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Unformatted text preview: 1 WORKING WITH NONSTANDARD MODELS
Harvey M. Friedman
friedman@math.ohiostate.edu
http://www.math.ohiostate.edu/~friedman/ July 31, 2003
Most of the research in foundations of mathematics that I
do in some way or another involves the use of nonstandard
models. I will give a few examples, and indicate what is
involved.
1. General algebra and measurable cardinals. An
unexpectedly direct connection.
2. Borel selection and higher set theory. A descriptive set
theoretic context extensively pursued by some functional
analysts.
3. Equational Boolean relation theory and Mahlo cardinals.
A discrete mathematical context.
4. We conclude with an adaptation of 1 to weak second order
logic.
1. GENERAL ALGEBRA AND MEASURABLE CARDINALS.
Some innocent looking statements in general algebra turn
out to be equivalent to the existence of a measurable
cardinal. In fact, the first measurable cardinal turns out
to be clearly identifiable in a basic context in general
algebra.
Here an algebra is just a relational structure based on
finitely many constant and function symbols and no relation
symbols.
PROPOSITION 1.1. Every algebra with a sufficiently large
domain has a proper extension with the same countable
subalgebras up to isomorphism.
PROPOSITION 1.2. Every algebra with a sufficiently large
domain has a proper extension with the same finitely
generated subalgebras up to isomorphism. 2
THEOREM 1.3. Propositions 1.1 and 1.2 are provably
equivalent to the existence of a measurable cardinal, over
ZFC.
THEOREM 1.4. The least cardinal k, if any, such that every
algebra of cardinality ≥ k has a proper extension with the
same countable (alternatively finitely generated)
subalgebras up to isomorphism, is the least measurable
cardinal.
Note that these Propositions and Theorems use the notion of
cardinality, which can be argued to be not strictly
algebraic. We give obvious reformulations which do not use
the notion of cardinality.
PROPOSITION 1.1’. Every algebra with a sufficiently
inclusive domain has a proper extension with the same
countable subalgebras up to isomorphism.
PROPOSITION 1.2’. Every algebra with a sufficiently
inclusive domain has a proper extension with the same
finitely
generated subalgebras up to isomorphism.
THEOREM 1.3’. Propositions 1.1’ and 1.2’ are provably
equivalent to the existence of a measurable cardinal, over
ZFC.
THEOREM 1.4’. Let D be a nonempty set. The following are
equivalent.
i) every algebra whose domain includes D has a proper
extension with the same countable subalgebras up to
isomorphism;
ii) every algebra whose domain includes D has a proper
extension with the same finitely generated subalgebras up
to isomorphism;
iii) there is a countably additive 0,1 valued measure on
all subsets of D, in which singletons have measure 0 and D
has measure 1.
We now give a proof of Theorems 1.3 and 1.4.
LEMMA 1.5. Let k be a measurable cardinal. Every algebra of
cardinality ≥ k has a proper extension with the same
countable subalgebras up to isomorphism. 3
Proof: Let k be measurable and A be an algebra of
cardinality ≥ k. We will assume that the domain of A is a
cardinal l ≥ k. Let j:V Æ M be an elementary embedding with
critical point k, where M is a transitive class containing
all ordinals. j is obtained by taking the ultrapower of V
via a kcomplete nonprincipal ultrafilter on k.
j(A) is an algebra with domain j(l), and jdom(A) is an
isomorphism from A into j(A).
Since k is the critical point of j, it is clear that k is
not in the range of j. Hence jdom(A) is a proper
isomorphism from A into j(A).
We will use the important fact that M is closed under w
sequences; i.e., every w sequence drawn from M is an
element of M. In fact, every k sequence drawn from M is an
element of M (see [Ka94], p. 50, Proposition 5.7 d).
Let B be a countable subalgebra of j(A). We wish to show
that B is isomorphic to a subalgebra of A. Since the domain
of j(A) is an ordinal, B Œ M. Let B* be any algebra with
domain Õ w which is isomorphic to B. Then B* Œ M and there
is an isomorphism from B* onto B that lies in M.
In particular, we have that
B* is embeddable in j(A)
holds in M. Since j is the identity on V(k), we have j(B*) =
B*, and so
B* is embeddable in A
holds in V.
We have thus shown that A and j(A) have the same countable
subalgebras up to isomorphism. Now j(A) is not really an
extension of A. But obviously j(A) is isomorphic to an
extension A’ of A since A is embeddable into j(A). Clearly
A and A’ have the same countable subalgebras up to
isomorphism. QED
The reversal involves the interaction of a well founded
model and another model which may not be well founded i.e., nonstandard. Of course, the use of nonstandard models
here is not intense, but is still significant. 4 We fix a cardinal k such that every algebra of cardinality ≥
k has a proper extension with the same finitely generated
subalgebras up to isomorphism. Our aim is to show that
there is a measurable cardinal £ k.
LEMMA 1.6. k is uncountable.
