Quick question. The metric $\displaystyle d(x,y)=|x-y|$ does not metrize $\displaystyle \mathbb{R}\cup\left\{-\infty,\infty\right\}$ right? Specifically because it fails the triangle inequality.
I would define $\displaystyle d(\infty,a)=\infty$ for any $\displaystyle a\in\mathbb{R}$. I believe that is the appropriate extension of the metric defined on $\displaystyle \mathbb{R}$, but then that implies that normal metic for the reals does not metrize the extended reals. Does that make sense?
P.S. Sorry about putting this in the wrong forum, the question arose in a topology setting but I can see how its more of an analysis question.
Extended real number line - Wikipedia, the free encyclopedia
But the sentence before can be interesting too. However, homeomorphisms represent a rude language to me So I'll leave it to you to understand that !There is no metric which is an extension of the ordinary metric on R.