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Compactness; show inf is achieved

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data
    ra19.JPG


    2. Relevant equations
    N/A
    3. The attempt at a solution
    Since we're in Rn, from the given information, we can say that X is compact.
    But I don't really understand the hint in part a(actually I'm really confused). HOW can we find such sequences xn and yn? And how can we find a subsequence xn_k that converges?

    Any help is appreciated!


    [note: also under discussion in math help forum]
     
  2. jcsd
  3. Feb 22, 2010 #2
    Ok I think the best way to approach this is by starting component wise.
    By thinking of xn as the cross product of a whole lot of closed bounded intevals on R, you can consider each individual interval seperately. Call the interval of interest [tex] I_{k}[/tex]

    pick [tex] y_{n},x_{n}[/tex] to converge to a point (a,b) on it's interval product.
    now consider a subsequence of [tex] x_{n} [/tex] such that for every [tex] y_{n}[/tex]
    [tex]d(x_{n_{k}},y_{n})[/tex] is a minimum
     
  4. Feb 22, 2010 #3
    then by picking y_(n_(k_m))) in such a way that d(y(j),x(nk)) is a min, for all n_k.
    it'll be an inf for d(xn,yn) since x was picked so that it was the shortest distance for every point yn, and y was picked so that it was the shortest distance for every point of the newely refurbished series xnk. and since it was picked to converge, the subsequence converges to the same point.
     
  5. Feb 22, 2010 #4
    the hint was worthless though I wont lie.
     
  6. Feb 23, 2010 #5
    I don't quite understand your point, so maybe I should go back to the hint...

    I am not sure how to begin because I don't know how to FIND the xn and yn in the hint in the first place? Do we need to produce an actual concrete example for which d(xn,yn)->d(X,Y)? Or can we just say such sequences exist?

    Also, since X is compact, and by definition we know there EXISTS a convergence subsequence xnk whose limit lies in X. But then the hint tells me to FIND such a subsequence, why??

    Can someone kindly explain this?
    Thanks a million!
     
  7. Feb 23, 2010 #6
    You pick a pair of sequences converging to a point. You pick a subsequence so that d(y1,xn1) is a min, d(y2,xn2) is min.. Etc
    and from their, you can make it even smaller by consider the values of a sequence ynkm where
    d(ynk1,xn1) is min, d(ynk2,xn2) is a min... These sequence are guarenteed by compactness .
     
  8. Feb 23, 2010 #7
    You basically say that their are 2 sequences that satisfy it, and you show why it's true
     
  9. Feb 23, 2010 #8
    Does anyone have any idea about part b??
     
  10. Feb 23, 2010 #9
    Any countable set (in [tex]\mathbb{R}^n[/tex]) is closed. So just choose two sequences Which converge to the same point from different sides.

    E.g., let
    [tex]X = \{ \frac{1}{n} : n \in \mathbb{N} \}[/tex]
    [tex]Y = \{ -\frac{1}{n} : n \in \mathbb{N} \}[/tex]
     
  11. Feb 23, 2010 #10
    But I thought for part b we're supposed to find an example where X and Y are unbounded, and the infinmum is never achieved, no?

    In your example, both X and Y are bounded...Also, I don't get why the infinmum is NOT achieved in your example. How can we prove it?

    Thank you!
     
  12. Feb 23, 2010 #11
    Oh sorry! Let me correct myself!

    In the metric space [tex]\mathbb{R}^2[/tex], let

    [tex]
    X = \{ (\ n,\ 1/n) : n \in \mathbb{N} \}
    [/tex]
    [tex]
    Y = \{ (\ n,-1/n) : n \in \mathbb{N} \}
    [/tex]

    i.e., the graphs of what I told you before
     
  13. Feb 23, 2010 #12
    Hi, but still your X and Y are bounded...And I don't understand why the infinmum is NOT achieved in your example. How to justify it?

    thanks.
     
  14. Feb 23, 2010 #13
    I just edited it look again... I The infimum is 0. But there exist no points in X, Y that satisfy this.

    (This is just the graph of 1/n and -1/n going from 0 to infinity - both converge to 0, but for never actually take that value.)
     
  15. Feb 23, 2010 #14
    But how can we PROVE that d(X, Y) = inf{d(x, y) : x E X; y E Y} = 0 ?
    I have no idea how to prove this...:(
     
  16. Feb 23, 2010 #15
    To show d(X, Y) = 0:
    1. note that d(X,Y) is bounded below by zero
    2. show for an arbitrary epsilon > 0, there exists elements in X and Y that are closer than epsilon from each other
    It follows from that.

    To show show that for all x in X, y in Y that d(x,y) > 0, note the intersection of the closures of X and Y is empty (both are closed themselves and clearly disjoint). Apply the definition of the limit points of a set and it follows.

    I need to go out for an hr, if you have another question, i'll check later. See ya.
     
  17. Feb 23, 2010 #16
    OK, so 0 is a lower bound.
    We have to prove that 0 is the GREATEST lower bound. But it seems like you're talking about convergence, and in R2 (unlike in R), we don't have the monotone convergence theorem which links the greatest lower bound with convergence...
    So how can we prove that 0 is the GREATEST lower bound?

    To show that d(x,y) > 0, can we instead say that since d(x,y) is a metric, d(x,y)=0 <=> x=y
    And in your example, x E X and y E Y are never equal, so d(x,y)>0??


    Thanks!
     
  18. Feb 23, 2010 #17
    The set [tex] A = \{ D(x,y) : x \in X, y \in Y \}[/tex] is a subset of [tex]\mathbb{R}[/tex]. If for an arbitrary [tex]\epsilon > 0 [/tex] you can show (and you can, easily) that it is not a lower bound for [tex]A[/tex] it follows that [tex]inf A = 0[/tex] (as we know 0 is a lower bound).

    Yes, I think that's fine. Maybe when you write it up add a one line explanation as to why X and Y are disjoint (don't know how picky your marker is).
     
  19. Feb 24, 2010 #18
    I know it's easy for you, but not for me:(
    If you don't mind, can you explain how to prove that ANYTHING >0 is not a lower bound.

    d(x,y) = ||x-y|| = (n-n)2 + [1/n-(-1/n)]2 = 4/n2 < ε

    Is that how you prove that the greatest lower bound is 0?

    Thanks!
     
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