TITLE: Examples of non symmetric distances
QUESTION [31 upvotes]: It is well known that the symmetric property is $d(x,y)=d(y,x)$ is not necessary in the definition of distance if the triangle inequality is carefully stated. On the other hand there are examples of functions satifying
(1) $d(x,y)\geq 0$ and $d(x,y)=0$ if and only if $x=y$
(2) $d(x,y)\leq d(x,z)+d(z,y)$
which are not symmetric: in the three point space $(a,b,c)$ take the non-zero values of $d$ as $1=d(a,b)=d(c,b)$, $2=d(b,a)=d(b,c)=d(a,c)=d(c,a)$.
Do you know other examples of "non symmetric distances"? Are there examples on the real numbers, etc.? Are there examples of spaces were every function satisfying (1) and (2) is symmetric?

REPLY [5 votes]: I came here looking for something else and was surprised that nobody has mentioned distances induced by convex bodies (these are classically called Minkowski distances, and "convex distance functions" in the Computational Geometry literature). 
Let $K$ be an arbitrary convex body (= compact convex set with non-empty interior) in $\mathbb R^d$ with the origin in its interior. You can define the $K$-norm $||v||_K$ of a vector $v$ to be the unique positive $\lambda\in [0,\infty)$ with $\lambda v \in \partial K$, and the $K$-distance between $a$ and $b$ to be $||b-a||_K$. 
If $K$ is centrally symmetric then these are a norm and a distance in the usual sense (the Euclidean distance and all the $L_p$ distances are examples). 
If $K$ is not centrally symmetric then this distance is not symmetric ($||-v||\ne -||v||$), but it still satisfies the asymmetric triangle inequality.