Axiom:Lower Dimensional Axiom

Axiom
Let $a,b,c,\ldots,x,y,z$ be points.

Let $\mathsf{B}$ be the relation of betweenness.

Let $\equiv$ be the relation of equidistance.

Let $=$ be the relation of equality.

1 Dimension
The lower $1$-dimensional axiom is the assertion:


 * $\exists a,b: \neg \left({a = b}\right)$

Intuition
There are two points, hence the space is at least 1-dimensional.

2 Dimensions
The lower $2$-dimensional axiom is the assertion:


 * $\exists a,b,c: \neg\mathsf{B}abc \lor \neg\mathsf{B}bca \lor \neg\mathsf{B}cab$

Intuition
There are three points that are not collinear.

It follows that the space is at least 2-dimensional.

$n$ Dimensions
Let $n \in \N, n \ge 3$. The lower $n$-dimensional axiom is the assertion:


 * $\exists a,b,c,p: \left({\displaystyle \bigwedge_{1 \le i < j < n} \neg\left({p_i = p_j}\right) \land \bigwedge_{i=2}^{n-1} ap_1 \equiv ap_i \land \bigwedge_{i=2}^{n-1} bp_1 \equiv bp_i \land \bigwedge_{i=2}^{n-1} cp_1 \equiv cp_i}\right)$


 * $\land \left({\neg\mathsf{B}abc \land \neg\mathsf{B}bca \land \neg\mathsf{B}cab}\right)$

Intuition
There exist $n-1$ distinct points.

There are also three points $a,b,c$.

It's possible to set up these points such that all of $a,b,c$ are equidistant from the $n-1$ points and yet $a,b,c$ are not collinear.

In other words, the set of all points equidistant from of $n-1$ distinct points is not a line.

These axioms effectively give a lower bound on the dimension of the space considered.

Also see

 * Upper Dimensional Axiom