This preview shows pages 1–3. Sign up to view the full content.
1
P
0
1
INCOMPLETENESS
by
Harvey M. Friedman
December 4, 2005
‘Beautiful’ is a word used by mathematicians with a semi
rigorous meaning.
We give an ‘arguably beautiful’ explicitly
P
0
1
sentence
independent of ZFC. See Proposition A(4,3) from section 1.
1.
P
0
1
INDEPENDENT STATEMENT.
We use [1,n] for the discrete interval {1,.
..,n}.
Let A [1,n]
k
. We write A’ = [1,n]
k
\A. This treats [1,n]
k
as the ambient space.
Let R [1,n]
2k
. We define
RA = R[A] = {y [1,n]
k
: (
$
x A)(R(x,y))}.
We say that R is strictly dominating if and only if for all
x,y [1,n]
k
, if R(x,y) then max(x) < max(y).
We start with a basic finite fixed point theorem.
THEOREM 1.1. For all k,n
≥
1 and strictly dominating R
[1,n]
2k
, there exists A [1,n]
k
such that R[A’] = A.
Furthermore, A [1,n]
k
is unique.
We can obviously take complements, obtaining what we call
the ‘complementation theorem’ for RA.
THEOREM 1.2. For all k,n
≥
1 and strictly dominating R
[1,n]
2k
, there exists A [1,n]
k
such that RA = A’.
Furthermore, A [1,n]
k
is unique.
Here is a trivial modification of Theorem 1.1 with the same
proof. We call this the ‘complementation theorem’ for
R[A\{t}
k
].
THEOREM 1.3. For all k,n,t
≥
1 and strictly dominating R
[1,n]
2k
, there exists A [1,n]
k
such that R[A\{t}
k
] = A’.
Furthermore, A [1,n]
k
is unique.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2
We now incorporate the antichain concept.
Let R [1,n]
2k
. We say that A is an antichain for R if and
only if A [1,n]
k
and A,RA are disjoint. I.e., RA A’.
The reader will recall the following familiar ‘maximal
antichain’ theorem.
THEOREM 1.4. For all k,n
≥
1, every R [1,n]
2k
has a
maximal antichain.
Note that the equation in Theorems 1.2, RA = A’, asserts
that A is a very strong kind of antichain. We refer to the
following trivial restatement of Theorem 1.2 as a ‘complete
antichain’ theorem.
THEOREM 1.5. For all k,n
≥
1, every strictly dominating R
[1,n]
2k
has an antichain A such that RA = A’. Furthermore, A
is unique.
Note that we could equally well write RA
⊇
A’ instead of RA
= A’, since we are also asserting A is an antichain for R.
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 JOSHUA
 Math

Click to edit the document details