Theorem. Let $f: X \rightarrow Y$ be a function and $S \subseteq X$ s.t.
$$ f \: \text{1-1} \leftrightarrow S = f^{-1}(f(S)). $$ Proof. $$ f \: \text{1-1} \leftrightarrow \underbrace{S = f^{-1}(f(S))}_{s_0} \leftrightarrow ( (\underbrace{f \: \text{1-1} \rightarrow s_0}_{(i)}) \wedge (\underbrace{f \: \text{1-1} \leftarrow s_0}_{(ii)}) ) $$ (i) $$ f \: \text{1-1} \rightarrow ( \underbrace{S \subseteq f^{-1}(f(S))}_{(i.1)} ) \wedge ( \underbrace{S \supseteq f^{-1}(f(S))}_{(i.2)} ). $$ (i.1) (contrapositive) $$ x \notin f^{-1}(f(S)) \leftrightarrow x \notin \{ x \in X : f(x) \in f(S) \} \leftrightarrow (\forall z \in S)f(x) \neq f(z) \xrightarrow{f:X \rightarrow Y} x \notin S. $$ $\Box$

(i.2) (contrapositive): Given $x \notin S $ and $z \in S \xrightarrow{(i.1)} z \in f^{-1}(f(S))$, we have, since $f$ is 1-1, $x \neq z \rightarrow f(x) \neq f(z)$ and thereby $$ x \in \{ x \in X : f(x) \notin f(S) \} \xrightarrow{f:X \rightarrow Y} x \notin \{ x \in X : f(x) \in f(S) \} \leftrightarrow x \notin f^{-1}(f(S)). $$ $\Box$

(ii) Given $ x \notin S $ and $z \in S$. From (i.1) we can tell that $z \in f^{-1}(f(S))$, while (i.2) implies that $ x \notin f^{-1}(f(S)) $.
This is equivalent to saying $f(z) \in f(S)$ and $f(x) \notin f(S)$, or $f(x) \neq f(z)$. Since $S$ is an arbitrary subset of $X$, the case when $S = \{x_0\}$ is included, where $x_0$ is an arbitrary element. This legitimates $$ (\forall x,z \in X) x \neq z \rightarrow f(x) \neq f(z). $$ Which is the definition of $f$ is 1-1. $\Box$

2014 DEC 26 (v1.1)
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