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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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- $\begingroup$ The tag elementary-functions is for questions about elementary functions, not about functions in general. $\endgroup$Martin Sleziak– Martin Sleziak2013-01-24 07:20:15 +00:00Commented Jan 24, 2013 at 7:20
- $\begingroup$ sorry about the duplicate, thanks for the heads up $\endgroup$Peej Gerard– Peej Gerard2013-01-24 17:50:53 +00:00Commented Jan 24, 2013 at 17:50
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2 Answers
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0 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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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