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I have already looked at previous posts 1 and 2, which are related to my question, but mine is a combination of the two.

My question relates to defining a metric on the extended real line $$\bar{\mathbb{R}} : = [-\infty, \infty] := \mathbb{R} \cup \{\infty, -\infty\}.$$

In 1 they talk about not being able to define a metric on the extended real line with the usual distance metric on $\mathbb{R}$. I understand that this is not possible if we require that metrics are finite valued functions (i.e., with range $[0,\infty))$. However, in 2 they talk about allowing metrics to actually take the value $\infty$ (i.e., with range $[0,\infty]:= [0,\infty)\cup \{\infty\}$). This all made me wonder if one can actually define a metric taking infinite values on $\bar{\mathbb{R}}$ that is consistent with the usual distance metric on $\mathbb{R}$ (i.e., when we restrict to computing the distance between two (finite-valued) real numbers).

This certainly seems to be possible. Indeed, define a function $d:\bar{\mathbb{R}}\times \bar{\mathbb{R}}\rightarrow [0,\infty]$ by \begin{align*} &d(x,y) = \vert x-y\vert,\quad x,y\in\mathbb{R},\\ &d(x,\infty) = d(\infty, x) = d(x,-\infty) = d(-\infty, x)=\infty,\quad x\in\mathbb{R},\\ &d(-\infty,\infty)= d(\infty,-\infty)=\infty,\\ &d(\infty,\infty) = d(-\infty,-\infty) =0, \end{align*} By definition, $d$ is symmetric and the triangle inequality holds, with the addition operator extended to $[0,\infty]$ by requiring that \begin{align*} a+\infty = \infty + a = \infty,\quad 0\le a\le \infty, \end{align*} and where we take $\le$ to be the usual order on $\bar{\mathbb{R}}$.

Thus, the pair $(\bar{\mathbb{R}}, d)$ seems to be a valid metric space, if we extend our definition of metric spaces to include "metrics" that actually take infinite values, and if we extend the addition operator to $[0,\infty]$ from $[0,\infty)$, and extend the order relation $\le$ to $[0,\infty]$ from $[0,\infty)$.

I have not thought too much past this, but was hoping for some insights, comments, etc. I have the following follow-up questions:

  1. Is this all valid? And if so, does doing any of this useful in any way? Or at least have some mathematical meaning? Do people work with these objects? Et cetera. I think you get the point.
  2. Does this infinite metric (or any infinite metric in general) give rise to topologies on the set of interest? If so, in this example of the extended real line, how does the induced "metric" topology compare with the usual topology on $\bar{\mathbb{R}}$ with basis elements $(a,b), (a,\infty], [-\infty , a)$, $a,b,\in \mathbb{R}$ (i.e., the topology induced by the order relation on $\bar{\mathbb{R}}$)?

I can't think of other questions, but any other insights would be appreciated!

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  • $\begingroup$ The reason that a "standard" metric on $\overline {\Bbb R}$ must be bounded is that the topology is compact.... The A by pompelle raises an "issue" that won't go away. $\endgroup$ Apr 9 '18 at 22:56
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The problem is that that metric doesn't give you the usual topology in which neighborhoods of $\infty$ are sets of the form $(a,\infty]$. Instead the set $\{\infty\}$ turns out to be clopen, and so is $\{-\infty\}$.

An open ball of radius $r$ and center $\infty$ would be $\{x\in\overline{\mathbb{R}}:\ d(x,\infty)<r\}=\{\infty\}$. And for the same reason $\{-\infty\}$ is also open. They are closed because $\{\infty\}=(\{-\infty\}\cup\mathbb{R})^c$.

So, the topology induced by this metric is the disjoint union of $\mathbb{R}$ and a discrete set with two elements $\{-\infty,\infty\}$.

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