Neighbourhoods in a quasi-distance induced topology A quasi-distance on a set $X$ is a non-negative function $\delta(x,y)$ defined on $X\times X$ such that:


*

*$\delta(x,y)=0$ if and only if $x=y$

*$\delta(x,y)=\delta(y,x)$ for all $x,y\in X$

*there exist a constant $K>0$ such that for every $x,y,z\in X$ it holds $$\delta(x,y)\leq K(\delta(x,z)+\delta(y,z))$$
It's not to difficult to see that $\delta$ induces a uniform structure by the sets $\{(x,y):\delta(x,y)<r\}$. And then we have a topology in $X$ induced by $\delta$.
For what I have read, a $\delta$-ball centered at $x$  should be a neighborhood for such $x$. I'm looking for a proof of this fact.
 A: It would be enough to show that every $\delta$ ball is itself an open set, since any open set contains itself. To do this, let $\epsilon > 0$ arbitrary, and denote $B_{\delta} (x, \epsilon)$ as the $\delta$ ball of radius $\epsilon$ around $x$. Then, choose $y \in B_{\delta} (x, \epsilon)$ and consider $B_{\delta} (y , \epsilon/K - \delta (x,y) )$, $K>0$ is the constant defined in your post. If we can show that $B_{\delta} (y , \epsilon/K - \delta (x,y) ) \subset B_{\delta} (x, \epsilon)$, then we are done, since then $B_{\delta} (x, \epsilon)$ contains an open ball around all of its points. One issue here is the condition on $K$: If $K \leq 1$, then $\epsilon/K - \delta(x,y) >0$, and we have the following: 
Suppose $z \in B_{\delta} (y , \epsilon/K - \delta (x,y) )$. Then,
$$\delta (x,z) \leq K (\delta (x,y) + \delta (y,z)) < K (\epsilon/K) = \epsilon$$
So $z \in B_{\delta} (x, \epsilon) \implies B_{\delta} (y , \epsilon/K - \delta (x,y) ) \subset B_{\delta} (x, \epsilon) $, as desired.
Now when $K>1$ it is less clear: We require that the $y$ we choose is such that $y \in B_{\delta} (x, \epsilon/K)$. From here we can do all of the above work and conclude the same thing, but we still have the additional restriction on the $y$ we chose. This is not necessarily an open set, but we can conclude that it properly contains another set containing $x$.
A: For $ K < 1$ quasi-distance is not defined, infact if you set $z = y$ and $x \ne y$ in generalized triangular inequality you get $\delta \left( {x,y} \right) \leqslant \delta \left( {x,y} \right)$, i.e. $K \geqslant 1$. For $K = 1$ you get an usual metric space, and in relative topology the balls are open and the family of balls is a base. In general case $ K > 1 $ the family of balls is instead a base of neighborhood and the balls are not 
necessarily closed.
