Theorem. Given a function $f:X \rightarrow Y, \: A,B \subseteq X.$ The following implication holds:
$$f \: \text{1-1} \rightarrow f(A \cap B) = f(A) \cap f(B).$$ Proof. $$f \: \text{1-1} \rightarrow ( \underbrace{f(A \cap B) \subseteq f(A) \cap f(B)}_{(i)} ) \wedge ( \underbrace{f(A \cap B) \supseteq f(A) \cap f(B)}_{(ii)} )$$ (i) (c-p) $$y \notin f(A) \cap f(B) \leftrightarrow y \notin f(A) \vee y \notin f(B).$$ Since $y$ is not in $f(A)$ it won't live in the subset $f(A \cap B)$ either: $$y \notin f(A) \supseteq f(A\cap B) \rightarrow y \notin f(A \cap B).$$ Same holds for $y \notin f(B)$: $$y \notin f(B) \supseteq f(A\cap B) \rightarrow y \notin f(A \cap B).$$ Putting the bits together: $$y \notin f(A) \cap f(B) \rightarrow y \notin f(A \cap B) \leftrightarrow y \in f(A \cap B) \rightarrow y \in f(A) \cap f(B) \leftrightarrow f(A \cap B) \subseteq f(A) \cap f(B).$$ $\Box$

(ii) $y \in f(A)$ and $y \in f(B)$ implies, since $(\forall x,z \in X) f(x) = f(z) \rightarrow x = z$, that $y \in \{ f(x) : x \in A \wedge x \in B \}$ , or $y \in f(A \cap B)$. In short $$(y \in f(A) \wedge y \in f(B) \rightarrow y \in f(A \cap B)) \leftrightarrow f(A \cap B) \supseteq f(A) \cap f(B).$$

2014 DEC 27 (v1.0)
Contact me: m.herrmann followed by an -at- followed by blaetterundsterne.org.

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