```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 <ma...@panix.com> 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.
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