Intersection of Image with Subset of Codomain
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Theorem
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $A \subseteq S$ and $B \subseteq T$.
Then:
- $f \sqbrk {A \cap f^{-1} \sqbrk B} = f \sqbrk A \cap B$
Proof
\(\ds f \sqbrk {A \cap f^{-1} \sqbrk B}\) | \(=\) | \(\ds \set {\map f x: x \in A \cap f^{-1} \sqbrk B}\) | Definition of Image of Subset under Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {\map f x: x \in A \land x \in f^{-1} \sqbrk B}\) | Definition of Set Intersection | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {\map f x: x \in A \land \map f x \in B}\) | Definition of Preimage of Subset under Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {\map f x: x \in A} \cap \set {\map f x: \map f x \in B}\) | Definition of Set Intersection | |||||||||||
\(\ds \) | \(=\) | \(\ds f \sqbrk A \cap B\) | Definition of Image of Subset under Mapping |
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $4 \ \text{(d})$