A b-metric is defined similar to a metric in which the triangle inequality is replaced by the inequality $$d(x,z)\leq s\Big[d(x,y)+d(x,z)\Big]\quad\forall\ x,y,z$$ where $s\geq1$.
There is an example of a b-metric which is not continuous.
My question: If the b-metric space is complete, can we conclude that the b-metric is continuous?
Any help would be appreciated. Thanks!
Nope. Let $X$ be the set $\{0\} \cup \{\frac{1}{n}: n \in \mathbb{N}\} \subset \mathbb{R}$ together with one additional point $e$. The distance between two points neither of which is $e$ is just their usual distance in $\mathbb{R}$. Also set $d(e,e) = 0$, $d(e,0) = 2$, and $d(e,\frac{1}{n}) = 1$ for all $n$. This is a "b-metric" with $s = 2$ by inspection, and it is complete (any Cauchy sequence is either eventually constant or converges to $0$). But $\frac{1}{n} \to 0$ in the topology generated by $d$, while $d(e,\frac{1}{n}) \not\to d(e,0)$.
