Is "ternary metrizability" equivalent to pseudometrizability?

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 $\sigma\subseteq\tau$.

Solution 1:

First, for convenience, I will impose an additional axiom on my ternary metrics:

• If $a\neq b$, then there exists $c$ such that $t(a,b,c)>0$.

Note that if $t(a,b,c)=0$ for all $c$, then applying the tetrahedral inequality to the 4-tuples $(c,d,a,b)$ and $(c,d,b,a)$ gives $t(c,d,a)=t(c,d,b)$ for all $c$ and $d$. That is, $t$ cannot distinguish $a$ and $b$ (and neither can the topology induced by $t$). If we quotient out the equivalence relation that identifies all pairs $a,b$ with this property, we obtain a ternary metric which satisfies by axiom. Moreover, the topology of $t$ is then $T_0$ (in fact, $T_1$): if $t(a,b,c)>0$, then the set of $d$ such that $t(a,c,d)<t(a,b,c)$ is an open set that contains $a$ but not $b$.

So, a ternary metric without this axiom is just a ternary metric with this axiom that has indistinguishable copies of some points, and identifying those indistinguishable copies corresponds taking the $T_0$ quotient on the induced topologies. We thus lose no significant generality by imposing this axiom on ternary metrics (and comparing them to metrics, rather than to pseudometrics).

Every metric space $(X,d)$ with at least three points is ternary metrizable. Define $$t(x,y,z)=\min(d(x,y),d(x,z),d(y,z)).$$ I claim this is a ternary metric. The only nontrivial part is the tetrahedral inequality, so suppose $x,y,z,w\in X$ and we wish to show that $$t(x,y,z)\leq t(x,y,w)+t(x,z,w)+t(y,z,w).$$ Without loss of generality, assume that $t(x,y,z)=d(x,y)$. If $t(x,y,w)=d(x,y)$ as well we are done, so without loss of generality we may assume that instead $t(x,y,w)=d(x,w)$. If $t(y,z,w)=d(y,w)$ we are now done by the triangle inequality. If $t(y,z,w)=d(y,z)$ we are also done since we know $d(y,z)\geq t(x,y,z)$. Finally, if $t(y,z,w)=d(z,w)$ we are done again by the triangle inequality since $d(x,w)+d(z,w)\geq d(x,z)\geq t(x,y,z)$.

Now I claim that $t$ induces the same topology as $d$. It is clear that every $t$-subbasic open set is $d$-open. Conversely, let $x\in X$ and $\epsilon>0$; we wish to show the ball $B_d(x,\epsilon)$ contains a $t$-neighborhood of $x$. By hypothesis, $X$ has at least three points, so pick two other points $y$ and $z$. Without loss of generality, we may assume that $t(x,y,z)\geq2\epsilon$ (if not, shrink $\epsilon$). Now suppose $w$ is such that $t(x,y,w)<\epsilon$ and $t(x,z,w)<\epsilon$ (the set of such $w$ is an intersection of two $t$-subbasic open neighborhoods of $x$). This implies that either $d(x,w)<\epsilon$ or else both $d(y,w)<\epsilon$ and $d(z,w)<\epsilon$, since $d(x,y)$ and $d(x,z)$ are both at least $2\epsilon$. But now the triangle inequality gives $d(y,z)<2\epsilon$ in the latter case, which is impossible. So we must have $d(x,w)<\epsilon$, and our $t$-neighborhood of $x$ is contained in $B_d(x,\epsilon)$.

On the other hand, here is a ternary metric space that is not metrizable. Let $X$ be an infinite set and pick an element $a\in X$. Partition $X\setminus\{a\}$ into pairs; call pairs that are terms of this partition good pairs. Define $t(x,y,z)$ to be $1$ if $x,y,z$ are all distinct and contain a good pair, and $0$ otherwise. To verify that this satisfies the tetrahedral inequality, suppose $(x,y,z,w)$ is a 4-tuple of elements of $X$; we wish to show $$t(x,y,z)\leq t(x,y,w)+t(x,z,w)+t(y,z,w).$$ If $t(x,y,z)=0$ this is trivial, so we may assume $x,y$ are a good pair and $z$ is distinct from them. Unless $w$ is equal to either $x$ or $y$, we then have $t(x,y,w)=1$ and are done. On the other hand, if $w$ is equal to $x$ or $y$, then $t(y,z,w)$ or $t(x,z,w)$ is $1$ and we are again done.

To verify that $t$ satisfies my positivity axiom, suppose $x,y\in X$ are distinct. If $x,y$ are a good pair, then $t(x,y,z)>0$ for any $z$ distinct from them. Otherwise, at least one of $x$ and $y$ is different from $a$, so there is an element $z$ which forms a good pair with it. This $z$ is then distinct from $x$ and $y$ so $t(x,y,z)>0$.

Thus $t$ is indeed a ternary metric. Now let us consider the topology induced by this ternary metric. The only nontrivial subbasic open sets to consider are those of the form $U_{x,y}=\{z:t(x,y,z)<1\}$. If $x$ and $y$ are a good pair, then $U_{x,y}=\{x,y\}$. Since the topology is $T_1$, this means every point of $X$ except for $a$ is isolated. On the other hand, if $a\in U_{x,y}$ then $x,y$ is not a good pair, which implies $U_{x,y}$ is cofinite (if $z\not\in U_{x,y}$ then $z$ can only be either the good pair partner of $x$ or the good pair partner of $y$). Thus every neighborhood of $a$ is cofinite.

We thus conclude that $X\setminus\{a\}$ is discrete and $X$ is its one-point compactification with $a$ as the point at infinity. This space is not metrizable if $X$ is uncountable.

Finally, here are some things I can say about the topology of ternary metrizable spaces.

Lemma: Let $(X,t)$ be a ternary metric space, let $a\in X$, and let $(x_i)$ be a net in $X$. Then the following are equivalent.

1. $(x_i)$ converges to $a$.
2. For all $b\in X$, $t(a,b,x_i)$ converges to $0$.
3. For all $b,c\in X$, $t(b,c,x_i)$ converges to $t(b,c,a)$.

Proof: The implications $(1\Rightarrow 2)$ and $(3\Rightarrow 1)$ are immediate from the definition of the topology of $t$, so all that remains is to prove $(2\Rightarrow 3)$. Now the tetrahedral axiom tells us $$t(b,c,x_i)\leq t(b,c,a)+t(a,b,x_i)+t(a,c,x_i).$$ Assuming (2), the final two terms converge to $0$, so we conclude that $\limsup t(b,c,x_i)\leq t(b,c,a)$. Reversing the roles of $a$ and $x_i$ in the tetrahedral inequality, we also conclude that $t(b,c,a)\leq\liminf t(b,c,x_i)$. Thus $t(b,c,x_i)$ converges to $t(b,c,a)$, as desired. $\blacksquare$

Corollary: Any ternary metrizable space is completely regular. Any countable ternary metrizable space is metrizable.

Proof: The equivalence of (1) and (3) says the topology of $t$ is the coarsest topology which makes the real-valued function $x\mapsto t(b,c,x)$ continuous for each $b,c\in X$. When $X$ is countable, these are only countably many functions so we can conclude $X$ is metrizable. $\blacksquare$

It seems plausible that the arguments used in this Corollary are the only obstructions to ternary metrizability. That is, it seems plausible that a $T_0$ space $X$ with at least three points is ternary metrizable iff it embeds in $[0,1]^X$ (i.e., its topology is induced by a set of $|X|$ real-valued functions).

Solution 2:

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.