TITLE: What is straight line?
QUESTION [10 upvotes]: I have found the definition of line in metric space.
It is general but has two problems. Considering about $\mathbb R^2$ equipped rectilinear distance, every line by this definition contains a rectangle and is not a string. Besides, it is possible 3 points in a line not collinear.

Here is what I have thought.
Let $<X,d>$ be any metric space.
Definition 0 Point $b$ is said between $a,c$ iff $d(a,b)+d(b,c)=d(a,c)$
Definition 1 A subspace $S$ of $X$ is collinear iff $\forall x\forall y \forall z[\mbox{exist one between the other two}]$  
Definition 2 A subspace $L$ is called a line iff it is a maximal collinear subspace in $X$
Definition of collinear set is same as the previous if $|S|\le 3$ but different in otherwise. By this definition a collinear set is a collinear set by the previous definition but the converse is not valid.
According to this definition, it has 2 theorems.
Theorem 1
Every subset of a collinear set is collinear.
Theorem 2
Every collinear set can be extended to a line.
1 holds by definition whereas it does not hold by the previous definition.
2 also holds by the previous definition but requires Zorn's lemma by the current definition.
Hence a Corollary 3 Every set is collinear iff it is included in a line 
Finally, by the current definition every line in $\mathbb R^2$ equipped rectilinear distance is actually a curve， no longer contains proper rectangles. 
It looks better now. My question:Is it a precise definition?

REPLY [2 votes]: The definition is valid. However, there are still some “very unusual” lines in the space ${\mathbb R}^2$ equipped with the rectilinear distance. For example, the set $\{(0,0),(1,0),(0,1),(1,1)\}$ is a line according to this definition (it is a maximal collinear set).
We can make your definition stronger as follows.

Definition 1. Let us say that a subspace $S$ of a metric space $(X, d)$ satisfies the $k$-collinear condition if for every points $x_1, \dots, x_k$ in $S$ there exist a permutation $\pi$ such that
  $$\sum_{i=1}^{k-1}d(x_{\pi(i)},x_{\pi(i+1)}) = d(x_{\pi(1)},x_{\pi(k)}).$$

Trivially, every set $S$ satisfies the 2-collinear condition. A set $S$ satisfies the 3-collinear condition precisely when it is collinear according to your definition. Clearly, if a set satisfies the $k$ collinear condition than it also satisfies the $k+1$-collinear condition (we can just let $x_{k+1} = x_k$). However, the set $\{(0,0),(1,0),(0,1),(1,1)\}$ satisfies the 3-collinear condition but not the 4-collinear condition. So in general the 4-collinear condition is strictly stronger than the 3-collinear condition. 
Question. Can we get stronger and stronger conditions by increasing $k$? E.g., is 5-collinear condition even stronger than 4-collinear condition? 
It turns out that $4$-collinear condition implies $k$-collinear conditions for all $k$. This, in particular, follows from the four-point characterization of tree metrics. This result can be restated as follows:

A set $S$ satisfies the 4-collinear condition if and only if there is an isometric embedding 
  $$\phi:S \hookrightarrow{\mathbb R},$$ 
  i.e. there is a map $\phi:S\to\mathbb R$ s.t. $d(x,y) = |\phi(x) - \phi(y)|$ for every $x$ and $y$ in $S$.

Similarly to your definition, we give the following definition of a line.

Definition 2. A subspace $S$ of a metric space $(X, d)$ is a line if it is a maximal subspace of $(X,d)$ satisfying the 4-collinear condition.

Now every line $S$ in a Banach space (in particular, in ${\mathbb R}^2$ equipped with the rectilinear distance) is a curve. Moreover, there is a natural parametrization $\gamma(t)$ of $S$ ($\gamma:{\mathbb R} \to S$) such that $d(\gamma(s), \gamma(t)) = |s-t|$ for every $s,t\in \mathbb R$.