Ordinals

**chimp** · October 11th 2009, 09:19 AM

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

Does anyone have any suggestions?

**aliceinwonderland** · October 12th 2009, 07:50 PM

Originally Posted by chimp

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

Does anyone have any suggestions?

[Trichotomy of ordinals]
Let $\alpha, \beta, \gamma$ be ordinals. Then, it satisfies one of the alternatives, $\alpha \in \beta, \alpha=\beta, \beta \in \alpha$ .

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

If $\bigcap X = \emptyset$ , we are done. It is an ordinal number. Otherwise, it is a member of some ordinal number $\gamma$ in X. By lemma 1, it is an ordinal number.

We show that $x = \bigcap X$ is the least element of X. If $x = \emptyset$ , we are done.

If $x \neq \emptyset$ , we claim that x is the least element (w.r.t $\epsilon$ -image) in X.
Take any $y \in X$ . If $x \in y$ , we are done again. Otherwise, if $x \notin y$ , then $y \in x$ or y=x by trichotomy of ordinals. This implies that $y \subseteq x$ . Since y is an element of X, $x \subseteq y$ . This forces y=x. Thus x is the least element of X.

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