Is the infinity function continuous? Define the function $f: \mathbb{R} \to \mathbb{R}\cup\{\infty\}$ as $f(x) = \infty$ for all $x \in \mathbb{R}$. 
Is this function continuous? 
Intuitively it seems like it is, but I cannot prove this function is continuous in the usual $\varepsilon-\delta$ way because the difference $|f(x)-f(y)|$ is undefined for all $x,y \in \mathbb{R}$. 
This problem came up when I was trying to determine if a continuous function $g: \mathbb{R} \to \mathbb{R}\cup\{-\infty\}\cup\{\infty\}$ must be finite a.e. Any insight is appreciated. 

You are introducing the set $\bar{\mathbb R}:={\mathbb R}\cup\{-\infty,\infty\}$ as codomain of functions $f:\>{\mathbb R}\to\bar{\mathbb R}$. If you want to discuss continuity of such functions you have to introduce a topology on $\bar{\mathbb R}$. The usual topology is obtained by letting the sets $\ ]M,\infty]$, $M>0$, form a neighborhood base of $\infty$, and similarly for $-\infty$. In this way $\bar{\mathbb R}$ becomes homeomorphic to the interval $[-1,1]$.
Under this setup the function $f(x):=\infty$ is admissible, and is continuous. It is a totally different matter when we are asking whether this function is integrable, even on a finite interval $[a,b]$. Of course this is not the case.