Proof: Obviously k is not finite. Suppose k = w. We claim
that (w,0,S) does not have any proper extension with the
same finitely generated subalgebras up to isomorphism. This
is because in any proper extension, there will be a
finitely generated substructure with an element that is not
generated from 0, whereas no such substructure exists in
(w,0,S). In fact, all substructures of (w,0,S) are (w,0,S).
QED
We begin with the structure (V(k+1),Œ). Replace the binary
relation Œ with the binary function symbol Œ* which is the
characteristic function of Œ. Recall that Œ* is 0,1 valued.
Also add the constants 0,1 representing the ordinals 0,1.
Thus we have formed (V(k+1),Œ*,0,1).
We fix K to be a generous finite set of prenex formulas in
Œ*,0,1,=, which are closed under subformulas. We add Skolem
functions for the formulas in K in a standard way.
Specifically, take each of the formulas in K and add a
Skolem function for the outermost quantifier alternation,
whose arity is equaled to the number of outermost universal
quantifiers. In particular, if the formula begins with an
existential quantifier, then we use a Skolem constant. As
usual, these Skolem functions are only required to provide
witnesses if a witness exists.
We write L* for the language based on Œ*,0,1,=, and the
finitely many Skolem functions introduced in the previous
paragraph.
Note that the conjunction of the assertions that each of
these Skolem functions are actually Skolem functions for
the intended formula is naturally expressed as a universal
sentence in L*.
It is convenient to also add the unary function symbol S
for the successor function on w, which is 0 off of w. We
write the resulting algebra as (V(k+1),Œ*,0,1,S,...). Note 5
that w Õ V(k+1) since k is an uncountable cardinal. In
fact, w Œ V(k+1).
We fix an algebra B = (E,Œ*’,0,1,S’,...) which properly
extends (V(k+1),Œ*,0,1,S,...), where the two algebras have
the same finitely generated substructures up to
isomorphism.
LEMMA 1.7. (V(k+1),Œ*,0,1,S,...) and (E,Œ*’,0,1,S’,...)
satisfy the same purely universal sentences in L*. The
former is an elementary substructure of the latter with
respect to the formulas in L*.
Proof: Suppose the latter satisfies an existential sentence
j. Then j holds in some finitely generated subalgebra of
the latter, by simply taking the subalgebra generated by
witnesses for j. Hence j holds in some finitely generated
subalgebra of the former. Hence j holds in the former.
For the second claim, note that since the two algebras
satisfy the same purely universal sentences in L*, the
Skolem functions used for the former are also Skolem
functions for the latter. The claim then follows by
induction on the complexity of the formulas involved. QED
LEMMA 1.8. For all x Œ E\w, S’(x) = 0.
Proof: Let x Œ E\w, S’(x) ≠ 0. Look at the subalgebra of
(E,Œ*’,S’,0,l,...) generated by x. This subalgebra has an
element, x, with S’(x) ≠ 0, that is not generated from 0 and
S’. However, in any subalgebra of (V(k+1),Œ*,S,0,1,...),
every element, y, with S(y) ≠ 0, must be generated from 0
and S. So (E,Œ*’,S’,0,1,...) and (V(k+1),Œ*,S,0,1,...) do
not have the same finitely generated subalgebras up to
isomorphism. QED
At this point, we revert to a single binary relation. Thus
we use (V(k+1),Œ) instead of (V(k+1),Œ*,S,0,1,...), and
(E,Œ’) instead of (E,Œ*’,S’,0,1,...). We have x Œ’ y ´
Œ*(x,y) = 1, and x Œ’ y ´ E*’(x,y) = 1.
Since (V(k+1),Œ) is a standard model of the cumulative
hierarchy on a successor ordinal, we see that (E,Œ’) is a
possibly nonstandard model of the cumulative hierarchy on a
successor “ordinal”. The “ordinals” of (E,Œ’) may be
nonstandard. We will be using the “generosity” of K, and
Lemma 1.7. 6 In (V(k+1),Œ), w is the first limit ordinal, and its Œ
predecessors are exactly the x such that S(x) ≠ 0. Hence in
(E,Œ*’,S’,0,1,...), w is the first limit ordinal, and its
Œ’ predecessors are exactly the x such that S’(x) ≠ 0.
LEMMA 1.9. For all x, x Œ’ w if and only if x Œ w.
Proof: Let x Œ’ w. Since “("y Œ w)(S(y) = 1)” holds in
(V(k+1),Œ), it holds in (E,Œ’). Hence S’(x) = 1. By Lemma
1.8, x œ E\w. Therefore x Œ w. QED
We summarize where we are right now. (V(k+1),Œ) is a
substructure of (E,Œ’). In fact, it is an elementary
substructure of (E,Œ’) for the generous finite set of
formulas K closed under subformulas. Furthermore, k is an
uncountable cardinal, and the two structures have the same
w, and the same predecessors of w. Thus in the appropriate
sense, the first structure is standard (well founded), but
the second structure is an w model that may be nonstandard.
LEMMA 1.10. Suppose a < k+1 and ("x Œ’ a)(x Œ a). Then ("x
Œ’ V(a))(x Œ V(a)). I.e., if the ordinals in (E,Œ’) are
standard below the ordinal a, then the cumulative hierarchy
in (E,Œ’) on a is also standard.
