[FOM] order types a proof

Dave Marker marker at math.uic.edu
Mon Mar 7 08:29:59 EST 2005

Joe asks for an explicit example of two nonisomorphic omega_1-like
orders (say both with bottom element)

Let Q be the rationals and let I= 1+Q or the rationals in [0,1)

Let A be the order \omega_1 x I

Let B be the order 1+\omega_1 x Q  (that is \omega_1 x Q with a bottom
element added).

Then A and B are nonisomorphic.

The proof is realatively concrete (but uses \aleph_1 is regular so is
not choiceless--indeed if \aleph_1 is singular they are isomorphic).

Suppose f is an isomorphism. Show that there is an \alpha
such that f maps \alpha x I onto 1+\alpha\times Q
(let \alpha_0 = 0, let \alpha_{n+1} be least such that every element
of \alpha_n x I maps into \alpha_{n+1} x Q and
every element of \alpha_n x Q maps into \alpha_{n+1} x I,
then let \alpha=\sup\alpha_n).

But then we have a contradiction because in A there is a least
element greater that \alpha x I, but in B there is no
least element greater than f(\alpha x I).


More information about the FOM mailing list