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In euclidean metric space $\mathbb{R}^k$ with its usual metric, let $E$ be a compact nonempty subset of $\mathbb{R}^k$, and $\delta = \sup\{d(x,y) | x,y \in E\}$. Show that there exist $x_0, y_0 \in E$ such that $d(x_0,y_0) = \delta.$

I am not sure what approach will work. Actually, using basic figure like circle in $\mathbb{R}^2$, the point giving maximum of distance should be on its boundary. I try to use sequences :

For each $n \in \mathbb{N}$, there exists $u_n = d(x_n,y_n)$ such that $$|u_n - \delta| < \frac{1}{n}.$$ So $u_n \rightarrow \delta.$ I try to use the property of closed set that it contains every limits of its sequence. But I do not know whether $(x_n), (y_n)$ are convergent. Using their convergent subsequences seems do not work since the index might not be the same so I cannot use the face $|d(x_{n_l},y_{n_k}) - \delta| < \epsilon$.

Any suggestion or approach I should use to solve this problem ?

user117375
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  • related: http://math.stackexchange.com/questions/208943/the-diameter-of-a-compact-set – Closure Feb 28 '15 at 06:09
  • Note that $(x,y)\mapsto d(x,y)$ is a continuous function from ${\mathbb R}^k$ to $[0,\infty)$. Since $E$ is compact nonempty set this function has a maximum on it. – Janko Bracic Feb 28 '15 at 06:52
  • I cannot use continuity now. It is not defined generally in metric space yet. If I can use continuity, then, like you point out, I then can use extreme value thm on compact set in metric space. But it is not what is valid now. – user117375 Feb 28 '15 at 13:15

2 Answers2

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It is actually not so hard. The trick is that you can first chose $y_{n_k}$ converges and then chose further $x_{n_{k_l}}$ converges.

Brian Ding
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In $\mathbb{R}^n$, $E$ being compact is equivalent to the fact that for any sequence $(x_n)$ in $E$, there is a subsequence $(y_n)$ that converges. Let's exploit this fact to get the result.

Take $(x_n,y_n)$ such that $d(x_n,y_n)\to \delta$. Then we may choose $(n_k)$ a subsequence of indices such that $(x_{n_k})$ converges. We still can't say that $(y_{n_k})$ converges. But $(y_{n_k})$ is a sequence in $E$, so we can find a subsequence $(n_{k_{\ell}})$ of $(n_{k})$ such that $(y_{n_{k_\ell}})$ converges. Now note that $(x_{n_{k_\ell}})$ is a subsequence of the convergent subsequence $(x_{n_k})$, so in fact both $(x_{n_{k_\ell}})$ and $(y_{n_{k_\ell}})$ converge, say to $x,y$ respectively. We know $x,y \in E$ since $E$ is closed (compact implies closed in $\mathbb{R}^n$.) Continuity of $d(\cdot,\cdot)$ then implies $$ d(x,y) = \lim_\ell d(x_{n_{k_\ell}},y_{n_{k_\ell}}) = \delta. $$

