[FOM] 486:Naturalness Issues

Harvey Friedman friedman.8 at asc.ohio-state.edu
Sat Mar 17 00:44:11 EDT 2012

Z+up(3/2,1) = (3/2,1). Z+up(x) is always order equivalent to x. 

Z+up(x) is the result of adding 1 to all coordinates greater than all coordinates not in Z^+. 

Yes, (x,y) is the concatenation of x,y if x,y are in Q*.

I have a complete, or almost complete, treatment of what invariance conditions can be used for the maximal square.

I have given one form of this with uniform transformations. This will be strengthened when the fundamental equivalence relations on Q* are worked out.

This almost appears to make the statement mathematically inevitable.

I.e., I would like to claim that Invariant Maximality is inevitable.

Harvey Friedman

On Mar 15, 2012, at 7:59 PM, David Roberts wrote:

> Dear Harvey,
> On 15 March 2012 01:20, Harvey Friedman <friedman at math.ohio-state.edu> wrote:
> Will do!
> (snip)
>> We say that T:Q* into Q* is a UNIFORM TRANSFORMATION (with respect to ~) if
>> and only if for all x in Q*, (x,Tx) ~ (Tx,TTx).
>> Note that Z+up:Q* into Q* is a uniform transformation. We can show that all
>> uniform transformations are very much like Z+up.
> I like this abstraction very much. Z+up, while perhaps nice in hindsight,
> and certainly very concrete, always looked to me to be a tailor-made function
> in order to arrive at the theorems. I have two questions
> 1) Are we stuck with the definition of ~ as given, or can that be altered
> as well?
> 2) What is the definition of (-,-)? I guessed it was concatenation of strings,
> but someone pointed out to me that this would mean x order equivalent
> to Tx, which is not true for Z+up. Can you please clarify? (consider, for
> example, Z+up(3/2,1) = (3/2,2))
> I have asked at MathOverflow [1] if people can come up with examples
> of uniform transformations, and if they can see a way to arrive at
> Z+up from some natural combinatorial problem.
> I can think of some nice ways to arrive at the condition (x,Tx) ~ (Tx,TTx)
> in a very natural way (given some relation ~), but it depends on the definition
> of (-,-).
> Best regards,
> David Roberts
> [1] http://mathoverflow.net/questions/91238/
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