Image of Union under Mapping/Proof 2
Jump to navigation
Jump to search
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
As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation:
- $\RR \sqbrk {A \cup B} = \RR \sqbrk A \cup \RR \sqbrk B$
$\blacksquare$
Sources
- 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Functions