analytic hierarchy
(Spector [1S])
A set S
of natural numbers)
is said to be "inductively defined with respect to a predicate
Q" iff, for each ordinal
(T1 r T 2 ,
where
'r, S
=
Q(x,Tl)) ->
&
(Spector [15])
Sc
=
9 = (L,<
If
S = Sc,
The ordinal
is a
A1
c
Sc+1
inductively defined with respect to a
Theorem 2.3:
Q(x,Sv
&
Q(x,T 2 ), and
being the least ordinal such that
Theorem 2.1:
x((3v)(v < T
c
of a set
predicate is
< w.
linear ordering, then
,(£) < o.
Proof:
Let
By IV-1.1,
we may assume that the field of
J(xy) x -
Sv
y}.
Q(x,T)
Q(x, T)
is
total
By IV-2.2, it will suffice, for
the proof of the theorem, to show that
defined with respect to a
I
A
S
predicate.
is inductively
(See I-1.1).
Let
be the following predicate:
x = J(y, y) V x e T V{ (3 s) (seq(s)
<->
(z)(K(x)-< z-< L(x)
V
L(x)-< z -< K(x)) ->
&
0)(O
< i < A(s)
(
J(z, s(i)) e T).
Q(x,T)
is
A
with respect to
Corollary 2.4:
<
,
S
and, by I-1.1,
Q.
If
Therefore
£
is a
;(9)
A
(10
- 66 -
is defined inductively
=
c < M1 .
I
scattered linear ordering, then
&
is a
Since
Z
is
(£-o) + 1.
If
>
(
Proof:
3(Z) = w,
then,
by 1-1.7, 2'((T.c)
+ 1)
Contradiction. I1
w.
=
scattered linear ordering, so
A
If
Lemma 2.5:
is a strict Boolean algebra with a scattered
B
B
base and the operations and relations of
are hyperarithmetic,
then;
<
6 (B)
(1)
(ii)
B s Dg
where
9 < w ,
(iii)
B ~~B'
where
B'
and relations of
Proof:
B'
By II-1.5,
1-2.12,
1
and
by 11-1.2.
H1
strict and the operations
are recursive.
B f D.,
B P
IV-1.5,
£
where
is hyperarithmetic.
DQ,
9 < WI.
where
We now wish to prove that if
I
By
IV-2.4, 6(B) < a)1 .
By 1-2.10, and
is scattered.
By L-2.14
is
is
7,
(iii) holds
<
<()
then
If we try to apply IV-2.2 directly, we run into the following
difficulty:
S
If
£
is a
,(r'defined in the proof of IV-2.3, need not necessarily be
1
defined with respect to a
the equivalence relations,
and, if
if
linear ordering, then the set
1
I
X
is
is
1
predicate.
This is so because
range over the field of
-- v
E , then its field need not be
and not
A,
then its field can't be
Therefore, we take the following indirect approach.
be
-
67
-
9,
(In fact,
iT
Let
,.)
"Scat"
e e
is the G6del number of a scattered
recursive linear ordering
Lemma 2.6:
(i) Scat e
(11)
TFV
Scat
E~ 11
e e Li
&
Proof:
(i) "e e Scat" <->
(z) ~({e)(f(x),z) = 0
Since
Li
&
(f)[(~3x)(ly)((e}(f(x),f(y)) = 0
(e}(z,f(y))
is arithmetic,
(ii) Suppose "Scat" e F .
0)].
"Scat" is
1
Scat e '
=A
T71.
We are going to
.
use the hyperarithmetic predicate, "e e Scat", to take a hyperarithmetic sum of all scattered recursive linear orderings.
Let
mth
pn
be the
member of Scat.
arithmetic function.
linear ordering, -<,
If Scat is
on
&
prime.
Let
A , then
L = {(m
Let
n
n (m.,n)eN 2
I
g(m)
be the
g
is a hyper-
.
We define a
as follows:
P=pm
x<y<->x
(n = n'
(n + 1)th
&t
&
=
(g(n)}(m,m')
&
(n <n'vy
= 0).
