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Unformatted text preview: 1 DOES NORMAL MATHEMATICS NEED NEW
AXIOMS?
by
Harvey M. Friedman*
Department of Mathematics
Ohio State University
friedman@math.ohiostate.edu http://www.math.ohiostate.edu/~friedman/ October 26, 2001
Lecture Notes
Abstract. We present a range of mathematical theorems whose
proofs require unexpectedly strong logical methods, which in
some cases go well beyond the usual axioms for mathematics.
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There are a variety of mathematical results that can only be
obtained by using more than the usual axioms for mathematics.
For several decades there has been a gradual accumulation of
such results that are more and more concrete, more and more
connected with standard mathematical contexts, and more and
more relevant to ongoing mathematical activity.
Probably the most well known mathematical problem that cannot
be proved or refuted with the usual axioms (ZFC) is the
continuum hypothesis  that every set of real numbers is
either countable or of cardinality the continuum (Kurt Gödel
and Paul Cohen).
But mathematicians have instinctively learned to hide from
this kind of problem by focusing on relatively “concrete”
subsets of complete separable metric spaces. In particular,
the Borel measurable sets and functions in and between
complete separable metric spaces proves to be a natural
boundary.
By way of illustration,
“every Borel set of real numbers is either countable or of
cardinality the continuum via a Borel measurable function”
is a well known theorem of descriptive set theory. 2
All problems discussed here live within this Borel measurable
universe. Some are even further down in sets and functions on
the integers.
1.
2.
3.
4.
5.
6.
7.
8. Exotic High School Math. Warmup.
Ordinals in Freshman Calculus. Warmup.
Borel Measurable Selection.
Thin Set Theorem. Warmup.
Complementation Theorem. Warmup.
Disjoint Covers.
Boolean Relation Theory.
A Sketch. 1. EXOTIC HIGH SCHOOL MATH.
This problem was used a few years ago in Paul Sally’s program
for gifted high school students at U Chicago.
THEOREM 1.1. There is a longest sequence of 1’s and 2’s in
which no block xi,...,x2i is a subsequence of a later block
xj,...,x2j. The longest length is 11.
Ex: 12221111111. The relevant blocks are 12, 222, 2211,
21111, 111111. None is a subsequence of any later one.
One of the students was able to give a correct proof of this.
THEOREM 1.2. There is a longest finite sequence of 1’s, 2’s,
3’s, in which no block xi,...,x2i is a subsequence of a later
block xj,...,x2j.
The students couldn’t prove this. The natural proof is very
infinitary, and the longest length is gigantic. More detailed
work gives a less infinitary proof, but all proofs must be
somewhat exotic. More generally:
THEOREM 1.3. For all k ≥ 1 there exists a longest finite
sequence from {1,...,k} in which no block xi,...,x2i is a
subsequence of any later block xj,...,x2j.
Theorem 1.3 can be proved using induction with 3 quantifiers
over the positive integers, but not using induction with just
2 quantifiers over the positive integers.
As for bounds, let n(k) be the longest length with k letters.
Then n(1) = 3, n(2) = 11. 3 n(3) > A7198(158386), where Ak(n) is the kth function in the
Ackermann hierarchy at n. A1 = doubling, A2 = base 2
exponentiation, A3 = base 2 superexponentiation, etcetera.
See [Fr01].
2. ORDINALS IN FRESHMAN CALCULUS.
Here is a theorem in what used to be freshman calculus that
is apparently new.
THEOREM 2.1. Let A,B be countable sets of real numbers. There
is a oneone continuous function from A into B or there is a
oneone continuous function from B into A.
This is the usual Œ/d pointwise continuity notion.
The proof uses a countable ordinal analysis of the topology
of countable sets of real numbers (or countable metric
spaces).
There are several ways to indicate that the use of countable
ordinals is necessary to prove Theorem 2.1.
THEOREM 2.2. There is no Borel function that takes a pair of
countable sets of reals, given as infinite sequences, to a
oneone continuous function from one into the other. There is
no Borel function F:¬• x ¬• Æ {0,1} such that if F(x,y) = 0
then there is a oneone continuous map from rng(x) into
rng(y), and if F(x,y) = 1 then there is a oneone continuous
map from rng(y) into rng(x).
For those familiar with reverse mathematics, Theorem 2.1 is
provably equivalent to ATR0 over RCA0.
See [Frxx].
3. BOREL MEASURABLE SELECTION.
Example: The function (1x2)1/2 on [1,1] is a selection for
the unit circle on the set [1,1]. It is also a selection for
the unit circle.
