Complete metric space in which every pair of points there is a distinct point satisfying the triangle equality is path connected

Question

Given $X$ a metric space in which for each $x,y\in X$ there is a $z\in X$ such that $x\neq z\neq y$ and the equality $d(x,y)=d(x,z)+d(z,y)$ holds then X must be path connected. My approach is that given $a,b\in X$to this is by using a form of transfinite induction to build a function $f$ from $[0,1]$ to X in which at each step I add a new $z$ between two points and set it's inverse image as $d(f(a),z)/d(f(a),f(b))$ then whenever a cluster point forms I can use completeness of $X$ to fill it in as needed. this is very messy and tedious and may also be wrong. Is there a more elegant approach to this problem?

Answer 1

So with every two points $a,b$ we can find a point $z$ "between" $a$, $b$. Let us start with the set $\{a,b\}$ and consider a (countable) set $Z_0$ of all the points we can get this way. That is

(**) $a,b\in Z_0$ and for every $z_1, z_2$ in $Z_0$ there is a point $z_3\in Z_0$ "between" $z_1, z_2$.

Consider the closure $\bar Z_0\subseteq X$. That set may no longer satisfy (**).

Then we can repeat the construction and construct sets $Z_1,...$ indexed by ordinals, for limit ordinals we take the unions of the previous $Z_i$. At the end we will get a closed set $Z$ of points from $X$ satisfying (**)

Let $T$ be sets of the points on $[0,1]$ corresponding to points in $Z$ (the correspondence is described in the OP). Then $0,1\in T$ and for every $t_1,t_2\in T$ there is $t_3\in T$, moreover $T$ is closed. Claim: $T=[0,1]$ and $Z$ is the image of a path connecting $a,b$. This is easy to prove and I leave it to the OP.