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 $\mathcal P \left({S}\right)$ be the power set of $S$.

Let $\mathbb S \subseteq \mathcal P \left({S}\right)$.


Then:

$\displaystyle f \left[{\bigcup \mathbb S}\right] = \bigcup_{X \mathop \in \mathbb S} f \left[{X}\right]$


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


Sources