Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
The extended real line is compact. Open neighborhoods of $-\infty$ contain a set $(-\infty, x)$ for some $x \in \mathbb{R}$ and similar for $+\infty$. –  WimC Feb 26 '12 at 22:58
    
Thanks WimC. :-) –  Anand Feb 26 '12 at 23:02

1 Answer 1

up vote 4 down vote accepted

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

$$h(x)=\begin{cases} \frac12+\frac1\pi\arctan x,&\text{if }x\in\mathbb{R}\\\\ 1,&\text{if }x=\infty\\ 0,&\text{if }x=-\infty\;. \end{cases}$$

share|improve this answer
    
Thanks Brian M. Scott. :-) –  Anand Feb 26 '12 at 23:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.