# 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

1. $$p(x,y)=0$$ iff $$x=y$$
2. $$p(x,y)=p(y,x)$$
3. $$p(x,z)\leq s[p(x,y)+p(y,z)]$$

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 :(

• Note the use of \text{} in MathJax, as in my edit to this question. May 28 at 2:04
• Thank you for the comment, I edited the question. May 28 at 2:42
• You will probably need to explain what a $b$-metric is in order to get an answer to your question May 31 at 10:35

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 $$d$$.

• Thank you so much for the answer. What I needed to show is that $d$ (a function from a metric space $X\times X$ to another metric space $\mathbb{R}$) is not continuous at $(\infty,1)$. And the definition you're using is the one I used, but I get confused when my prof asked what is the neighborhood around $(\infty,1) \in X \times X$. I know that in extended real system, the neighborhood of $\infty$ is a subset of $\mathbb{R}$ which contains all sufficiently large real numbers. But what about $X\times X$? What is the neighborhood of $(\infty,1)$? Jun 2 at 12:56
• @math-newbie there is no "the" neighborhood, there can be multiple neighborhoods. Jun 2 at 15:05