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Suppose that$(E_n)$$_{n \in \mathbb N}$ be a sequence of closed and bounded sets in complete space $M$ such that $ E_{n+1} \subseteq E_n$ for all $ n \in\mathbb N$. If $\lim \operatorname{diam} E_n $= $0$, prove that $E$ contains one point where $E = \bigcap_{n \in\mathbb N} E_n$. $(E_n)$ $_{n \in \mathbb N}$ are bounded and closed

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  • $\begingroup$ What's a prefect space? $\endgroup$ Commented Nov 27, 2014 at 16:37
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    $\begingroup$ If any Cauchy (note spelling and capital letter) sequence converges, then the space is complete. Perfect is something else ... unless you insist on using "perfect" ... or "prefect" ... in the above sense in your question. What is bounded? Is your space a metric space? $\endgroup$
    – Mirko
    Commented Nov 27, 2014 at 16:58
  • $\begingroup$ Yesi changed it. $\endgroup$ Commented Nov 27, 2014 at 17:10

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Hint. Pick $x_n\in E_n$. Given any $\varepsilon>0$ use that $\operatorname{diam} E_n\to 0$ to pick $N$ such that $\operatorname{diam} E_N<\varepsilon$. Note that if $n,m>N$ then $x_n,x_m\in E_N$, hence $d(x_n,x_m)<\varepsilon$. Hence the $x_n$ form a Cauchy sequence. Since $M$ is a complete (and you should have stated metric) space, this sequence converges to some $x$. Use that each $E_n$ contains a tail of this sequence and is closed, hence each $E_n$ contains the limit $x$.

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