# Example of a strong b-metric which is not a metric.

Definition: Let $$X$$ be an arbitrary set, $$d:X\times X\to [0,\infty)$$ be a mapping satisfying:

(a) $$\forall_{x,y\in X}\; d(x,y)=0\iff x=y$$;

(b) $$\forall_{x,y\in X}\; d(x,y)=d(y,x)$$;

(c) $$\exists_{s\geq 1}\;\forall_{x,y,z\in X}\; d(x,y)\leq d(y,z)+sd(x,z)$$.

Then $$d$$ is called a strong b-metric and $$(X,d)$$ is called a strong b-metric space.

Naturally, every metric space is a strong b-metric space as it fulfills (c) with $$s=1$$ (the classic triangle inequality).

Are there any natural examples of strong b-metric spaces which are not metric spaces? Note that every finite set $$X$$ equipped with a mapping $$d$$ fulfilling (a) and (b), fulfills (c) as well. Hence we are interested in examples on an infinite set $$X$$.

• Consider s < 1 for counter examples, especially s = 0 and s = -1. Commented Jun 9, 2019 at 21:36
• @WilliamElliot It says $\exists s \ge 1$ so those $s$ you mention are not allowed. Commented Jun 10, 2019 at 8:26
• where are these things from? an observation - If $d(x, y) \leq d(y, z)+s d(x, z)$ then also $$d(x,y)=d(y,x) \le d(x,z) + s d(y,z)$$ so that $$d(x,y) \leq \frac{1+s}2\left(d(x, z) + d(y, z)\right)$$ Commented Jun 13, 2019 at 13:14

## 1 Answer

As recognized by Calvin Khor in the comments, one has $$d(x,y) \le \frac{s+1}{2}(d(x,z)+d(z,y)).$$ Thus, we have a quasi-metric. I am not sure, if this name is used in the case of metric spaces. However for norms this is a well-known concept, called quasinorm. For example, the spaces $$L^p(\mu)$$ and $$l^p$$ for $$p \in (0,1)$$ are quasi-normed spaces and even quasi-banachspaces. Much simpler one can take $$d(x,y) = |x-y|^n$$ on $$\mathbb{R}$$ to get a simple example.

However, there are no examples of b-metric spaces which are not metric spaces. The proof is not really complicated, see the paper On the Macias-Segovia Metrization of quasi-metric spaces by R. Aimar, B. Iaffei , L. Nitti, published in Revista de la Unión Matemática Argentina 41(2) (1998).

• +1 I wasn't sure how to go in the reverse direction i.e. from the symmetric quasinorm condition to the asymmetric $b$-metric condition. Commented Jun 14, 2019 at 18:16