# Projective Plane of order n

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.

• 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? – Morgan Rodgers Mar 25 at 23:53
• "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 $|K| \geq 3$) – Morgan Rodgers Mar 25 at 23:56
• @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. – lj_growl Mar 26 at 0:16
• I don't understand your most recent comment at all. Is what $n+1+1$? What does it mean for $P$ to intersect $K$? – Morgan Rodgers Mar 26 at 3:10

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.