De Bruijn–Erdős theorem

finite plane

De Bruijn–Erdős theorem

Let $S$ be a set of $n\ne 0$ ‘points’, say $\{L_1,L_2,\mathrm{\dots }{L}_n\}$, and let ${\mathrm{\Lambda }}_1$, ${\mathrm{\Lambda }}_2$, … ${\mathrm{\Lambda }}_{\nu }$ be $\nu$ different subsets of $S$ of which any two have exactly one point in common. Now the De Bruijn–Erdős theorem says that $\nu ⩽n$, and that if $\nu =n$ then (up to renumbering) at least one of the following three must be true:

• ${\mathrm{\Lambda }}_i=\{L_i,L_n\}$ for $i\ne n$, and ${\mathrm{\Lambda }}_n=\{L_n\}$

• ${\mathrm{\Lambda }}_i=\{L_i,L_n\}$ for $i\ne n$, and ${\mathrm{\Lambda }}_n=\{L_1,L_2,\mathrm{\dots }{L}_{n-1}\}$

• $n$ is of the form $q^2+q+1$, each ${\mathrm{\Lambda }}_i$ contains $q+1$ points, and each point is contained in $q+1$ of the ${\mathrm{\Lambda }}_i$

and we recognise the last case as http://planetmath.org/node/6943finite projective planes of order $q$. For $n=1$ ($q=0$) the first and last case overlap, and for $n=3$ ($q=1$) the last two cases do. The second case is also known as a near-pencil. The last two cases together are examples of linear spaces.

To exclude the first two cases one usually defines projective planes to satisfy some non-triviality conditions; unfortunately that also excludes projective planes of order 0 and 1.

The formulation above was found in the literature [Cam94]. Naturally, if the $L_i$ are points we tend to interpret the subsets ${\mathrm{\Lambda }}_{\iota }$ as lines. However, interpreted in this way the condition that every two $\mathrm{\Lambda }$ share an $L$ says rather more than is needed for the structure to be a finite plane (http://planetmath.org/node/3509linear space) as it insists that no two lines are parallel. The absence of the dual condition, for two $L$ to share a $\mathrm{\Lambda }$, actually means we have too little for the structure to be a finite plane (linear space), as for two points we may not have a line through them. And indeed, while the second and third cases are finite planes without parallel lines, the first case is not a plane.

Erdős–De Bruijn theorem

A dual formulation (which we could whimsically call the Erdős–De Bruijn Theorem) remedies both under- and over-specification for a plane. Indeed, the conditions are now exactly tailored to make the structure a finite plane (with some parallel lines in the first of the three cases).

Let $\mathrm{\Sigma }$ be a set of $\nu \ne 0$ ‘points’, say $\{{\mathrm{\Lambda }}_1,{\mathrm{\Lambda }}_2,\mathrm{\dots }{\mathrm{\Lambda }}_{\nu }\}$, and let $L_1$, $L_2$, … $L_n$ be $n$ subsets of $\mathrm{\Sigma }$ (‘lines’) such that for any two points there is a unique $L_i$ that contains them both. We must now be careful about the points (the former lines) being “different” (this was easier to formulate in the previous version, in which form it was a simple incidence structure): this condition now takes the form that no two points must have the same collection of lines that are incident with them. Then $n⩾\nu$, and if $n=\nu$ then (up to renumbering) at least one of the following three must be true:

• $L_i=\{{\mathrm{\Lambda }}_i\}$ for $i\ne n$, and $L_n=\{{\mathrm{\Lambda }}_1,{\mathrm{\Lambda }}_2,\mathrm{\dots }{\mathrm{\Lambda }}_n\}$

• $L_i=\{{\mathrm{\Lambda }}_i,{\mathrm{\Lambda }}_n\}$ for $i\ne n$, and $L_n=\{{\mathrm{\Lambda }}_1,{\mathrm{\Lambda }}_2{\mathrm{\Lambda }}_{n-1}\}$

• $n$ is of the form $q^2+q+1$, each $L_i$ contains $q+1$ points, and each point is contained in $q+1$ of the $L_i$

For a proof of the theorem (in the former version) see e.g. Cameron [Cam94].

Title	De Bruijn–Erdős theorem
Canonical name	DeBruijnErdHosTheorem
Date of creation	2013-03-22 15:11:21
Last modified on	2013-03-22 15:11:21
Owner	marijke (8873)
Last modified by	marijke (8873)
Numerical id	7
Author	marijke (8873)
Entry type	Theorem
Classification	msc 05B25
Related topic	FiniteProjectivePlane4
Related topic	IncidenceStructures
Related topic	Geometry
Related topic	LinearSpace2