The de Bruijn-Erd\"os theorem in incidence geometry via Ph. Hall's marriage theorem

TL;DR

This paper presents a novel proof of the de Bruijn-Erd"os theorem in incidence geometry utilizing Ph. Hall's marriage theorem, offering a new perspective on the theorem's foundational principles.

Contribution

The paper introduces a proof of the de Bruijn-Erd"os theorem based on Hall's marriage theorem, connecting incidence geometry with combinatorial matching theory.

Findings

Proof of de Bruijn-Erd"os theorem using Hall's marriage theorem

Establishes a link between incidence geometry and combinatorial matching

Provides a new methodological approach to the theorem

Abstract

The paper is devoted to a proof of the de Bruijn-Erd\"os theorem in incidence geometry based on the Ph. Hall's marriage theorem (the theorem about the systems of distinct representatives).

Equations

$$\sum _{a\in P}{k}_a=\sum _{l\in \mathcal{L}}{s}_l.$$

$$s_l⩽k_a.$$

$${\bigcup }_{i\in J}{\phi }_i$$

$$\sum _{l\in \mathcal{L}}{s}_l⩽\sum _{l\in \mathcal{L}}{k}_{f(l)}⩽\sum _{a\in P}{k}_a.$$

$$⩾2+(s_l-1)(s_{l^{\prime }}-1)⩾2+2(s_l-1)=2s_l⩾s_l+s_{l^{\prime }}=n+1,$$

Full text

The  de  Bruijn–Erdös  theorem  in  incidence  geometry  via  Ph.  Hall’s  marriage  theorem

Nikolai  V.  Ivanov

††footnotetext: © Nikolai  V.  Ivanov,   2017.   Neither the work reported in this paper,  nor its preparation were supported by any governmental or non-governmental agency,  foundation,  or institution.   ††footnotetext: The author is grateful to F.  Petrov and  M.  Prokhorova for stimulating correspondence.

Theorem  (de Bruijn–Erdös).   Let  $P$  be a finite set consisting of $n$ elements,   and let $\mathcal{L}$ be a collection of $m$ proper subsets of  $P$.   Suppose that each  $l\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{L}$  contains at least two elements and for every  $x\hskip 0.50003pt,\hskip 3.00003pty\hskip 1.99997pt\in\hskip 1.99997ptP$,  $x\hskip 1.99997pt\neq\hskip 1.99997pty$  there exists exactly one subset  $l\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{L}$  such that $x\hskip 0.50003pt,\hskip 3.00003pty\hskip 1.99997pt\in\hskip 1.99997ptl$.   Then  $m\hskip 1.99997pt\geqslant\hskip 1.99997ptn$,   and  $m\hskip 1.99997pt=\hskip 1.99997ptn$  if and only if one of the following two cases occurs:

( i )i  *there exists  $z\hskip 1.99997pt\in\hskip 1.99997ptP$  such that $\mathcal{L}$ consists of  $P\hskip 1.00006pt\smallsetminus\{\hskip 1.00006ptz\hskip 1.00006pt\}$  and all the pairs  $\{\hskip 1.00006ptx\hskip 0.50003pt,\hskip 3.00003ptz\hskip 1.00006pt\}$  with  $x\hskip 1.99997pt\neq\hskip 1.99997ptz$;   *

( ii )  there exists a natural number $k$ such that $n\hskip 1.99997pt=\hskip 1.99997ptk\hskip 1.00006pt(k\hskip 1.99997pt-\hskip 1.99997pt1)\hskip 1.99997pt+\hskip 1.99997pt1$,  every element of  $\mathcal{L}$  is a set consisting of $k$ elements,   every element of  $P$  belongs to exactly $k$ elements of  $\mathcal{L}$,   and the intersection of every two elements of  $\mathcal{L}$  is non-empty.

The elements of $P$ are called points  and the elements of $\mathcal{L}$ lines.  For a point $a$ let $k_{\hskip 0.35002pta}$ be the number of lines containing $a$,  and for a line $l$  let  $s_{\hskip 0.35002ptl}$  be the number of points in $l$,  i.e. the cardinality of the set $l$.  Following de Bruijn–Erdös  [dB-E],  we start with two observations.  First ,  counting the pairs $(a\hskip 0.50003pt,\hskip 3.00003ptl\hskip 1.00006pt)\hskip 1.99997pt\in\hskip 1.99997ptP\times\mathcal{L}$  such that  $a\hskip 1.99997pt\in\hskip 1.99997ptl$  in two ways leads to the equality

