deletions | additions       diff --git a/subsection_Linear_Separability_begin_lem__.tex b/subsection_Linear_Separability_begin_lem__.tex  deleted file mode 100644  index d5b1bf3..0000000  --- a/subsection_Linear_Separability_begin_lem__.tex  +++ /dev/null 
...
\subsection{Linear Separability}  \begin{lem}  Two point sets in the plane are linearly separable IFF their convex hulls are linearly separable.  \end{lem}  \begin{proof}  ($\Rightarrow$) by contradiction. Linear separability implies we can draw a line in the plane such that all $p\in \P$ fall on one side and all $q\in \Q$ the other. Can our convex hulls ever cross this line?  Assume that $C(\P)$ crosses this line. This means one of the line-segments of $B(\P)$ crosses our line. All such line-segments are defined between two points in $V(\P)\subseteq \P$. 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 one side of a line, $\P \subset C(\P)$ is on this side also.  \end{proof}  \begin{lem}  Two convex hulls are linearly separable IFF they do not intersect.  \end{lem}  \begin{proof}  ($\Rightarrow$) our separation line splits the plane into two disjoint sets $A,B$. Each of these is a superset of one of our convex-hulls. WLOG let $C(\P)\subset A, C(\Q) \subset B$.   \begin{align*}  C(\P) \cap C(\Q) &\subseteq A\cap B \\  &= \emptyset  \end{align*}  RTP: if non-intersecting then linearly separable.   ($\Leftarrow$)Find $u\in B(\P),v\in B(\Q)$ with minimum distance.  \end{proof}  \begin{cor}  From Lemma 1 and 2 it follows immediately that   \end{cor}