# Image of Union under Mapping/Proof 1

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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:

- $\forall A, B \in \powerset S: \map {f^\to} {A \cup B} = \map {f^\to} A \cup \map {f^\to} B$

## Proof

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 |

$\blacksquare$

## Sources

- 1968: A.N. Kolmogorov and S.V. Fomin:
*Introductory Real Analysis*... (previous) ... (next): $\S 1.3$: Functions and mappings. Images and preimages: Theorem $3$ - 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(ii)}$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 21.4 \ \text{(ii)}$: The image of a subset of the domain; surjections

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