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UNPROVABLE THEOREMS IN DISCRETE MATHEMATICS
Harvey M. Friedman
Department of Mathematics
Ohio State University
friedman@math.ohiostate.edu
www.math.ohiostate.edu/~friedman/
April 26, 1999
An unprovable theorem is a mathematical result that cannot
be proved using the commonly accepted axioms for mathematics
(ZermeloFrankel plus the axiom of choice), but can be proved
by using the higher infinities known as large cardinals.
Large cardinal axioms have been the main proposal for new
axioms originating with Gödel.
Gödel's famous incompleteness theorems from the 1930's
demonstrate the possibility of unprovable theorems in discrete and finite mathematics.
Known examples in discrete and finite mathematics have been
hopelessly unsatisfactory from the point of view of a mathematician. (Disguised statments about formal set theories;
Diophantine equations too long to write down in base 10 in
1000 years, ad hoc concoctions, etc.).
This is not to be confused with various important
undecidability results concerning the impossibility of giving
algorithms for solving an infinite family of problems.
E.g., the algorithmic undecidability of the class of
Diophantine equations with a solution over the integers.
(Incidentally, this is open over the rationals).
Here we are talking about specific individual assertions of
substantive mathematical content, in discrete or finite
mathematics, which can only be proved by going well beyond
the usual axioms for mathematics.
Before getting into discrete and finite mathematics, I want
to discuss some of the original examples from the 1980’s
where large infinite sets are used in fairly concrete mathematics  at least fairly concrete for an analyst. The
startling point is that these uses of large infinities are
demonstrably necessary  we know that they simply cannot be
removed. 2 Cantor’s famous diagonalization theorem can be stated as
follows. For any x Œ I• there exists y Œ I which not a term
in x. We call such a y a diagonalizer for x. We want to
investigate how a diagonalizer for x can be obtained from x.
First of all, it is easy to see that a diagonalizer cannot be
obtained continuously, where we give I• the usual infinite
product topology (which is also separable).
The Borel measurable subsets of a topological space form the
least s algebra containing the open sets. The Borel measurable functions from one space into another are the
functions under which inverse images of open sets are Borel
measurable.
It is well known that we can obtain diagonalizers by a Borel
measurable function. I.e., there is a Borel measurable
function F such that for all x Œ I•, F(x) is a diagonalizer
for x.
In such a construction, the y chosen may depend not only on
the terms in x but the order in which they appear in x.
So it is natural to ask: can diagonalizers be chosen Borel
measurably and dependent only on the set of terms?
The answer is no (Borel diagonalization theorem).
THEOREM 1. Let F:I• Æ I be Borel measurable. Assume that for
all x,y Œ I•, if x and y have the same terms (are permutations of each other; are finite permutations of each
other) then F(x) = F(y). Then there exists x such that F(x)
is a term in x.
The following elegant variant is one of very many, and
assumes the general flavor of a fixed point theorem:
THEOREM 2. Let F:I• Æ I• be Borel measurable. Suppose for all
x,y Œ I•, if x and y are subsequences of each other then F(x)
and F(y) are subsequences of each other. Then there exists x
such that F(x) is a subsequence of x.
Theorem 1 is proved by a basic application of the Baire category theorem for the space I•, where this time I is given the
DISCRETE topology I. Any Borel measurable F:I• Æ I will be a 3
Borel measurable F:I• Æ I. By using an appropriate 01 law,
we see that F must be constant off of a meager set in the I•
topology. The conclusion follows immediately.
The Borel diagonalization theorem lives squarely in the realm
of discrete mathematics, and so one would not expect to be
using the normally useless discrete topology on I. Yet the
significant point is that one can suitably formalize
“separable mathematics” or “inherently countable mathematics” and show that all three forms of the Borel diagonalization theorem cannot be proved there. The proof of
Theorem 2 is trickier, but with the same underlying idea.
We mention two additional Borel diagonalization theorems that
share the same logical features  i.e., necessary use of
nonseparable mathematics.
Let K be the Cantor space {0,1}• with the usual compact
topology. Let s:K Æ K be the shift map, where s(x) is obtained from x by chopping off the first term of x. For x Œ K
let x(2) = (x1,x4,x9,x16...).
