Does the extended real line form a metric space such that we can define limits in the usual way?

I read on a blog that if we define $$d(x,y) = \left|\,\overline{\arctan}(x) - \overline{\arctan}(y)\right|\;,$$


$$\overline{\arctan}(x)= \begin{cases} \pi /2 & \text{if } x=\infty \\ -\pi/2 & \text{if } x= -\infty \\ \arctan(x) & \text{else} \end{cases} $$

This clearly gives a metric space, but I'm having trouble showing that limits in this metric space correspond to the usual ones. Basically, I'm trying to explain why $\lim_{x \rightarrow \infty} f(x)$ has a definition which is different from $\lim_{x \rightarrow p} f(x)$.

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    $\begingroup$ It has a different definition because $\infty$ is not an element of $\mathbb R$ $\endgroup$ – Ilya Mar 29 '12 at 17:18
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    $\begingroup$ Where are you having trouble? For finite $x$ the fact that $|x_n-x|\to 0$ iff $d(x_n,x)\to 0$ is basically just continuity of $\arctan$, and that $x_n\to\pm\infty$ iff $d(x_n,\pm\infty)\to 0$ follows pretty easily from the nature of $\arctan$. $\endgroup$ – Brian M. Scott Mar 29 '12 at 17:40

Expanding the comment by Brian M. Scott: $\tan:(-\pi/2,\pi/2)\to\mathbb R$ is a continuous function, and so is $\arctan:\mathbb R\to(-\pi/2,\pi/2)$. Therefore, for every $x\in\mathbb R$ $$x_n\to x \iff \arctan x_n\to \arctan x \tag1$$ Equivalence (1) expresses the fact that sequences with a finite limit converge (or fail to converge) equally well under both metrics.

There are differences too: introduction of new elements $\pm \infty$ in the space; the possibility that a sequence can converge to one of them.


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