Discontinuity of $b$-Metric Here I have a definition of $b$-metric on a set: Let $s \geq 1$, $X$ is any nonempty set, and $p:X \times X \rightarrow [0,\infty)$ that satisfied

*

*$p(x,y)=0$ iff $x=y$

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

*$p(x,z)\leq s[p(x,y)+p(y,z)]$
for all $x,y,z\in X$. The function $p$ is called $b$-metric on $X$.
I have showed that if I have $X = \mathbb{N}\cup\{\infty\}$ and $d: X \times X \rightarrow \mathbb{R}$ where
$$d(m,n) = \begin{cases}
  0&,\text{for }m = n\\
  \left|\dfrac{1}{m} - \dfrac{1}{n}\right|&, \text{ for } m\text{ and } n\text{ are both even or } m\text{ is even and } n=\infty\text{ or}\\ 
  &\ \ m=\infty\text{ and } n\text{ is even }\\
  8&, \text{ for } m\text{ and } n\text{ are both odd or } m\text{ is odd and } n=\infty \text{ or }\\ 
  &\ \ m=\infty\text{ and } n\text{ is odd }\\
  5&, \text{ others}.
  \end{cases}$$
then $d$ is $b$-metric on $X$ with $s=3$, I want to show that $d$ (as function on a metric space) discontinuous at $(\infty,1) \in X \times X$ and this was how I tried to show the discontinuity.
Let $f: X\rightarrow\mathbb{R}$ where $f(x) = d(2x,1)$ for all $x\in X$ (the metric I use for $X$ and $\mathbb{R}$ is usual metric). I choose $\varepsilon_0 = 2$. By Archimedean Properties, for all $N > 0$ we have $m_N \in \mathbb{N}$ such that $N \leq m_N$. Thus
\begin{align*}
  |f(2m_N) - f(\infty)| = {} & |d(2m_N,1) - d(\infty,1)|\\
  = {} & |5 - 8| = 3 > \varepsilon_0.
  \end{align*}
So, $f$ discontinuous at $\infty$. And intuitively, $d$ discontinuous at $(\infty,1) \in X \times X$.
My question is:
Is there any properties I can use such that my intuitive (the last line) is right?
Any help is appreciated :(
 A: What does it mean for a b-metric to be continuous?
I will assume the following definition:
$$
d(x_n,y_n)\to d(x,y)
\qquad\text{for all sequences }x_n,y_n
\text{ with } x_n\to x, y_n\to y.
$$
In your work with the function $f$ you have basically shown that
$d(2n,1)$ does not converge to $d(\infty,1)$.
Therefore, we choose $x_n=2n$, $x=\infty$, $y_n=1$, $y=1$ in the above definition.
It remains to show that $2n = x_n\to x=\infty$ with respect to the b-metric,
or $d(2n,\infty)\to0$.
This can be calculated and should not be too hard.
The convergence $y_n\to y$ is trivial.
In summary, this shows that the b-metric $d$ is not continuous with respect to the b-metric $d$.
about neighboorhoods of $(\infty,1)$:
Neighborhoods are sets that contain an $\varepsilon$-ball around the point.
Thus, we need to describe the open $\varepsilon$-ball
$B_\varepsilon((\infty,1))$ with respect to the b-metric $d$.
Let $\varepsilon\in (0,1)$. Then one can show that
$$
B_\varepsilon((\infty,1)) = 
$\{(x,y)\in X\times X\mid d(x,\infty)+d(y,1)<\varepsilon\}
=\{2k \mid \frac1{2k} < \varepsilon, k\in\Bbb N\}\times \{1\} \cup \{(\infty,1)\},
$$
when one uses the definition of $d$.
