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Let $x_n$ be a sequence in a Hilbert space such that $\left\Vert x_n \right\Vert=1$ and $ \langle x_n,\ x_m \rangle =0 $, for all $n \neq m$.

Let $ K= \{ x_n/ n : n \in \mathbb{N} \} \cup \{0\} $.

I need to show that $K$ is compact, $\operatorname{co}(K)$ is bounded, but not closed and finally find all the extreme points of $ \overline{\operatorname{co}(K)} $ .

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    $co(M)$ is the convex hull of $M$?2012-06-24
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    Yes, that's right.2012-06-24
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    Are you sure you stated this correctly? Is, possibly, $||x_n||=1/n$? Or are you supposed to show weak compactness?2012-06-24
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    Oh... I've made a mistake. Let me fix it!2012-06-24
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    Ok. Compact is easy then, since $$||x_n/n|| = ||x_n/n-0|| =1/n \rightarrow 0 \, (n\rightarrow \infty)$$2012-06-24
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    For the compactness, I am supposed to take an arbitrary bounded sequence $ y_n $ in K and show that it has a convergent subsequence.2012-06-24
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    Yes. Any subsequence of a convergent sequence converges.2012-06-24

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