Axiom:Lower Dimensional Axiom

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

Let $\mathsf B$ denote 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:


 * $\exists a, b, c: \neg \mathsf B abc \land \neg \mathsf B bca \land \neg \mathsf B cab$

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:
 * $\exists a, b, c, p: \left({\displaystyle \bigwedge_{1 \mathop \le i \mathop < j \mathop < n} \neg \left({p_i = p_j}\right) \land \bigwedge_{i \mathop = 2}^{n - 1} a p_1 \equiv a p_i \land \bigwedge_{i \mathop = 2}^{n - 1} b p_1 \equiv b p_i \land \bigwedge_{i \mathop = 2}^{n - 1} c p_1 \equiv c p_i}\right)$


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

where $a, b, c, p$ etc. are points.

Intuition
There exist $n-1$ (pairwise) distinct points.

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

It is 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

 * Axiom:Upper Dimensional Axiom