Intersection of Ordinals is Ordinal

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Theorem

Let $A$ be a non-empty class of ordinals.


Then $\bigcap A$ is an ordinal.


Proof

Let $i = \bigcap A$.


Set

By Intersection of Non-Empty Class is Set, $i$ is a set.

$\Box$


Transitive

Let $n \in i$ and let $m \in n$.

Let $a \in A$.

By the definition of intersection, $n \in a$.

Since $a$ is an ordinal, it is transitive.

Thus $m \in a$.

As this holds for all $a \in A$, $m \in i$.

Thus $i$ is transitive.

$\Box$


$\in$-connected

Let $x,y \in i$.

Since $A$ is non-empty, it has an element $a$.

By the definition of intersection, $x, y \in a$.

Since $a$ is an ordinal, it is $\in$-connected.

Thus $x \in y$ or $y \in x$.

As this holds for all $x, y \in i$, $i$ is $\in$-connected.

$\Box$


Well-founded

Let $b$ be a non-empty subset of $i$.

Let $a \in A$ (such exists because $A$ is non-empty).

By Intersection is Largest Subset, $b \subseteq a$.

Since $a$ is an ordinal it is well-founded.

Thus $b$ has an element $x$ such that $x \cap b = \varnothing$.

As this holds for all such $b$, $i$ is well-founded.

$\blacksquare$