Is every Cartesian biaffine plane affine?

Question

This question concerns the (synthetic) geometry of linear spaces.

Definition 1. A linear space is a pair $(P,\mathcal L)$ consisting of a set $P$ whose elements are called points and a family $\mathcal L$ of subsets of $P$ called lines such that the following axioms hold:

$\bullet$ every line $L\in \mathcal L$ contains at least three distinct points;

$\bullet$ for every distinct points $x,y\in P$ there exists a unique line $L\in\mathcal L$ containing those points;

$\bullet$ the set $P$ is not a line.

Two lines $L,\Lambda\in\mathcal L$ are called concurrent if $L\cap\Lambda$ is a singleton.

Definition 2. A linear space $(P,\mathcal L)$ is called a plane if there exist two concurrent lines $X,Y\in\mathcal L$ such that for every point $x\in P$ there exists a line $Z\in\mathcal L$ such that $x\in Z$ and $X\cap Z\ne\varnothing\ne Z\cap Y$.

Definition 3. A plane $(P,\mathcal L)$ is called

$\bullet$ projective if every distinct lines $L,\Lambda\in\mathcal L$ are concurrent;

$\bullet$ affine if for every line $L\in\mathcal L$ and point $p\in P\setminus L$, there exists a unique line $\Lambda\in\mathcal L$ such that $p\in \Lambda\subseteq P\setminus L$;

$\bullet$ semiaffine if for every line $L\in\mathcal L$ and point $p\in P\setminus L$, there exists at most one line $\Lambda\in\mathcal L$ such that $p\in \Lambda\subseteq P\setminus L$;

$\bullet$ a biaffine if for every disjoint lines $L,L'\in\mathcal L$ and point $p\in P\setminus L$, there exists a unique line $\Lambda\in\mathcal L$ such that $p\in \Lambda\subseteq P\setminus L$;

$\bullet$ Cartesian if there exists two concurrent lines $X,Y\in\mathcal L$ such that for every point $p\in P\setminus(X\cup Y)$ there exist unique lines $H,V\in\mathcal L$ such that $H\cap X=\varnothing\ne H\cap Y$, $V\cap X\ne\varnothing=V\cap Y$, and $p\in H\cap V$.

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It is easy to see that

Cartesian $\Leftarrow$ affine $\Rightarrow$ biaffine $\Rightarrow$ semiaffine $\Leftarrow$ projective

and none of these implications can be reversed.

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By the Kuiper-Dembowski Theorem (1962), every finite semiaffine plane is one of the following:

(i) a projective plane;

(ii) an affine plane;

(iii) a projective plane with a removed point;

(iv) an affine plane with an attached point at infinity.

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This classification of finite semiaffine planes implies the following characterization:

Theorem. A finite plane is affine if and only if it is biaffine and Cartesian.

Is this characterization still true for infinite planes?

Problem. Is every biaffine Cartesian plane affine?

Answer 1

The answer to this question is "No". A non-affine biaffine Cartesian plane can be constructed as follows.

First, we fix a suitable terminology. Every function $F:X\to Y$ is identified with its graph $\{(x,y)\in X\times Y:y=F(x)\}$. Two functions $F,G$ are called transversal if $|F\cap G|=1$. A family of functions $\mathcal F$ is called transversal if any two distinct functions $F,G\in\mathcal F$ are transversal. A bijective function $F:\omega\to\omega$ will be called an $\omega$-bijection.

Lemma. Let $\mathcal F$ be a finite transversal family of $\omega$-bijections. For any numbers $x,x'\in\omega$ and $y,y'\in\omega$ with $x\ne x'$ and $y\ne y'$ there exists an $\omega$-bijection $\Phi$ such that $\{(x,y),(x',y')\}\subseteq\Phi$ and the family $\mathcal F\cup\{\Phi\}$ is transversal.

Proof. If $\{(x,y),(x',y')\}\subseteq F$ for some $F\in\mathcal F$, then put $\Phi:=F$ and finish the proof. So, assume that $\{(x,y),(x',y')\}\not\subseteq F$ for every $F\in\mathcal F$. Let $m:=|\mathcal F|$ and $\{F_i\}_{i\in m}=\mathcal F$ be an enumeration of the set $\mathcal F$.

