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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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    thanks for the feedback. it sounds good to me so far. the limits of these subsequences then must also be in the intersection, E?2012-09-19
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    Yes, but it requires some proof $-$ it relies in the fact that a compact subset of a metric space is closed. See if you can do it; if not, I'll give you a prod in the right direction.2012-09-19
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    if the limit of (xj) was not in E, then there would be an N such that x would not be in E_N. Since the E_j's are nested, x would not be in E_m for all m >= N. Since E_j is closed for all j, E_j contains all of its limit points. Thus, there exists a ball B(x, d(x, E_M) for which no member of (x_j) is in and so x_j could not converge to x. Similarly, the limit y of y_j is also in E. (Excuse my lack of formatting)2012-09-19
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    I think that's correct, but it's quite convoluted. It suffices to say: for each $k$, the sequence $(x_{\sigma(j)})$ eventually lies in $E_k$. Since each $E_k$ is closed, its limit lies in $E_k$ for each $k$, and hence in $E$. Same goes for $(y_{\tau \sigma(j)})$. You're not done yet, though; you still have to prove that $\text{diam}(E) \ge p$.2012-09-19
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    However, I think you may need to use double sub-sequences... For instance, if x_j = (a,b,a,b..) and y_j = (b,a,b,a..) and d(a,b) >= p, then a subsequence x_j_k = (a,a,a,a...) and y_j_k = (a,a,a, ..) and then lim x_j_k = a = lim y_j_k and that would be wrong2012-09-19
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    so now we have that x = lim (xσ(j)) and y = lim (yτσ(j) are in E. i'm somewhat confused though on how to prove that d(x,y) >= p, especially given the fact that the subsequences could converge to the exact same point, as I mention in the comment right above2012-09-19
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    I have used double-subsequences; that's why we have $y_{\tau \sigma(j)}$ and not $y_{\sigma(j)}$. Then $x_{\tau \sigma(j)}$ is also convergent and by construction we have $d(x_{\tau \sigma(j)}, y_{\tau \sigma(j)}) \ge p$ for each $j$. Use the fact that $d$ is continuous.2012-09-19
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    Also your $a,b$ example doesn't work, since if your $j_k$s were all the same then (necessarily) if $x_{j_k} = a$ then $y_{j_k}=b$.2012-09-19
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    well, since d is continuous, taking the limit of both sides results in: lim j->infinity d(xτσ(j),yτσ(j)) = d(lim j->infinity xτσ(j),lim j->infinity yτσ(j)) = d(x,y) >= p . eh?2012-09-19
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    Pretty much, yes; the fact is that $(d(x_{\tau \sigma(j)},y_{\tau \sigma(j)}))$ is a convergent sequence in $\mathbb{R}$ lying entirely in $[p,\infty)$, which is closed, and hence $d(x,y) \in [p, \infty)$.2012-09-19
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    @Clive Newstead How do we know that E contains only the limits of the sequences?2018-10-11
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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