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Prove $f(S \cup T) = f(S) \cup f(T)$

I'm revisiting set theory and am troubled by this question.

Let $f:A \rightarrow B$, and $C \subset A$, $D \subset A$.

Prove that $f(C \cup D) = f(C) \cup f(D)$.

Any thoughts?

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marked as duplicate by Martin Sleziak, Stefan Hansen, Fabian, Arthur Fischer, Old John Jan 24 '13 at 10:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
The tag elementary-functions is for questions about elementary functions, not about functions in general. –  Martin Sleziak Jan 24 '13 at 7:20
    
sorry about the duplicate, thanks for the heads up –  Peej Gerard Jan 24 '13 at 17:50
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2 Answers 2

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I'll show $\subseteq$. Let $y\in f(C\cup D)$. Then there exists an $x\in C\cup D$ such that $f(x)=y$. This means $x\in C$ or $x\in D$, hence $f(x)\in f(C)$ or $f(x)\in f(D)$. This implies $f(x)\in f(C)\cup f(D)$ and we've established $$f(C\cup D)\subseteq f(C)\cup f(D).$$ Approach the other containment in a similar manner.

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thank you very much –  Peej Gerard Jan 24 '13 at 17:50
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if $f : A \to B$ and $E \subset A$ then by definition $\{ f(e) \in B | e \in E \}$.

So you want to prove

$$\{ f(e) \in B | e \in C \cap D \} = \{ f(e) \in B | e \in C \} \cap \{ f(e) \in B | e \in D \}$$

so

$$\{ f(e) \in B | e \in C \cap D \} = \{ f(e) \in B | e \in C \& e \in D \}$$

done

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