# A Borel algebra containing $\infty$

If I adjoint $\infty$ to the real numbers ($\overline{\mathbb{R}}=\mathbb{R}\cup\infty$) is there a reasonable way to define a $\sigma$-algebra "$\mathcal{B}_{\overline{\mathbb{R}}}$" such that $\mathcal{B}_{\overline{\mathbb{R}}}$ restricted to sets not containing $\infty$ is equal to the Borel algebra $\mathcal{B}_{\mathbb{R}}$ ? If so, what topology on $\overline{\mathbb{R}}$ would induce $\mathcal{B}_{\overline{\mathbb{R}}}$ ?

-
By the way, I believe that we are more interested in $R \cup \{ -\infty, +\infty\}$ than in $R \cup \{\infty\}$. –  user52350 Dec 9 '12 at 12:02

You can do it by considering the Borel sets of the unit circle in the complex plane. This is homeomorphic to the one-point compactification of the real numbers, which is how one should look at $\overline{\mathbb R}$.

It can be shown that the Borel subsets of the circle are simply Borel sets of the plane intersected with the circle. It also turns out that the sigma algebra is not different than that of the real numbers.

-
Equivalently, topologize $\overline{\mathbb R}$ as the one-point compactification of the line. It is a Polish space, and the theory of Borel sets in that setting is very nice. –  GEdgar Dec 7 '12 at 17:03

First note that to any topological space you can define the $\sigma$-algebra of its Borel sets.

In general, given any topological space $X$ and a subspace $Y$ of $X$, if $Y$ is a Borel subset of $X$, the the Borel subsets of $Y$ (taken as a topological space in its own right) are exactly the Borel subsets of $X$ which happen to be subsets of $Y$.

Usually, one takes $\overline{\mathbb{R}}$ to be the one-point compactifaction of $\mathbb{R}$ (as in Asaf's answer and GEdgar's comment to it). The open subsets of $\overline{\mathbb{R}}$ would then be the open subsets of $\mathbb{R}$, together with all sets of the form $U \cup \{ \infty \}$ where $\mathbb{R} \setminus U$ is compact (in $\mathbb{R}$). Then $\mathbb{R}$ is an open (hence Borel) subspace of $\overline{\mathbb{R}}$, and therefore the Borel subsets of $\overline{\mathbb{R}}$ which are subsets of $\mathbb{R}$ coincide with the Borel subsets of $\mathbb{R}$.

-