There is a typo in the question that makes it difficult to understand what is being asked, but if one replaces the $d$'s with $p$'s in the bottom part of the question, then an interesting question emerges. (@youssef: I think, but I'll have to check details, that the answer is no in general, but yes if $p$ is continuous.) I'm voting to reopen. — Will Brian Dec 6 at 17:50
4:54 PM
@WillBrian As a follow-up to your edit I have also corrected a few minor typos and explicitly added to the post that this is different from metric space. (Since this can be missed if somebody does not read carefully.) 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. — Martin Sleziak Dec 6 at 18:02
It seems that Wikipedia calls this a metametric. A reference given there is Väisälä, Jussi (2005), "Gromov hyperbolic spaces", Expositiones Mathematicae, 23 (3): 187–231, doi: 10.1016/j.exmath.2005.01.010. From this paper: "A metametric space is metrizable. In fact, a metametric $d$ can be changed to a metric $d_1$ simply by setting $d(x,x)=0$ and $d_1(x,y)=d{x,y}$ for $x\ne y$. Then $d$ and $d_1$ define the same topology." — Martin Sleziak Dec 7 at 1:00
@WillBrian Since it seems that you are a bit interested in the question (you mentioned that you voted that reopen), I wanted to let you know that I have posted the question (with some additional context) on another site: Does metric-like space generate a topology? — Martin Sleziak 2 days ago
@MartinSleziak I have cast the final vote to reopen, so that you can put some of the details from your earlier comments into an answer below — Yemon Choi 16 hours ago
You can find links to some basic info about comment replies, for example, here. (In particular, they do not work if you add space after the @.) — Martin Sleziak 1 min ago
-4
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} Th...
@YemonChoi I did post a CW-answer as you suggested. I did not add anything that was not already said in the linked math.SE post. But perhaps somebody will add something more now that the post is open again.
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