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

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

    This is the Question:

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

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

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

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

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

    And how do I state formally that $\displaystyle \mbox{\{x_n(t)\}}$ converge pointwise to $\displaystyle x(t)$???
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  2. #2
    Super Member girdav's Avatar
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    Re: Show {x_n} is NOT a compact set in C([0,1])

    In fact $\displaystyle \{x_n\}$ converges pointwise to the constant function equal to $\displaystyle 0$. If the set $\displaystyle \{x_n,n\in\Bbb N\}$ was compact, we would be able to extract a uniformly convergent sub-sequence $\displaystyle \{x_{n_k}\}$. This one should converge uniformly to $\displaystyle 0$. Do you see why it is not possible?
    Thanks from sweetadam
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    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 $\displaystyle Def^\underline{n}$ of uniform convergence:

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

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

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

    Is this right?
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  4. #4
    Super Member girdav's Avatar
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    Re: Show {x_n} is NOT a compact set in C([0,1])

    That right.
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