TITLE: A Borel algebra containing $\infty$
QUESTION [7 upvotes]: 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 $\mathcal{B}_{\overline{\mathbb{R}}}$
?

REPLY [4 votes]: 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 $\mathbb{R}$.