Date: Feb 1, 2013 3:22 AM
Author: William Elliot
Subject: Re: Order Isomorphic

On Wed, 30 Jan 2013, Butch Malahide wrote:
> On Jan 30, 11:01 pm, William Elliot <> wrote:

> > Is every infinite subset S of omega_0 with the inherited order, order
> > isomorphic to omega_0?

> Yes, of course. (And you don't need the subscript; omega without a
> subscript means omega_0, the first infinite ordinal.)

> > Yes.  S is an ordinal, a denumerable ordinal.
> > Let eta be the order type of S.

> No, S is *not* an ordinal, unless S = omega. S is a subset of an
> ordinal, so it's a well-ordered set, so it's isomorphic to an ordinal,
> so it's order type is an ordinal.

> > Does the same reasoning hold to show that an uncountable subset
> > of omega_1 with the inherited order is order isomorphic to omega_1.

> Yes, of course. The ordinal omega_1 is the unique ordinal such that
> (a) it's uncountable, and (b) each of its proper initial segments is
> countable. Any well-ordered set with those two properties is
> isomorphic to omega_1. If S is an uncountable subset of omega_1, then
> S (with the natural order) has those two properties.

Let eta be an ordinal with those two properties.
Since omega_1 is the first uncountable ordinal, omega_1 <= eta.
If omega_1 < eta, then omega_1 in eta and the initial segment
[0,omega_1] is uncountable. Thus omega_1 = eta.

> More generally (and just as trivially), if kappa is an initial
> ordinal, and if S is a subset of kappa which has the the same
> cardinality as kappa, then S is order-isomorphic to kappa.

Yes, it's simple.