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Equivalent characterization of uniform convergence

  1. May 7, 2007 #1
    Hi all,

    I'm learning some calculus theory and I found one point I don't fully understand:

    \mbox{Let M} \subset \mathbb{R} \mbox{ be non-empty set and let } f, f_{n}, n \in \mathbb{N} \mbox{ be functions defined on M. Then the following is true:}

    f_n \rightrightarrows f \mbox{ on M} \Leftrightarrow \lim_{n \rightarrow \infty} \sup \left\{\left|f_{n}(x) - f(x)\right|; x \in M \right\} = 0


    f_{n} \rightrightarrows f \mbox{ on M }

    \Leftrightarrow\ \forall \epsilon > 0\ \exists n_{0} \in \mathbb{N}\ \forall n \geq n_{0} \forall x \in M\ : |f_{n}(x) - f(x)| < \epsilon

    \Leftrightarrow \forall \epsilon > 0\ \exists n_{0} \in \mathbb{N}\ \forall n \geq n_{0} : \sup_{x \in M} \left\{ | f_{n}(x) - f(x) | \leq \epsilon \right\}

    \Leftrightarrow \lim_{n \rightarrow \infty} \left( \sup_{x \in M} |f_{n}(x) - f(x)| \right) = 0

    I don't get why in the last but one condition in the proof there is [itex]\leq \epsilon[/tex] instead of [itex]< \epsilon[/tex].

    Could you please tell me the reason?

    Thank you very much

    Best regards,
    Last edited: May 7, 2007
  2. jcsd
  3. May 7, 2007 #2

    matt grime

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    This case it is the standard result that if S is a set of real number and, for all s in S, s<K, then sup(s)<=K. E.g. take S =(0,1), every s in S is strictly less than 1, but the sup is 1.
  4. May 7, 2007 #3
    Thank you matt! I see it now.
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