A11 linear
ordering, and every scattered recursive linear ordering is
Let
£ = (L,-<}.
is a scattered,
T
isomorphic to a segment of
£.
-
68 -
Therefore, for any
: < o ,
&
o
is isomorphic to a segment of
for any
c
and,
wo,
_<
If
B
Thus,
> M,
(X.)
3(T) > ol.
therefore,
This
H|
contradicts Corollary 2.4.
Lemma 2.7:
£.
is a strict Boolean algebra with a scattered
basis and the field,
relations,
and operations of
B
are
E,
then:
6(B) <
(i)
(ii)
Proof:
(i)
B ; D., for some ordinal
Suppose that
IV-2.7, and
6(B) > o .
definition of
is
9 < w1 .
satisfies the hypothesis of
If
e e Li,
hell
let
[e)(x,y) = 0).
= {N,-<),
We give the following
Scat:
"e eScat" <->
f
B
(x)(y)(x-< y <->
where
E
:
&
(e e L
(3f)(f:D 1+1 e|+1 --
> B
&
a Boolean monomorphism. )
First of all, we claim that (1) is, in fact, a definition
of
"scat.".
If such a monomorphism
1 + ||ell + 1
is scattered, and, hence,
Conversely, suppose
Then
IV-2. 4,
(1 +
where
< o
9'
f, exists, then, by 1-2.13,
|1ell
.
is scattered.
is scattered.
1 + 1|ell + 1
l|e I
Hjell
is scattered and recursive.
+ 1) < l.
By 1-2.14,
By IV-2. 5,
1-2.10,
Di+ ll ell+ 1
and IV-1.5,
B ~
By
~D~9
D,
A
where
@ >wo1
By 1-2.13, there is a monomorphism, f:D,,
- 69 -
D
We def ine
by means of the following diagram:
f
I
DG
->
D1 +1 lel 1+1
Thus,
suffices,
it
1.
that (1)
--
Dg
>B
- -
for the proof of
IV-2.7, to show
To do this, we just write it out.
(See
11-1.3)
Let
6
(1)
"le e Scat"
->
x e B)
&
B.
be the zero element of
({g
2
(af)(x)(y)(z)(({g 1 (e))(x) =
&
(e e Li
(e)) (x,y)
=
0
([g5 (e)}(x,y)
=
=
0
0
->
&
(({gl(e))(x) = 0
-
70
x U y
->
({g3 (e)}(x,y,z)
((g4 (e)}(xy,z)
x < y)
->
-
=
x r y
->
x
=
f(x) =
y)
6
)
&
->
x
=
"x
U y
"x
z",
=
y
=
"x < y",
"x e B",
Noting that
are
"x n y = z"
and bringing the function
,
quantifiers of (1) to the front of (1), we see that (1) is
.*
in
(ii)
Lemma 2.8:
<
Proof:
This follows directly from (i).
If
is a scattered,
.
E
f
linear ordering, then
VD._
Follows directly from 11-1.2, and
Z = (L,-<}
If
Theorem 2.9:
is a E
IV-2.7.
linear ordering then
B(s) <co .
Proof:
£=
By I-1.11,
,
£r
reil
Ar
=
there is some
. such that
r
a
By 1-1.1, 3(1 i[a,b)) > o1 . However,
1.
and El; i.e. x << y <->
a-< x,< b
Contradiction.
|
Theorem 2.10:
If
B
Suppose
6(B) > w .
By
B
is scattered,
b.
3(1) > co,
then
There must be two
ria,b)={t,<<}.
Let
£r[a,b)
is
& a-< y-< b
scattered
& x-< y.
is a strict Boolean algebra whose field,
relation and operations are
Proof:
If
; (X) > W .
r
such that
Zr
r
(1(r) re).
I(.)
L.u.b.
and where
elements a, b e I
where each
1-
El , then
8(B) < w.
satisfies the hypothesis of
II-1.1, and
IV-2.l0, and
1-2.7, there is an isomorphism
-71-
f:B
D
->
Let
? (X'[a,b)) > w.
scattered and
such that
a, b e I
there are two elements
the proof of
As in
> W .
