Theorem. Given a function $f:X \rightarrow Y, \: A,B \subseteq X$ the following two sets are equal:
$$ f(A \cup B) = f(A) \cup f(B).$$
Proof.
(i) $ f(A \cup B) \subseteq f(A) \cup f(B) $:
$$ f(x) = y \in f(A \cup B) \leftrightarrow y \in f(A \cup B) \wedge (x \in A \vee x \notin A) $$
$$ (
\underbrace{y \in f(A \cup B) \wedge x \in A}_{(i.1)}
)
\vee
(
\underbrace{y \in f(A \cup B) \wedge x \notin A}_{(i.2)}
) $$
If (i.1) holds $y \in f(A).$
If (i.2) holds $y \in f(B). \: \Box $
(ii) $ f(A \cup B) \supseteq f(A) \cup f(B) $:
$$ y \in f(A) \rightarrow y \in \{ f(x) : x \in A \vee \text{whatsoever} \} $$
Since $\textit{whatsoever}$ is arbitrary we may substitute it with a special case:
$$ y \in f(A) \rightarrow y \in \{ f(x) : x \in A \vee x \in B \} \leftrightarrow y \in f(A \cup B). $$
If $y \in f(B)$ the same implication follows. $\Box$
2014 DEC 29 (v1.0)
Contact me: m.herrmann followed by an -at- followed by blaetterundsterne.org.
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