Let S be 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 for all x Œ A, 4
(x,f(x)) Œ S. We say that f is a selection for S iff f is a
selection for S on {x: ($y)((x,y) Œ S)} = dom(S).
We use ¬ for the reals with the usual topology.
THEOREM 3.1. Let S Õ ¬2 be Borel and symmetric about the line
y = x. Then S has a Borel selection on ¬ or ¬2\S has a Borel
selection on ¬.
This is closely related to an earlier theorem of D.A. Martin
called Borel determinacy, involving infinite zero sum games.
Theorem 3.1 can be proved using uncountably many iterations
of the power set operation, but not using only countably many
iterations of the power set operation. See [Fr81].
Recent developments in Borel selection come from work of two
analysts at University of Paris:
THEOREM 3.2. Let S Õ ¬2 be Borel and E Õ ¬ be Borel. If
there is a Borel selection for S on every compact subset of
E, then there is a Borel selection for S on E.
By Debs/Saint Raymond, Theorem 3.2 can be proved using large
cardinal axioms. We proved that Theorem 3.2 cannot be proved
in ZFC. There is a rather weak form of Theorem 3.2 that
cannot be proved in ZFC.
THEOREM 3.3. Let S Õ ¬2 be Borel. If there is a Borel
selection for S on every compact set of irrationals, then
there is a Borel selection for S on the irrationals.
Here is a continuous form which can be proved using
uncountably many iterations of the power set operation, but
not using only countably many iterations of the power set
operation.
THEOREM 3.4. Let S Õ ¬2 be Borel. If there is a continuous
selection for S on every compact set of irrationals, then
there is a continuous selection for S on the irrationals.
See [Fryy].
4. THIN SET THEOREM. 5
THEOREM 4.1. Thin Set Theorem. Let f be a multivariate
function from Z into Z. There exists infinite A Õ Z such that
fA ≠ Z.
Here fA is the set of all values of f at arguments from A.
If you know the right branch of combinatorics, then this is
not hard. Otherwise, you pretty much have to reinvent it.
We know that Theorem 4.1 cannot be proved constructively,
even in dimension 2.
One precise way of saying this is the following.
THEOREM 4.2. There exists a computable f:Z2 Æ Z such that for
all infinite computable A Õ Z, fA = Z.
For those who know reverse math, TST is provable in ACA but
not in ACA0. TST in dimension 2 is not provable in WKL0.
Notice the logical form of TST. For any function of a certain
kind there is a set of a certain kind such that some Boolean
relation holds between the set and the image of the set under
the function.
In fact, we do not even use the set. For any function of a
certain kind there is a set of a certain kind such that some
Boolean relation holds of the image of the set under the
function. See [FS00] and [CGH].
5. THE COMPLEMENTATION THEOREM.
This can be viewed as a fixed point theorem. Let f be a
multivariate function from Z into Z. We say that f is
strictly dominating if and only if for all x Œ dom(f), f(x)
> x. Here   is the sup norm.
THEOREM 5.1. Complementation Theorem. Let f be a strictly
dominating multivariate function from Z into Z. There exists
infinite A Õ Z such that fA = Z\A. Furthermore, there is a
unique A Õ Z such that fA = Z\A.
Suppose we have determined membership in A of all x < n. 6
For x = n, we inquire whether x Œ f[A] thus far. If so,
then x œ A. If not, then x Œ A. Since f is strictly dominating, we never change our mind about x Œ f[A].
There is obviously no leeway in this construction. Hence the
uniqueness.
The structure of these unique fixed points is unclear, even
in the case of one dimensional affine transformations, let
alone higher dimensional transformations.
Note the logical form of the complementation theorem: For any
function of a certain kind there is a set of a certain kind
such that some Boolean relation holds between the set and the
image of the set under the function.
6. DISJOINT COVERS.
The disjoint cover relation turns out to be an interesting
Boolean relation between sets. We way that A is disjointly
covered by B,C if and only if
i) A Õ B » C;
ii) B « C = ∅.
We write this as A/B,C.
Using this notation, we can restate the complementation
theorem as follows.
COMPLEMENTATION THEOREM. Let f be a strictly dominating
multivariate function from Z into Z. There exists infinite A
Õ Z such that Z/A,fA.
We say that A Õ Z is biinfinite if and only if A has
infinitely many positive and infinitely many negative
elements.
COMPLEMENTATION THEOREM? Let f be a strictly dominating
multivariate function from Z into Z. There exists biinfinite
A Õ Z such that Z/A,fA.
