# Blocking set in a projective plane.

Let S be any nontrivial blocking set in a projective plane of order q, such that S not containing any line. Let A be a set of points in the same projective plane of order q, raging over all these S.

Is it true that if A$\cap$S != $\phi$ then |A| >= q+1 and equality exists only if A is a line?

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p contains a line so it cannot be a blocking set. –  jeff Aug 14 '10 at 13:30
I'm sorry, the definition is for ANY blocking set S, therefore you cannot take a concrete one. –  jeff Aug 14 '10 at 13:48
I've mentioned that for each blocking set S that doesn't contain a line, and according to Bruen, |S| >= q + sqrt(q) + 1 so how what singeltons have to do here? So i want A to intersect every blocking set S that doesn't contain a line. –  jeff Aug 14 '10 at 14:13
Then i suppose i'm using the wrong term for blocking set. What i mean is: Set of points such that each line from the projective plane consists at least one of these points. –  jeff Aug 14 '10 at 14:23
So in my question: S = ANY of these sets such that there is no line in them –  jeff Aug 14 '10 at 14:25

Let $\ell_1,\ell_2,\ell_3$ be three lines in the plane that do not all contain the same point. The triangle formed by $\ell_1,\ell_2,\ell_3$ is the set obtained from $\ell_1 \cup \ell_2 \cup \ell_3$ by removing the three pairwise intersections of the lines. Clearly, any triangle is a blocking set: every line in the plane meets each line $\ell_1,\ell_2,\ell_3$, and cannot contain all the three points that we are removing from the triangle. Triangles are the only blocking sets we consider.

We say that $A$ has property (*) if the intersection of $A$ with every triangle is not empty. Note that if a subset of the plane intersects non-trivially every blocking set, then it has property (*). We prove the (a priori) stronger statement that a set with property (*) and at most $q+1$ elements consists of all the points contained in a line.

Lemma 1. Let $A$ be a set of at least three points in the plane such that any line intersects $A$ in at most two points; then $A$ cannot have property (*).

Proof. Let $A' \subset A$ be a subset with three elements, and note that the points in $A'$ are not collinear. Let $\ell_1,\ell_2,\ell_3$ be the three lines each containing two of the points of $A'$. By assumption $\ell_1 \cap A , \ell_2 \cap A , \ell_3 \cap A \subset A'$; thus, the triangle formed by $\ell_1 , \ell_2 , \ell_3$ is a blocking set disjoint from $A$. $\square$

Lemma 2. Let $A$ be a set of at most $q+1$ points in the plane that are not collinear and let $\ell$ be a line in the plane such that $|A \cap \ell| \geq 3$; then $A$ cannot have property (*).

Proof. Let $a$ be a point of $\ell \setminus A$: such a point exists, since the points of $A$ are not collinear and $\ell$ contains $q+1$ points. There are $q$ lines different from $\ell$ through $a$ and at most $q-2$ points in $A \setminus \ell$; thus there are two distinct lines $\ell_1 , \ell_2$ through $a$ disjoint from $A$. Let $b'$ be any point of $A$ not contained in $\ell$ and let $b$ be any point not contained in $A$ on a line $\ell'$ joining $b'$ to a point of $A$ contained in $\ell$; there are $q$ lines through $b$ different from $\ell'$ and only $q-1$ points in $A \setminus \ell'$; we deduce that there is a line $\ell_3$ through $b$ disjoint from $A$. Thus the triangle formed by $\ell_1,\ell_2,\ell_3$ is a blocking set disjoint from $A$. $\square$

Corollary. Let $A$ be a set of at most $q+1$ points in the plane having property (*); then $A$ consists of all the points contained in a line.

Proof. By Lemma 1 we know that there must be a line containing three points of $A$. If the points of $A$ were not collinear, then Lemma 2 would imply that $A$ does not have property (*). Thus the points of $A$ are collinear. To conclude it suffices to show that if $A$ misses a point of the line containing $A$, then $A$ does not have property (*). This is easy: let $a$ be a point of $A$ and let $\ell_1,\ell_2$ be distinct lines through $a$; let also $b$ be a point on the line containing $A$ not in $A$, and let $\ell_3$ be any line through $b$ not containing $A$. The triangle formed by $\ell_1,\ell_2,\ell_3$ is a blocking set disjoint from $A$. $\square$

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Looks nice, although i'm not sure why you are using a conics in your proof, Can't you just define a set of points in the projective plane? BTW, is it possible to conclude from this about equality? thanks. –  jeff Aug 14 '10 at 19:30
The version above is completely different and should answer the full question. –  damiano Aug 14 '10 at 22:52
The case $q=2$ requires a little extra care, but is not difficult. The reason the above argument does not apply is that triangles contain lines in the case $q=2$. –  damiano Aug 19 '10 at 11:11