Proof: By induction on a. This is trivial for a = 0 since
V(0) = ∅.
Let a = b+1 and ("x Œ’ a)(x Œ a). Then ("x Œ’ b)(x Œ b ⁄ x
= b). Hence ("x Œ’ b)(x Œ b). By the induction hypothesis,
("x Œ’V(b))(x Œ V(b)). Let y Œ’ V(a). Then in (E,Œ’), we
have a = b+1 and y Õ V(b). Hence ("x Œ’ y)(x Œ V(b)).
Therefore y has the same Œ’ predecessors as an element of
V(a). Since (E,Œ’) satisfies extensionality, y Œ V(a).
Let a be a limit ordinal and ("x Œ’a)(x Œ a). Let b < a.
Then ("x Œ’ b)(x Œ a). Hence ("x Œ’ b)(x Œ b).
By induction hypothesis, ("b < a)("x Œ’ V(b))(x Œ V(b)).
Let x Œ’ V(a). We want x Œ V(a). Since (E,Œ’) satisfies “a
is a limit ordinal and V(a) is the union of the V(b), b <
a”, we let b Œ’ a be such that x Œ’ V(b) holds in (E,Œ’).
Then b < a, and x Œ’ V(b). Hence x Œ V(b). QED
LEMMA 1.11. There is an ordinal in the sense of (E,Œ’) that
is not an ordinal < k+1. 7 Proof: Suppose that every ordinal in the sense of (E,Œ’) is
an ordinal < k+1. Then the hypothesis of Lemma 1.10 holds
for every a < k+1. Hence by Lemma 1.10, for all a < k+1,
("x Œ’ V(a))(x Œ V(a)). Hence for all ordinals a in (E,Œ’),
("x Œ’ V(a))(x Œ V(a)).
Now ("x)($a)(x Œ V(a)) holds in (V(k+1),Œ), and therefore
in (E,Œ’). Hence ("x Œ E)($a)(x Œ V(a)). Therefore V(k+1) =
E, which is a contradiction. QED
LEMMA 1.12. There exists x Œ’ k, x œ V(k+1).
Proof: Since “k is the largest ordinal” holds in
(V(k+1),Œ), it also holds in (E,Œ’). So any ordinal x in
the sense of (E,Œ’) has x Œ’ k. Apply Lemma 1.11. QED
By Lemma 1.12, we now fix a to be least ordinal such that
there exists x Œ’ a, x œ V(k+1). Clearly a £ k. By Lemma
1.10, a is also the least ordinal such that there exists x
Œ’ a such that x œ a.
LEMMA 1.13. a is a limit ordinal.
Proof: (E,Œ’) satisfies “a is nonempty” and so (V(k+1),Œ)
also satisfies “a is nonempty”. Hence a > 0. Suppose a =
b+1. Let x Œ’ a, x œ V(k+1). Clearly (E,Œ’) satisfies “a =
b+1”. So x Œ’ b or x = b. Both alternatives violate x œ
V(k+1). QED
We now fix x Œ’ a, x œ V(k+1).
We define a nonprincipal ultrafilter W over a. I.e.,
i) W Õ P(a);
ii) for all y Õ a, y Œ W ⁄ a\y Œ W;
iii) for all y,z Œ W, y « z Œ W;
iv) for all y Œ W, y ≥ 2.
Take W to be the set of all A Õ a such that x Œ’ A.
LEMMA 1.14. W is a nonprincipal ultrafilter over a.
Proof: i) is immediate. Let y Õ a. Let z = a\y. “y » z = a”
holds in (V(k+1),Œ), and so in (E,Œ’). Since x Œ’ a, we
have x Œ’ y ⁄ x Œ’ a\y. Hence y Œ W ⁄ a\y Œ W. 8
Let y,z Œ W, and u = y « z. “y « z = u” holds in
(V(k+1),Œ), and so in (E,Œ’). Since x Œ’ y,z, we have x Œ’
u. Hence u Œ W.
Let y Œ W. Then x Œ’ y. So “y is nonempty” holds in (E,Œ’),
and hence in (V(k+1),Œ). Now suppose y = {z}. Then “every
element of y is z” holds in (V(k+1),Œ), and hence in
(E,Œ’). Hence “x = z” holds in (E,Œ’). Therefore x = z Œ a,
which is a contradiction. QED
We now show that W is countably complete; i.e., the
intersection of any infinite sequence of elements of W is
an element of W.
LEMMA 1.15. W is a countably complete nonprincipal
ultrafilter over a.
Proof: Let D0,D1,... all lie in W. Then x Œ’ D0,D1,... . Let
D be the intersection of the D’s, and f = {<n,Dn>: n Œ w}.
Obviously “f is a function with domain w” holds in
(V(k+1),Œ). Hence “f is a function with domain w” holds in
(E,Œ’). Also “D is the intersection of the values of f”
holds in (V(k+1),Œ). Hence “D is the intersection of the
values of f” holds in (E,Œ’).
