# Can the strict triangle inequality hold in a length space?

Let $(M, d)$ be a metric space where

• (0) for relevance: set $M$ contains at least three distinct elements and

distance function $d : M \times M \to \mathbb R$ explicitly satisfies

• (1) non-negativity: $d[ \, x, y \, ] \ge 0$ (where $x, y \in M$ are not necessarily distinct),

• (2) identity of indiscernibles: $d[ \, x, y \, ] = 0 \, \iff \, x \equiv y$,

• (3) symmetry: $d[ \, y, x \, ] = d[ \, x, y \, ]$, and

• (4) the general triangle inequality (a.k.a. inclusive subadditivity): $d[ \, x, z \, ] \le d[ \, x, y \, ] + d[ \, y, z \, ]$ for any three (not necessarily all distinct) $x, y, z \in M$.

Now consider the following two additional conditions:

• (a) the strict triangle inequality (a.k.a. exclusive subadditivity): $d[ \, x, z \, ] \lt d[ \, x, y \, ] + [ \, y, z \, ] \iff (y \not\equiv x \hbox{ and } y \not\equiv z)$,
which includes the case that $x, y, z$ are all pairwise distinct;

• (b) metric space $(M, d)$ is a length space;
which shall be defined here explicitly and suitably generally as follows:
$\,$
$\forall \, x, z \in M \, | \, x \not\equiv z : (\exists \, y \in M \, | \, d[ \, x, y \, ] \lt d[ \, x, z \, ] \hbox{ and } d[ \, y, z \, ] \lt d[ \, x, z \, ]) \implies$
$\,$
$d[ \, x, z \, ] = \text{inf}_{\{\forall \, C \subset M | \, \exists \, q \, \in \, C \, : \, q \not\equiv x \text{ and } q \not\equiv z\}}{\Large[} \, \text{sup}_{\{ n \in \mathbb N \}}{\large[} \,$
${\left\{ \text{inf}_{ \{ \forall \text{ permutations } p \, : n \rightarrow C \} } \!\! \left[ \sum_{k = 1}^n d[ \, p_{(k - 1)}, p_{(k)} \, ] \right] | {x \equiv p_{(0)} \hbox{ and } z \equiv p_{(n)} \hbox{ and } (\forall \, k \le n : p_{(k)} \in C)} \right\}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad {\large]} \, {\Large]},$
where the expression part
$\text{sup}_{\{ n \in \mathbb N \}} \left[ \, {\left\{ \text{inf}_{ \{ \forall \text{ permutations } p \, : n \rightarrow C \} } \!\! \left[ \sum_{k = 1}^n d[ \, p_{(k - 1)}, p_{(k)} \, ] \right] | {x \equiv p_{(0)} \hbox{ and } z \equiv p_{(n)} \hbox{ and } (\forall \, k \le n : p_{(k)} \in C)} \right\}} \right]$
is to be regarded as (general, possibly polygonal) length of chain $C$ between $x$ and $z$.

Surely both these conditions (a) as well as (b) are separately consistent with conditions (0) through (4) stated above, describing a corresponding metric space.

My question:
Are these conditions (a) and (b) consistent with each other?
Or in other words: Can the strict triangle inequality hold in a length space?

In particular, such a space contains no nonconstant minimizing geodesics. Examples of complete metric spaces without any nonconstant minimizing geodesics are constructed in this mathoverflow answer. You should check these examples to see if they satisfy your condition (you may have to modify the construction though).

Edit. Here is a construction of what you are asking for. I will leave you to fill in the details.

Definition. A point $m$ in a metric space $(X,d)$ is an $\epsilon$-midpoint between distinct points $x, y\in M$ if $$\frac{1}{2} d(x,y)< \min(d(m,x), d(m,y))\le \max( d(m,x), d(m,y) ) < \frac{1}{2} d(x,y) + \epsilon.$$

Definition. A metric space $(X,d)$ is strict if any triple of distinct points in $X$ satisfies strict triangle inequalities.

Lemma. Given a countable strict metric space $(X,d)$, $\epsilon>0$ and two distinct points $x,y\in M$, there exists a strict metric $d'$ on $X':= X \sqcup \{m\}$ extending the metric $d$ on $X$, such that $m$ is an $\epsilon$-midpoint between $x, y$.

Now, here is a construction of a (countable) strict length metric space $(X,d)$.

Start with a 2-point set $X_1=\{x,y\}$, $d_1(x,y)=1$. Then, using Lemma, inductively build finite strict metric spaces $(X_n,d_n)$ by adding $\epsilon$-midpoints (where each $\epsilon$ is of the form $\frac{1}{i}$ for some $i\in {\mathbb N}$), in such a way that for each $n$ and every pair of points $x,y\in X_n$ and every $i\in {\mathbb N}$, there exists $k=k(n,x,y,m)$ such that on the $k$-th stage of the construction, we add to $X_{k}$ an $\frac{1}{i}$-midpoint between $x$ and $y$. (This requires a certain enumeration of ${\mathbb N}^3$ where the first two coordinates in ${\mathbb N}^3$ represent points in our finite metric spaces and the last coordinate corresponds to the index $i$ as above.) Lastly, take $$X:= \bigcup_{n} X_n$$ where the metric $d$ on $X$ is defined as the direct limit of the metrics $d_n$, i.e. it restricts to the metric $d_n$ on $X_n$ for each $n$. Then $(X,d)$ is a strict length metric space.

Note that $(X,d)$ will be incomplete and will not be a path-metric space.

• Moishe Cohen: "In particular, such a space contains no nonconstant minimizing geodesics." -- But is such a metric space that you describe even called "length space"?? (And even: what exactly is the definition of "nonconstant minimizing geodesic" in suitable generality?) "Examples [...] are constructed in this mathoverflow answer. You should check these" -- Too bad I got, for all intents and purposes, kicked out off mathoverflow; due to having asked questions pretty much like the one above ... (Still: +1 for your interest and effort.) – user12262 Apr 21 '17 at 4:42
• @user12262: Yes, it is a length space (in the standard sense), which is what makes this example so amazing! I do not have much time to think about this, but read the example and see if it satisfies the stronger property you are asking. – Moishe Kohan Apr 21 '17 at 12:26
• Moishe Cohen: Well, I see that the question there relates to mine here, but doesn't quite match it: A space 'with intrinsic metric' (as defined there) is indeed an example of a length space; but in turn: a space satisfying the strict triangle inequality (of my definition) surely is an example of a space which 'has no nontrivial min.zing geodesics.' So the example given over there doesn't necessarily make an affirmative answer here quite yet. Moreover, I'm stymied by its jargon or assumptions: '$c_0$'? 'distance element $\|dx\|_{\infty}$'? '$m$-th position'?? – user12262 Apr 21 '17 at 19:20