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1) Every compact space is locally compact.

2) Every metric space is a Hausdorff space.

3) $\mathbb{R}^n$ is a locally compact space.

Proof:

1) Suppose $X$ - compact space. Taking $p\in X$ we see that $X$-open set containing $p$ since $X\in \tau$. Also $\overline{X}=X$ - compact. Hence $X$ is locally compact.

2) Suppose $(X,d)$ - metric space. Taking $p,q\in X$ with $p\neq q$ then neighborhoods of $p$ and $q$, $N_{\delta}(p)$ and $N_{\delta}(q)$, respectively with $\delta=\frac{1}{4}d(p,q)$ have empty intersection.

3) I have some problems with that point. I understand that there I have to use that compact subsets in $\mathbb{R}^n$ are exactly bounded and closed.

Can anyone please help with this point?

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    $\begingroup$ You need to show that for each point of $\mathbb{R}^n$, it has a compact neighborhood. That is, a compact set containing an open neighborhood of the point. Then the closure of the open neighborhood, as a closed subset of a compact set, is compact. $\endgroup$ – Qiyu Wen Jun 18 '16 at 4:56
  • $\begingroup$ @QiyuWen, I have the following idea: Taking $x\in \mathbb{R}^n$. Then $N_{\varepsilon}(x)$ is open set containing $x$ since $N_{\varepsilon}(x)\in \tau$. But $\overline{N_{\varepsilon}(x)}$ is a closed ball (because we are dealing with usual euclidean METRIC). Hence $\overline{N_{\varepsilon}(x)}$ is closed and bounded $\Rightarrow$ compact! Am I right? $\endgroup$ – ZFR Jun 18 '16 at 5:04
  • $\begingroup$ Yes, you are right. $\endgroup$ – Qiyu Wen Jun 18 '16 at 5:07
  • $\begingroup$ @QiyuWen, Thanks for hint! $\endgroup$ – ZFR Jun 18 '16 at 5:07
  • $\begingroup$ @RFZ Well,you don't need to use the closure for this,but that's certainly another way to go! $\endgroup$ – Mathemagician1234 Jun 18 '16 at 5:13
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By the Heine-Borel theorem, every closed and bounded subset of $\mathbb R^n$ is compact. So let B be a closed ball in $\mathbb R^n$ centered at x of radius r.Consider the open ball A centered at x of radius r/2. Clearly, $A\subset B$. Also,since B is a compact subset of $\mathbb R^n$, every open cover of B has a finite open subcover-that is,there is a family of open balls B ={$B_r$'(x) | r'$\leq r$} such that B$\subseteq\cup$B and there exists a finite collection S$\subseteq$B such that B B$\subseteq\cup$B . Since $A\subset B$, then A$\subseteq\cup$B and A$\subseteq\cup$B. Since A is an arbitrary open subset of B, we're done!

You can also make the argument more detailed by expressing the open sets in B and S explicitly as open balls with chosen radii. But I was too lazy to do that.....lol

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  • $\begingroup$ So you want to say that my approach in above comment is wrong? $\endgroup$ – ZFR Jun 18 '16 at 5:15
  • $\begingroup$ @RFZ NO,it looks correct. It's just different from mine. $\endgroup$ – Mathemagician1234 Jun 18 '16 at 5:25

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