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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.

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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?

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yes I think. Thanks, I was not sure. I will seek an independent demonstration of the topology :) – Hiperion Jul 11 '12 at 16:30
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

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.

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