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the diameter of nested compact sequence

Let $(E_j)$ be a nested sequence of compact subsets of some metric space; $E_{j+1} \subseteq E_j$ for each $j$. Let $p > 0$, and suppose that each $E_j$ has diameter $\ge p$ . Prove that $E = \bigcap_{j=1}^{\infty} E_j$ also has diameter $\ge p$.

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    Hi Marcus, welcome to Math.SE. This looks like a homework question; for such questions we expect you to follow [these guidelines from the FAQ](http://meta.math.stackexchange.com/questions/1803/how-to-ask-a-homework-question). Please edit your question accordingly.2012-09-19

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For each $j$ pick two points $x_j, y_j \in E_j$ such that $d(x_j,y_j) \ge p$. Since $x_j \in E_1$ for all $j$, and $E_1$ is compact, the sequence $(x_j)$ has a convergent subsequence $(x_{\sigma(j)})$ say, and likewise $(y_{\sigma(j)})$ has a convergent subsequence $(y_{\tau \sigma(j)})$.

What can you say about the limits of these subsequences?

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    @ASlowLearner: It's more than six years since I posted this answer! In any case, I'm not sure why you'd need $E$ to contain only the limits of the sequences in order to show it has diameter $\ge p$, since even if it contains other points, they could only increase its diameter.2018-10-11