Can betweenness be defined in terms of collinearity?

Question

Consider the space $\mathbb{R}^n$ for $n \geq 2$. We can consider two structures on that space. The first is betweenness, which is a ternary predicate that tells us when one point is strictly between two other points. The second is collinearity, which is a ternary predicate that applies precisely when three points are distinct and collinear. Now, certainly, collinearity can be defined in terms of betweenness. Can betweenness be defined in terms of collinearity?

Answer 1

Yes. The idea is this: Using the colinearity relation, we can define the relation $R(A,B,Z)$ which says "$Z$ lies on the ray which begins at $A$ and extends through $B$". Then the betweenness relation $B(A,B,Z)$ which says "$Z$ is strictly between $A$ and $B$" is defined by $Z\neq A\land Z\neq B\land R(A,B,Z)\land R(B,A,Z)$.

To define $R$, we note that for any two points $A$ and $B$, the line $L_{AB}$ through $A$ and $B$ can be viewed (by a linear transformation) as a copy of the real line, with $A = 0$ and $B = 1$, so the ray extending from $A$ through $B$ is the positive real line. Then multiplication on this copy of $\mathbb{R}$ can be carried out by a "straightedge construction", every step of which is definable using colinearity. For this construction, we need a third point $C$ not on the line through $A$ and $B$, as an extra parameter. But any $C$ not on the line will do, so we can quantify out this parameter (incidentally, this explains why the construction doesn't work in $\mathbb{R}^1$). Because a real number is $\geq 0$ if and only it is a square, a point $Z$ satisfies $R(A,B,Z)$ if and only if it is a square for this definable multiplication operation.

Here are the details:

• Let $C(x,y,z)$ be the colinearity relation.

• Given distinct points $A$ and $B$, we can define the unique line $L_{AB}$ containing $A$ and $B$ by $L(A,B,x): C(A,B,x)\lor (x = A)\lor (x = B)$.

• Given four coplanar points $A$, $B$, $A'$, $B'$ with $A\neq B$ and $A'\neq B'$, we can define when the lines $L_{AB}$ and $L_{A'B'}$ are parallel by $P(A,B,A'B'): (L(A,B,A')\land L(A,B,B'))\lor \lnot\exists x\, (L(A,B,x)\land L(A',B',x))$.

• Given five coplanar points $A,B,A',B',C$ with $A\neq B$ and $A'\neq B'$ and such that $L_{AB}$ and $L_{A'B'}$ are not parallel, we can define the (unique) point $D$ on $L_{AB}$ such that $L_{CD}$ is parallel to $L_{A'B'}$ by $I(A,B,A',B',C,x): L(A,B,x)\land P(C,x,A',B')$.

• Given distinct points $A,B$, a point $X$ on $L_{AB}$, and a point $C$ not on $L_{AB}$, note that $A,B,C,X$ are coplanar. We define $Y$ to be the point on $L_{AC}$ such that $L_{XY}$ is parallel to $L_{BC}$, and we define $Z$ to be the point on $L_{AB}$ such that $L_{YZ}$ is parallel to $L_{XC}$. $Z$ is definable by $S(A,B,C,X,z): \exists y (I(A,C,B,C,X,y)\land I(A,B,X,C,y,z))$.

• Now the point is that if $A,B,C,X,Y,Z$ are as in the above construction, then the triangle $ABC$ is similar to the triangle $AXY$, and under this similarity, the point $X$ corresponds to the point $Z$. Let's normalize to assume that the distance $AB$ from point $A$ to point $B$ is equal to $1$. Then, for example, $AX > 0$ if $X$ lies along the ray extending from $A$ through $B$, and $AX < 0$ if $X$ lies along the ray extending from $A$ away from $B$.

• It follows from the similarity that $\frac{AX}{AB} = \frac{AZ}{AX}$, so $AZ = (AX)^2$. In particular, $AZ \geq 0$. And conversely, for any point $Z$ with $AZ\geq 0$, we can obtain $Z$ from the above construction by picking $X$ so that $AX = \sqrt{AZ}$.

• Given $A\neq B$ and $C$ not on $L_{AB}$, the above discussion implies that $Z$ lies on the ray extending from $A$ through $B$ if and only if there exists $X$ on $L_{AB}$ such that $S(A,B,C,X,Z)$. But now we can observe that this condition is independent of the choice of $C$, so we can quantify it out. It follows that $R(A,B,Z)$ is definable by $A\neq B\land \exists C(\lnot L(A,B,C)\land \exists X\, S(A,B,C,X,Z))$.