Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By girdav

Math Help - Show {x_n} is NOT a compact set in C([0,1])

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    20
    Thanks
    1

    Show {x_n} is NOT a compact set in C([0,1])

    This is the Question:

    Let x_n(t) = \begin{cases}  nt  & 0 \leq t \leq \frac{1}{n} \\ 2-nt & \frac{1}{n}<t\leq\frac{2}{n} \\0 & \frac{2}{n}<t\leq1 \end{cases}

    Show that  \{x_n\} is NOT compact set in C([0,1])

    My idea is to show that  \mbox{\{x_n(t)\}} converge pointwise to x(t), and x(t) \notin C([0,1])

    Where x(t) = \lim_{n\to \infty}x_n(t) = \begin{cases}1&t=0\\0&0<t\leq1\end{cases}

    Is this enough to show  \{x_n\} is NOT compact set in C([0,1])???

    And how do I state formally that  \mbox{\{x_n(t)\}} converge pointwise to x(t)???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32

    Re: Show {x_n} is NOT a compact set in C([0,1])

    In fact \{x_n\} converges pointwise to the constant function equal to 0. If the set \{x_n,n\in\Bbb N\} was compact, we would be able to extract a uniformly convergent sub-sequence \{x_{n_k}\}. This one should converge uniformly to 0. Do you see why it is not possible?
    Thanks from sweetadam
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    20
    Thanks
    1

    Re: Show {x_n} is NOT a compact set in C([0,1])

    Thank you for fast reply

    I'm really new to these concepts, sorry if my answer is silly lol

    from the Def^\underline{n} of uniform convergence:

    for any x_m, m\in n_k, and any \epsilon < 1:

    \left|{x_m(\frac{1}{n}) - x(\frac{1}{n})\right|= 1-0 > \epsilon

    therefore no sub-sequence converge to 0 uniformly, hance \{x_n,n\in\Bbb N\} is not compact

    Is this right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32

    Re: Show {x_n} is NOT a compact set in C([0,1])

    That right.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: November 19th 2011, 06:32 AM
  2. Finite union of compact sets is compact
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: April 8th 2011, 07:43 PM
  3. Show a set is compact
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: January 22nd 2011, 10:32 AM
  4. show operator not compact
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 1st 2010, 08:07 AM
  5. sequentially compact? Show bounded?
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 11th 2010, 09:11 PM

Search Tags


/mathhelpforum @mathhelpforum