We claim that for all y Œ E, if “y is a value of f” holds
in (E,Œ’), then y is a value of f.
To see this, assume “y is a value of f” holds in (E,Œ’),
and let n Œ E be such that “y = f(n)” holds in (E,Œ’).
Since “f is a function with domain w” holds in (E,Œ’), we
see that n Œ’ w. By the crucial Lemma 1.9, n Œ w. Now “v =
f(n)” holds in (V(k+1),Œ), where v is f(n). Hence “v =
f(n)” holds in (E,Œ’), and hence y = v = f(n).
We now claim that “x lies in every value of f” holds in
(E,Œ’). Let y be such that “y is a value of f” holds in
(E,Œ’). Then y is a value of f, and hence x Œ’ y.
Obviously “D is the intersection of the values of f” holds
in (E,Œ’) since it holds in (V(k+1),Œ). Hence x Œ’ D by the
previous claim. Therefore D Œ W. QED
LEMMA 1.16. There is a measurable cardinal £ k. 9
Proof: By Lemma 1.15, there is a countably complete
nonprincipal ultrafilter over the ordinal a £ k. Let b be
the least ordinal such that there is a countably complete
nonprincipal ultrafilter over b. Then b £ a is also the
least cardinal such that there is a countably complete
nonprincipal ultrafilter over b. Hence b is a measurable
cardinal £ k (see [Ka94], p. 23). QED
THEOREM 1.3. Propositions 1.1 and 1.2 are provably
equivalent to the existence of a measurable cardinal, over
ZFC.
Proof: By Lemmas 1.5 and 1.16. QED
THEOREM 1.4. The least cardinal k, if any, such that every
algebra of cardinality ≥ k has a proper extension with the
same countable (alternatively finitely generated)
subalgebras up to isomorphism, is the least measurable
cardinal.
Proof: Let k be the least cardinal such that every algebra
of cardinality ≥ k has a proper extension with the same
countable subalgebras up to isomorphism. By Lemma 1.16,
there is a measurable cardinal £ k. Let l be a measurable
cardinal £ k. By Lemma 1.5, every algebra of cardinality ≥ l
has a proper extension with the same countable subalgebras
up to isomorphism. Since k is least with this property, we
have l = k. Hence there is no measurable cardinal < k. Since
there is a measurable cardinal £ k, k is the least
measurable cardinal.
The argument can be repeated using “finitely generated” in
place of “countable”, since Lemma 1.16 was proved using
“finitely generated”. QED
For an announcement of many further results along these
lines, see [Fr03a].
2. BOREL SELECTION AND HIGHER SET THEORY.
We extract material from [Fr03] to give the flavor of the
applications of nonstandard models. Here these applications
are much more intense than in the previous section,
although the models are still slightly standard  they are
wmodels. 10
Let S a set of ordered pairs, and A be a set. We say that f
is a selection for S on A iff dom(f) = A and ("x Œ A),
(x,f(x)) Œ S.
f is a selection for S iff f is a selection for S on dom(S)
= {x:($y)((x,y) Œ S)}.
THEOREM 2.1. Let S Õ NN x NN be Borel and E Õ NN be Borel.
If there exists a continuous selection for S on every
compact subset of E, then there exists a continuous
selection for S on E.
Theorem 2.1 is due to Debs and Saint Raymond of Paris VII,
[DSR01X], using Borel Determinacy. We showed that it cannot
be proved using just countably many iterations of the power
set operation.
The same results apply to the simpler
THEOREM 2.2. Let S Õ NN x NN be Borel. If there is a
constant selection for S on every compact set, then there
is a Borel selection for S.
Theorem 2.2 is also due to Debs and Saint Raymond, in
[DSR99] and [DSR01X], using that the set of all Borel
selectors of a Borel relation can be coded by a P11 set of
reals.
PROPOSTION 2.3. Let S Õ NN x NN be Borel. If there is a
Borel selection for S on every compact set, then there is a
Borel selection for S.
Proposition 2.3 turns out to be independent of ZFC despite
being only a Borel statement. In fact, we have shown that
this is equivalent to “some function strictly dominates all
functions constructible in any given function”.
DOM: ("f)($g)("h Œ L[f])(g eventually dominates h).
PROPOSTION 2.4. Let S Õ NN x NN be Borel. If there is a
Borel selection for S on every compact set, then there is a
Borel selection for S.
There is a way of proving Proposition 2.4 using Borel
determinacy if one can bound the levels of the Borel
selections for S over compact sets. Using DOM and various
absoluteness arguments, we obtain such a bound in [Fr03]. 11 We use nonstandard models to show that Theorem 2.2 cannot
be proved with just countably many iterations of the power
set operation.
More specifically, in [Fr03] we show that ZFC\P + V = L +
Theorem 2.2 proves that for all a < w1 there is an wmodel
of ZFC\P + V(a) exists. We sketch the ideas.
We work in ZFC\P + V = L + a < w1 + for all countable b,
there is a set of rank £ a lying in L(b+1)\L(b). Our job is
to refute Theorem 2.2.
