[FOM] 254:Pi01 Progress/more
Harvey Friedman
friedman at math.ohio-state.edu
Thu Nov 10 04:37:34 EST 2005
Pi01 INCOMPLETENESS 1
by
Harvey M. Friedman
November 10, 2005
"Beautiful" is a word used by mathematicians with a semi rigorous meaning.
We give an "arguably beautiful" explicitly Pi01 sentence independent of ZFC.
See Proposition A below.
For A containedin [1,n]k, we write A' = [1,n]k\A. This treats [1,n]k as the
ambient space.
We focus on relations R containedin [1,n]3k x [1,n]k. Here we use R in two
ways. Firstly, as a subset of [1,n]4k. Secondly, and most importantly, as an
operator whose arguments and values are subsets of [1,n]k.
In particular, for all A containedin [1,n]k, we define RA = R<A> = R[A3] =
{w in [1,n]k: (therexists x,y,z in A)(R(x,y,z,w))}. For elegance, we will
omit the brackets < > and write RA whenever this does not lead to ambiguity.
Note that in the above, [ ] is the standard forward image notation for R
containedin [1,n]3k x [1,n]k = [1,n]4k.
We say that R is strictly dominating if and only if for all x,y,z,w in
[1,n]k, if R(x,y,z,w) then max(x),max(y),max(z) < max(w). This definition is
the reason why we prefer to write [1,n]3k x [1,n]k rather than [1,n]4k.
We start with the basic unique fixed point theorem for the operator R<A'>.
THEOREM 1.1. For all k,n >= 1 and strictly dominating R containedin ]1,n]3k
x [1,n]k, there exists A containedin [1,n]k such that R<A'> = A.
Furthermore, A containedin [1,n]k is unique.
For A containedin [1,n]k and t >= 1, we write A\t = {x in A: t is not a
coordinate of x} = "A with t omitted".
The "omission" operator R<A'\(8k)!> also has a unique fixed point.
THEOREM 1.2. For all k,n >= 1 and strictly dominating R containedin [1,n]3k
x [1,n]k, there exists A containedin [1,n]k such that R<A'\(8k)!> = A.
Furthermore, A containedin [1,n]k is unique.
However, we can't get a COMMON fixed point
R<A'\(8k)!> = R<A'> = A.
THEOREM 1.3. The following is false. For all k,n >= 1 and strictly
dominating R containedin [1,n]3k x [1,n]k, there exists A containedin [1,n]k
such that R<A'\(8k)!> = R<A'> = A.
We will weaken this three term common fixed point equation so that it is
attainable. The weakening involves replacing = both with containedin and
with "having the same elements of a certain kind".
Our development depends heavily on a very strong regularity condition on R.
We say that R containedin [1,n]3k x [1,n]k is order invariant if and only if
for all x,y in [1,n]4k of the same order type, R(x) iff R(y). The number of
such R is bounded by an exponential expression in k that does not depend on
n.
The powers of t are the numbers t^i, i >= 0.
THEOREM 1.4. For all k,n >= 1 and strictly dominating order invariant R
containedin [1,n]3k x [1,n]k, there exists A containedin [1,n]k such that
the three sets R<A'\(8k)!> containedin R<A'> containedin A contain the same
k tuples of powers of (8k)!+1.
What happens if we apply R to the expressions in the three term tower of
Theorem 1.4?
This is in rough analogy with standard mathematical contexts where one
passes from degree 1 equations to degree 2 equations and higher. Things can
get much more difficult.
Obviously
RR<A'\(8k)!> containedin RR<A'> containedin RA
follows immediately from
R<A'\(8k)!> containedin R<A'> containedin A.
However,
"the first tower of sets above contain the same k tuples of powers of
(8k)!+1"
does not follow from
"the second tower of sets above contain the same k tuples of powers of
(8k)!+1."
PROPOSITION A. For all k,n >= 1 and strictly dominating order invariant R
containedin [1,n]3k x [1,n]k, there exists A containedin [1,n]k such that
the three sets RR<A'\(8k)!> containedin RR<A'> containedin RA contain the
same k tuples of powers of (8k)!+1.
Proposition A is obviously an explicitly Pi01 sentence. It is independent of
ZFC. Here is much more precise information.
Let MAH = ZFC + {there exists a strongly n-Mahlo cardinal}_n. Let MAH+ = ZFC
+ "for all n there exists a strongly n-Mahlo cardinal".
THEOREM 1.5. MAH+ proves Proposition A. However, Proposition A is not
provable in any consistent fragment of MAH that derives Z = Zermelo set
theory. In particular, Proposition A is not provable in ZFC, provided ZFC is
consistent. These facts are provable in RCA0.
THEOREM 1.6. It is provable in ACA that Proposition A is equivalent to
Con(MAH).
*************************************
I use http://www.math.ohio-state.edu/%7Efriedman/ for downloadable
manuscripts. This is the 254th in a series of self contained numbered
postings to FOM covering a wide range of topics in f.o.m. The list of
previous numbered postings #1-249 can be found at
http://www.cs.nyu.edu/pipermail/fom/2005-June/008999.html in the FOM
archives, 6/15/05, 9:18PM.
250. Extreme Cardinals/Pi01 7/31/05 8:34PM
251. Embedding Axioms 8/1/05 10:40AM
252. Pi01 Revisited 10/25/05 10:35PM
253. Pi01 Progress 10/26/05 6:32AM
Harvey Friedman
More information about the FOM
mailing list