[TABLE]

Second,  if $(a\hskip 0.50003pt,\hskip 3.00003ptl\hskip 1.00006pt)\hskip 1.99997pt\in\hskip 1.99997ptP\times\mathcal{L}$  and  $a\hskip 1.99997pt\not\in\hskip 1.99997ptl$,  then for every $x\hskip 1.99997pt\in\hskip 1.99997ptl$  there is a line containing $\{\hskip 0.50003pta\hskip 0.50003pt,\hskip 3.00003ptx\hskip 1.00006pt\}$ and these lines are distinct .   It follows that in this case

[TABLE]

Systems of distinct representatives.   A  (finite)  family of subsets of a set $S$ is a map  $\varphi\colon I\longrightarrow 2^{S}$ from a finite set $I$ to the power set $2^{S}$.  Usually $\varphi\hskip 0.50003pt(i)$ is denoted by $\varphi_{i}$.  A  system of distinct representatives  for $\varphi$ is an  injective  map  $f\colon I\longrightarrow S$  such that  $f(i)\hskip 1.99997pt\in\hskip 1.99997pt\varphi_{i}$  for all  $i\hskip 1.99997pt\in\hskip 1.99997ptI$.  We will denote by $|\hskip 1.00006ptX\hskip 1.00006pt|$ the number of elements of a set $X$.

Theorem  (Ph.  Hall).   A system of distinct representatives for $\varphi$ exists if and only if  the union

[TABLE]

*contains  $\geqslant\hskip 1.99997pt|\hskip 1.00006ptJ\hskip 1.00006pt|$  elements for every subset  $J\hskip 1.99997pt\subset\hskip 1.99997ptI$.   *

Corollary.   Suppose that for every non-empty subset  $J\hskip 1.99997pt\subset\hskip 1.99997ptI$  not equal to $I$ itself the union  (3)  contains  $>\hskip 1.99997pt|\hskip 1.00006ptJ\hskip 1.00006pt|$  elements.  Then for every  $j\hskip 1.99997pt\in\hskip 1.99997ptI$  and every element $a\hskip 1.99997pt\in\hskip 1.99997pt\varphi_{j}$ there exists a system of distinct representatives $f\colon I\longrightarrow S$ for $\varphi$ such that  $f(j)\hskip 3.99994pt=\hskip 3.99994pta$.

Proof**.**     Let  $K\hskip 1.99997pt=\hskip 1.99997ptI\hskip 1.00006pt\smallsetminus\hskip 1.00006pt\{\hskip 1.00006ptj\hskip 1.00006pt\}$,  and let  $\psi_{i}\hskip 3.99994pt=\hskip 3.99994pt\varphi_{i}\hskip 1.00006pt\smallsetminus\hskip 1.00006pt\{\hskip 1.00006pta\hskip 1.00006pt\}$  for every  $i\hskip 1.99997pt\in\hskip 1.99997ptK$.  Then $\psi$ is a family of subsets of $S$.  The theorem applies to $\psi$ and implies that there exists a system of distinct representatives  $g\colon K\longrightarrow S$  for $\psi$.   The map  $f\colon I\longrightarrow S$  defined by  $f(i)\hskip 3.99994pt=\hskip 3.99994ptg(i)$  for  $i\hskip 1.99997pt\in\hskip 1.99997ptK$  and  $f(j)\hskip 3.99994pt=\hskip 3.99994pta$  is the desired system of distinct representatives for $\varphi$.    $\blacksquare$

In fact ,  a standard proof  of  the  Ph.  Hall’s  theorem using an induction by the number of elements of $I$ includes a proof of this Corollary.  See,  for example,  the textbook  [C]  by  P.  Cameron.

The family of complements to lines.   Let  $I\hskip 1.99997pt=\hskip 1.99997pt\mathcal{L}$  and  $\varphi_{l}\hskip 1.99997pt=\hskip 1.99997ptP\hskip 1.00006pt\smallsetminus\hskip 1.00006ptl$  for  $l\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{L}$.   Then $\varphi$ is family of subsets of $P$.   If  $l\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{L}$,   then  $l\hskip 1.99997pt\neq\hskip 1.99997ptP$  and hence $\varphi_{l}\hskip 1.99997pt\neq\hskip 1.99997pt\emptyset$.   If  $l\hskip 0.50003pt,\hskip 3.00003ptl^{\prime}\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{L}$  and  $l\hskip 1.99997pt\neq\hskip 1.99997ptl^{\prime}$,   then  $l\hskip 1.00006pt\cap\hskip 1.00006ptl^{\prime}$ contains no more than one point and hence  $\varphi_{l}\hskip 1.00006pt\cup\hskip 1.00006pt\varphi_{l^{\prime}}$  is equal either to $P$,   or to $P$ with one point removed.   It follows that the same is true for the union  (3)  if  $J$  contains at least two lines,   and hence in this case  (3)  contains  $\geqslant\hskip 1.99997ptn\hskip 1.99997pt-\hskip 1.99997pt1$  points.