THEOREM 3. Let F:K Æ K be Borel measurable. Assume that for
all x Œ K, F(sx) = F(x). Then there exists x such that F(x) =
x(2).
Let T be the usual circle group.
THEOREM 4. There is a continuous f:T Æ T which agrees
somewhere with: every Borel measurable F:T Æ T that satisfies F(2x) = F(x).
EMERGING GENERAL THEME: For certain interesting classes of
functions X and Y, some element of X agrees somewhere
with every element of Y.
Not yet systematically explored.
I now want to mention a Borel theorem whose proof necessarily employs much larger sets. We say that S Õ IxI is
symmetric if and only if (x,y) Œ S ´ (y,x) Œ S.
THEOREM 5. Every symmetric Borel measurable subset of the
unit square (or KxK) contains or is disjoint from the graph
of a Borel measurable (or continuous) function. 4
The proof (which relies heavily on work of D.A. Martin on
infinite games) necessary uses uncountably many uncountable
cardinalities! However monstrous such cardinalities are, they
are still well within ZFC.
But now we come to a Borel theorem that can only be proved by
using infinities going way beyond ZFC.
UNPROVABLE THEOREM 6. Let F be a Borel measurable function
from the space of sequences of finitely generated groups into
finitely generated groups which is isomorphically invariant.
Then F maps all of the infinite subsequences of some argument
into a value that is embeddable into a term of that argument.
We now turn to discrete and finite mathematics.
The connections with large infinities are more delicate and
have taken longer to develop.
The next Theorem captures the essence of recursion on N.
Let N be the set of nonnegative integers. F:Nk Æ N is
strictly dominating iff for all x1,...,xk Œ N, F(x1,...,xn) >
x1,...,xn. For A Õ N, we write F[A] for the forward image of F
on Ak.
THEOREM 7. Let k ≥ 1 and F:Nk Æ N be strictly dominating.
There exists A Õ N such that A = N\F[A] (or equivalently,
F[A] = N\A). Furthermore A is unique and A is infinite.
We construct A by induction. Suppose we have determined
membership in A for all i Œ [0,n). We place n in A if and
only if n is not the value of F at integers already placed in
A. By the strict dominance of F, these placements remain
correct later on.
Obviously there is no leeway in how we can construct A, so
that A is unique. And clearly A must be infinite since if A
were finite, both A and F[A] would be finite, contradicting
that their union is all of N.
Write A D B for A\B » B\A. We can restate Theorem 7 (without
uniqueness) as follows:
THEOREM 7’. Let k ≥ 1 and F:Nk Æ N be strictly dominating.
There exists an infinite A Õ N such that N Õ A D F[A]. 5
We now present an elaboration of Theorem 7’. We write A+A for
{x+y: x,y Œ A}.
THEOREM 7’’. Let k,n ≥ 1 and F:Nk Æ N be strictly dominating. There exists infinite sets A1 Õ ... Õ An Õ N such that
for all 1 ≤ i < n, Ai+Ai Õ Ai+1 D F[Ai+1].
Theorem 7’’ is a completely trivial weakening of Theorem 7’;
just set all of the A’s to be the A from Theorem 7’. However,
take a look at this:
UNPROVABLE THEOREM 8. Let k,n ≥ 1 and F:Nk Æ N be strictly
dominating. There exists infinite sets A1 Õ ... Õ An Õ N such
that for all 1 £ i < n, Ai+Ai Õ (Ai+1 D F[Ai+1])\A1.
We have shown that 8 can only be proved by using axioms going
well beyond the usual axioms of mathematics (ZFC). These
axioms are called large cardinal axioms, and in this case
they are what are called “Mahlo cardinals of finite order.”
These cardinals are far larger than strongly inaccessible
cardinals  which correspond to Grothendieck’s “universes,”
which Grothendieck and other algebraists sometimes postulate
for convenience. But in down to earth algebraic settings, it
is well known that they are a luxury that can be easily
dispensed. with. In contrast, here we show the demonstrable
unremovability of such monsters  in fact monsters that are
far more imposing than even Grothendieck’s “universes” or
strongly inaccessible cardinals.
Of course, 8 is rather specialized, and needs to be placed
in a thematic context. We have been proposing a theory of
solvability of Boolean equations. The idea is to embed 8 into
a general class of problems; in particular, to view it as
simply one instance of solvability of a general kind of
Boolean equation.
