[FOM] Order types: a proof
Jeremy Clark
jeremy.clark at wanadoo.fr
Mon Mar 7 05:31:22 EST 2005
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
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