[FOM] Correction to problem on order types

[Jeremy Clark]

I don't know if this is the error Joe had in mind, but it did occur to 
me after I submitted an answer that there are in fact many more order 
types satisfying this condition. Take a set S \subset \omega_1. Then 
define A_x to be Q (order type of rationals) if x \in S and 1+Q 
otherwise. Let A_S = sum A_x for x \in \omega_1. I *think* one can show 
something like: if S \setminus T is stationary in \omega_1 (meets every 
closed unbounded set in \omega_1) then A_S cannot be order isomorphic 
to A_T. It is possible to define 2^\omega_1 sets such that any two of 
them have the property that their setwise difference is stationary. So 
it follows that there are 2^\omega_1 sets satisfying Joe's condition.

regards,

Jeremy Clark


On Mar 6, 2005, at 7:05 am, Robert M. Solovay wrote:

> Joe,
>
> 	Can you elaborate what the error is? I got eleven as well. The
> empty set; the one point set and then 3 times 3 where the possibilities
> are open, closed, or "the long line" for each end.
>
> 	--Bob Solovay
>
>
>
> On Sat, 5 Mar 2005 JoeShipman at aol.com wrote:
>
>> In the statement of the problem, I should require that every open 
>> interval (a,b) with a<b is isomophic to the real numbers, not 
>> rational numbers.
>>
>> That was the form I had originally come up with the problem, but then 
>> I got too clever.  Dave Marker claims that changing "reals" to 
>> "rationals" allows more solutions than I had contemplated, and 
>> pointed out why my comtemplated proof for "rationals" fails.
>>
>> -- JS
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