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Re: Order Isomorphic
Posted:
Jan 31, 2013 10:42 AM
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On Wed, 30 Jan 2013 21:01:37 -0800, William Elliot <marsh@panix.com>
wrote:
>Is every infinite subset S of omega_0 with the inherited order,
>order isomorphic to omega_0?
>
>Yes. S is an ordinal, a denumerable ordinal.
???
>Let eta be the order type of S.
>
>Since S is a subset of omega_0, eta <= omega_0.
>Since omega_0 is the smallest infinite ordinal, omega_0 <= eta.
>Thus S and omega_0 are order isomorphic.
>
>Does the same reasoning hold to show that an uncountable subset
>of omega_1 with the inherited order is order isomorphic to omega_1.
>
>It seems intuitive that since S is a subset of omega_1, that
>order type S = eta <= omega_1. How could that be rigorously
>shown?
Have you thought about this even a little bit?
Let s_0 be the smallest element of S. Let s_1 be the
smallest element of S minus s_0. Etc. Since S is infinite,
you never run out of elements, you can always find
s_{n+1}. It's easy to show that S = {s_n}:
Say s is in S. Since s has only finitely many predecessors,
there must be a stage in the construction when s is
the smalllest remaining element of S, and it becomes
s_n at that point.
Now n <-> s_n is your isomorphism.
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