Image of Union under Mapping/Family of Sets

Theorem

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 $\family {S_i}_{i \mathop \in I}$.


This can be expressed in the language and notation of direct image mappings as:

$\ds \map {f^\to} {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \map {f^\to} {S_i}$


Proof

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

$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {S_i}$

$\blacksquare$


Sources

  • 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 10$: Inverses and Composites
  • 1968: A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis ... (previous) ... (next): $\S 1.3$: Functions and mappings. Images and preimages: Remark $2$
  • 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.6 \ \text{(a)}$
  • 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Exercise $1$
  • 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $3 \ \text{(a})$
  • 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): Notation and Terminology
  • 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): Appendix $\text{A}.3$: Functions: Problem $\text{A}.3.1$