[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).
Dave
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