To describe this promising approach, let us return to Theorem
7’.
THEOREM 7’. Let k ≥ 1 and F:Nk Æ N be strictly dominating.
There exists an infinite A Õ N such that N Õ A D F[A].
We can view this as follows:
GIVEN: Let R(A,B) be a formal Boolean relation between two
subsets A,B of N. 6
DECIDE: For all k ≥ 1 and strictly dominating F:Nk Æ N, there
exists an infinite A Õ N such that R(A,F[A]).
The point is that N Õ A D B counts as a formal Boolean
relation between A,B Õ N. Specifically, a Boolean relation
is taken as the assertion that some combination of union,
intersection, and complementation is N.
All Boolean relations can be written as a finite list of one
or more of these basic Boolean relations:
i) inclusions of the form a Õ b, where a is an intersection
of one or more variables, and b is the union of one or more
variables;
ii) the union of one or more variables = N;
iii) the intersection of one or more variables = ∅.
We can give a complete anlysis of which R  in list form makes the statement true.
In an appropriate sense, all true problem instances follow
formally from Theorem 7’. And if we were to use all F, and
not just the strictly dominating F, then all true problem
instances would be formally trivial.
GIVEN: Let n ≥ 1 and R(A1,...,A2n) be a formal Boolean
relation between subsets of N.
DECIDE: For all k ≥ 1 and strictly dominating F:Nk Æ N, there
exists infinite A1 Õ ... Õ An Õ N such that
R(A1,...,An,F[A1],...,F[An]).
We again give a complete analysis of which R in list form
makes this statement true.
And again, in an appropriate sense, all true problem instances follow formally from Theorem 7’. Again, if we were to
use all F, then all true problem instances would be formally
trivial.
GIVEN: Let n ≥ 1 and R(A1,...,A3n) be a formal Boolean
relation between subsets of N.
DECIDE: For all k ≥ 1 and strictly dominating F:Nk Æ N, there
exists infinite A1 Õ ... Õ An Õ N where R(A1,..,An,F[A1]
,...,F[An],A1+A1,...,An+An). 7
We conjecture that there is an exponential time algorithm
which, with the help of Mahlo cardinals of finite order,
provably solves this problem.
We already know that there is no algorithm which, without the
help of such cardinals, provably solves this problem.
We make the following finite obstruction conjecture. This
conjecture asserts that any R that makes the following finite
problem true also makes the previous infinite problem true.
GIVEN: Let n ≥ 1 and R(A1,...,A3n) be a formal Boolean
relation between subsets of N.
DECIDE: For all k,r ≥ 1 and strictly dominating F:Nk Æ N,
there exists A1 Õ ... Õ An Õ N, each with at least r
elements, such that R(A1,..,An, F[A1],...,F[An],A1+A1,..,An+An).
Specifically, we conjecture that the finite obstruction
conjecture can be proved with the help of Mahlo cardinals of
finite order, but not without. We already know that the
finite obstruction conjecture cannot be proved without the
help of Mahlo cardinals of finite order. But can it be proved
with their help?
We now turn to the “greedy” construction of finite graphs.
The term “greedy” is from the theory of algorithms, where so
many constructions proceed so that at each stage, some aspect
is optimized among all possible choices.
A digraph G is a pair (V,E), where V = V(G) is a finite set
and E = E(G) Õ VxV. V is the set of vertices and E is the set
of edges. (No multiple edges allowed).
If (x,y) is an edge, then we say that y is a target of x.
We say that G’ is a point extension of G if and only if
i)V(G’)\V(G) = {x} for some x;
ii) the edges in G are exactly the edges in G’ that connect
vertices in G;
iii) in G’, x is not a target of any vertex.
Thus in G’, all of the new edges point from the new vertex x
to a vertex in G. 8
We write G’/G for the new part of G’ over G, whose vertices V
are x together with the targets of x, and whose edges are
those edges in G’ that connect these vertices.
A digraph construction sequence (dgc) is a finite or
infinite sequence of digraphs G1,G2,... such that each Gi+1 is
a point extension of Gi.
For any set X let DG(X) be the set of digraphs all of whose
vertices lie in X. We consider “weight” functions w:DG(X)Æ N.
