Prove that if X is a nonempty set of ordinals, then is an ordinal. Moreover, is the least element of X.

Does anyone have any suggestions?

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- October 11th 2009, 09:19 AMchimpOrdinals
Prove that if X is a nonempty set of ordinals, then is an ordinal. Moreover, is the least element of X.

Does anyone have any suggestions? - October 12th 2009, 07:50 PMaliceinwonderland
[Trichotomy of ordinals]

Let be ordinals. Then, it satisfies one of the alternatives, .

Lemma 1. Let be an ordinal. Then, any member of is itself an ordinal number.

If , we are done. It is an ordinal number. Otherwise, it is a member of some ordinal number in X. By lemma 1, it is an ordinal number.

We show that is the least element of X. If , we are done.

If , we claim that x is the least element (w.r.t -image) in X.

Take any . If , we are done again. Otherwise, if , then or y=x by trichotomy of ordinals. This implies that . Since y is an element of X, . This forces y=x. Thus x is the least element of X.