Let $x_0:= x$, $x_1:= x'$, $y_0:= y$, $y_1:= y'$. For every $i\in m$ choose a pair $(x_{i+2},y_{i+2})\in F_i\setminus(\bigcup_{j<i}F_j)$ such that $x_i\notin\{x_j:j<i\}$ and $y_i\notin\{y_j:j<i\}$. The choice of the pair $(x_{i+2},y_{i+2})$ is always possible because the family $\mathcal F$ is transversal and hence the set $\bigcup_{j<i}(F_i\cap F_j)$ is finite and the set $F_i\setminus\bigcup_{j<i}F_j$ is infinite. After completing the inductive construction, we obtain two sequences $(x_i)_{i\in m+2}$ and $(y_i)_{i\le m+2}$ such that $\varphi:=\{(x_i,y_i)\}_{i\in m+2}$ is an injective function such that $\{(x,y),(x',y')\}\subseteq \varphi$ and $|\varphi\cap F|=1$ for every $F\in\mathcal F$.

Next, choose a sequence $(x_i)_{i=m+2}^\infty$ of pairwise distinct elements of $\omega$ such that $\{x_i:i\ge m+2\}=\omega\setminus\{x_i\}_{i<m+2}$. For every number $i\ge m+2$ let $y_i$ be the smallest number in the infinite set $\omega\setminus(\{y_j:j<i\}\cup\{F(x_i):F\in\mathcal F\})$. Then $\Phi:=\{(x_i,y_i)\}_{i\in\omega}$ is an injective function from $\omega$ to $\omega$ such that $\varphi\subseteq\Phi$ and $|\Phi\cap F|=|\varphi\cap F|=1$ for every $F\in\mathcal F$. It remains to show that the function $\Phi$ is bijective. In the opposite case, the set $\omega\setminus\Phi[\omega]$ is not empty and we can consider its smallest element $s:=\min(\omega\setminus\Phi[\omega])$. Since the finite set $\{0,\dots,s-1\}$ is contained in $\Phi[\omega]$, there exists a number $k\ge m+2$ such that $\{0,\dots,s-1\}\subseteq \{\Phi(x_i):i<k\}$. For every $i\ge k$, the choice of $s\ne y_i=\min(\omega\setminus(\{y_j:j<i\}\cup\{F(x_i):F\in\mathcal F\})$ ensures that $s=\{F(x_i):F\in\mathcal F\}$ and hence $s\in \Psi_i(x_i)$ for some $\Phi_i\in\mathcal F$. Since $|\mathcal F|<\omega=|\omega\setminus k|$, there exist two distinct numbers $i,j\in \omega\setminus k=\omega\setminus\{0,1,\dots,k-1\}$ such that $\Psi_i=\Psi_j$ and hence $\Psi_i(x_i)=s=\Psi_j(x_j)=\Psi_i(x_j)$, which implies that the function $\Psi_i=\Psi_j\in\mathcal F$ is not injective, which contradicts the choice of the family $\mathcal F$. This contradiction shows that the function $\Phi:\omega\to \omega$ is bijective. $\qquad\square$

Let $\mathcal B$ be a set of injective functions $B$ such that $|B|=2$ and $dom[B]\cup rng[B]\subseteq \omega$. Every function $B\in\mathcal B$ is a set of the form $\{(x,y),(x',y')\}$ for some numbers $x,x',y,y'\in\omega$ with $x\ne x'$ and $y\ne y'$. Let $\{B_n\}_{n\in\omega}$ be an enumeration of the countable set $\mathcal B$ such that $B_n\ne B_m$ for any distinct numbers $n,m\in\omega$. Using Lemma, choose a sequence of $\omega$-bijections $(F_n)_{n\in\omega}$ such that for every $n\in\omega$, the family $\{F_i\}_{i\le n}$ is transversal and $B_n\subseteq F_n$.

Now consider the linear space $X:=\omega\times\omega$ endowed with the family of lines $$\mathcal L:=\{\omega\times\{y\}:y\in\omega\}\cup\{\{x\}\times \omega:x\in\omega\}\cup\{F_n:n\in\omega\}.$$ The transversality of the family $\{F_n\}_{n\in\omega}$ ensures that any distinct lines $L,\Lambda\in\{F_n:n\in\omega\}$ are concurent. The linear space $X$ has only two families of parallel lines: $\{\omega\times\{y\}:y\in\omega\}$ and $\{\{x\}\times\omega:x\in\omega\}$, witnessing that $X$ is biaffine and not affine.