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INCOMPLETENESS: finite set equations
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
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sentence
independent of ZFC. See Proposition A from section 1.
1.
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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]
3k
[1,n]
k
. We define
R<A> = {y [1,n]
k
: (
$
x A
3
)(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]
3k
[1,n]
k
, 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 R<A>.
THEOREM 1.2. For all k,n
≥
1 and strictly dominating R
[1,n]
3k
[1,n]
k
, there exists A [1,n]
k
such that R<A> =
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]
3k
[1,n]
k
, there exists A [1,n]
k
such that R<A\{t}
k
>
= A’. Furthermore, A [1,n]
k
is unique.