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How do you show that there is always a straight line bisecting a finite set of points on a plane?
(1) for all lines l in L, p is not on l is nonempty
I'm not sure what this means.
First, what do you mean by "bisecting a finite set of points"? In particular, if the line passes through some of the points and has the same number of points on each side, is that "bisecting"?
Yeah, that sentence should have read:
"Then L is finite, so the set T consisting of all points p such that (1) for all lines l in L, p is not on l, and (2) p is below and to the left of B (that is, [tex] p \in (-\infty, -a] \times (-\infty, -a] [/tex] ) is nonempty." In other words,
[tex] \{p \; : \; \ell \in L \Rightarrow p \notin \ell \; \textrm{and} \; p \in (-\infty, -a] \times (-\infty, -a] \} \neq \emptyset [/tex].
It's actually a very easy proof; my post above just got a bit bogged down in formalism. The idea is that you select a point p far to the left and below all your points (again, call the set of all your points S), such that no line runs through p and any two points in S, and imagine rotating a line slowly through p. You then keep track of how many times the line has "hit" a point in S. The crucial idea is that you can arrange things so that this count always increases by just one at a time (this is why we chose p so that it's not collinear with any two points in S). Furthermore, at some point, the line has "hit" all the points in S, so the final count is |S|. There is therefore a time at which the line has hit exactly half of the points in S.
I guess that separation lemma could work, but what if there's an odd number of points? Then the bisecting line must pass through an odd number of points. Not sure if that's always possible.