nullUser
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  • Ok, I get how to choose appropriate subsequences now. But, I cannot use continuity in the final part. Actually, we do not define rigorously what continuity in metric space is, which include $\mathbb{R}^k$. – user117375 Feb 28 '15 at 13:03
  • I think it might be like this : There exists $N \in \mathbb{B}$ such that $$|d(x_{n_{k_l}},x)| + |d(y_{n_{k_l}},y)| + |d(x_{n_{k_l}},y_{n_{k_l}}) - \delta| < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon$$ – user117375 Feb 28 '15 at 13:11
  • However, I suspect that is it true that: $$|d(x,y) - \delta| \leq d(x_{n_{k_l}},x) + d(y_{n_{k_l}},y) +|d(x_{n_{k_l}},y_{n_{k_l}}) - \delta| < \epsilon.$$ – user117375 Feb 28 '15 at 15:13
  • Anymore suggestion avoiding continuity ? – user117375 Feb 28 '15 at 15:20
  • Recall that in $\mathbb{R}^n, d(x,y) = |x-y| = \sqrt{\sum(x_i-y_i)^2}$ is just the square root of a polynomial. Are you sure you can't use continuity. Continuity is exactly the property you need to conclude this, so avoiding continuity amounts to using continuity in a convoluted way. – nullUser Feb 28 '15 at 19:24
  • Actually, the rigorous definition of continuity in metric space is not defined yet, I am afraid that that include $\mathbb{R}^k$. Anyway, I come up with a new idea as follows : I have a subsequence $u {n{k_l}} = d(x_{n {k_l}}, y{n_{k_l}})$ which satisfies 1) $u_{n_{k_l}} \rightarrow \delta = \sup {d(x,y)}$ 2. $x_{n_{k_l}} \rightarrow x, y_{n_{k_l}} \rightarrow y$ for some $x, y \in E$. I claim that $u_{n_{k_l}} \rightarrow d(x,y)$, then bu uniqueness of limit, $\delta = d(x,y)$. – user117375 Mar 01 '15 at 20:15
  • @user117375, you don't need continuity in an arbitrary metric space, you just need continuity for function $d:\mathbb{R}^{2k}\to \mathbb{R}$ which you certainly do know about. In any case, what you have said in your previous comment is true but doesn't help you. "I claim that $u_{n_{k_l}}\to d(x,y)$" is exactly the statement $d(x_{n_{k_l}},y_{n_{k_l}})\to d(x,y)$, i.e. you are using that $d$ is continuous at $(x,y)$. Calling it $u_{n_{k_l}}$ instead of $d(x_{n_{k_l}},y_{n_{k_l}})$ doesn't tell you anything you didn't already know. There is no way to get around continuity here. – nullUser Mar 01 '15 at 20:21
  • I see that $u_{n_{k_l}} \rightarrow d(x,y)$ might be the same as using continuity, like you said. But I view $u_{n_{k_l}}$ as a sequence in $\mathbb{R}$ and I use the definition ($\delta - \epsilon)$ for sequence in $ \mathbb{R}$ to show my claim. I think that it does not involve, in my approach, any continuity in general metric space. I think this way is valid in my situation that general continuity does not state yet. – user117375 Mar 01 '15 at 20:31
  • @user117375, $u_{n_{k_l}}$ is just another name for $d(x_{n_{k_l}},y_{n_{k_l}})$, indexed by $l$ they are both (the same!) sequence in $\mathbb{R}$. I am not using any fact about metric spaces. I am literally only using the definition of continuity on $\mathbb{R}$. Calling a function $d$ doesn't mean you are using metric space theory. $d$ is a function defined on $\mathbb{R}^{2k}$ taking values in $\mathbb{R}$. I am using the definition of continuity for function $f:\mathbb{R}^{2k}\to\mathbb{R}$. Nothing I am doing uses metric spaces. – nullUser Mar 01 '15 at 20:35
  • You should know the following facts about continuity on $\mathbb{R}$: 1) polynomials are continuous, 2) the square root function on $[0,\infty]$ is continuous, 3) the sum of finitely many continuous functions is continuous, 4) the composition of continuous functions is continuous. Therefore $d(x,y):= \sqrt{\sum_{i=1}^k (x_i-y_i)^2}$ is continuous. All of this requires only knowledge of continuity on $\mathbb{R}$. – nullUser Mar 01 '15 at 20:38
  • Ok, I see your point. Sorry for being so silly. But I already use $(\delta - \epsilon)$ notion for claiming that the sequence $u_{n_{k_l}} \rightarrow d(x,y)$, and use uniqueness of limit to conclude $\delta = d(x,y)$. I think this way works too, though it is harder than what you do. – user117375 Mar 01 '15 at 21:00