Borel sigma algebra on Extended Real Line

Question

I am trying to understand the Borel Sigma Algebra on the extended real line. In order to do this, I must first understand the topology of the extended real line. In particular, I was wondering if $\{(a,b): a,b \in \mathbb{R} \}$ can generate (as in, use $\cup$ and complement of these to get all open sets) the extended real line topology. Since this set generates $\mathbb{R}$, we can find the complement of $\mathbb{R}$ and get $\{-\infty, \infty \}$, but I am not sure how to get the neighbourhoods of $\infty$ for example, $(a, \infty ]$.

Answer 1

As you have discovered, the finite open intervals are not enough to generate the usual topology of the extended real line.

In order to generate the topology you need to explicitly declare $[-\infty,a)$ and $(a,\infty]$ (for $a\in\mathbb R$ or even just for $a\in\mathbb Q$ or $a\in \mathbb Z$) to be open.

(Remember that when you're generating a topology, taking complements is not a valid operation because complement is not supposed to preserve openness).

If all you're interested in is generate the Borel algebra, less can do it -- but you still need some way to distinguish between $-\infty$ and $\infty$. However including just, for example $\{-\infty\}$ or $[-\infty,0)$ among your generators would do the trick.

Answer 2

The extended real line has a linear order, where $-\infty < x < \infty$ for all $x \in \mathbb{R}$,so also $-\infty< \infty$ and for $x,y$ reals $x < y$ means the same. This is just the "order completion" of $\mathbb{R}$.

Any linearly ordered set $X$ has an order topology, with as a subbase all sets of the form $L_a = \{x \in X: x < a\}$ and $U_a = \{x \in X: x > a\}$, the lower and upper sets w.r.t. all $a \in X$. We then have all open intervals $(a,b)$ as $L_b \cap U_a$ in the generated base, and all sets $[m, a) = L_a$ if $X$ has a minimum $m$ and $(a,M] = U_a$ if $M$ is the maximum for $X$ (if these exist).

So we get $[-\infty, x)$ as an open set of the extended reals, because we take the order topology on it, just as we do on $\mathbb{R}$.