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diff --git a/vorlesungen/slides/9/pf/trennung.tex b/vorlesungen/slides/9/pf/trennung.tex
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+%
+% trennung.tex -- slide template
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\bgroup
+\definecolor{darkgreen}{rgb}{0,0.6,0}
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Trennung}
+\vspace{-20pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+\begin{center}
+\begin{tikzpicture}[>=latex,thick]
+
+\coordinate (u) at (3.5,4.5);
+\coordinate (v) at (2.5,2);
+\coordinate (va) at ({(3.5/2.5)*2.5},{(3.5/2.5)*2});
+
+\uncover<3->{
+\fill[color=darkgreen!20] (0,0) rectangle (5.3,5.3);
+\node[color=darkgreen] at (1.5,4.9) {$u\not\ge w$};
+\node[color=darkgreen] at (4.4,0.6) {$u\not\ge w$};
+}
+
+\uncover<5->{
+\begin{scope}
+\clip (0,0) rectangle (5.3,5.3);
+\draw[color=darkgreen] (0,0) -- ($3*(v)$);
+\end{scope}
+
+\node[color=darkgreen] at ($1.2*(va)$)
+	[below,rotate={atan(2/2.5)}] {$(1+\mu)v$};
+}
+
+\uncover<2->{
+	\fill[color=red!20] (0,0) rectangle (u);
+}
+
+\fill[color=red] (u) circle[radius=0.08];
+\node[color=red] at (u) [above right] {$u$};
+
+\uncover<4->{
+	\fill[color=blue!40,opacity=0.5] (0,0) rectangle (v);
+}
+
+\uncover<2->{
+	\fill[color=blue] (v) circle[radius=0.08];
+	\node[color=blue] at (v) [above] {$v$};
+}
+
+\uncover<4->{
+	\draw[color=blue] (0,0) -- (va);
+
+	\fill[color=blue] (va) circle[radius=0.08];
+	\node[color=blue] at (va) [above left] {$(1+\varepsilon)v$};
+}
+
+\draw[->] (-0.1,0) -- (5.5,0) coordinate[label={$x_1$}];
+\draw[->] (0,-0.1) -- (0,5.5) coordinate[label={right:$x_2$}];
+
+\uncover<2->{
+	\draw[->,color=red] (3.0,-0.2) -- (3.0,1.5);
+	\node[color=red] at (3.0,-0.2) [below]
+		{$\{w\in\mathbb{R}^n\;|\; w<u\}$};
+}
+
+\end{tikzpicture}
+\end{center}
+\end{column}
+\begin{column}{0.48\textwidth}
+\begin{block}{Satz}
+$u>v\ge 0$\uncover<4->{, dann gibt es $\varepsilon>0$ mit
+\[
+u\ge (1+\varepsilon)v
+\]}%
+\uncover<5->{und für $\mu>\varepsilon$ ist 
+\[
+u \not\ge (1+\mu)v 
+\]}
+\uncover<6->{%
+\begin{proof}[Beweis]
+\begin{itemize}
+\item<7->
+$u>v$ $\Rightarrow$ $u_i/v_i>1$ falls $v_i>0$
+\item<8->
+\[
+\vartheta = \min_{v_i\ne 0} \frac{u_i}{v_i} > 1
+\]
+\uncover<9->{$\varepsilon = \vartheta - 1$}
+\end{itemize}
+\end{proof}}
+\end{block}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup