Are metric segments convex?

For a metric space $$(X,d)$$ and points $$x,y \in X$$ we define the metric segment between them as the following set:

$$\left [ x,y \right ] = \left \{ z \in X : d(x,z)+d(z,y)=d(x,y)\right \}$$

Can we say that metric segments are convex? That is, for an arbitrary metric space $$(X,d)$$ and points $$x,y,u,v \in X$$, does $$u,v \in \left [ x,y \right ]$$ imply $$\left [ u,v \right ] \subseteq \left [ x,y \right ]$$?

• The trouble with your definition of convexity is ambiguity of the notation [u,v] since segments are not uniquely determined by their end points. Please, clarify. Commented Jun 5, 2022 at 22:35
• @MoisheKohan, I don't see this ambiguity. The formula is strict. They are uniquely determined. Commented Jun 5, 2022 at 22:45
• While it is true that two different endpoints can define the same segment, if two endpoints are given, the segment is unique and strictly defined. Commented Jun 5, 2022 at 22:50
• Oh, nevermind, I was misreading your definition. Commented Jun 5, 2022 at 23:48
• For example, in $\mathbb R^2$ with the norm $\|(x,y)\|_\infty = \max\{|x|,|y|\}$, the "metric segment" between $(0,0)$ and $(1,0)$ is the filled-in square with vertices $(0,0),(1/2,1/2),(1/2,-1/2), (1,0)$. Commented Jul 20, 2022 at 21:33

2 Answers

No. Here's my example:

Let $$a=(0,0),\ b=(1,0),\ c=(1,1),\ d=(0,1)$$. Let $$X$$ be a union of segments $$ab,bc,cd,da,bd$$. Distance between two points is an euclidean length of a shortest path between them, for example $$\rho(a,c)=2$$. Then $$b,d\in [a,c]$$ and $$(0.5,0.5)\in [b,d]$$ but $$(0.5,0.5)\notin [a,c]$$.

Edit (after the request on the comment):

The fact that it's a metric: Intrinsic metric.

The example of the metric on the whole $$\Bbb R^2$$. Consider the infinite countryside $$\Bbb R^2$$ with a mud in the rhomb abcd, where $$a=(1,0),\ b=(0,1),\ c=(-1,0),\ d=(0,-1)$$. In mud you go much slower (e.g. five times slower) then on the grass (outside mud). The distance is given by the time you have to get from one place to another (infimum over all paths joining two points). Consider two points: $$e=(0,-5)$$, $$f=(0,5)$$. The shortest paths go through points $$a$$ or $$c$$ and therefore $$[e,f]$$ is a union of line segments $$ea,af,ec,cf$$. It's far from being convex.

• Thank you. This seems to be a valid counterexample to my question. However, I would prefer a metric on the whole R/R2. Could you provide such a counterexample? Commented Jun 5, 2022 at 22:51
• I added the example. Commented Jun 5, 2022 at 23:23
• Do you perhaps mean $ea,\color{red}{a}f,ec,cf$ in the last paragraph?
– Ruy
Commented Jun 6, 2022 at 0:06
• @Ruy Yes, thank you. Corrected. Commented Jun 6, 2022 at 0:29
• @BrianTung, thank you, I meant $\rho(a,c)$. This is because I changed the labels during writing. Commented Jun 6, 2022 at 1:02

@Mateo
A metric segment is not convex because, informally, it may include multiple paths between the endpoints.

Actually a restricted notion of metric segment can be derived, and is convex. Not knowing if it has a name already, let's call it metric thread.

First define an order relation $$\underset {[x, y]} \preceq$$ between points that belong to the metric segment $$[x, y]$$, by: $$a \underset {[x, y]} \preceq b \iff d(x, y) = d(x, a) + d(a, b) + d (b, y)$$. It means "$$a$$ and $$b$$ belong to a same thread between $$x$$ and $$y$$, in that direction". (And we'll just write $$\preceq$$ without $$_{[x, y]}$$ when there is no ambiguity).

Note that, if $$a \preceq b$$, as $$d(x,a) + d(a,b) \ge d(x, b) \text { and } d(x, b) + d(b, y) >= d(x, y)$$ by triangular inequality, the two inequalities are equalities, so $$d(x, a) + d(a, b) = d(x, b)$$, $$a$$ belongs to the metric segment $$[x, b]$$. And similarly, $$b$$ belongs to the metric segment $$[a, y]$$.

Relation $$\preceq$$ is a partial order:

• Reflexivity: $$d(a, a) = 0$$,
so $$d(x, a) + d(a, a) + d(a, y) = d(x, a) + d(a, y)$$
$$= d(x, y)$$ because $$a \in [x, y]$$.
So $$a \preceq a$$.
• Antisymmetry: if $$a \preceq b$$ and $$b \preceq a$$,
then $$d(x, y) = d(x, a) + d(a, b) + d (b, y) = d(x, b) + d(b, a) + d (a, y)$$.
As $$a, b \in [x, y]$$ we have also: $$d(x, y) = d(x, a) + d(a, y) = d(x, b) + d(b, y)$$.
So $$0 = d(a, b) + d(b, y) - d(a, y) = d(b, a) + d(a, y) - d(b, y)$$.
As $$d(a, b) = d(b, a)$$,
$$d(b, y) - d(a, y) = d(a, y) - d(b, y)$$,
so $$d(a, y) = d(b, y)$$.
We have seen above that $$b \in [a, y]$$, so $$d(a, y) = d(a, b) + d(b, y)$$,
$$d(a, b) = 0, a=b$$.
• Transitivity: if $$a \preceq b$$ and $$b \preceq c$$, $$d(x, y) = d(x, a) + d(a, b) + d(b, y) = d(x, b) + d(b, c) + d (c, y)$$.
$$d(a, c) \le d(a, b) + d(b, c) = d(x, y) - d(x, a) - d(b, y) + d(x, y) - d(x, b) - d(c, y)$$
$$= 2 d(x, y) - d(x, a) - d(x, y) - d(c, y)$$
$$= d(x, y) - d(x, a) - d(c, y)$$.
By triangular inequality we also have $$d(x, y) \le d(x, a) + d(a, c) + d(c, y)$$, i.e. inverse inequality.
So there is equality, i.e. $$a \preceq c$$.

Then we define a metric thread of $$[x, y]$$ as any totally ordered subset of $$[x, y]$$.

Let's prove now that it is convex: if $$u \underset {[x, y]} \preceq v, \text { and } a \underset {[u, v]} \preceq b$$, then $$a \underset {[x, y]} \preceq b$$.

$$u \underset {[x, y]} \preceq v$$: $$u, v \in [x, y] \text {, i.e. } d(x, y) = d(x, u) + d(u, y) = d(x, v) + d(v, y)$$, $$\text {and } d(x, y) = d(x, u) + d(u, v) + d(v, y)$$

$$a \underset {[u, v]} \preceq b$$: $$a, b \in [u, v] \text {, i.e. } d(u, v) = d(u, a) + d(a, v) = d(u, b) + d(b, v)$$, $$\text {and } d(u, v) = d(u, a) + d(a, b) + d(b, v)$$

Then $$d(x, y) - d(x, a) - d(a, b) - d(b, y)$$
$$= d(x, u) + d(u, v) + d(v, y) - d(x, a) - d(a, b) - d(b, y)$$
$$= d(x, u) + d(u, a) + d(a, b) + d(b, v) + d(v, y) - d(x, a) - d(a, b) - d(b, y)$$
$$= d(x, u) + d(u, a) + d(b, v) + d(v, y) - d(x, a) - d(b, y)$$
which is $$\ge 0$$ by triangular inequalities
($$d(x, u) + d(u, a) \ge d(x, a)$$, and $$d(b, v) + d(v, y) \ge d(b, y)$$.
But by triangular inequality, $$d(x, y) - d(x, a) - d(a, b) - d(b, y) \le 0$$
So $$d(x, y) - d(x, a) - d(a, b) - d(b, y) = 0$$,
i.e. $$a \underset {[x, y]} \preceq b$$.

Note that metric threads do not make a partition of the metric segment, i.e. a point can belong to many metric threads of the same segment. In the case of the taxicab metric, the metric segment is a full rectangle; any subset of points in this rectangle, whose coordinates are strictly growing, is included into an infinity of threads.