[FOM] Order types: a proof

[Jeremy Clark]

Joe Shipman wrote:

> Can anyone provide an EXPLICIT (choiceless) construction of
> an example not isomorphic to one of the 11 obtained from
> the "real" examples above by replacing the reals with the
> rationals in the obvious way?  It's nice to show that there
> are 2^aleph-1 of them with a stationary set argument, but it
> ought to be possible to get just one more without going
> through such advanced combinatorics.


Well, yes: just take \omega_1 copies of (1+Q). That is to say, the 
proof for our large collection of stationary sets is still going to 
work if we specify a stationary set (in this case \omega_1 itself). 
Suppose that this is isomorphic to 1 + (\omega_1 times Q) (the only 
contender in your list that it could be isomorphic to). You need 
(countable) choice to derive a contradiction, but the construction of 
the set itself is choiceless.

Or you can construct (choiceless) Brouwerian counterexamples to your 
theorem (trivial ones: subsets of a singleton set, for example, which 
cannot be shown to be either empty or non-empty: such a set, trivially 
ordered, would be a counter-example to your theorem), but I don't think 
that is what you are asking for.

regards,

Jeremy Clark