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Let $H$ be a Hilbert space. Discuss the validity of the following statement: If ${K_n}$ is a decreasing sequence of nonempty, bounded, closed convex sets in $H$, then $\bigcap_n K_n \neq ∅$.

My work:

Let $L_n=\inf\{\|x\|:x\in K_n\}$. Then there is a unique $x_n\in K_n$ such that $\|x_n\|=L_n$. Now I wanted to prove that the sequence $\{x_n\}$ is cauchy but stuck. I found a similar solution here

Intersection of nested closed bounded convex sets in Euclidean space

but was not helpful because as I think it has some mistakes. Can anybody please help me?

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Extremal
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    What mistakes? This looks like a duplicate. – Jonas Meyer May 04 '15 at 21:14
  • Ok then in order to show that $(x_n+x_m)/2 \in E_m$ by convexity we need to have $x_n\in E_m$. But $m>n$. So how do we know that $x_n\in E_m$? – Extremal May 04 '15 at 21:17
  • And what is $R$ in that solution? – Extremal May 04 '15 at 21:19
  • I guess $E_n$ was intended, and $R=R_*$, but I haven't checked closely the details. – Jonas Meyer May 04 '15 at 21:29
  • That is why I mentioned that it has mistakes. And the last statement does not seems to be a contradiction as $R$ is the limit of an increasing sequence. – Extremal May 04 '15 at 21:32
  • For now I've deleted my answer. It said this: (BEGIN QUOTE) Let $\ell^2$ be the Hilbert space of sequences $(c_1,c_2,c_3,\ldots)$ of complex numbers such that $\sum_k |c_k|^2<\infty$. Let $$ K_n = \Big{ (,\underbrace{0,\ldots,0}{n-1}, c_n, c{n+1},c_{n+2},,\ldots) : \forall k\ge n,\ c_k\ge 0\text{ and } \sum_{k=n+1} c_k=1 \Big}. $$ Then $\ell^2\supseteq K_1\supseteq K_2\supseteq\cdots$ and these are closed, convex, and bounded, and $\bigcap_{n=1}^\infty K_n = \varnothing$. (END QUOTE) However, now I'm not sure these sets are closed. ${}\qquad{}$ – Michael Hardy May 04 '15 at 22:28
  • They are closed if the function that takes a sequence of non-negative numbers to its sum is continuous on the set of sequences of non-negative numbers whose sum is finite. That's unproblematic if one looks only at sequences of length $N$ for some integer $N$. ${}\qquad{}$ – Michael Hardy May 04 '15 at 22:32
  • $(,\underbrace{0,\ldots,0}_{n-1}, ,\underbrace{1/n, \ldots,1/n}_n,,0,0,0,\ldots)\to0$ as $n\to\infty$, because the norm of that vector is $1/n$. Hence the subsets I described are not actually closed. ${}\qquad{}$ – Michael Hardy May 05 '15 at 04:54

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