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, Logic, Set Theory, Quantification, Russell's paradox, Ernst Zermelo

Click to edit the document details