$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\scrM}{\mathscr{M}}$
$\newcommand{\scrO}{\mathscr{O}}$
At a point in my measure theory class, we defined a Hausdorff space and my professor stated that evidently the extended real line is Hausdorff. I want to confirm this for myself. Here is my definition of Hausdorff,
Definition: A topological space $(X,\scrM)$ is called Hausdorff if for any two points $p,q \in X$ with $p \neq q$ there are open sets $U,V \in \scrM$ for which $p \in U$, $q \in V$ and $U \cap V = \varnothing$.
To define the Borel $\sigma$-algebra on the extended reals $\R^*$ we first defined basic open sets; sets of the form $[-\infty, a), (a,b) , (a, \infty]$ and from there we set $\scrO^*$ to be the minimal topology generated by the basic open sets of $\R^*$. From this we defined the Borel $\sigma$-algebra in $\R^*$ to be $\sigma(\scrO^*)$; which is the minimal $\sigma$-algebra generated by $\scrO^*$.
I wanted to prove to myself that $\R^*$ was in fact Hausdorff, but I'm having some conceptual issues. When we mean $\R^*$ is Hausdorff, we do in fact mean that the topology on $\R^*$ is the Borel $\sigma$-algebra on $\R^*$ right? So that the topological space itself is really $(\R^*, \sigma(\scrO^*))$? So I would need to show that for any two points $p,q \in \R^*$ that we can sequester them into disjoint Borel sets? If that's the case, does this attempt seem valid?
Attempt: Let $p,q \in \R^*$ and assume without loss of generality that $p< q$. Then the sets $[-\infty, p]$ and $[q,\infty]$ are certainly Borel sets since $[-\infty, q)^c = [q,\infty]$ and $[p,\infty)^c = [-\infty, p)$; i.e they are complements of basic open sets in $\R^*$. Also evidently $p \in [-\infty, p]$ and $q \in [q,\infty]$ but since $p<q$ it follows that $[-\infty,p] \cap [q, \infty] = \varnothing$ and so $\R^*$ is in fact Hausdorff.
Thanks in advance your help. If any clarifications are needed let me know, I'll be happy to edit the post.
Edit: It occurs to me that $\sigma(\scrO^*)$ is not necessarily a topology since $\sigma$-algebras are only closed under countable rather than arbitrary unions. Is it the case that the Borel $\sigma$-algebra here is in fact a topology on $\R^*$?
Best Answer
$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\scrM}{\mathscr{M}}$ $\newcommand{\scrO}{\mathscr{O}}$ The issue here is that I was confusing the measurability of functions $f: \Omega \rightarrow \R^*$ with what it means to have a topology.
With the definitions given in lecture, if we set $BO$ to be the collection of all basic open sets in $\R^*$, then we set $\scrO^*$ to be the minimal topology on $\R^*$ containing BO. This is a topology on $\R^*$ since $$ \scrO^* = \bigcap_{BO \subseteq T} T $$ and the arbitrary intersection of topologies on $\R^*$ is again a topology on $\R^*$. Elements of $\scrO^*$ are called open sets in $\R^*$. Then from this the pair $(\R^*, \scrO^*)$ constitutes a topological space.
To define the Borel $\sigma$-algebra on $\R^*$ we use the topology $\scrO^*$, but they are distinct notions. The collection $\sigma(\scrO^*)$ is the minimal $\sigma$-algebra generated by the open sets in $\R^*$ and the pair $(\R^*, \sigma(\scrO^*))$ constitutes a measurable space, not a topological space.
I had confused this with the fact that maps $f:\Omega \rightarrow \R^*$ are measurable if $f^{-1}(B) \in \mathscr{F}$ for any $B \in \sigma(\scrO^*)$, i.e the preimage of Borel sets are measurable.
So in summary, $\R^*$ is a topological space under the minimal topology containing all basic open sets, denoted $\scrO^*$. And it is also a measure space with the Borel $\sigma$-field $\sigma(\scrO^*)$ but one has to do with topology while one is related to measurability.