Topology on generalized metric space and metric space Let $X$ be a nonempty set and $d: X\times X\to R$ be a function such that for all $x,y\in X$ and all distinct $u, v\in X$ each of which is different from $x$ and $y$
(1) $ d(x,y)\geq 0$ ;
(2) $d(x,y)=0$ if and only if $x=y$;
(3) $d(x,y)=d(y,x)$;
(4) $d(x,y)\leq d(x,u)+d(u,v)+d(v,y)$.
Then $d$ is generalized metric on $X$ and $(X,d)$ is called generalized metric space. 
Question: Please describe the topology of Generalized metric space and show that it is different of topology induced by a classic metric $d$.
 A: Remark: Let $X=\{1,2,3,4\}$ and $d(i,i)=0$ for all $i \in X$. And $d(i,j)=1$ for all $i,j \in X$ such that $i\neq j$ and $(i,j)\neq (1,2)$ and $(i,j)\neq (2,1)$. And $d(1,2)=d(2,1)=3$.
$d$ is a generalized metric. But not a classic metric because $d(1,2)>d(1,3)+d(3,2)$.
Answer: Let $X$ a generalized metric space. Let $x \in X$, we have:
either, (1) $x$ is an isolated point: $\exists \eta>0$ such that $d(x,a)<\eta \implies a=x$
either, (2) for all $\epsilon>0$, $\exists \eta>0$, such that $d(a,x)<\eta$ and $d(b,x)< \eta \implies d(a,b)< \epsilon$;
or (3) $\exists \ell_x>0, \forall \epsilon>0$, $\exists \eta>0$, such that $d(a,x)<\eta$ and $d(b,x)< \eta \implies |d(a,b)-\ell_x|< \epsilon$
Indeed, if $x$ is not an isolated point, let $a\neq b \neq c \neq e \neq x \in B(x,\eta)$. 
$d(a,b)<d(a,x)+d(x,c)+d(c,b)$. So $d(a,b)-d(c,b)<2\eta$. 
$d(c,b)<d(c,e)+d(e,x)+d(x,b)$. So $d(c,b)-d(c,e)<2\eta$. 
So, $d(a,b)-d(c,e)<4\eta$.
And $|d(a,b)-d(c,e)|<4 \eta$. So $d(a,b)$ has a limit $\ell$ when $a,b$ tends to $x$. If $\ell=0$, it's the case (2). If $\ell\neq 0$, it's the case (3).
We show that $d(x,z)\leq d(x,y)+d(y,z)$ when $x,y$ or $z$ are points of the case (2): if $a,b$ are near $x$, $d(x,z)\leq d(x,a)+d(a,y)+d(y,z)$  and $d(a,y)<d(a,b)+d(b,x)+d(x,y)$. $d(a,b), d(x,a)$ and $d(b,x)$  tends to $0$ when $a,b$ tend to $x$, because $x$ is in  the case (2). So $d(x,z)\leq 3\epsilon+d(x,y)+d(y,z)$, for all $\epsilon>0$.
Idem, if $y$ or $z$ are in the case (2).
If $x$ is a point  of case (3), the points near $x$ are case (1).
So, we can choose a classic metric $D$ such that, for all $y \in X$, for $\epsilon$ small enough $B_d(y, \epsilon)=B_D(y,\epsilon)$.
And, if the open set of the topology of the generalized metric are the set $O$ such that for all $x\in O$, $\exists \epsilon>0$, such that $B_d(x, \epsilon) \subset O$, we have that the topology of $d$ is the same that the classical topology of $D$.
A: If a function $d: X\times  X\to \mathbb{R}$ satisfyies the condition $(1)-(4)$ then $d$ is a metric in clasical sense. Indeed it is enough to show the triangle inequality, but this is obvious if we put $u=v$ in $(4)$ and use $(2).$
