1. Not finding help here? Sign up for a free 30min 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!

Convergent Sequences on l infinity

  1. Jun 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Define [itex] R^\infty_f = \{ (t^{(1}),t^{(2}), \ldots, ) |\; t^{(i}) \in \mathbb{R}\; \forall i, \; \exists k_0 \text{ such that } t^{(k})=0 \; \forall k\geq k_0 \} [/itex]

    Define [itex] l^\infty = \{ (t^{(1}),t^{(2}), \ldots, ) |\; t^{(i}) \in \mathbb{R}\; \forall i, \; \sup_{k\geq 1} | t^{(k})| < \infty \} [/itex]

    Observe that [itex] R^\infty_f [/itex] is a linear subspace of [itex] l^\infty[/itex]. Show that [itex] R^\infty_f [/itex] is not closed in [itex] l^\infty[/itex], then show that the closure of [itex] R^\infty_f [/itex] is the space [itex] c_0 [/itex];

    2. Relevant equations

    The space c_0 is the set of all sequences converging to zero


    3. The attempt at a solution

    It's not too hard to show that this set is not closed. It suffices to show that there is a convergent sequence in [itex] l^\infty[/itex] such that every term is in [itex] R^\infty_f [/itex], but whose limit is not in [itex] R^\infty_f [/itex]. I constructed the following sequence

    [itex] x_1 = (1, 0, \ldots, ) [/itex]
    [itex] x_2 = (1, \frac{1}{2}, 0 , \ldots, ) [/itex]
    [itex] \vdots [/itex]
    [itex] x_n = (1, \ldots, \frac{1}{n}, 0, \ldots} ) [/itex]

    which converges to the point [itex] a = (1, \frac{1}{2}, \ldots, \frac{1}{n-1}, \frac{1}{n}, \frac{1}{n+1}, \ldots ) [/itex]

    It's the closure part that I'm worried about. I'm not terribly sure how I would go about showing that...
     
  2. jcsd
  3. Jun 21, 2008 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    It will suffice to show that every sequence in c_0 is a limit of sequences in R_f. Do you agree?
     
  4. Jun 21, 2008 #3
    Yes, since every point in c_0 will necessarily be the limit of some sequence in R_f. Though I think that this only shows that c_0 is a subset of the closure - not necessarily the whole closure.
     
  5. Jun 21, 2008 #4

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Yes, but on the other hand, R_f clearly sits in c_0 (and c_0 is closed!).
     
  6. Jun 21, 2008 #5
    True enough.

    So I need to show that every sequence in c_0 is a limit of sequences in R_f.

    How do I show that every sequence that converges to zero is the limit of a sequence. Indeed, what does it mean for a sequence to be a limit of another sequence?
     
  7. Jun 21, 2008 #6

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Think of this as a problem set in an abstract normed space. What does it mean for a sequence {x_n} to converge to x?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?