3(9)
where
IV-2.9,
is
I£[a,b)
- f~
We
([a,b)).
define the following Boolean algebra
(x A y = z <->
and
y,
x,
For every
(ii)
(x
is
(i) The field of
z e A,
x n y = z)
ZDZ[a,b] .
B
x <
0
y
z <->
(x
=
iff
(x
and
&
x U y = z)
x
is strict and scattered and its
field,
operations and relations are easily seen to be
Since
)(xI[a,b))
UT1
initial segment
and
.
= W()
mean
x e
is a
1TR
and
71
O
Let
y e f
be the Gandy ordering [1,6] with
of order type o . 9 is recursive
1*
~
=x(x e L & x e 01 ). Let x << y
and
x-< y,
and let
linear ordering of order type
linear ordering of order type
> M 1.
6(D
71
This contradicts IV-2.8. 11
8(A) > o1 .
> og,
9 = (L, -)
Let
Example:
n z).
n
=
By 11-1.2,
co
2.3(
= (t, <<).
w *
-2)
is a
.2
>
.
is a strict Boolean algebra
D
.2
02
whose relation and operations are
iT1.
D
By 111-2.5,
can't be isomorphic to any strict Boolean algebra whose
operations, field and relations are
Corollary 2.11:
7
.
There is a strict Boolean algebra
-
72
-
B
with a
scattered basis such that the operations, field and relation
TI
isn't isomorphic to any strict Boolean
11
algebra whose field, operations and relations are 7.
Now we turn to an interesting parallel between strict
of
B
are
B
and
Boolean algebras whose field, operations, and relations are
E(
TR
11
algebras of
and Lindenbaum
- axiomatizable theories.
This parallel is expressed in the following theorem.
).
pp.'1.A-'
Chapter II,
Theorem 2.12:
If
B
is the Lindenbaum algebra of a
axiomatizable theory, T, then 6(B) < o .
of
5 , and
class of
a
generated by
Claim:
be the language of
b
Let
Proof:
I
-
in
C.
is an ideal, let
B
B/I.
If
C r- B, let
Let
A
B
be the
The following predicates are
(i)
(ii)
= <
a _< a A~
I(C)
i71
is a sentence
be the equivalence
be the ideal
set of axioms.
in
a, @ and
C:
a
IMl(c) < 0;
|HI
(c)
Proof of claim: (i)
the predicate
Jai,
a
-
WI(c)'
(v) |=1I(c)
(vi)
If
T.
F$
0;
(iii)IMIl(c)
(iv)
(See
"r eA"
is an atom;
is a finite union of atoms.
iff = -AP.
<
The only occurrence of
in the proof theoretic definition
"-
Of
-
73
-
I
posiinV
(ii)
"There
<->
such that
is
Ia'I(c)
a sequence
<
c
p
V.
TI
...
.
Tn
.V
(vi)
<)
> 1
<-
Q(Cx:,C)
Q and
9
Let
Example:
-T.
is
aI(c) = 0
A
(I
L
11
"
is a
II(c)
y"
HI
LI
= 0.
is inductively defined with respect
Therefore, by
6(B) <
IV-2.2,
be the Gandy ordering.
{L,--<}
=
an atom
()is
) ->
= 0
aII(c)
(see 1-2.3)
Q
-
C
as follows:
finite union of atoms
IB)
I(c)
Follows directly from (v).
Q(c,C)
We define
to
.
|x|I(c)
of elements of
,T
is a sentence of
(T)(T
<->
Special case of (i).(iii)
assume without loss of generality that
L = N.
be a countable set of propositional letters.
set of all Boolean combinations of
{P }.
Let
.
We can
? ,29'
Let
Let Q;
3
be the
A - &be
the
-im
following set of axioms.
(~pn
nn
< n'
a
E
set.
and let
B
A
is
(n' < n
Let
T
&
n'
01)}.
be the theory whose axioms are
be the Lindenbaum algebra of
ordered basis for
Thus
V
8(B) > w.
