Starting with the smallest infinite ordinal omega, we can look for
points where there's
a jump in complexity.  Omega + 1  is a a successor ordinal; so it's
ordinary.  Same
for omega + n,  n<omega.  Next, omega+omega seems no more complex than
omega.
So, after a while, we get to omega^2  .  It's a limit ordinal, so maybe
it's more
complex than omega.  But omega+omega is also a limit ordinal, and I feel
it's
no more complex than omega.  So if  alpha is some infinite ordinal,
say we define
(a) an alpha half-interval as:
       any non-empty subset of alpha of the form:  { delta in alpha such
that delta>= beta}
       for some  beta<alpha.

For omega^2,  and a beta<omega^2,  { beta, beta+1, .... w^30, ...
w^10000000  ... }
is an example of a half interval, a bit like  the half-line [3, oo[  in
the set R of real
numbers.

(b) an alpha interval (? "proper interval" ? ) is:
     any non-empty subset of alpha of the form:
       { delta in alpha such that  gamma>delta>= beta}, some beta, gamma
< alpha.

For omega^2, we have proper intervals:
[ w, w+w[     :=  { delta: omega<=delta< omega+omega},
[ w*n,  w*(n+1)[      or  [omega*n, omega*(n+1) [  for some integer n>1 .

Does there exist a half open interval A of omega^2 and a
//    property J
proper interval B of omega^2 such that A and B are order-isomorphic ?

I don't think so: B will always be made up of a finite number of copies of
omega followed by nothing, or a positive integer.

It seems on the other hand A will always
(%%)
be order-isomorphic to omega^2 (cf. definition (a) ).         (%%)

Omega^2 fails to have order-isomorphic A, B as asked for next to
"property J" above.  J is for jump, a jump in complexity. So
we say omega^2 has property J.

Next, after omega^2, I think the next ordinal with "property J" is
omega^3 .  After that,  omega^4,  omega^5, omega^6, omega^7 and so on ...
After all the omega^n for n  = 1, 2, 3  ...  [positive integers ],  I
guess we're
landed at omega^omega as the next smallest ordinal with "property J".

It seems intuitively that we now have to sort of "think out of the box"
and come up with something new ...

Does  omega^(omega^2)  have "property J"  ?
If this depends only on the "half-intervals", as in (%%)  above, that
simplifies things somewhat.

I would imagine that logicians who work in Proof theory, who
are used to large countable ordinals, might have thought
about similar problems.  Or maybe those in recursion theory...

David Bernier

The recursive ordinals are less complex than the non-recursive ordinals.

----

In article <68Qsk.6119$fA6.16...@wagner.videotron.net>,
David Bernier  <david...@videotron.ca> wrote:

Check out the definition of a "regular ordinal," e.g., see

  http://en.wikipedia.org/wiki/Cofinality

This is not exactly the same as your definition, but it's close.
--
Tim Chow       tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth.  ---Galileo, Dialogues Concerning Two New Sciences

On Aug 26, 3:36 am, David Bernier <david...@videotron.ca> wrote:
|Does there exist a half open interval A of omega^2 and a
|//    property J
|proper interval B of omega^2 such that A and B are order-isomorphic ?

If I understand you right, an ordinal alpha has property J if
it can be written as beta+gamma and also as
delta+gamma+theta for ordinals beta, gamma, delta, and
theta.

That's equivalent to being expressible as beta+gamma
for gamma<alpha. (That beta<alpha is automatic.)
For, if gamma<alpha, we can always write alpha as
gamma+theta by taking away an initial segment of
order-type gamma. If you want an interval really in
the middle and alpha is infinite, you can always
write instead 1+gamma+theta. :-) So one simple
way to describe property J is, "is not a sum of
smaller ordinals".

The condition is equivalent to alpha being a power of
omega. If you write alpha in Cantor normal form,

   omega^rho1 * n1 + ... + omega^rho_k * n_k,

where n_i are natural numbers and rho1>...>rho_k,
when any of the n_i are >1, or there is more than
one term, you can lop off the initial term and get a
tail that is smaller than alpha.

If on the other hand alpha is omega^rho, then any
tail has order type alpha. This follows again from
Cantor normal form, although this may be overkill.
Any ordinals <omega^rho have Cantor normal form
with all the powers being <rho. The sum has then
also Cantor normal form with the powers being
<rho, hence again is <omega^rho.

Keith Ramsay