This is refutable. A way to fix this is to weaken Z/A,fA by
using something smaller than Z. The natural way is to bring
in a second function. 7
THEOREM? Let f,g be strictly dominating multivariate
functions from Z into Z. There exists biinfinite A Õ Z such
that fA/A,gA.
This is also refutable. The problem is that A is on the left
and right sides of /. This suggests bringing in a second set.
THEOREM. Let f,g be strictly dominating multivariate functions from Z into Z. There exist biinfinite A,B Õ Z such
that fA/B,gB.
The natural extension to three sets is this.
THEOREM. Let f,g be strictly dominating multivariate
functions from Z into Z. There exist biinfinite A,B,C Õ Z
such that fA/B,gB and fB/C,gC.
Does this make a difference?
THEOREM? Let f,g be strictly dominating multivariate functions from Z into Z. There exist biinfinite A,B,C Õ Z such
that fA/C,gB and fB/C,gC.
This turns out to be refutable. However, we can fix this by
modifying the strict domination condition.
We say that f is expansively trapped if and only if there
exists p,q > 1 such that
px £ f(x) £ qx
holds for all x Œ dom(f).
PROPOSITION. Let f,g be expansively trapped multivariate
functions from Z into Z. There exist biinfinite A,B,C Õ Z
such that fA/C,gB and fB/C,gC.
This Proposition is provable, but the proof uses large
cardinals (Mahlo cardinals of finite order). It is not
provable in ZFC.
Note the logical form of the Proposition:
For all multivariate functions f,g of a certain kind there
exists sets A,B,C of a certain kind such that some Boolean 8
relation holds between the three sets and their six images
under the two functions.
7. BOOLEAN RELATION THEORY.
We have noted the logical form of the thin set theorem and
the complementation theorem (and also the Proposition). Here
are two more examples of this form.
a. Let f be a multivariate continuous function from ¬ into
¬. There exists an unbounded open set A Õ ¬ such that fA ≠
¬. (True).
b. Every bounded linear operator on Hilbert space maps some
nontrivial closed subspace into itself. (Open problem).
We now give a formal presentation of Boolean relation theory.
If f is a multivariate function of arity n and A is a set
then fA = {f(x1,...,xn): x1,...,xn Œ A}.
A BRT setting is a pair (V,K) where V is a set of
multivariate functions, and K is a set of sets.
Ex: Let V be the set of all multivariate functions from N
into N, and K be the set of all infinite subsets of N.
Ex: Let V be the set of all strictly dominating multivariate
functions from N into N, and K be as above.
We now describe equational Boolean relation theory in (V,K).
In its most elemental from, it consists of analyzing all
statements of the form
For all f Œ V there exists A Œ K such that a given Boolean
equation holds between A,fA.
A Boolean equation between elements of K is an equation
between Boolean terms – i.e., using the Boolean operations of
intersection, union, and complementation.
Complementation is defined using
the set of all elements
of sets in K and
values of functions in V 9
as the universal set. It can be readily seen that there are
16 Boolean equations between two sets that are formally
distinct.
More generally, we can attempt to analyze all statements of
the form
For all f1,...,fk Œ V there exists A1,...,An Œ K such that a
given Boolean equation holds between the (k+1)n sets
A1,...,An
f1A1,...,f1An
...
fkA1,...,fkAn.
It can be readily seen that there are 2^2(k+1)n such Boolean
equations that are formally distinct.
It appears that equational Boolean relation theory is
interesting for any interesting pair (V,K) – and frequently
quite nontrivial even for just a single function and a single
set. There are a great many such interesting pairs (V,K)
throughout mathematics.
The Thin Set Theorem is an example of inequational Boolean
relation theory. Here inequations play the exact role that
equations play in equational Boolean relation theory, where
an inequation is simply the negation of an equation.
We can go further and consider propositional combinations of
equations, and even compound terms like f(A » gB). We have
not investigated these more involved situations.
Let ET(Z) be the
functions from Z
infinite subsets
infinite subsets set of all expansively trapped multivariate
into Z. Let INF(Z) be the set of all
of Z. Let BINF(Z) be the set of all biof Z. The Proposition is a statement in equational BRT on (ET(Z),
BINF(Z)) with two functions and three sets.
CONJECTURE 1. Every one of the 2512 statements in equational
BRT on (ET(Z),BINF(Z)) with two functions and three sets is
provable or refutable using Mahlo cardinals of finite order. 10
Because of the Proposition, clearly ZFC does not suffice.
