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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