# 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?

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

$\displaystyle \emph{Proof. }$ Let $\displaystyle a \in f^{-1}(\overline{f(S)})$. Then $\displaystyle f(a) = b$ for some $\displaystyle b \in \overline{f(S)}$. So we have that $\displaystyle b \notin f(S)$. Now $\displaystyle b \notin f(S)$ implies that, for every $\displaystyle k \in A$ such that $\displaystyle f(k) = b$, $\displaystyle k \notin S$, in particular, $\displaystyle x$. Thus $\displaystyle x \notin S$, hence $\displaystyle x \in \overline{S}$ so that $\displaystyle f^{-1}(\overline{f(S)}) \subseteq \overline{S}$
• Apr 29th 2010, 03:12 PM
Plato
Quote:

Originally Posted by james121515
Can you please check this basic set theory/function proof?
$\displaystyle \textbf{Theorem. }f^{-1}((\overline{f(S)}) \subseteq \overline{S}$
$\displaystyle \emph{Proof. }$ Let $\displaystyle a \in f^{-1}(\overline{f(S)})$. Then $\displaystyle f(a) = b$ for some $\displaystyle b \in \overline{f(S)}$. So we have that $\displaystyle b \notin f(S)$. Now $\displaystyle b \notin f(S)$ implies that, for every $\displaystyle k \in A$ such that $\displaystyle f(k) = b$, $\displaystyle k \notin S$, in particular, $\displaystyle x$. Thus $\displaystyle x \notin S$, hence $\displaystyle x \in \overline{S}$ so that $\displaystyle f^{-1}(\overline{f(S)}) \subseteq \overline{S}$

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