Image of Union under Mapping/Proof 1
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:
| \(\ds A\) | \(\subseteq\) | \(\ds A \cup B\) | 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 |
$\blacksquare$
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