[FOM] 528: More Perfect Pi01
Harvey Friedman
hmflogic at gmail.com
Sat Aug 16 05:19:00 EDT 2014
I have upgraded the extended abstract #82
https://u.osu.edu/friedman.8/foundational-adventures/downloadable-manuscripts/
with a new statement involving step maximal cliques.
Step maximal cliques are perfectly natural and have the following
advantages. It allows me to use Q^k instead of Q[0,n]^k. It allows me to
use only sections at (1,...,n),(2,...,n+1).
The matter is simple enough for me to explain it here. First recall this
Pi01 independent statement:
For all k,n, every order invariant graph on Q[0,n]^k has a maximal clique
whose sections at any (i,...,n-1),(i+1,...,n) agree below i.
The new Pi01 independent statement is:
For all k,n, every order invariant graph on Q^k has a step maximal clique
whose sections at (1,...,n),(2,...,n+1) agree below 1.
What is step maximality? Let G be a graph on Q^k. S is a step maximal
clique for G if and only if for all m, S|<=m is a maximal clique in G|<=m.
I don't know whether the following are independent.
For all k,n, every order invariant graph on Q[0,n]^k has a maximal clique
whose sections at (1,...,n-1),(2,...,n) agree below 1.
For all k,n, every order invariant graph on Q^k has a maximal clique whose
sections at any (i,...,n-1),(i+1,...,n) agree below i.
For all k,n, every order invariant graph on Q^k has a maximal clique whose
sections at (1,...,n),(2,...,n+1) agree below 1.
I am going to try to prove that
For all k,n, every order invariant graph on Q^4 has a step maximal clique
whose sections at (1,2,3,4),(2,3,4,5) agree below 1.
is independent of ZFC while writing the book Order Invariant Relations and
Incompleteness.
****************************************
My website is at https://u.osu.edu/friedman.8/ and my youtube site is at
https://www.youtube.com/channel/UCdRdeExwKiWndBl4YOxBTEQ
This is the 528th 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 #1-527 can be found at the FOM posting
immediately preceding this posting.
Harvey Friedman
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