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Let $mathcal{M}=(M,0,+,<)$ be a linearly ordered divisible abelian group. Let's define an $mathcal{M}$-metric on $M$ to be a map $d:Mtimes M rightarrow M$ such that

(1) $forall x,yin M,, d(x,y)=0 Leftrightarrow x=y$,

(2) $forall x,yin M,, d(x,y)=d(y,x)$,

(3) $forall x, y, zin M, , d(x,y) leq d(x,z) + d(z,y)$.

Note that one such map defines a topology on $M$ in the usual way that metrics do.

Let $tau$ be the right half-open interval topology in $M$. Namely the topology with basis ${ [x,x+varepsilon) : , x, varepsilonin M, varepsilon>0}$.

In there some $mathcal{M}$-metric on M that generates $tau$?

If $mathcal{M}=(mathbb{R},0,+,<)$ then the answer is no since the topological space described would be separable and not second countable, hence not metrizable.

Any ideas/intuitions regarding the question, even if they are not full answers, are apreciated.

To manage expectations, let me start with a simple observation. $M$ does not admit a translation-invariant $mathcal{M}$-metric. If $rho$ is a translation-invariant $mathcal{M}$-metric, then $rho(x, 0) = rho(-x, 0)$
for all $x$, which means that any open ball about $0$ is symmetric. Hence
the open set $[0, rightarrow)$ cannot contain such a ball. This suggests
that even if a metric exists, it may not be very useful.

The same reasoning applied to $mathbb{R}$ can be applied to many other cases. If $M$ admits an $mathcal{M}$-metric, then $d(M) = |M|$, i.e. $M$ does not have a dense subset of lower cardinality. Since the balls with centre and radius in a dense subset form a base for the topology, we have
$w(M) le d(M)d(M) = d(M)$. On the other hand we have $w(M) ge |M|$ for
the same reason as in $mathbb{R}$. (Note that here we do not have to distinguish between order-dense, dense in the
order topology and dense in the half-open interval topology, as they are
equivalent for a dense linear order.)

This necessary condition can be sharpened by an observation that is useful
for non-Archimedean groups. If $mathcal{N}$ is a nontrivial convex
subgroup of $mathcal{M}$, the following are equivalent:

1. $M$ admits an $mathcal{M}$-metric


2. $N$ admits an $mathcal{M}$-metric


3. $N$ admits an $mathcal{N}$-metric


4. $M$ admits an $mathcal{N}$-metric

That 4. implies 1. is trivial and 1. implies 2. because the restriction of a metric to a subspace induces the subspace topology.
The implication $2. implies 3.$ can be obtained by taking a positive $b inmathcal{N}$ and defining $sigma(x, y) = min{rho(x,y), b}$ as usual. Since $N$ is open in $M$, the quotient $M/N$ is discrete and
$M$ is homeomorphic to $(M/N) times N$. Hence, given an
$mathcal{N}$-metric on $N$, we can make $M$ locally isometric to $N$ to
arrive at 4.

Giving a useful necessary and sufficient condition is probably very hard,
but there is a sufficient condition that is not too difficult. If $mathcal{M}$ has a nontrivial countable convex subgroup, $M$ admits an
$mathcal{M}$-metric. By the previous result, we need only consider the
case $|M| = aleph_0$. Let $B$ be the set of real numbers between 0 and 1
whose binary expansions have finitely many ones. This is also a countable
dense linearly ordered set with no minimum or maximum, so it is order-isomorphic to $M$. It will suffice then to find an $mathcal{M}$-metric
that induces the half-open interval topology on $B$.

Since $M$ is countable, there is a strictly decreasing
sequence ${a_n}$ of positive elements with $inf_n a_n = 0$. Define
$rho(x, y) = 0$ when $x=y$ and otherwise $rho(x, y) = a_m$, where $m$ is the largest integer such that the binary expansions of $x$ and $y$ are
equal to $m$ places. Symmetry and positivity of $rho$ are obvious and it
is easily seen to satisfy the ultrametric triangle inequality
$rho(x, z) le max{rho(x, y),rho(y,z)}$.

To see that the topology induced by $rho$ is finer that the half-open interval topology, note that in the latter a point $x$ has a base of
neighbourhoods of the form $[x, y)$ where $y>x$. There is an $n$ such that
the binary expansions of $x$ and $y$ contain no ones after the $n$th place.
For such $n$, the $rho$-ball of radius $a_{n+1}$ about $x$ is contained in
$[x, y)$. Similarly, the topology induced by $rho$ is coarser than the
half-open interval topology, since every $rho$-ball about $x$ contains a
half-open interval starting at $x$. Thus $rho$ induces the half-open
interval topology on $B$.

Note that under the same hypothesis, $M$ also admits a real-valued metric,
by Urysohn's metrization theorem.

Side note: For topological purposes, it usually makes no difference
whether a ordered abelian group is divisible or just densely ordered. On one hand, in a densely ordered group there is for every $b > 0$ and every
positive integer $n$ an $a > 0$ such that $na <= b$. On the other hand,
in a divisible group there may be no $a > 0$ such that
$lim_{ntoinfty} a/n = 0$.