TITLE: Is "ternary metrizability" equivalent to pseudometrizability?
QUESTION [12 upvotes]: Below, $X$ is always a set with at least three elements to avoid triviality.

Say that a ternary metric on a set $X$ is a map $t:X^3\rightarrow\mathbb{R}$ with the following properties:

Non-negativity: $t(x_1,x_2,x_3)\ge 0$ and $t(x_1,x_1,x_2)=0$. (However, we may have $t(x_1,x_2,x_3)=0$ even if the $x_i$s are distinct.)

Symmetry: $t(x_1,x_2,x_3)=t(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)})$ for each permutation $\sigma\in S_3$.

Tetrahedral inequality: for all $x_1,x_2,x_3,y$ we have $$t(x_1,x_2,x_3)\le t(x_1,x_2,y)+t(x_1,x_3,y)+t(x_2,x_3,y).$$


The motivating example is $X=\mathbb{R}^{n\ge 2}$ and $t(x,y,z)=$ the area of the triangle with vertices $x,y,z$. Now say that a topological space $(X,\tau)$ is ternary metrizable iff there is a ternary metric $t$ on $X$ such that $\tau$ is generated by the family of sets $$\{\{y: t(x_1,x_2,y)<\epsilon\}:\epsilon\in\mathbb{R}, x_1,x_2\in X\}.$$For example, in $\mathbb{R}^3$ the "area ternary metric" mentioned above induces the usual topology: the (sub)basic open sets are infinite tubes.
My question is:

What is the relationship between pseudometrizability and ternary metrizability?

The most natural guess is that they coincide exactly, and I do suspect that this is the case. However, I can't prove it (or indeed either of the implications involved) at the moment. It's not even obvious to me that for every metrizable space $(X,\tau)$ there is a ternary metric on $X$ inducing a topology $\sigma\subseteq\tau$.

REPLY [3 votes]: This is a follow-up to Eric Wofsey's answer with a partial resolution of the question of which topologies can be induced by a ternary metric.


Proposition. Every completely regular space $X$ is homeomorphic to a clopen subspace of a ternary metric space.

Proof. Let $C$ be the collection of all continuous functions from $X$ to $[0,1]$. It is a well-known fact that the topology on $X$ is the initial topology on $X$ with regards to the family $C$.
If $X$ is a singleton, then we can obviously do this with a metric space and the ternary metric Eric defined, so assume that $X$ has at least two elements.
Consider the set $Y=X \sqcup C$. Define a symmetric function $t: Y^3 \to \mathbb{R}$ as follows.

$t(x,y,z) = 0$ for $x,y,z \in X$.
$t(x,y,f) = |f(x)-f(y)|$ for $x,y \in X$ and $f \in C$.
$t(x,f,g) = 1$ and $t(x,f,f) = 0$ for $x \in X$ and $f,g \in C$ with $f \neq g$.
$t(f,g,h) = 1$ and $t(f,g,g) = 0$ for $f,g,h \in C$ pairwise distinct.

This clearly satisfies the non-negativity and symmetry properties, so we just need to verify the tetrahedral inequality and the fact that $X$ is a clopen subspace with the original topology on $X$.
For an inequality of the form $t(x,y,z) \leq \dots$, there is nothing to show. Similarly, we may always assume that the three arguments of $t$ are pairwise distinct. This leaves six cases. Let $x,y,z \in X$ and $f,g,h,i \in C$.
Case 1. $t(x,y,f) \leq t(x,y,z) + t(x,z,f) + t(z,y,f)$ follows from the fact that $|f(x)-f(y)| \leq |f(x)-f(z)| + |f(z) - f(y)|$.
Case 2. $t(x,y,f) \leq t(x,y,i) + t(x,i,f) + t(i,y,f)$ follows from the fact that

if $f=i$, then $t(x,y,f) \leq t(x,y,i)$ and
if $f\neq i$, then $|f(x) - f(y)| \leq t(x,i,f) = 1$.

Case 3. $t(x,f,i) \leq t(x,z,y) + t(x,z,i) + t(z,f,i)$ follows from the fact that $t(x,f,i) = t(z,f,i) = 1$.
Case 4. $t(x,f,i) \leq t(x,f,h) + t(x,h,i) + t(h,f,i)$ follows from the fact that

if $f=i$, then $t(x,f,i) = t(x,i,g) = 1$,
if $g=i$, then $t(x,f,g) = t(x,f,i) = 1$, and
$t(x,f,g) = t(i,f,g) = 1$ otherwise.

Case 5. $t(f,g,h) \leq t(f,g,z) + t(f,z,h) + t(z,g,h)$ follows from the fact that $t(f,g,h) = t(f,g,z) = 1$.
Case 6. $t(f,g,h) \leq t(f,g,i) + t(f,i,h) + t(i,g,h)$ follows from the fact that at least one of the right-hand terms is $1$.
To see that $X$ is a clopen subspace of $Y$, let $f \in C$ be a constant function, and pick $x \in X$. We obviously have that $t(a,x,f) = 0$ if $a \in X \cup \{f\}$ and $t(a,x,f) = 1$ if $a \in C \setminus \{f\}$. Therefore $X \cup \{f\}$ is clopen for each constant function $f \in C$. Since there is more than one constant function on $X$, we have that $X$ is the intersection of two clopen sets and is therefore clopen.
The topology on $X$ is generated by functions from $X$ to $[0,1]$ that actually attain $0$ for some $y$. For any such function $f$, $f(x) = |f(x) - f(y)|= t(x,y,f)$, so we have that the topology induced on $X$ by $t$ is finer than the original topology. On the other hand, any function of the form $x\mapsto t(x,a,b)$ for $a,b \in Y$ is in $C$, so the topology induced by $t$ must be coarser than the original topology as well. Hence they are the same. $\square$

This result makes it seem like a precise characterization might be non-local. It's not clear to me, for example, that every compact Hausdorff space is a ternary metric space.