Quasimetric induces two different topologies A function $d:X\times X\to \mathbb R^+$ is called quasimetric on $X$ if:

*

*$d(x,y)=0$ if and only if $x=y$;

*$d(x,z)\leq d(x,y)+d(y,z)$.

One can define a subset $A\subset (X,d)$ to be open if for each $a\in A $ there is $\epsilon>0$ such that $\{y\in X:d(a,y)<\epsilon\}$ is contained in $A$. It can be easily checked, collection of these open sets define a topology $\tau$.
Similarly, one can say a subset $A\subset (X,d)$ is open if  for each $a\in A $ there is $\epsilon>0$ such that $\{y\in X:d(y,a)<\epsilon\}$ is contained in $A$. Collection of these open sets define a topology $\tau'$.
Thus, a quasimetric induces two different topologies. Is it so?
 A: They do.
Orient the unit circle $S^1 \subseteq \mathbb{C}$ `anti-clockwise', so that $d(x,y)$ is the anti-clockwise distance from $x$ to $y$.
The anti-clockwise `interval' $[1,i)$ of points $e^{i\theta}$, $0\leq \theta < \frac{\pi}{2}$ is open in $\tau$, but not in $\tau'$.

A: Let's call the set the first set $$B_l(a,r) = \{x \in X\mid d(a,x) < r\}$$ because the centre is left in the condition, and likewise
$$B_r(a,r) = \{x \in X\mid d(x,a) < r\}$$ when the centre is on the right.
Both $$\tau_l = \{A\subseteq X: \forall a \in A: \exists r>0: B_l(a,r) \subseteq A\}$$ and
$$\tau_r = \{A\subseteq X: \forall a \in A: \exists r>0: B_r(a,r) \subseteq A\}$$ are topologies on $X$; the intersection axiom is shown on the basis of $B(a,r)\cap B(a,s)=B(a,\min(r,s))$ for both types (left or right) of balls, while the union and first axiom are trivial to see..
They can indeed be different.
Take $\Bbb R$ in the quasimetric $d(x,y)=\max(y-x,0)$ and in that case $B_l(a,r) = (-\infty,a+r)$ so we induce the so-called left-order topology, while $B_r(a,r) = (a-r,+\infty)$ and we induce the right-order topology, which are different (but homeomorphic, and the meet of these topologies is the trivial topology, while their join is the Euclidean topology).
I believe taking the left-balls is more common to induce the quasimetric topology. But I'm no expert. I believe we could also induce the Sorgenfrey topologies on $\Bbb R$ this way, using $d(x,y)=y-x$ for $y\ge x$ and $d(x,y)=1$ for $y<x$ and then $B_l(a,\frac12) = [a,a+\frac12)$ e.g. so we get the usual right facing Sorgenfrey topology as $\tau_l$. The $\tau_r$ is then be the left facing one. These topologies have meet the Euclidean one and as join the discrete one. So they also have nice places in the lattice of topologies on $\Bbb R$..
