Let $\mathbb{\overline R}$ denote the extended real line. In a course on topology, I have heard people say that compact Hausdorff space is a space which topologists love and $\mathbb{\overline R}$ seems to carry a lot of those good properties.
So that’s why I don’t quite understand why we insist working with $\mathbb{R}$ instead of $\mathbb{\overline R}$? In undergrad analysis and topology, you almost never heard about $\mathbb{\overline R}$. What you hear instead is you hear your professor telling you that $\infty$ is not a number. But when we start talking about measure theory, which I had a hell of a time with, it seems to be standard to say that $m(\mathbb{R}) = +\infty$.
To me it would seem to ease a lot of the troubles with the reals by simply capping it off at the end. Maybe I exaggerated, the real has its problems (for example our computer cannot represent every real, the physical world is quantized but we use the reals to represent physical quantities anyways…), but at least this seems to me be the logical step towards having a more “comfortable” space to work with. So can someone give a strong reason why $\mathbb{\overline R}$ do not enjoy as widespread of use as $\mathbb{R}$?
With respect to $\mathbb{R}$, the extension $\bar{\mathbb{R}}$ lacks properties like: being a group, a ring, and a field.
There are two meaningful extension of $\mathbb R$: one with a single $\infty$ point and one with two points: $+\infty$ and $-\infty$. Sometimes one is better than the other… so there is not a clear choice here.
As pointed out in comments and answers, $\mathbb{R}$ has some nice properties that $\mathbb{\bar{R}}$ doesn’t. There is also the hyperreals that are a very neat construction – they contain infinite and infinitesimal numbers, and form a field. You can do analysis very nicely in them. For example, in the hyperreals, the definition of a derivative is roughly $f'(x)=\dfrac{f(x+dx)-f(x)}{dx}$ where $dx$ is any infinitesimal (specifically, you have to get rid of the hyperreal part and get $f'(x)=Re(\dfrac{f(x+dx)-f(x)}{dx})$ loosely speaking).
In day to day use, neither of these objects justify the added complexity over $\mathbb{R}$. And especially for $\mathbb{\bar{R}}$, it’s very easy to say “consider the one/two-point compactification of $\mathbb{R}$, and…”, as indeed I have done a few times myself.