Image of Union under Mapping/Proof 2

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