# Set Theory Function Proof

• Apr 29th 2010, 02:31 PM
james121515
Set Theory Function Proof
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 $f^{-1}(\overline{f(S)}) \subseteq \overline{S}$
• Apr 29th 2010, 03:12 PM
Plato
Quote:

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 $a\in \overline{S}$.