
Re: Order Isomorphic
Posted:
Feb 1, 2013 3:22 AM


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 wellordered 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 wellordered 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 orderisomorphic to kappa. > Yes, it's simple.