Let KP’ = KP with infinity. Let W = class of all (w,R) such
that
1. (w,R) satisfies KP’ + V = L.
2. There is an internal ordinal in (w,R) whose set of
predecessors is of type a.
3. (w,R) satisfies “for all b, there is a set of rank £ a
lying in L(b+1)\L(b).”
W must be plentiful since L(w1) satisfies these things, and
we are working within ZFC\P. Obviously W is Borel.
Let (w,R) Œ W. At each level of the constructible hierarchy
in (w,R), we can consider the family of sets of rank ≤ a
“present” there. This attaches a transitive set of rank £
a+1 to each “ordinal” of (w,R). These are strictly
increasing under inclusion because of condition 3 on
elements of W.
Write (w,R)’ for the union of these sets. I.e., the set of
all sets of rank £ a “present” in (w,R).
We say that (w,R) is special iff it lies in W and every
element of (w,R) is definable in (w,R).
We also say that L(b) is special iff every (w,R) isomorphic
to L(b) is special.
Define L(b)’ = (w,R)’, where (w,R) is isomorphic to L(b).
LEMMA 2.5. Let L(b) satisfy KP’, a < b < w1, where a new
real appears in L(b+1). Then L(b) is special. There are
arbitrarily large countable b such that L(b) is special. 12
Let (w,R) Œ W be special. We need a special code (w,R)* for
(w,R). We use T = set of true sentences in (w,R), T Õ N.
(w,R)* is the first real recursive in the double jump of T
such that
1. (w,R)* strictly dominates every real recursive in T.
2. For all n, (w,R)*(n) is even iff n Œ T.
Note that given a special code, we can recover an
underlying special (w,R) uniquely up to isomorphism, quite
effectively.
The set of special codes is a Borel set.
We are now ready to define the Borel set S Õ NN x NN.
Let x,y Œ NN. (x,y) Œ S iff either x is not a special code,
or x,y are both special codes and
#) Let x = (w,R)*, y = (w,Y)*, effectively chosen. Let U =
maximum common initial segment of (w,R)’ and (w,Y)’. Then U
is the set of strict predecessors of some element of
(w,Y)’.
In the case that the maximum common initial segment of
(w,R)’ and (w,Y)’ demonstrate the ill foundedness of both
(w,R)’ and (w,Y)’, we take (x,y) œ S  but it doesn’t make
any difference what we do in this case.
LEMMA 2.6. There is no Borel selection for S.
LEMMA 2.7. Let V Õ NN be compact. There is a countable
ordinal bound to the lengths of the maximum well ordered
initial segments of the internal ordinals of every (w,R) Œ
W such that (w,R)* lies in V.
LEMMA 2.8. There is a constant Borel selection for S on
every compact subset of NN.
We now have the following.
THEOREM 2.9. ZFC\P + V = L + Theorem 2.2 proves that for
all a < w1 there is an wmodel of ZFC\P + V(a) exists.
As discussed in [Fr03], we see that using only countably
many iterations of the power set operation is insufficient 13
to prove Theorem 2.2. However, using all countable
iterations of the power set operation is sufficient to
prove Theorem 2.2, and even Theorem 2.1.
As a second, related application of nonstandard models, in
[Fr03] we derive DOM from the following.
PROPOSTION 2.3. Let S Õ NN x NN be Borel. If there is a
Borel selection for S on every compact set, then there is a
Borel selection for S.
Since DOM is derivable from Proposition 2.3, we see that
Proposition 2.3 is not provable in ZFC. The derivation of
Proposition 2.3 in ZFC + DOM appears in [Fr03].
We work in ZFC\P + ÿDOM. We wish to refute Proposition 2.3.
To simplify the notation in the sketch, we will assume that
no real dominates every constructible real.
Let X be the class of all (w,R) satisfying KP’ + V = L. X
is Borel.
We say (w,R) is unusual iff every element is definable.
We say L(a) unusual iff L(a) satisfies KP’ and every
element is definable.
Let (w,R) Œ X be unusual. The unusual code (w,R)* is
defined as follows.
Let T be the theory of (w,R). (w,R)* is the first sequence
recursive in the double jump of T such that
1. (w,R)* eventually strictly dominates every sequence
recursive in T.
2. for each n, (w,R)*(n) is even iff n Œ T. The set of all
unusual codes is Borel.
Now define Borel S’ Õ NN x NN as follows.
Let x,y Œ NN. (x,y) Œ S’ iff
1. x is not an unusual code, or
2. x is an unusual code and y is an infinite backward R
chain, where (w,R) comes from x; or 14
3. x,y are unusual codes and x is internal to (w,Y), where
Y is obtained from y.
Note that S’ is arithmetic.
LEMMA 2.10. If L(a) satisfies KP’, a < w1, and L(a+1)\L(a)
meets NN, then L(a) is unusual. There are arbitrarily large
a < w1 such that L(a) is unusual.
LEMMA 2.11. There is no Borel selection for S’.
LEMMA 2.12. Let V be compact. There exists a countable
ordinal bound to the lengths of the maximum well ordered
initial segments of the internal ordinals of the (w,R) Œ X
such that (w,R)* lies in V.
