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Let $P=(S,L)$ be a projective plane of order $n$. Let $K$ be a nonempty subset of $S$ with the property that no three points belonging to $K$ are collinear. Prove that $|K|$ is less than or equal to $n+2$.

I began by staying that since $P$ is of order $n$, there are exactly $n^2+n+1$ points in $S$ and since $K$ has the stipulation that no three points are collinear, the points in $K$ must be at least on two lines. Now here is where I'm stuck. I don't know how to go from there to less than or equal to $n+2$. Any help would be appreciated.

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  • $\begingroup$ Each line of $P$ meets $K$ in either $0$, $1$, or $2$ points. There are a total of $n^{2}+n+1$ lines. Each point is on $n+1$ lines. Does this help you count? $\endgroup$ – Morgan Rodgers Mar 25 at 23:53
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    $\begingroup$ "the points in $K$ must be at least on two lines": what would this mean? That at least two lines intersect $K$? (note that this would be true as long as $|K|\geq1$) That you need at least two lines to cover all the points of $K$? (Note that this is only true if $|K| \geq 3$) $\endgroup$ – Morgan Rodgers Mar 25 at 23:56
  • $\begingroup$ @MorganRodgers So, is it n+1+1? So, n+2? Because if P intersects K at two points, you would have n+1 points plus one more? I don't know that I'm fully understanding the count. $\endgroup$ – lj_growl Mar 26 at 0:16
  • $\begingroup$ I don't understand your most recent comment at all. Is what $n+1+1$? What does it mean for $P$ to intersect $K$? $\endgroup$ – Morgan Rodgers Mar 26 at 3:10
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Suppose $K$ is a set of $n+2$ points, no three of which are collinear. Let $x$ be any point which is not in $K$. Every point in $K$ is on some line passing through $x$. But there are only $n+1$ lines passing through $x$, so two points of $K$ must be on the same line through $x$. Thus $K\cup\{x\}$ does contain three collinear points. Since $x$ was an arbitrary point not in $K$, this shows that $K$ cannot be enlarged beyond $n+2$ points.

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