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The following is a question about the answer given here:

I have been trying to prove that if $X$ is an infinite dimensional Banach space and $O\subseteq X$ is an open set such that its closure $\overline{O}$ is compact then contradiction. But I just can't do it.

Please could somebody help me prove it?

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    Can you show that the closed unit ball is not compact? $\overline O$ contains closed balls. – Jonas Meyer Sep 03 '14 at 05:23
  • http://math.stackexchange.com/q/253081/ – Jonas Meyer Sep 03 '14 at 05:26
  • @JonasMeyer Thank you for your comments. Yes, for example I know that the space is finite dimensional if and only if the closed unit ball is compact. But I can't argue like this: if a set $S$ contains a non-compact then $S$ is itself non-compact. A counter example is $(0,1)\subseteq [0,1]$. So how to use that $\overline{O}$ contains non-compact balls? I proceed with reading the linked post. –  Sep 03 '14 at 05:35
  • If a space has a closed subset that is not compact, then the space is not compact. (Think of the contrapositive.) – Jonas Meyer Sep 03 '14 at 05:37
  • @JonasMeyer Thank you, it's now so obvious that I don't know why I didn't figure it out myself. –  Sep 03 '14 at 06:18

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The Riesz Lemma: Let $X$ be an infinite-dimensional normed linear space. Let $\alpha \in (0,1)$ be given. Then there exists $\{ x_{n} \}_{n=1}^{\infty} \subset X$ such that (a) $\|x_{n}\| =1$ and (b) $\|x_{n}-x_{m}\| \ge \alpha$ for all $n \ne m$. So, $\{ \frac{r}{2} x_{n} + x \}_{n=1}^{\infty} \subseteq B_{r}(x)$ for any $r > 0$ and any $x \in X$, which spoils compactness for any set in an infinite-dimensional normed space containing an open ball. http://en.wikipedia.org/wiki/Riesz%27s_lemma

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