Intersection of Ordinals is Smallest
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Theorem
Let $A$ be a non-empty set or class of ordinals.
Let $m = \bigcap A$ be the intersection of all the elements of $A$.
Then $m$ is the smallest element of $A$.
Proof
By Intersection of Ordinals is Ordinal, $m$ is an ordinal.
By Intersection is Largest Subset, $m \subseteq a$ for each $a \in A$.
It remains to show that $m \in A$.
Let $m^+ = m \cup \left\{{m}\right\}$ be the successor of $m$.
By Relation between Two Ordinals:
- for each $a \in A$, either $m^+ \subseteq a$ or $a \in m^+$.
Let $m^+ \subseteq a$ for all $a \in A$.
Then by Intersection is Largest Subset $m^+ \subseteq m$.
This contradicts the fact that Ordinal is not Element of Itself.
Thus there is an $a \in A$ such that $a \in m^+$.
Thus $a \in m$ or $a = m$.
Let $a \in m$.
Then $a \subsetneqq m$.
This contradicts the fact that $m \subseteq a$.
Thus $a = m$.
Thus it follows that $m \in A$.
$\blacksquare$
