Base of topology for metric-like space Let $X$ be a nonempty set and $p:X\times X\rightarrow\mathbb{R}^+ $ be a function satisfying the following conditions for all $x,y,z\in X$: \begin{align} &1)\enspace p(x,y)=0\implies x=y \\ &2)\enspace p(x,y)=p(y,x)\hspace{1,2cm}\\ \hspace{0,2cm}&3)\enspace p(x,z)\leq p(x,y)+p(y,z) \end{align}
Then the pair $(X,p)$
is said to be a metric-like space.
I want to show please that each metric-like $p$
on $X$ generates a topology $τ_p$ on $X$ whose base is the family of open-balls  $$B(x,\varepsilon)=\{y\in X:|p(x,y)-p(x,x)|<\varepsilon\}.$$
Thank you.
 A: Based on Yemon Choi's suggestion I am posting here an answer. So far it is mostly a summary of stuff which is was said in a post on another site and in the comments above. But if you have something to add (and it is not enough for a separate answer), feel free to edit this. (After all, this is community wiki.) 
A quick Google search leads to the paper A. Amini-Harand: Metric-like spaces, partial metric spaces and fixed points, doi.org/10.1186/1687-1812-2012-204. This paper contains a definition of metric-like space in the same way as given in the question and contains a claim that the open balls, defined as above, indeed give a topology (without a proof).
Counterexamples


*

*Does a metric-like space generate a topology if open balls are defined as $B_\sigma(X,\varepsilon)=\{y\in X; |\sigma(x,y)-\sigma(x,x)|<\varepsilon\}$? posted on Mathematics Stack Exchange contains a counter-example of a metric-like space on 3-points such that the open balls defined in this way do not give a topology.

*mlk's answer to the same question gives a wider class of counter-examples. If we choose any function $m\colon X\times X\to\mathbb R$ that is bounded, symmetric and fulfills $m(x,x)=0$, then the function given by $p(x,y)=m(x,y)+2M_0$ for $M_0=\sup |m(x,y)|$ is metric-like. Moreover, the open balls are exactly the sets $\{y\in X; |m(x,y)|<\varepsilon\}$. It is not difficult to find $m$ such that this does not give a topology.


Related notions


*

*As mlk points out in their answer, there is a related notion of patrial metric, which is also mentioned in Amini-Harand's paper. One of the reason for the problems might be that a different version of triangle inequality is needed. The definition of partial metric requires:


*

*$x=y$ iff $p(x,y)=p(x,x)=p(x,y)$

*$0\le p(x,x) \le p(x,y)$ 

*$p(x,y)=p(y,x)$

*$p(x,z) \le p(x,y)+p(y,z)-p(x,x)$


*The Wikipedia article on metric (current revision) contains a definition of metametric which is exactly the same as the above definition of metric-like function. The reference given there is: Väisälä, Jussi (2005), "Gromov hyperbolic spaces", Expositiones Mathematicae, 23 (3): 187–231, doi: 10.1016/j.exmath.2005.01.010. However, the topology in this paper is defined differently. (For example, a point $x$ is isolated whenever $p(x,x)>0$.)

*The Wikipedia article on metric also defines the notion of premetric (current revision) where only conditions $d(x,x)=0$ and $d(x,y)\ge0$ are required. (Including the warning that this is not a standard term and terminology can vary). Clearly, $d(x,y)=|p(x,y)-p(x,x)|$ is a premetric. According to the Wikipedia article, every premetric gives a topology but in this way: A set $U$ is open if for every $x\in U$ there exist some ball $B(x,\varepsilon)=\{y\in X; d(x,y)<\varepsilon\}$ with $B(x,\varepsilon)\subseteq U$. It is explicitly mentioned that these balls are not necessarily open. (So the balls described here are not necessarily a base. And the topology is obtained from these balls in a different way than described in the linked paper.) This way of obtaining a topology is analogous to the way a topology is obtained from a metametric in Väisälä's paper.