B.
The order type of
II
-
74
-
T.
(pn)neN
{pn~neN
is
A,
is an
Ml
+ 1.
I
To justify IV-2.12 which might, at first glance
Remark:
seern somewhat artificial, we make the following observations.
In order for the proof of IV-2.ll to go through it
that
be a Boolean algebra such that
B
f1 (x,y), f2
that there exist hyp. functions
(i)
f 1 (xy)-
x U y;
(ii)
f 2 (xy)
x n y;
TT
B
f3
is isomorphic to the Lindenbaum
axiomatizable theory.
TT1-axiomatizable
baum algebra of some
and
y
and
is isomorphic to an algebra which satisfies
B
these conditions iff
T
T,
f3 (x)
However
algebra of a
is
x, y e B:
such that for every
(iii)
<"
sufficeD
f1 ,f2 'f3
If
B
is the Linden-
theory then
<
is
are the propositional connectives
On the other hand, if
B
satisfies the conditions mentioned
above, then we can use the standard proof that every Boolean
algebra is isomorphic to the quotient of a free algebra and a
filter, to show that
B
is isomorphic to the Lindenbaum
algebra of a FI-axiomatizable theory.
-
75
-
BIBLIOGRAPHY
1.
P. Erd6s and A. Hajnal, A classification of the denumerable
order types, Fund. Math., Vol. 51 (1962),
117-129.
2.
A Mostowski and A. Tarski, Boolesche Ringe mit geordneter
Basis, Fund. Math., 32 (1939), 6986.
3.
S. Mazurkievicz and W. Sierpinski, Contribution a' la
Topologie des ensembles denumbrables
Fund. Math., 1, 1920, 17-27.
4.
R. D. Mayer and R. S. Pierce, Boolean Algebra with Ordered
Bases, Pac. J. Math., 10 (1960) 925-94F2.
5.
S. C. Kleene, Arithmetical Predicates and Function
Quantifiers, Trans. of the Amer. Math. Soc.
(1955), 312-339.
6.
S.
7.
H. Rogers, Computing Degrees of Unsolvability, Math.
Analen, 130 (195-9T 129-140.
8.
S. C. Kleene, Introduction to Metamathematics, New York
and Toronto (vai'Nostrand). 1952.
9.
M. Davis, Computalibility and Unsolvability, McGraw-Hill,
195b, x+210 pp.
10.
S.
C. Kleene, Recursive Predicate and Quantifiers,
of the Amer. Math. Soc., 5313)71-43.
C. Kleene,
3
An Notation for Ordinal Numbers,
J.
Trans.
S.
L.
T1935)
150-15.
11.
R. Tarski, Arithmetical Classes and Types of Boolean
Algebras, Bull Amer.~ith. Soc., 55 (1979)
12.
A. Ehrenfeucht, Application of the Theory of Games to the
Completeness Problem for Formalized Theories,
Fund-Math., 50, (1961T-129-141.
-
76 -
--
-
E---
M
- - --
13.
S. C. Kleene, On the Forms of Predicates in the Theory
of Constructive Ordinals, Amer~JIur. of Math.
-6_, (1944) 41-58.
14.
H. Rogers, Recursive Functions Over Partial Well 0rderin.,
Proc. of Amer. Math. Soc., lO (1959), 647-053.
15.
C. Spector, Inductively Defined Sets of Natural Numbers.
Warsaw Symposium on Infinitistic Methods,
1959, 98-102.
16.
R. Gandy, Proof of Mostowski's Conjecture, Bull. Acad.
Polon. Sci. Ser. Sci. Math. Astronom. Phys.
Vol. 9, (1960), 577-582.
-
77 -
-Vw
BIOGRAPHY
Lawrence Feiner was born in Bronx, New York,
May 9,
1942.
He was graduated from the Massachusetts
Institute of Technology with a B.S. in June 1964 and
continued study there as a graduate student.
78I
I-