Let $l$ be a line and let  $x\hskip 0.50003pt,\hskip 3.00003pty\hskip 1.99997pt\in\hskip 1.99997ptl$  and  $x\hskip 1.99997pt\neq\hskip 1.99997pty$.  Let $a$ be a point not in $l$.  Then the two lines $l_{x}\hskip 0.50003pt,\hskip 3.00003ptl_{y}$ containing $a$ and $x\hskip 0.50003pt,\hskip 3.00003pty$ respectively intersect only in $a$.  It follows that the intersection of three lines $l\hskip 0.50003pt,\hskip 3.00003ptl_{x}\hskip 0.50003pt,\hskip 3.00003ptl_{y}$ is empty,  and hence the union of their complements is equal to $P$.  It follows that for  $J\hskip 1.99997pt=\hskip 1.99997ptI\hskip 1.99997pt=\hskip 1.99997pt\mathcal{L}$  the union  (3)  is equal to $P$ and hence contains $n$ elements.

Proof of the inequality  $m\hskip 1.99997pt\geqslant\hskip 1.99997ptn$****.    Suppose that $m\hskip 1.99997pt\leqslant\hskip 1.99997ptn$.   Let  $J\hskip 1.99997pt\subset\hskip 1.99997pt\mathcal{L}$.   If  $|\hskip 1.00006ptJ\hskip 1.00006pt|\hskip 1.99997pt=\hskip 1.99997pt1$,  then  (3)  is the complement of a line and hence contains  $\geqslant\hskip 1.99997pt1$  points.  If  $2\hskip 1.99997pt\leqslant\hskip 1.99997pt|\hskip 1.00006ptJ\hskip 1.00006pt|\hskip 1.99997pt\leqslant\hskip 1.99997ptm\hskip 1.99997pt-\hskip 1.99997pt1$,  then  (3)  contains  $\geqslant\hskip 1.99997ptn\hskip 1.99997pt-\hskip 1.99997pt1\hskip 1.99997pt\geqslant\hskip 1.99997ptm\hskip 1.99997pt-\hskip 1.99997pt1$  points,  and hence contains  $\geqslant\hskip 1.99997pt|\hskip 1.00006ptJ\hskip 1.00006pt|$  elements.  If  $|\hskip 1.00006ptJ\hskip 1.00006pt|\hskip 1.99997pt=\hskip 1.99997ptm$,  then  (3)  contains  $n\hskip 1.99997pt\geqslant\hskip 1.99997ptm\hskip 1.99997pt=\hskip 1.99997pt|\hskip 1.00006ptJ\hskip 1.00006pt|$  elements.  Therefore,   by  Hall’s theorem there exists a system of distinct representatives for $\varphi$.  Such a system of distinct representatives is an injective map  $f\colon\mathcal{L}\longrightarrow P$  such that  $f(l\hskip 0.50003pt)\hskip 3.99994pt\not\in\hskip 3.99994ptl$  for every  $l\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{L}$.  By the inequality  (2),  this implies that  $\displaystyle s_{l}\hskip 1.99997pt\leqslant\hskip 1.99997ptk_{\hskip 0.35002ptf(l\hskip 0.35002pt)}$  for every  $l\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{L}$.  By summing all these inequalities and using the injectivity of $f$ we see that

[TABLE]

Moreover ,  the second inequality is strict unless  $m\hskip 1.99997pt=\hskip 1.99997ptn$  (otherwise the last sum has an additional summand  $k_{a}\hskip 1.99997pt>\hskip 1.99997pt0$  compared to the previous one).  But  (1)  implies that both inequalities in  (4)  should be actually equalities.   It follows that  $m\hskip 1.99997pt=\hskip 1.99997ptn$.   $\blacksquare$

