Set Theory Function Proof

**james121515** · April 29th 2010, 02:31 PM

Hi guys,

Can you please check this basic set theory/function proof?

$\textbf{Theorem. }f^{-1}((\overline{f(S)}) \subseteq \overline{S}$

$\emph{Proof. }$ Let $a \in f^{-1}(\overline{f(S)})$ . Then $f(a) = b$ for some $b \in \overline{f(S)}$ . So we have that $b \notin f(S)$ . Now $b \notin f(S)$ implies that, for every $k \in A$ such that $f(k) = b$ , $k \notin S$ , in particular, $x$ . Thus $x \notin S$ , hence $x \in \overline{S}$ so that $f^{-1}(\overline{f(S)}) \subseteq \overline{S}$

**Plato** · April 29th 2010, 03:12 PM

Originally Posted by james121515

Can you please check this basic set theory/function proof?
$\textbf{Theorem. }f^{-1}((\overline{f(S)}) \subseteq \overline{S}$
$\emph{Proof. }$ Let $a \in f^{-1}(\overline{f(S)})$ . Then $f(a) = b$ for some $b \in \overline{f(S)}$ . So we have that $b \notin f(S)$ . Now $b \notin f(S)$ implies that, for every $k \in A$ such that $f(k) = b$ , $k \notin S$ , in particular, $x$ . Thus $x \notin S$ , hence $x \in \overline{S}$ so that $f^{-1}(\overline{f(S)}) \subseteq \overline{S}$

I think it can be shorter.
$a \in f^{-1}(\overline{f(S)})$ implies that $f(a)\notin f(S)$
So that $a\notin S$ or $a\in \overline{S}$ .

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