Preimage of Intersection under Mapping/Family of Sets
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Theorem
Let $S$ and $T$ be sets.
Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.
Let $f: S \to T$ be a mapping.
Then:
- $\ds f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
where:
- $\ds \bigcap_{i \mathop \in I} T_i$ denotes the intersection of $\family {T_i}_{i \mathop \in I}$.
- $f^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $f$.
Proof 1
As $f$ is a mapping, it is by definition also a many-to-one relation.
It follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation.
Thus Image of Intersection under One-to-Many Relation: Family of Sets can be applied for $\RR = f^{-1}$:
- $\ds \RR \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {T_i}$
where $\RR \sqbrk {T_i}$ denotes the image of $T_i$ under $\RR$.
$\blacksquare$
Proof 2
\(\ds x\) | \(\in\) | \(\ds f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i}\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map f x\) | \(\in\) | \(\ds \bigcap_{i \mathop \in I} T_i\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall i \in I: \, \) | \(\ds \map f x\) | \(\in\) | \(\ds T_i\) | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall i \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds f^{-1} \sqbrk {T_i}\) | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}\) |
$\blacksquare$
Sources
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- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.6 \ \text{(c)}$
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Exercise $1$
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