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Complete Metric Subspaces

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine whether the following metric subspaces are complete:
    a) the set E of sequences containing only entries 0 & 1 in [tex](m,||\cdot||_{\infty})[/tex]
    b) the unit sphere in any Banach Space


    2. Relevant equations
    a) for [tex]x=\{\lambda_1,\lambda_2,\ldots,\lambda_n,\ldots \}[/tex]
    [tex]||x||_{\infty}=sup\{|\lambda_n|:n=1,2,\ldots\}[/tex]

    b)[tex]\{x\in X:||x-x_0||=1\}[/tex]

    3. The attempt at a solution
    I think:
    A complete space is one in which all Cauchy sequences converges to a sequence (of points) in the space

    a) it seems that if I construct whatever sequence I construct will always have zeros, but my limit will be a sequence of only 1's, so it will not be in the space.
    That is, [tex]||x||_{\infty}=sup\{|\lambda_n|:n=1,2,\ldots\}=1[/tex]
    If this is correct, how do I show that?

    b) It seems this space is complete by the same reasoning above, but again, how do I show that?
     
  2. jcsd
  3. Feb 26, 2010 #2

    Dick

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    You are thinking about the space E wrong. Your sequence is a single point in E. To discuss convergence you need a sequence of points in E, i.e. a sequence of sequences.
     
  4. Feb 26, 2010 #3
    I was thinking of:
    [tex]x_1=(1,0,\ldots) [/tex]
    [tex]x_2=(1,1,0,\ldots) [/tex]
    [tex]\vdots [/tex]
    [tex]x_n=(1,1,\ldots,1,0,\ldots)[/tex]
     
  5. Feb 26, 2010 #4

    Dick

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    Is that Cauchy?
     
  6. Feb 26, 2010 #5
    I don't think so, since it doesn't have the property:
    [tex]|x_n-x_{n+1}|\rightarrow0 \text{ as }n\rightarrow\infty[/tex]
     
    Last edited: Feb 26, 2010
  7. Feb 26, 2010 #6

    Dick

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    Right. It's not. Look at the definition of Cauchy in terms of epsilon and think about what happens if you put epsilon=1/2.
     
  8. Feb 26, 2010 #7
    That would mean:
    [tex]||x_n-x_m||_{\infty}<\epsilon=1/2[/tex]
     
    Last edited: Feb 26, 2010
  9. Feb 26, 2010 #8
    I'm still confused, and we haven't even started talking about B yet.
    known knowns:
    [tex](m,||\cdot||_{\infty})[/tex] is complete since it contains all bounded sequences.
     
    Last edited: Feb 26, 2010
  10. Feb 26, 2010 #9

    Dick

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    If x_n and x_m are in E and |x_n-x_m|<1/2, doesn't that mean x_n=x_m?
     
  11. Feb 26, 2010 #10
    Yes it does. So, are you saying that my sequence I thought of is not in E or that it's in E and E it's not Cauchy so therefore E is not complete?
     
  12. Feb 26, 2010 #11

    Dick

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    Your sequence is in E. But it's not Cauchy and it doesn't converge. So it doesn't tell you anything about whether E is complete or not. Having thought about the epsilon=1/2 thing, if {a_n} is a Cauchy sequence in E, what can you say about the sequence?
     
  13. Feb 26, 2010 #12
    You can say that [tex]||a_n-a_m||_{\infty}<\epsilon[/tex] is true
     
  14. Feb 26, 2010 #13

    Dick

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    Apply what we decided in posts 9 and 10 to that.
     
  15. Feb 26, 2010 #14
    I don't understand where you are heading with this. Are you saying that we cannot construct a Cauchy sequence in E?
     
  16. Feb 26, 2010 #15

    Dick

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    No. I'm saying if you put the definition of a Cauchy sequence together with the fact that "if |x_n-x_m|<1/2 then x_n=x_m", then it's easy to show all Cauchy sequences converge. Look up the definition of Cauchy sequence again and think about that.
     
  17. Feb 26, 2010 #16
    In this case:

    A sequence [tex]x_n[/tex] in [tex]\left(X,||\cdot||_{\infty}\right)[/tex] is Cauchy iff

    [tex]\forall \epsilon >0\quad \exists N\quad \ni \quad ||x_n-x_m||_{\infty}<\epsilon \quad \forall m,n\geq N [/tex]

    You are saying if we let [tex]\epsilon = 1/2 [/tex] then:
    [tex]||x_n-x_m||_{\infty}<1/2 \quad \Rightarrow \quad x_n=x_m[/tex]

    So, what insight is this providing in E (the set of sequences containing only entries 0 & 1)?
     
  18. Feb 26, 2010 #17
    In this case:

    A sequence [tex]x_n[/tex] in [tex]\left(X,||\cdot||_{\infty}\right)[/tex] is Cauchy iff

    [tex]\forall \epsilon >0\quad \exists N\quad \ni \quad ||x_n-x_m||_{\infty}<\epsilon \quad \forall m,n\geq N [/tex]

    You are saying if we let [tex]\epsilon = 1/2 [/tex] then:
    [tex]||x_n-x_m||_{\infty}<1/2 \quad \Rightarrow \quad x_n=x_m[/tex]

    So, what insight is this providing in E (the set of sequences containing only entries 0 & 1)?
     
  19. Feb 26, 2010 #18

    Dick

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    Ok, so that says that if {x_n} is Cauchy, then there exists an N such that for all n,m>N, x_n and x_m are the SAME SEQUENCE. Doesn't it?? The sequence {x_n} has to eventually be constant. Call that constant sequence 'c'. What's the limit of the sequence {x_n}??
     
    Last edited: Feb 26, 2010
  20. Feb 26, 2010 #19
    It would be the sequence containing only "c"?
     
  21. Feb 26, 2010 #20

    Dick

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    Hmmm. I'm not sure what that means. I don't think I'm getting through. All of the elements in the sequence {x_n} are equal to x_(N+1) for n>N, right?
     
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