deletions | additions       diff --git a/begin_lem_Two_point_sets__.tex b/begin_lem_Two_point_sets__.tex  index 605daa8..618776d 100644  --- a/begin_lem_Two_point_sets__.tex  +++ b/begin_lem_Two_point_sets__.tex 
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WLOG let's assume that $C(\P)$ crosses this line. This means one of the line-segments of $C(\P)$'s boundary crosses our line. All such line-segments are defined between two points of $C(\P)$. As such we have an element of $\P$ on either side of our separability line, a contradiction.  ($\Leftarrow$) follows immediately from the fact that $\P,\Q$ are subsets of their respective convex-hulls. If all of $C(\P)$ is on the side of a line, $\P \subset C(\P)$ is on the same side.   \end{proof}  \end{proof} \begin{lem}  Two convex hulls are linearly separable implies they do not intersect.  \end{lem}