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Analysis - Metric space proof (prove max exists)

  1. Sep 29, 2012 #1
    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.

    Thank you in advanced, Linda
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 29, 2012 #2

    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 [itex]E \times E[/itex] is compact and that [itex]\rho : E \times E \rightarrow \mathbb{R}[/itex] is a continuous function.
     
  4. Sep 29, 2012 #3

    jbunniii

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    Another option is to recognize that by definition of the supremum, you can find sequences [itex](x_n)[/itex] and [itex](y_n)[/itex] such that [itex]M - 1/n \leq \rho(x_n, y_n) \leq M[/itex], where [itex]M[/itex] is the supremum. What does the fact that [itex]E[/itex] is compact imply about the sequences [itex](x_n)[/itex] and [itex](y_n)[/itex]?
     
  5. Sep 29, 2012 #4
    Thanks for replying jbunniii!

    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:
    [itex]{(x_n)_n}[/itex] and [itex]{(y_n)_n}[/itex] such that [itex]K - 1/n \leq \rho((x_n)_n, (y_n)_n) \leq K[/itex] where K is in E.

    Is that right?
     
  6. Sep 29, 2012 #5

    jbunniii

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    So far so good, although you should use a different letter to denote subsequence, such as [itex](x_{n_k})[/itex], not [itex](x_{n_n})[/itex].

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

    OK, so the subsequences are convergent. That means there are points [itex]x,y \in E[/itex] such that [itex]x_{n_k} \rightarrow x[/itex] and [itex]y_{n_k} \rightarrow y[/itex]. Now what can you say about [itex]\rho(x,y)[/itex]?
     
  7. Sep 30, 2012 #6
    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?
     
  8. Sep 30, 2012 #7

    jbunniii

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    It is true, but you have to explain/prove why this works.

    I don't think this is necessary, but perhaps the author is thinking of a different proof.
     
  9. Sep 30, 2012 #8

    jgens

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    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.
     
  10. Sep 30, 2012 #9
    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?
     
  11. Sep 30, 2012 #10

    jbunniii

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    Oops, yes, you're right of course. This is the only way to get both subsequences to share the same indices.
     
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