We seek to minimize weights of new parts. A wminimal dgc is
a dgc G1,G2,... from DG(X) where each Gi+1 is such that for no
point extension Gi+1’ of Gi with the same vertices as Gi+1, is
w(Gi+1’/Gi) < w(Gi+1/Gi).
The following is obvious:
THEOREM 9. Let G Œ DG(X) and v1,v2,... be a sequence of distinct elements of X\V(G) of length 0 ≤ n ≤ •. Let w:DG(X) Æ
N. There exists a wminimal dgc of length 1+n starting with
G, where the new vertices are v1,v2,... .
We need to consider some structural properties of digraphs.
Let [N]k, k ≥ 1, be the set of all k element subsets of N. Let
G Œ DG([N]k). We say that x is a summit in G if and only if x
Œ V(G) and every vertex y that x points to in G has
max(y) ≤ max(x).
Let x,y Œ [N]k. We say that x is entirely lower than y if and
only if every element of x is < every element of y.
THEOREM 10. Let k,p ≥ 1 and w:DG([N]k) Æ N have finite range.
There exists a wminimal dgc of finite length, starting with
any element of DG([N]k), such that in the final digraph, all k
element subsets of some p element set appear as summits with
the same number of targets.
We prove Theorem 10 with the help of infinitely many
uncountable cardinals. We conjecture that these are required.
Now, look at this:
UNPROVABLE THEOREM 11. Let k,p ≥ 1 and w:DG([N]k) Æ N have
finite range. There exists a finite wminimal dgc of finite
length, starting with any element of DG([N]k), such that in 9
the final graph, all k element subsets of some p element set
appear as summits with the same entirely lower targets.
It is necessary and sufficient to use subtle cardinals of
finite order to prove 11. These cardinals are somewhat bigger
than the ones we have been talking about earlier.
There is a more abstract way to state the conclusion of 11.
UNPROVABLE THEOREM 11'. Let k,p ≥ 1 and w:DG([N]k) Æ N have
finite range. There exists a wminimal dgc of finite length,
starting with any element of DG([N]k), such that in the final
digraph, there are p summits of any collective order type
with the same entirely lower targets.
In fact, this requires the same large cardinals even if p is
set to 2.
Finally, we discuss the issue of removing all mention of
infinite objects from these unprovable theorems. This
corresponds to going from discrete mathematics to finite
mathematics.
For 11 (or 11') this is simply a matter of introducing finite
parameters in the most straightforward way:
UNPROVABLE THEOREM 12. Let n >> k,p,r ≥ 1 and w:DG([1,n]k) Æ
[r]. There exists a wminimal dgc of finite length, starting
with any element of DG([1,n]k), such that in the final
digraph, all k element subsets of some p element set are
summits with the same entirely lower targets.
Here the wminimal dgc are defined as before except that all
terms must lie in dom(w) = DG([1,n]k). By a compactness or
Konig tree argument, 12 is equivalent to 11.
An appropriate elimination of infinite objects in 8 is more
delicate. Recall 8:
UNPROVABLE THEOREM 8. Let k,n ≥ 1 and F:Nk Æ N be strictly
dminating. There exists infinite sets A1 Õ ... Õ An Õ N such
that for all 1 ≤ i < n, Ai+Ai Õ (Ai+1 D F[Ai+1])\A1.
The approach we take is to strengthen the inclusion relation.
Thus for A,B Õ N, we write A Õd B if and only if A Õ B and
for all n ≥ 0, B « [1,n] ≤ A « [1,n]d.
UNPROVABLE THEOREM 8'. Let t >> k,n,r ≥ 1 and F:[1,t]k Æ N be
strictly dminating. There exists finite sets A1 Õk^2 ... Õk^2 An 10
Õ N, with at least r elements, such that for all 1 ≤ i < n,
Ai+Ai Õ (Ai+1 D F[Ai+1])\A1.
We can show that it is necessary and sufficient to use subtle
cardinals of finite order in order to prove 8'.
We can even go further and bound t as a function of k,n,r in
terms of iterated exponentiation. The resulting assertion is
purely universal and provably equivalent to 8'. ...
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 Fall '08
 JOSHUA
 Math

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