**The case  $m\hskip 1.99997pt=\hskip 1.99997ptn$.   I **.    Suppose that there is a point  $a\hskip 1.99997pt\in\hskip 1.99997ptP$  contained in  $m\hskip 1.99997pt-\hskip 1.99997pt1$  lines.  Each of these lines contains at least one point in addition to $a$.   Since  $m\hskip 1.99997pt=\hskip 1.99997ptn$,  there are no other points.  On the other hand,  there is only one more line.  This line should contain all the points different from $a$.  It follows that we are in the case  ( i )  of the de Bruijn-Erdös theorem.   $\square$

**The case  $m\hskip 1.99997pt=\hskip 1.99997ptn$.   II **.    Suppose that  $P\hskip 1.99997pt=\hskip 1.99997ptl\hskip 1.00006pt\cup\hskip 1.00006ptl^{\prime}$  for some  $l\hskip 0.50003pt,\hskip 3.00003ptl^{\prime}\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{L}$  and  $l\hskip 1.00006pt\cap\hskip 1.00006ptl^{\prime}\hskip 1.99997pt\neq\hskip 1.99997pt\emptyset$.   Then the intersection  $l\hskip 1.00006pt\cap\hskip 1.00006ptl^{\prime}$  consists of exactly one point ,   say,  the point $z$.  It follows that the number of points  $n\hskip 3.99994pt=\hskip 3.99994pts_{l}\hskip 1.99997pt+\hskip 1.99997pts_{l^{\prime}}\hskip 1.99997pt-\hskip 1.99997pt1$.  Let us count the lines.

In addition to the lines  $l\hskip 0.50003pt,\hskip 3.00003ptl^{\prime}$,   for every two points  $b\hskip 1.99997pt\in\hskip 1.99997ptl\hskip 1.99997pt\smallsetminus\hskip 1.00006pt\ \{\hskip 1.00006ptz\hskip 1.00006pt\}$  and  $b^{\prime}\hskip 1.99997pt\in\hskip 1.99997ptl^{\prime}\hskip 1.99997pt\smallsetminus\hskip 1.00006pt\ \{\hskip 1.00006ptz\hskip 1.00006pt\}$  there is a line containing  $b\hskip 0.50003pt,\hskip 3.00003ptb^{\prime}$.   All these lines are distinct ,  and hence there are  $\geqslant\hskip 1.99997pt2\hskip 1.99997pt+\hskip 1.99997pt(s_{l}\hskip 1.99997pt-\hskip 1.99997pt1)(s_{l^{\prime}}\hskip 1.99997pt-\hskip 1.99997pt1)$  lines.   We may assume that  $s_{l}\hskip 1.99997pt\geqslant\hskip 1.99997pts_{l^{\prime}}$.   If  $s_{l^{\prime}}\hskip 1.99997pt\geqslant\hskip 1.99997pt3$,   then the number $m$ of  lines  is

[TABLE]

contrary to the assumption  $m\hskip 1.99997pt=\hskip 1.99997ptn$.  It follows that  $s_{l^{\prime}}\hskip 1.99997pt=\hskip 1.99997pt2$  and hence all points,  except the only point in  $l^{\prime}\hskip 1.00006pt\smallsetminus\hskip 1.00006ptl$,  belong to the line $l$.   Let $a$ be this point ,   so  $\{\hskip 0.50003pta\hskip 1.49994pt\}\hskip 1.99997pt=\hskip 1.99997ptl^{\prime}\hskip 1.00006pt\smallsetminus\hskip 1.00006ptl$.   In this case  $s_{l}\hskip 3.99994pt=\hskip 3.99994ptn\hskip 1.99997pt-\hskip 1.99997pt1\hskip 3.99994pt=\hskip 3.99994ptm\hskip 1.99997pt-\hskip 1.99997pt1$  and $a$ belongs to  $\geqslant\hskip 3.99994pts_{l}\hskip 3.99994pt=\hskip 3.99994ptm\hskip 1.99997pt-\hskip 1.99997pt1$  lines.   Therefore,  we are in the situation of the case  I  again,  and hence in the case  ( i )  of the de Bruijn-Erdös theorem.   $\square$

