# Image of Union under Mapping

## 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:

$\RR \sqbrk {A \cup B} = \RR \sqbrk A \cup \RR \sqbrk B$

$\blacksquare$