deletions | additions       diff --git a/begin_lem_Two_point_sets__.tex b/begin_lem_Two_point_sets__.tex  index 016fd51..63c63ec 100644  --- a/begin_lem_Two_point_sets__.tex  +++ b/begin_lem_Two_point_sets__.tex 
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\begin{proof}  ($\Rightarrow$) by contradiction: Linear separability implies we can draw a line in the plane such that all $p\in \P$ are on one side, and all $q\in \Q$ the other. Can our convex hulls ever cross this line?  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 elements 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.