Minimal Element of an Ordinal

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Theorem

The minimal element of any nonempty ordinal is the empty set.

That is, if $S$ is a nonempty ordinal, $\bigcap S = \O$

Proof

Let $S$ be an ordinal.

Let the minimal element of $S$ be $s_0$.

This exists by dint of an ordinal being a woset.

From Ordering on Ordinal is Subset Relation, $S$ is well-ordered by $\subseteq$.

So, by definition of an ordinal:

$s_0 = \set {s \in S: s \subset s_0}$

But as $s_0$ is minimal, there are no elements of $S$ which are a subset of it.

So:

$\set {s \in S: s \subset s_0} = \O$

Hence the result.

$\blacksquare$

Sources

• 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Exercise $1.7.1$