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I have been attempting to solve this HW problem, from Rosenlicht's Introduction to real analysis (pg. 92, 15th problem):

Given a non-empty compact metric space $E$, show that $\max\{d(x,y) \mid x,y \in E \}$ exists.

There was a hint provided with the problem, but I am not sure how to utilize it (something along the lines of trying to find sequences $p_n,q_n$ such the $\lim \;d(p_n,q_n) = \sup\{d(p,q) \mid p,q \in E \}$.

I guess showing that $\max\{d(x,y) \mid x,y \in E\}$ exists is equivalent to showing that $\{d(x,y) \mid x,y \in E\}$ is compact. I tried to do this by defining function $f_{p_0} = d(x,p_0)$ for some point $p_0 \in E$. Since $E$ is compact, $f_{p_0}(E)$ will be compact, therefore closed and bounded and will have a maximum. I can do this over every point in $E$. But, $E$ could be uncountable, so I will end up with uncountably many functions all of whose images would be compact but the maximum I am looking for would be in the union of all the images (closed and bounded), which need not be closed or bounded. So, I am not sure how to procced at this point. Any suggestions?

Thiagarajan
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2 Answers2

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Follow the hint, as a bounded set, $\sup\{d(x,y) \mid x,y \in E \}$ exists, say $l=\sup\{d(x,y) \mid x,y \in E \}$. By a property of the supremum, there are $p_n,q_n$ sequences in $E$ such that $d(p_n,q_n) \to l$. Since $p_n$ and $q_n$ are sequences in a compact space, $p_n$ and $q_n$ have convergent subsequences, say $p_{n_j}$ and $q_{n_j}$, approaching $p$ and $q$, respectively. Of course $d(p_{n_j},q_{n_j})\to l$ and, by compactness, $p,q \in E$.

Srivatsan
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Vinicius M.
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  • Oh ok,I see now.I don't have to construct the sequence.But I have to prove the fact that the metric is bounded,which can be proved using compactness.Thank you,that helped. – Thiagarajan Dec 01 '11 at 01:51
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    Note that you cannot just take any arbitrary subsequences $(p_{n_j})$ and $(q_{m_{j}})$ of $(p_n)$ and $(q_n)$, because it is important that they have the same indices, and that takes a little more work. – Jonas Meyer Dec 01 '11 at 01:52
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    yeah, sorry, one should construct a subsequence of $(p_n,q_n)$, but still works since $E^2$ is also compact – Vinicius M. Dec 01 '11 at 01:54
  • @ViniciusM So,it is necessary to prove that ExE is compact before I proceed to the solution?? – Thiagarajan Dec 01 '11 at 01:58
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    no, let $p_{n_j}$ be a convergent subsequence of $p_n$. Now consider the sequence $q_{n_j}$, since $q_{n_j} \in E$ and $E$ is compact, there's a convergent subsequence $q_{n_{j_k}}$ of $q_{n_j}$. You can work now with $(p_{n_{j_k}}, q_{n_{j_k}})$ – Vinicius M. Dec 01 '11 at 02:10
  • Ok.Got it.Thanks.:) – Thiagarajan Dec 01 '11 at 02:25
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Another approach is to show that $E \times E$ is compact, and that $d : E \times E \to \mathbb{R}$ is continuous. Then $d(E \times E)$ is compact, being the continuous image of a compact set, and so must contain its maximum. (This is basically the extreme value theorem.)

Nate Eldredge
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