# Axiom:Upper Dimensional Axiom

## Contents

## 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.

## 0 Dimensions

The **upper $0$-dimensional axiom** is the assertion:

- $\forall a,b: a = b$

### Intuition

There is only one point, hence the space is at most 0-dimensional.

## 1 Dimension

The **upper $1$-dimensional axiom** is the assertion:

- $\forall a,b,c: \mathsf{B}abc \lor \mathsf{B}bca \lor \mathsf{B}cab$

### Intuition

Any three points are collinear.

It follows that the space is at most 1-dimensional.

Be aware that a 0-dimensional space satisfies this axiom.

## $n$ Dimensions

Let $n \in \N, n \ge 2$.

The **upper $n$-dimensional axiom** is the assertion:

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

- $\implies \left({\mathsf B abc \lor \mathsf B bca \lor \mathsf B cab}\right)$

where:

- $a, b, c, p_i$ are points
- $\displaystyle \bigwedge$ denotes the general conjunction operator.

### Intuition

Any three points equidistant from $n$ distinct points are collinear.

In other words, the set of all points equidistant from $n$ distinct points forms a line.

For $n=2$, this might look like:

and for $n=3$, this might look like:

As was the case with the upper $1$-dimensional axiom, if $m < n$, then an $m$-dimensional space satisfies the upper $n$-dimensional axiom.

Hence the name **upper dimensional axioms**, as the axioms effectively give an upper bound on the dimension of the space considered.

## Also see

## Sources

- June 1999: Alfred Tarski and Steven Givant:
*Tarski's System of Geometry*(*The Bulletin of Symbolic Logic***Vol. 5**,*no. 2*: 175 – 214) : Page 181, 182 : Axiom $9$

Illustration courtesy of Steven Givant.