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Non-convergence written with sets

  1. Aug 10, 2010 #1
    Hey, everyone. I'm trying to prove the following:

    [tex] f_n [/tex] and [tex] f_n [/tex] are real-valued function in [tex] \Omega [/tex]

    [tex] \{\omega: f_n(\omega) \nrightarrow f(\omega) \} = \\
    \bigcup^{\infty}_{k=1} \bigcap^{\infty}_{N=1} \bigcup^{\infty}_{n=1}
    \{ \omega : | f_n(\omega) - f(\omega) | \geq 1/k \}


    I am convinced by the proof I've made up, but it isn't formal, so I would appreciate if you could help me give it more formality.
    Let's call the left side of the equality L and the right side R.
    L can be written:

    [tex] \exists k \in \mathbb{N} \quad \forall N \quad \exists n \geq N \quad | f_n(\omega) \nrightarrow f(\omega) | \geq 1/k

    On the other hand, the last part of R is
    [tex] \bigcap^{\infty}_{N=1} \bigcup^{\infty}_{n=1}
    \{ \omega : | f_n(\omega) - f(\omega) | \geq 1/k \} [/tex]

    which basically takes all the [tex] \omega [/tex] that [tex] \forall N [/tex] have
    at least one [tex] n \geq N [/tex] that makes the absolute difference bigger than 1/k
    If you take the union for all k, then you have the definition for being in L.
    Thanks in advance,

    Last edited: Aug 10, 2010
  2. jcsd
  3. Sep 26, 2010 #2


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    Science Advisor

    There's something wrong with your R. You take an intersection over N=1 to infty, but the index N does not appear in the collection over which the intersection is taken. You probably meant

    [tex]\bigcup^{\infty}_{k=1} \bigcap^{\infty}_{N=1} \bigcup^{\infty}_{n=N}
    \{ \omega : | f_n(\omega) - f(\omega) | \geq 1/k \}[/tex]

    and then you are already done, since the condition for omega to belong to this one and to to LHS are the same, namely

    [tex] \exists k \in \mathbb{N} \quad \forall N \quad \exists n \geq N \quad | f_n(\omega) \nrightarrow f(\omega) | \geq 1/k
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