**The case  $m\hskip 1.99997pt=\hskip 1.99997ptn$.   III **.    Suppose that we are neither in the case  I,  nor in the case  II.   In particular ,  no point is contained in  $m\hskip 1.99997pt-\hskip 1.99997pt1$  lines.  In this case,  if  $J$  is a set of  $m\hskip 1.99997pt-\hskip 1.99997pt1$ lines,  then  (3)  contains  $n\hskip 1.99997pt>\hskip 1.99997ptm\hskip 1.99997pt-\hskip 1.99997pt1$  points.  If  $J$  is a set of  $p$  lines and  $p\hskip 1.99997pt<\hskip 1.99997ptm\hskip 1.99997pt-\hskip 1.99997pt1$,  then  (3)  contains  $\geqslant\hskip 1.99997ptn\hskip 1.99997pt-\hskip 1.99997pt1\hskip 1.99997pt=\hskip 1.99997ptm\hskip 1.99997pt-\hskip 1.99997pt1\hskip 1.99997pt>\hskip 1.99997ptp$  points.  If  $p\hskip 1.99997pt=\hskip 1.99997pt1$  and  (3)  contains only $1$ point ,  then this point is contained in  $m\hskip 1.99997pt-\hskip 1.99997pt1$  lines and we are in the case  I,  contrary to the assumption.  Therefore,  if  $p\hskip 1.99997pt=\hskip 1.99997pt1$,  then  (3)  contains  $\geqslant\hskip 1.99997pt2\hskip 1.99997pt>\hskip 1.99997ptp$  points.  It follows that in this case the Corollary applies.

Let $f$ be a system of distinct representatives for $\varphi$.  Since $f$ is injective and  $m\hskip 1.99997pt=\hskip 1.99997ptn$,  the map $f$ is actually a bijection.  Therefore,  the inequalities  $\displaystyle s_{l}\hskip 3.99994pt\leqslant\hskip 3.99994ptk_{\hskip 0.35002ptf(l\hskip 0.35002pt)}$  together with  (1)  imply that  $s_{l}\hskip 3.99994pt=\hskip 3.99994ptk_{\hskip 0.35002ptf(l\hskip 0.35002pt)}$  for every  $l\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{L}$.   By the Corollary,  given a line $l$ and a point  $a\hskip 1.99997pt\not\in\hskip 1.99997ptl$,  one can choose $f$ is such a way that  $f(l\hskip 0.50003pt)\hskip 1.99997pt=\hskip 1.99997pta$.   It follows that  $s_{l}\hskip 3.99994pt=\hskip 3.99994ptk_{a}$  if  $a\hskip 1.99997pt\not\in\hskip 1.99997ptl$.   This implies,  in particular ,  that  if  $a\hskip 1.99997pt\not\in\hskip 1.99997ptl$,  then every line containing $a$ intersects $l$.   In turn,  this implies that every two lines intersect .

Suppose that  $l\hskip 0.50003pt,\hskip 3.00003ptl^{\prime}\hskip 1.99997pt\in\hskip 1.99997pt\mathcal{L}$  and  $l\hskip 1.99997pt\neq\hskip 1.99997ptl^{\prime}$.   By the previous paragraph  $l\hskip 1.00006pt\cap\hskip 1.00006ptl^{\prime}\hskip 1.99997pt\neq\hskip 1.99997pt\emptyset$  and hence if  $P\hskip 1.99997pt=\hskip 1.99997ptl\hskip 1.00006pt\cup\hskip 1.00006ptl^{\prime}$,  then we are in the case  II,  contrary to the assumption.   Therefore  $P\hskip 1.99997pt\neq\hskip 1.99997ptl\hskip 1.00006pt\cup\hskip 1.00006ptl^{\prime}$.   If  $a\hskip 1.99997pt\not\in\hskip 1.99997ptl\hskip 1.00006pt\cup\hskip 1.00006ptl^{\prime}$,   then  $s_{l}\hskip 1.99997pt=\hskip 1.99997ptk_{a}\hskip 1.99997pt=\hskip 1.99997pts_{l^{\prime}}$.   It follows that all the numbers  $s_{l}\hskip 0.50003pt,\hskip 3.00003ptk_{a}$  are equal.  Let $k$ be the common value of these numbers  $s_{l}\hskip 0.50003pt,\hskip 3.00003ptk_{a}$.   Let  $b\hskip 1.99997pt\in\hskip 1.99997ptP$.   There are $k$ lines containing $b$ and each of the  $n\hskip 1.99997pt-\hskip 1.99997pt1$  points not equal to $b$ is contained in one and only one of these lines.   Moreover ,  each of these lines consists of $b$ and exactly  $k\hskip 1.99997pt-\hskip 1.99997pt1$  points  not equal to $b$.   It follows that  $n\hskip 3.99994pt=\hskip 3.99994ptk\hskip 1.00006pt(k\hskip 1.99997pt-\hskip 1.99997pt1)\hskip 1.99997pt+\hskip 1.99997pt1$.   Since by the previous paragraph every two lines intersect ,  we are in the case  ( ii )  of the de Bruijn-Erdös theorem.    $\square$ $\blacksquare$