LEMMA 2.13. There is a Borel selection for S’ on every
compact subset of NN.
So we now have that Proposition 2.3 implies DOM over ZFC\P.
In particular, Proposition 2.3 is independent of ZFC.
3. EQUATIONAL BOOLEAN RELATION THEORY AND MAHLO CARDINALS.
Equational Boolean Relation Theory (equational BRT)
concerns the Boolean equations between sets and their
images under multivariate functions.
In [Fr02], we present equational BRT, and give a complete
proof of a crucial instance of equational BRT using the
existence of Mahlo cardinals of every finite order.
The proof that ZFC plus the existence of a Mahlo cardinal
of any specific finite order is not sufficient uses
nonstandard models in an extremely intensive way, where
there are nonstandard integers. That proof is not presently
publicly available. However, we will give an oversimplified
sketch of some key ideas.
Rather than repeat the rather full discussion of equational
BRT in [Fr02], we move directly to the special corner of
equational BRT where we have obtained rather complete
information.
We use the space MF(N) of all multivariate functions form N
into N, which are functions whose domain is a Cartesian 15
power of N (finite exponent) and whose range is a subset of
N.
Let f Œ MF(N) and A Õ N. Define fA to be the set of all
values of f at arguments from A.
We say that f Œ MF(N) is of expansive linear growth if and
only if there exist c,d > 1 such that for all but finitely
many x Œ dom(f),
cx £ f(x) £ dx
where x is the maximum coordinate of the tuple x.
Let ELG(N) be the set of all f Œ MF(N) of expansive linear
growth. Let INF(N) be the set of all infinite subsets of N.
We use X ». Y for X » Y together with the commitment that
X,Y are disjoint. For example,
X ». Y Õ Z ». W
means
X » Y Õ Z » W Ÿ X « Y = ∅ Ÿ Z « W = ∅.
Another way of explaining this convention is to think of ».
as a partially defined binary operation on sets, which is
defined if and only if the two sets are disjoint. The
additional rather standard convention is that if a
logically atomic statement is true then all expressions
within it must be defined.
PROPOSITION 3.1. For all f,g Œ ELG(N) there exist A,B,C Œ
INF(N) such that
A ». fA Õ C ». gB
A ». fB Õ C ». gC.
A cardinal k is 0Mahlo if and only if k is strongly
inaccessible.
A cardinal k is (n+1)Mahlo if and only if it is nMahlo and
every closed and unbounded subset of k has an element that
is nMahlo.
Let MAH = ZFC + {there exists an nMahlo cardinal}n. Let
MAH+ = ZFC + "for all n there exists an nMahlo cardinal". 16 THEOREM 3.2. MAH+ proves Proposition 3.1.
The proof of Theorem 3.2 is presented in detail in [Fr02].
Nonstandard models are used in an essential way in order to
prove the following.
THEOREM 3.3. MAH does not prove Proposition 3.1. ACA +
Proposition 3.1 proves Con(MAH).
The overarching idea is to work in ACA + Proposition 3.1,
and construct, in a series of many stages, a non wmodel of
MAH. Slowly but surely one builds countable structures that
more and more resemble a model of set theory with large
cardinals.
The two inclusions in Proposition 3.1 are of course rather
special, and so this immediately leads to consideration of
an arbitrary pattern of letters A,B,C in the two
conclusions. There are 8 positions for A,B,C, for a total
of 38 = 6561 statements.
I.e., we are looking at all 6561 statements of the
following form.
PROPOSITION. For all f,g Œ ELG(N) there exist A,B,C Œ
INF(N) such that
X ». fY Õ Z ». gW
S ». fT Õ U ». gV.
Here X,Y,Z,W,S,T,U,V are among the three letters A,B,C.
We have shown the following.
THEOREM 3.4. Proposition 3.1 and its other 11 symmetric
equivalent forms (total of 12) are the only ones among the
6561 that cannot be settled in RCA0. They are provably
equivalent, over ACA, to the 1consistency of MAH.
We now give some idea of how we construct a non wmodel of
MAH using Proposition 3.1.
We need only use Proposition 3.1 for multivariate functions
f from N into N which can be defined by finitely many
cases, where 17
i) each case given by a finite set of polynomial
inequalities with integer coefficients;
ii) f is defined within each case by a polynomial with
integer coefficients;
iii) for all x Œ dom(f), f(x) ≥ 2x.
Let V0 be the set of all such f. As a first step, we derive
the following in ACA + Proposition 3.1.
LEMMA 3.5. For all functions f,g Œ V0, there exist infinite
A Õ B Õ C Õ N such that
i) fA Õ B » gB;
ii) fB Õ C » gC;
iii) C « gC = ∅;
iv) A « fB = ∅.
Here is the next step.
LEMMA 3.6. For all k ≥ 1 and functions f,g Œ V0, there exist
infinite A1 Õ ... Õ Ak Õ N such that for all 1 £ i £ k1,
i) fAi Õ Ai+1 » gAi+1;
ii) Ak « gAk = ∅;
iii) A1 « fAk = ∅.
