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.

Heine-Borel theorem say: A subset in $\mathbb{R}^n$ is compact if and only if it is closed and bounded.

Is this theorem independent of the topology in $\mathbb{R}^n$?

If the answer is no, which is a counterexample?

I have seen a demonstration but uses the usual topology for $\mathbb{R}^n$.

Thank you for your help.

share|improve this question
add comment

2 Answers

up vote 5 down vote accepted

It is very dependent of the usual topology in $\,\Bbb R^n\,$ , of course. For example, if you take the discrete topology on this same space, then a subset is compact iff it is finite (meaning: it contains a finite number of elements.) Can you prove this?

share|improve this answer
    
yes I think. Thanks, I was not sure. I will seek an independent demonstration of the topology :) –  Hiperion Jul 11 '12 at 16:30
1  
Notice that often it is not interesting to introduce another topology: there exists only one separated topology compatible with the vector space structure of $\mathbb{R}^n$. –  Seirios Jul 11 '12 at 16:48
    
Ok thanks. I was trying to solve an exercise, I needed to be sure about my previous question. –  Hiperion Jul 11 '12 at 17:21
add comment

The Heine-Borel theorem doesn't just rely on the usual topology of $\mathbb{R}^n$, but it also relies on the usual, Euclidean, metric $d(x,y)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$. The usual metric is compatible with the metric $d'(x,y)=\frac{d(x,y)}{1+d(x,y)}$, which is bounded by 1 for all $x,y$. Now in the metric space $(\mathbb{R}^n,d')$, the set $[0,\infty)$ is closed, since it is closed in $(\mathbb{R}^n,d)$, and bounded since every subset is bounded. But, it is not compact, as the open cover $\{(-1,n)\}_{n\in\mathbb{N}}$ has no finite subcover.

share|improve this answer
add comment

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.