Intersection of Inductive Sets

Theorem
Let $\mathbb S$ be a non-empty indexed family of inductive sets.

Then $\bigcap \mathbb S$ is an inductive set.

Proof
From definition, a set $X$ is an inductive set both the following holds:
 * $\O \in X$
 * $x \in X \implies x^+ \in X$

For all $S \in \mathbb S$, $S$ is an inductive set.

From definition of an inductive set, $\O \in S$.

By definition of set intersection, $\O \in \bigcap \mathbb S$.

Now suppose $x \in \bigcap \mathbb S$.

By definition of set intersection:
 * $\forall S \in \mathbb S: x \in S$

Hence from definition of an inductive set, $x^+ \in S$.

By definition of set intersection, $x^+ \in \bigcap \mathbb S$.

Therefore $\bigcap \mathbb S$ is an inductive set.