The next step is to apply the compactness theorem so that
we have an infinite tower within a single structure.
LEMMA 3.7. There exists a countable model M of IS0 and A1 Õ
A2 Õ ... Õ dom(M) such that
i) A1 is unbounded in dom(M) and forms a set of
indiscernibles for bounded existential statements whose
existential quantifiers range over A2, with parameters from
the Ai# below the indiscernibles;
ii) for any bounded existential j(x), there exists a linear
function f(x) with standard coefficients, such that for all
x Œ Ai#, j(x) holds over Ai+1 if and only if f(x) œ Ai+1;
iii) for all x Œ A1, nothing in the A’s lies in (x/2,x).
Here E# is the closure of E » {0,1} under +,•.
The next step is to take the submodel of M generated by
0,1,+,•, and the union of the A’s. We inherit the < of M.
The elements of A1 become limit points and appropriate
indiscernibles. We also get an appropriate enumeration
theorem. 18
LEMMA 3.8. There exists a countable ordered commutative
semiring with unit, M = (D,<,0,1,+,•), and c1 < c2 < ...
unbounded in D, such that
i) the c’s are indiscernibles for bounded formulas using
parameters below indiscernibles;
ii) for any ci, every subset of [0,ci] defined by a formula
bounded by ci with parameters from [0,ci] is defined by some
formula bounded by ci+1 with a fixed number of symbols, with
a single parameter allowed from [0,ci+1];
iii) the c’s are limit points.
With such a structure at hand, we can internally build the
constructible universe, and show that there exists, for
each k, an internally kMahlo cardinal below c1. The first
limit ordinal serves as the w, and will of course be
nonstandard.
We again warn the reader that this is a simplified sketch
which should not be taken too literally.
4. WEAK SECOND ORDER LOGIC AND MEASURABLE CARDINALS.
Here we adapt the proof in section 1 to the case of
elementary equivalence and elementary extension in weak
second order logic.
WSOL (weak second order logic) is the same as first order
predicate calculus with equality, except that we adjoin the
quantifier “there are at most finitely many”. This has the
same effect as adding “there is at least one but at most
finitely many”, or adding the quantifier “there exist
infinitely many”.
Since we are moving to a language formulation of the
results of section 1, we will use relational structures
instead of algebras, but still require that they be in a
finite relational type.
However, the results would be the same if we just used
algebras as in section 1. In fact, when we reverse
Proposition 4.1 below, we will only use algebras, as in
section 1.
We also remark that all results in sections 1 and 4 would
remain unchanged if we, for example, used countable
relational types instead of just finite relational types. 19
We have the following analogous Propositions and Theorems.
PROPOSITION 4.1. Every structure with a sufficiently large
domain has a proper WSOL elementary extension.
PROPOSITION 4.2. Every structure with a sufficiently large
domain has a proper WSOL equivalent extension.
THEOREM 4.3. Propositions 4.1 and 4.2 are provably
equivalent to the existence of a measurable cardinal, over
ZFC.
THEOREM 4.4. The least cardinal k, if any, such that every
structure of cardinality ≥ k has a proper WSOL elementary
(alternatively equivalent) extension, is the least
measurable cardinal.
Here are the obvious reformulations which do not use the
notion of cardinality.
PROPOSITION 4.1’. Every structure with a sufficiently
inclusive domain has a proper WSOL elementary extension.
PROPOSITION 4.2’. Every structure with a sufficiently
inclusive domain has a proper WSOL equivalent extension.
THEOREM 4.3’. Propositions 1.1’ and 1.2’ are provably
equivalent to the existence of a measurable cardinal, over
ZFC.
THEOREM 4.4’. Let D be a nonempty set. The following are
equivalent.
i) every structure whose domain includes D has a proper
WSOL elementary extension;
ii) every structure whose domain includes D has a proper
WSOL equivalent extension;
ii) there is a countably additive 0,1 valued measure on all
subsets of D, in which singletons have measure 0 and D has
measure 1.
LEMMA 4.5. Let k be a measurable cardinal. Every algebra of
cardinality ≥ k has a proper WSOL elementary extension.
Proof: Let k be measurable and A be a structure of
cardinality ≥ k. We will assume that the domain of A is a
cardinal l ≥ k. Let j:V Æ M be an elementary embedding with
critical point k, where M is a transitive class containing 20
all ordinals. j is obtained by taking the ultrapower of V
via a kcomplete nonprincipal ultrafilter on k.
j(A) is an algebra with domain j(l), and jdom(A) is an
isomorphism from A into j(A).
Since k is the critical point of j, it is clear that k is
not in the range of j. Hence jdom(A) is a proper
isomorphism from A into j(A).
Let j(x1,...,xk) be a formula of WSOL with at most the free
variables shown. Let a1,...,ak Œ dom(A) = l. By the
elementarity of j, we have
V satisfies “A satisfies j(a1,...,ak)”
if and only if
M satisfies “j(A) satisfies j(j(a1,)..,j(ak))”.
