Intersection of Inductive Sets
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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