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

Does anyone have any suggestions?

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

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

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

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

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

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

If $\displaystyle x \neq \emptyset$, we claim that x is the least element (w.r.t $\displaystyle \epsilon$-image) in X.

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