Now satisfaction for WSOL is easily seen to be very
absolute. Therefore
A satisfies j(a1,...,ak)
if and only if
j(A) satisfies j(j(a1,)..,j(ak)).
Hence jdom(A) is a proper WSOL elementary embedding from A
into j(A). By a standard construction, there is a proper
WSOL elementary extension of A. QED
We fix a cardinal k such that every algebra of cardinality ≥
k has a proper WSOL equivalent extension. Our aim is to show
that there is a measurable cardinal £ k.
LEMMA 4.6. k is uncountable.
Proof: Obviously k is not finite. Suppose k = w. We claim
that (w,0,S,<) does not have any proper WSOL equivalent
extension. Here S is the successor operation on w.
To see this, let (A,0,S’,<’) be a proper WSOL equivalent
extension.
First note that in (w,0,S,<) 21 i) < is a linear ordering of w;
ii) 0 is the < least element;
iii) S is the successor function for <.
Hence
i) <’ is a linear ordering of w;
ii) 0 is the < least element;
iii) S’ is the successor function for <’.
Since S Õ S’, it is easy to see that (A,<’) must be a
proper end extension of (w,<).
We now use WSOL. Note that (w,<) satisfies
("x)(there are at most (standardly) finitely many y < x)
Hence (A,<) satisfies
("x)(there are at most (standardly) finitely many y < x).
This contradicts that (A,<’) is a proper end extension of
(w,<). QED
We now let (V(k+1),Œ*,0,1,S,...) be the algebra defined
after Lemma 1.6 and before 1.7 in section 1. We fix an
algebra B = (E,Œ*’,0,1,S’,...) which is a proper WSOL
equivalent extension of (V(k+1),Œ*,0,1,S,...).
We now follow the proof that there is a measurable cardinal
£ k given in section 1, which is Lemma 1.16. However, Lemma
1.16 was proved under the assumption that
“every algebra of cardinality ≥ k has a proper extension
with the same finitely generated subalgebras up to
isomorphism”
whereas we are operating under the different assumption
that
“every algebra of cardinality ≥ k has a proper WSOL
equivalent extension.”
LEMMA 4.7. (V(k+1),Œ*,0,1,S,...) and (E,Œ*’,0,1,S’,...)
satisfy the same purely universal sentences in L*. The 22
former is an elementary substructure of the latter with
respect to the formulas in L*.
Proof: The first claim is by assumption, even for WSOL (for
L*). As in the proof of Lemma 1.7, the second claim follows
immediately from the first claim. QED
LEMAM 4.8. For all x Œ E\w, S’(x) = 0.
Proof: Note that
("x)(x Œ w ´ S(x) ≠ 0)
holds in (V(k+1),Œ*,0,1,S,...).
By Lemma 4.7,
("x)(x Œ w ´ S’(x) ≠ 0)
holds in (E,Œ*’,0,1,S’,...).
Also by Lemma 4.7, w is satisfied to be the first limit
ordinal in (E,Œ*’,0,1,S’,...).
It suffices to show that, in (E,Œ*’,0,1,S’,...), the
elements of w are exactly the actual elements of w.
We now use WSOL. Clearly
“every element of w has at most (standardly) finitely many
elements”
holds in (V(k+1),Œ*,0,1,S,...). Hence
“every element of w has at most (standardly) finitely many
elements”
holds in (E,Œ*’,0,1,S’,...).
Hence every (E,Œ*’,0,1,S’,...) element of w is an ordinal
of (E,Œ*’,0,1,S’,...) with (standardly) finitely many
(E,Œ*’,0,1,S’,...) elements. Therefore every
(E,Œ*’,0,1,S’,...), element of w is an actual element of w.
QED
LEMMA 4.9. There is a measurable cardinal £ k. 23
Proof: The remainder of the proof that there is a
measurable cardinal £ k is identical to the proof of Lemma
1.16 in starting right after the proof of Lemma 1.8. This
is because in the proof of Lemma 1.16, the hypothesis in
section 1 on k is not used after the proof of Lemma 1.8.
THEOREM 4.3. Propositions 1.1 and 1.2 are provably
equivalent to the existence of a measurable cardinal, over
ZFC.
Proof: By Lemmas 4.5 and 4.9. QED
THEOREM 4.4. The least cardinal k, if any, such that every
structure of cardinality ≥ k has a proper WSOL equivalent
(alternatively elementary) extension, is the least
measurable cardinal.
Proof: Let k be the least cardinal such that every algebra
of cardinality ≥ k has a proper WSOL elementary extension.
By Lemma 4.9, there is a measurable cardinal £ k. Let l be a
measurable cardinal £ k. By Lemma 4.5, every algebra of
cardinality ≥ l has a proper WSOL equivalent extension.
Since k is least with this property, we have l = k. Hence
there is no measurable cardinal < k. Since there is a
measurable cardinal £ k, k is the least measurable cardinal.
The argument can be repeated using WSOL equivalence in
place of WSOL elementary, since Lemma 4.9 was proved using
WSOL equivalence. QED
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[Fr02] Equational Boolean Relation Theory,
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