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The question is in the title:

Is there a (extended) metric on the extended reals which yields regular and infinite limits?

but in particular I want know the explicit construction of said metric.

Of course, by "yields regular and infinite limits" I mean: Let $\overline{\mathbb{R}} = \mathbb{R} \cup \{\pm \infty\}$. The (extended) metric on $\overline{\mathbb{R}}$ is so that: Given a sequence $a : \mathbb{N} \to \overline{\mathbb{R}}$ , with only real numbers in its image, we have: $$\lim a = a_\infty \in \mathbb{R}\Leftrightarrow \forall \epsilon > 0 : \exists N\in \mathbb{N} : \forall n \in \mathbb{N}, n\geq N : |a_\infty - a_n| < \epsilon$$ $$\lim a = \infty \Leftrightarrow \forall c > 0 : \exists N \in \mathbb{N} : \forall n \in \mathbb{N}, n\geq N : a_n > c$$ $$\lim a = -\infty \Leftrightarrow \forall c > 0 : \exists N \in \mathbb{N} : \forall n \in \mathbb{N}, n\geq N : a_n < c$$

Here $\lim a$ refers to he limit in the aforementioned metric space.

This is hinted here, but I still do not know how to find the metric in question.

Community
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Stefan Perko
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1 Answers1

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As hinted in your link, there is a bijection $f:\overline{\Bbb R} \to \left[-\frac \pi 2,\frac \pi 2\right]$ given by $$f(x)=\begin {cases} -\frac \pi 2 & x=-\infty\\\arctan x & -\infty \lt x \lt \infty \\\frac \pi 2 & x=+\infty \end {cases}$$ Now your metric on $\overline{\Bbb R}$ is $d(x,y)=|f(x)-f(y)|$

Ross Millikan
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  • Let me add that, somewhat importantly, this map induces a homeomorphism between $(-\frac \pi 2, \frac \pi 2)$ and $\mathbb R$. This ensures that the convergence behavior we induce by this metric on $\mathbb R$ is indeed the same as the one provided by the usual metric. Also, a small argument is needed to justify that „converging to ∞“ in the sense of this metric is precisely the same as divergence to infinity (which is a good exercise if it's not immediately clear). – Lukas Juhrich Jan 14 '22 at 20:46