Intersection of Inductive Set as Subset of Real Numbers is Inductive Set
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Theorem
Let $\AA$ be a set of inductive sets defined as subsets of real numbers.
Then their intersection is an inductive set.
Proof
By definition of inductive set:
- $\forall S \in \AA: 1 \in S$
Thus by definition of set intersection:
- $\ds 1 \in \bigcap_{S \mathop \in \AA} S$
Also by definition of inductive set:
- $\forall S \in \AA: x \in S \implies x + 1 \in S$
So:
\(\ds x\) | \(\in\) | \(\ds \bigcap_{S \mathop \in \AA} S\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall S \in \AA: \, \) | \(\ds x\) | \(\in\) | \(\ds S\) | Definition of Intersection of Set of Sets | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall S \in \AA: \, \) | \(\ds x + 1\) | \(\in\) | \(\ds S\) | Definition of Inductive Set as Subset of Real Numbers | |||||||||
\(\ds \leadsto \ \ \) | \(\ds x + 1\) | \(\in\) | \(\ds \bigcap_{S \mathop \in \AA} S\) | Definition of Intersection of Set of Sets |
Hence the result, by definition of inductive set.
$\blacksquare$
Sources
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $3 \ \text{(a)}$