# 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:

- $\displaystyle f \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} f \sqbrk X$

### 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:

- $\displaystyle f \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} f \sqbrk {S_i}$

where $\displaystyle \bigcup_{i \mathop \in I} S_i$ denotes the union of $\family {S_i}_{i \mathop \in I}$.

## Proof 1

First we have:

\(\displaystyle A\) | \(\subseteq\) | \(\displaystyle A \cup B\) | Set is Subset of Union | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle f \sqbrk A\) | \(\subseteq\) | \(\displaystyle f \sqbrk {A \cup B}\) | Image of Subset under Mapping is Subset of Image |

\(\displaystyle B\) | \(\subseteq\) | \(\displaystyle A \cup B\) | Set is Subset of Union | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle f \sqbrk B\) | \(\subseteq\) | \(\displaystyle f \sqbrk {A \cup B}\) | Image of Subset under Mapping is Subset of Image |

\(\displaystyle \leadsto \ \ \) | \(\displaystyle f \sqbrk A \cup f \sqbrk B\) | \(\subseteq\) | \(\displaystyle f \sqbrk {A \cup B}\) | Union is Smallest Superset |

$\Box$

Then:

\(\displaystyle y\) | \(\in\) | \(\displaystyle f \sqbrk {A \cup B}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \exists x \in A \cup B: y\) | \(=\) | \(\displaystyle \map f x\) | Definition of Image of Subset under Relation | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \exists x: x \in A \lor x \in B: y\) | \(=\) | \(\displaystyle \map f x\) | Definition of Set Union | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle y\) | \(\in\) | \(\displaystyle f \sqbrk A \lor y \in f \sqbrk B\) | Definition of Image of Subset under Relation | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle y\) | \(\in\) | \(\displaystyle f \sqbrk A \cup f \sqbrk B\) | Definition of Set Union | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle f \sqbrk {A \cup B}\) | \(\subseteq\) | \(\displaystyle f \sqbrk A \cup f \sqbrk B\) | Definition of Subset |

$\Box$

Thus we have:

\(\displaystyle f \sqbrk A \cup f \sqbrk B\) | \(\subseteq\) | \(\displaystyle f \sqbrk {A \cup B}\) | |||||||||||

\(\displaystyle f \sqbrk {A \cup B}\) | \(\subseteq\) | \(\displaystyle f \sqbrk A \cup f \sqbrk B\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle f \sqbrk A \cup f \sqbrk B\) | \(=\) | \(\displaystyle f \sqbrk {A \cup B}\) | Definition of Set Equality |

$\blacksquare$

## Proof 2

As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation:

- $\mathcal R \sqbrk {A \cup B} = \mathcal R \sqbrk A \cup \mathcal R \sqbrk B$

$\blacksquare$

## 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): 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 $2.2 \ \text{(f)}$

- 1955: John L. Kelley:
*General Topology*... (previous) ... (next): Chapter $0$: Functions - 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $\S 2$