About the extreme value theorem over the extended real line This problem might seem very elementary to someone. I am using the following arguments. I would like to make sure that it is right. 

Let $f$ be a continuous function defined on $R$. If $\lim_{x\rightarrow\pm\infty}f(x)$ exists, then $f$ is a continuous function over the extended real line $R\cup\{\pm\infty\}$. Then we can say that there exists $x\in R\cup\{\pm\infty\}$ such that
$$
f(y) \le f(x), \qquad \forall y\in R.
$$

The usual extreme value theorem is stated over a compact set $[a,b]$; see the wikipedia.
Thank you very much! 
Anand

The argument works, because the extended real line is compact: $\mathbb{R}\cup\{\infty,-\infty\}$ with the natural order topology is homeomorphic to $[0,1]$. An explicit homeomorphism is
$$h(x)=\begin{cases}
\frac12+\frac1\pi\arctan x,&\text{if }x\in\mathbb{R}\\\\
1,&\text{if }x=\infty\\
0,&\text{if }x=-\infty\;.
\end{cases}$$
