In trying to understand what actually constitutes a "geometry" I came across many definitions of Euclidean spaces and geometries. Euclidean space is defined as an affine space with an inner product space acting on it. I was wondering if it could be defined in an equivalent and more natural manner, by relying only on the fundamental notion of distance, without the need for an inner product.
A set $E$ is an Euclidean space iff there is a function $d:E\times E\to \mathbb R$ that satisfies the following axioms:
(1) $d(a,b)+d(c,b)\ge d(a,c)$, for every $a,b,c\in E$.
(2) $d(a,b)=d(b,a)$, for every $a,b\in E$.
(2') $d(a,b)=0$ iff $a=b$.
(3) For every $p_1,p_2$ in $E$ there always exists a unique set $P$ of points that contains $p_1,p_2$ such that for any points $a,b,c\in P$ if $d(b,c) <= d(a, c)= >d(a,b) $ then $d(a,c)=d(a,b)+d(b,c)$.
(4) For any such set $P$ and for any point $p\notin P$ there is always a unique set $P_2$ (for which (3) holds) and that contains $p$, such that for every pair $(p_1,p_2)$ where $p_1\in P$ and $p_2\in P_2$, $D \le d(p_1, p_2)$, and for every $p_1 \in P$ there exists $p_2 \in P_2$ such that $d(p_1, p_2) =D$, $D \in \mathbb{R}$.
With the variation of this last property geometry should become non-Euclidean.
First two axioms define a usual metric, third defines geodesics, and last defines parallel geodesics.
For continuity there could be a requirement that for every geodesic $P$, for any real number $r$, there always exists a pair of points $p_1, p_2$ on $P$ such that $d(p_1, p_2) =r$.
Edit : I appreciate all the comments and suggestions, I would just like to say that I wouldn't refer to this as "my axioms" foe geometry as it was not my intention to just randomly come up with some generic new axioms.
It seemed reasonable to me to ask whether various geometries could be interpreted as sets on which there is a quantitative distance relation between points that can define geodesics through triangle equality. In this case Euclidean geometry.
- for any geodesic P and p in P and any r in R, there are p1, p2 such that d(p1, p) =d(p2, p) =r
After i posted the question i wanted to add this right away but then I didn't want to change the OG question too much.