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I failed to find a nice proof of the following statement: $\{x\in X: f(x)\in A\cup B\} = \{x\in X: f(x)\in A\} \cup \{x\in X: f(x)\in B\}$. I tried to use $\{x\in X: f(x)\in A\lor f(x)\in B\}$, and it is clear on the intuitive level, but I'd like to know what set theory axioms/results can I use to "rigurously" prove that.

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$y \in \{x\in X: f(x)\in A\cup B\}$

$\Longleftrightarrow f(y) \in A\cup B$

$\Longleftrightarrow f(y)\in A \hskip5pt \text{ or } \hskip5pt f(y) \in B$

$\Longleftrightarrow y \in \{x\in X: f(x)\in A\} \hskip5pt \text{ or } \hskip5pt y \in \{x\in X: f(x)\in B\}$

$\Longleftrightarrow y \in \{x\in X: f(x)\in A\} \cup \{x\in X: f(x)\in B\}.$

So, $\{x\in X: f(x)\in A\cup B\}=\{x\in X: f(x)\in A\} \cup \{x\in X: f(x)\in B\}.$

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  • $\begingroup$ Ok, that's a nice approach! I always forget about it. Thanks, Twink! $\endgroup$ – Igor Sep 29 '13 at 18:39

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