1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Metric Spaces of Bounded Sequences

  1. Sep 13, 2012 #1
    I was attempting to find a counterexample to the problem below. I think I may have, but was ultimately left with more questions than answers.

    Consider the space, [itex]L[/itex], of all bounded sequences with the metric [itex]\rho_1[/itex]

    [itex]\displaystyle \rho_1(x,y)=\sum\limits_{t=1}^{\infty}2^{-t}|x_t-y_t|[/itex]

    Show that a sequence that converges to [itex]X^*[/itex] in [itex](L,\rho_1)[/itex] does not necessarily converge to [itex]X^*[/itex] in [itex](L,\rho_\infty)[/itex]

    Where, [itex]\rho_\infty(x,y)=sup_t|x_t-y_t|[/itex].

    I believe convergence means:

    X converges to Y if [itex] \forall \epsilon>0 \exists N[/itex] s.t. [itex] \sum\limits_{t=N}^{\infty}2^{-t}|x_t-y_t|<\epsilon [/itex]

    I believe that [itex]x_n=sin (n)[/itex] converges to the 0 sequence in [itex]\rho_1[/itex], but not in [itex]\rho_\infty[/itex]. However, it seems like [itex]x_n=sin (n)[/itex] could also be shown to converge to {1,1,1,1,...} under [itex]\rho_1[/itex] which shouldn't be allowed.

    Could anyone help me out?
    Last edited: Sep 13, 2012
  2. jcsd
  3. Sep 13, 2012 #2
    I doubt that this is correct. It should be [itex]\rho_\infty(x,y)=\sup_t |x_t-y_t|[/itex].

    This is not what convergence is. First of all you must have a sequence of elements in L. That is, you have to start from a sequence of sequences. Let [itex]X_k=(x_k^n)_n[/itex], then we can say that [itex]X_k\rightarrow Y[/itex] if for each [itex]\varepsilon>0[/itex], there exists an N such that for all [itex]k\geq N[/itex] holds that [itex]\rho_1(X_k,Y)<\varepsilon[/itex]. That last inequality is of course the same as

    [tex]\sum_{t=1}^{+\infty} 2^{-t}|x_k^t-y^t|<\varepsilon[/tex]

    This is of course incorrect, since [itex]x_n=\sin(n)[/itex] is only one sequence. Again: you must have a sequence of sequences.
  4. Sep 13, 2012 #3
    Thanks, that was very helpful. It makes sense that I need to be thinking of a sequence of sequences. I was just having trouble wrapping my head around this L space.

    And you are of course right about my typo in the problem set-up, I fixed it.
  5. Sep 13, 2012 #4
    The counterexample you're looking for is very easy. Think about sequences with only 0's and 1's.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook