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At here, Example $8.2$, there is this statement:

Consider any countable and dense subset $\{ x_n : n \in \mathbb{N} \}$ of the unit ball of $X$ and let $K = \{ \frac{x_n}{n} : n \in \mathbb{N} \} \cup \{ 0 \}$. Plainly, $K$ is (weakly) compact and $X = \overline{span(K)}$.

Question: Why $K$ is compact and and $X = \overline{span(K)}$?

Idonknow
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1 Answers1

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To see that $X=\overline{span(K)}$, it's enough to show that $span(K)$ is dense in $X$ - so it's enough to show that $\{x_n: n\in\mathbb{N}\}\subseteq span(K)$. But this is easy, since $x_n={x_n\over n}+. . . +{x_n\over n}$ ($n$ times).

To show that $K$ is weakly compact: suppose I have an open cover $\mathcal{U}$ of $K$. Some open set $U_i\in\mathcal{U}$ contains $0$, since $0\in K$. What can you tell me about $K\setminus U_i$?

Idonknow
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Noah Schweber
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  • Why is it enough to show that ${ x_n : n \in \mathbb{N} } \subseteq span(K)$? Is it because after taking closure on both sides, we obtain $B_X \subseteq \overline{span(K)}$? – Idonknow Oct 26 '15 at 03:37
  • Do you understand why the closure of a dense subset of $X$ contains $X$? – Noah Schweber Oct 26 '15 at 05:08
  • If $D$ is a dense subset of $X$, then $\overline{D} = X$. In particular, we have $\overline{D} \supset X$. Am I right? – Idonknow Oct 26 '15 at 08:23
  • If ${ x_n: n \in \mathbb{N} }$ is a dense subset of $X$, then I can understand why it is enough to show that ${ x_n: n \in \mathbb{N} } \subseteq span(K)$. But in this case ${ x_n: n \in \mathbb{N} }$ is a dense subset of a unit ball of $X$... – Idonknow Oct 28 '15 at 01:52
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    If you contain a dense subset of the unit ball, and you're a subspace, then you contain a dense subset of the whole space. This is a good exercise. – Noah Schweber Oct 28 '15 at 01:57