# Analysis - Metric space proof (prove max exists)

http://imageshack.us/a/img12/8381/37753570.jpg [Broken]

I am having trouble with this question, like I do with most analysis questions haha.

It seems like I must show that the maximum exists.

So E is compact -> E is closed

To me having E closed seems like it is clear that a maximum distance exists in the metric space but I know that more work is required.

I think the way I am supposed to solve it is by using a similar proof to the Bolzano-Weierstrass theorem, as in picking a subsequence of a subsequence but I am really not sure how to begin and really apply this to the proof.

I was thinking maybe you show that the distance between a sequence and its sub-sub sequence is in E but I am really not confident that this is correct.

Could anyone help point me in the right direction and hopefully help me gain some intuition about this stuff.

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jbunniii
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If you know that a real-valued continuous function whose domain is a compact set achieves a maximum, then one way to approach this would be to prove that $E \times E$ is compact and that $\rho : E \times E \rightarrow \mathbb{R}$ is a continuous function.

jbunniii
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Another option is to recognize that by definition of the supremum, you can find sequences $(x_n)$ and $(y_n)$ such that $M - 1/n \leq \rho(x_n, y_n) \leq M$, where $M$ is the supremum. What does the fact that $E$ is compact imply about the sequences $(x_n)$ and $(y_n)$?

E being compact implies that the sequences {xn} and {yn} have a convergent subsequence in E.

So by taking the subsequence's of {xn} and {yn} this shows that you can find:
${(x_n)_n}$ and ${(y_n)_n}$ such that $K - 1/n \leq \rho((x_n)_n, (y_n)_n) \leq K$ where K is in E.

Is that right?

jbunniii
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E being compact implies that the sequences {xn} and {yn} have a convergent subsequence in E.

So by taking the subsequence's of {xn} and {yn} this shows that you can find:
${(x_n)_n}$ and ${(y_n)_n}$ such that $K - 1/n \leq \rho((x_n)_n, (y_n)_n) \leq K$ where K is in E.

Is that right?
So far so good, although you should use a different letter to denote subsequence, such as $(x_{n_k})$, not $(x_{n_n})$.

Also, by $K$, I assume you mean what I was calling $M$, namely $M = \sup_{u,v \in E} \rho(u,v)$. Note that $M$ is a real number, not an element of $E$.

OK, so the subsequences are convergent. That means there are points $x,y \in E$ such that $x_{n_k} \rightarrow x$ and $y_{n_k} \rightarrow y$. Now what can you say about $\rho(x,y)$?

Is it that you can say M <= p(x,y) <= M since the 1/n -> 0 as n->infinity

Then you get what you need!

p(x,y) = sup p(u,v)

The notes I found the question it kind of suggested that you have to take a subsequence of a subsequence, how come you think they suggested that?

jbunniii
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Is it that you can say M <= p(x,y) <= M since the 1/n -> 0 as n->infinity

Then you get what you need!

p(x,y) = sup p(u,v)
It is true, but you have to explain/prove why this works.

The notes I found the question it kind of suggested that you have to take a subsequence of a subsequence, how come you think they suggested that?
I don't think this is necessary, but perhaps the author is thinking of a different proof.

jgens
Gold Member
I don't think this is necessary, but perhaps the author is thinking of a different proof.

It's the same proof and there isn't really any way around this step if the prof wants all of the details. You use sequential compactness on {xn} to get a convergent subsequence {xnk}. Then you use sequential compactness on {ynk} to find a convergent subsequence {ynkj}. Then both {xnkj} and {ynkj} are convergent sequences.

Thanks guys!

Do you mean you have to prove that the limit as n,k-> infinity of p((x_n)_k, (y_n)_k) = p(x,y)?

Is there some kind of difference between using subsequence and sub-sub sequences?
I picture the subsequences having the same properties as sub-sub sequences, is there a deeper meaning to why you would pick sub-sub sequences?

jbunniii