$(\alpha, \infty]$ as a union of open sets. I would like to show that $(\alpha, \infty]$ is a Borel set using the definition of the Borel $\sigma$-algebra. I'm having trouble with including the infinity, because $\bigcup_n (\alpha, n) = (\alpha, \infty)$. If I work with the complements of $[-\infty, \alpha]$, I'm having trouble approaching infinity too. 
I know this is trivial, I just need a hint.
Thanks!
 A: By the definition of the topology on the extended real line $[-\infty,+\infty]$, a set of the form $(\alpha,+\infty]$ is open, hence it is a Borel set.
A common way to define the topology on $[-\infty,+\infty]$ is by specifying a bijection to a compact interval $\subset \mathbb{R}$, e.g. $h\colon [-1,1] \to [-\infty,+\infty];\quad h(x) = \frac{x}{1-\lvert x\rvert}$ for $-1 < x < 1$ and $h(-1) = -\infty;\; h(1) = +\infty$, and positing that the bijection be a homeomorphism, i.e. declaring a subset $U \subset [-\infty,+\infty]$ open if and only if $h^{-1}(U)$ is open in $[-1,1]$. Familiarity with the interval gives (hopefully) a good help dealing with $[-\infty,+\infty]$ as a topological space.
Another common way is to define neighbourhoods, then the points of $\mathbb{R}\subset [-\infty,+\infty]$ have their usual neighbourhood bases, and neighbourhood bases of $\pm\infty$ are given by sets of the form $(\alpha,+\infty]$ resp. $[-\infty, \alpha)$. It is a good exercise to verify that this construction yields the same topology as the other one.
