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 :(

  • 1
    $\begingroup$ Note the use of \text{} in MathJax, as in my edit to this question. $\endgroup$ May 28 at 2:04
  • $\begingroup$ Thank you for the comment, I edited the question. $\endgroup$ May 28 at 2:42
  • $\begingroup$ You will probably need to explain what a $b$-metric is in order to get an answer to your question $\endgroup$
    – postmortes
    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 $d$.

  • $\begingroup$ 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 $(\infty,1)$? $\endgroup$ Jun 2 at 12:56
  • $\begingroup$ @math-newbie there is no "the" neighborhood, there can be multiple neighborhoods. $\endgroup$
    – supinf
    Jun 2 at 15:05

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