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 if and only if 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.

$\blacksquare$

Sources

• 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 11$: Numbers