Intersection of Ordinals is Smallest

From ProofWiki
Jump to navigation Jump to search


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 $A$.


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 $m^+ = m \cup \set m$ 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 $m \in A$.