[FOM] joe shipman's question on ordertypes

Dave Marker marker at math.uic.edu
Sun Mar 6 17:35:45 EST 2005

Joe Shipman asked how many order types are there for linear orders
where if a<b then (a,b) is isomorphic to the rationals

We call an order \aleph_1-like, if every initial segment is countable.
Any \aleph_1-like dense linear order without endpoints would be
a suitable example.

There are 2^{\aleph_1} nonisomorphic \aleph_1-like dense linear orders
without endpoints. (This is Excercise 5.5.10 in my Model Theory text)

Here's the basic idea: For each stationary subset S of \omega_1
we build an \aleph_1-like dense linear order A^S as follows:

if \alpha\in S let A^S_\alpha = Q  (the rationals), while

if \alpha\not\in S let A^S_\alpha = 1+Q (the rationals with a least

(also let A^S_{-1} =Q)

Let A^S ={(\alpha,x): -1\le \alpha, x\in A^S_\alpha}
(this is the order Q+A^S_0+\dots+A^S_\alpha).

Suppose S and T are stationary sets and the symmetric difference
is also stationary. Then A^S and A^T are not isomorphic.
Suppose for contradication, f is an isomorphism.
The set of \alpha such that f maps the elements of A^S of the
form (\beta,x), \beta<\alpha onto similar elements of A^T is
closed unbounded. Thus we can find such an \alpha in the
symmetric difference of S and T. But in one of A^S and A^T these
elements have a least upper bound and in the other they don't.
Thus f is not an isomorphism.

There are standard arguments to find 2^\aleph_1 stationary subsets
of \aleph_1 such that any two have a stationary symmetric difference.

This type of argument is useful in many contexts (like building
nonfree groups of size \aleph_1 where every countable subgroups is free,
or in many of Shelah's many-models arguments).

Note these problems go away in the long lines where
intervals are isomorphic to the reals since we always have
least upperbounds for countable sets.

Dave Marker

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