What about (ET(Z),INF(Z))?
PROPOSITION’. Let f,g be expansively trapped multivariate
functions from Z into Z. There exists infinite A,B,C Õ Z such
that fA/C,gB, fB/C,gC, and A « fC = ∅.
This is provable using Mahlo cardinals of finite order but
not in ZFC.
CONJECTURE 2. Every one of the 2512 statements in equational
BRT on (ET(Z),INF(Z)) with two functions and three sets is
provable or refutable using Mahlo cardinals of finite order.
We have given a complete analysis of equational and
inequational BRT on (ET(Z),INF(Z)) and (ET(Z),BINF(Z)) with
only one function and one set. Here it suffices to use only
the very weak system ACA to prove or refute all statements
that arise.
Note that the Proposition consists of two disjoint cover
conditions in (ET(Z),BINF(Z)) with two functions and three
sets.
In a slightly different context we have succeeded in
analyzing all such statements with two disjoint cover
conditions, using Mahlo cardinals of finite order.
8. A SKETCH.
PROPOSITION*. Let f,g be integral piecewise polynomials
obeying f(x),g(x) ≥ 2x. There exist biinfinite A,B,C
such that fA/C,gB and fB/C,gC.
We have shown that this also cannot be proved in ZFC.
Begin by blowing up the discrete ordered ring Z by a well
ordered set of indeterminates of large cardinal length. This
results in an enormous discrete ordered ring R.
Verify that the relation x > 2y in R is well founded.
Transfinite recursion is then applied to this well founded
relation in order to construct the unique set A Õ R such that
R/A,gA. This is the same as the proof of the Complementation
Theorem, only here we are in a transfinite context rather
than Z. 11 It is easily seen that the indeterminates and their negatives
all lie in A.
We then form the chain indexed by natural numbers
±I = I0 Õ I1 Õ ... Õ A Õ R
‘minimally’, where each fIj/Ij+1,gIj+1.
For any choice of ±I’ Õ ±I, we can canonically form the chain
indexed by natural numbers
±I’ = I0’ Õ I1’ ... Õ A Õ R
‘minimally’, where each fIj’/ Ij+1’,gIj+1’, and each Ij’ Õ Ij.
We carefully choose I’ Õ I of order type w according to large
cardinal combinatorics. Each resulting Ij’ will have order
type Z.
Much more is true. For all j, the indeterminates appearing in
Ij' have order type w, and the degrees and integer
coefficients of elements of Ij' are bounded.
It then follows that we can stick the chain
±I' = I0' Õ I1' Õ I2'
back into Z and we are done. In fact, we get rather concrete
biinfinite sets A Õ B Õ C Õ Z such that fA/B,gB, fA/B,gC,
fA/C,gB, fA/C,gC, fB/C,gC.
9. A BIG PLAN: VARYING PARAMETERS.
THE PLAN: Let T be any interesting simply stated mathematical
theorem. Place T in a simple logical form, after identifying
appropriate concepts. Then choose parameters in the logical
form to vary, so as to create a classification problem with
only finitely many instances.
The resulting classification problem is expected to be deep,
interesting, and, with some frequency, doable by and only by
going far beyond the usual axioms for mathematics. 12
Boolean relation theory is just the special case of varying
the parameter: Boolean form. We look forward to varying
sensible arithmetic, algebraic, geometric, and analytic
parameters. REFERENCES
[CGH] P. Cholak, M. Giusto, J. Hirst, Free Sets and Reverse
Mathematics, http://www.nd.edu/~cholak/papers/vitae.html.
[Fr81] On the necessary use of abstract set theory, Advances
in Mathematics 41 (1981), 209280.
[Fr01] H. Friedman, Long Finite Sequences, Journal of
Combinatorial Theory, Series A 95, 102144 (2001).
[Frxx] H. Friedman, Metamathematics of comparability,
http://www.math.ohiostate.edu/~friedman/ [Fryy] H. Friedman, Selection for Borel relations,
http://www.math.ohiostate.edu/~friedman/ [Frzz] H. Friedman, Boolean relation theory notes,
http://www.math.ohiostate.edu/~friedman/ [Frww] H. Friedman, Lecture notes on baby Boolean relation
theory, h ttp://www.math.ohiostate.edu/~friedman/
[FS00] H. Friedman, S.G. Simpson, Issues and problems in
reverse mathematics, in: Computability Theory and Its
Applications, Contemporary Mathematics, volume 257, 2000,
127144.
*This research was partially supported by NSF Grant DMS9970459. ...
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