Image of Union under Mapping
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Theorem
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $A$ and $B$ be subsets of $S$.
Then:
- $f \sqbrk {A \cup B} = f \sqbrk A \cup f \sqbrk B$
This can be expressed in the language and notation of direct image mappings as:
- $\forall A, B \in \powerset S: \map {f^\to} {A \cup B} = \map {f^\to} A \cup \map {f^\to} B$
General Result
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
- $\ds f \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} f \sqbrk X$
Family of Sets
Let $S$ and $T$ be sets.
Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.
Let $f: S \to T$ be a mapping.
Then:
- $\ds f \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} f \sqbrk {S_i}$
where $\ds \bigcup_{i \mathop \in I} S_i$ denotes the union of $\family {S_i}_{i \mathop \in I}$.
Proof 1
First we have:
\(\ds A\) | \(\subseteq\) | \(\ds A \cup B\) | Set is Subset of Union | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds f \sqbrk A\) | \(\subseteq\) | \(\ds f \sqbrk {A \cup B}\) | Image of Subset under Mapping is Subset of Image |
\(\ds B\) | \(\subseteq\) | \(\ds A \cup B\) | Set is Subset of Union | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds f \sqbrk B\) | \(\subseteq\) | \(\ds f \sqbrk {A \cup B}\) | Image of Subset under Mapping is Subset of Image |
\(\ds \leadsto \ \ \) | \(\ds f \sqbrk A \cup f \sqbrk B\) | \(\subseteq\) | \(\ds f \sqbrk {A \cup B}\) | Union is Smallest Superset |
$\Box$
Then:
\(\ds y\) | \(\in\) | \(\ds f \sqbrk {A \cup B}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in A \cup B: \, \) | \(\ds y\) | \(=\) | \(\ds \map f x\) | Definition of Image of Subset under Relation | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x: x \in A \lor x \in B: \, \) | \(\ds y\) | \(=\) | \(\ds \map f x\) | Definition of Set Union | |||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds f \sqbrk A \lor y \in f \sqbrk B\) | Definition of Image of Subset under Relation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds f \sqbrk A \cup f \sqbrk B\) | Definition of Set Union | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds f \sqbrk {A \cup B}\) | \(\subseteq\) | \(\ds f \sqbrk A \cup f \sqbrk B\) | Definition of Subset |
$\Box$
Thus we have:
\(\ds f \sqbrk A \cup f \sqbrk B\) | \(\subseteq\) | \(\ds f \sqbrk {A \cup B}\) | ||||||||||||
\(\ds f \sqbrk {A \cup B}\) | \(\subseteq\) | \(\ds f \sqbrk A \cup f \sqbrk B\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds f \sqbrk A \cup f \sqbrk B\) | \(=\) | \(\ds f \sqbrk {A \cup B}\) | Definition of Set Equality |
$\blacksquare$
Proof 2
As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation:
- $\RR \sqbrk {A \cup B} = \RR \sqbrk A \cup \RR \sqbrk B$
$\blacksquare$
Also see
- Preimage of Union under Mapping
- Image of Intersection under Mapping
- Preimage of Intersection under Mapping
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.13 \ \text{(a)}$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 12 \alpha$
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.10$: Functions: Exercise $5 \ \text{(e)}$
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(f)}$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$