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Quick notation+statement verification

  1. Jul 11, 2005 #1
    Do you agree that, [tex] \forall k \in \left[ {a,b} \right]\;{\text{where}}\;\left( {a,b,k} \right) \in \mathbb{Q}^3 [/tex],
    [tex] \exists \,\varepsilon > 0{\text{ such that}}\;\forall n \in \mathbb{N},\;\left( {\left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} - a} \right) \subseteq \varepsilon \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{b - a}}{\varepsilon }} \right\rfloor } \right\} [/tex]

    |*Is this True or False ?
     
    Last edited: Jul 12, 2005
  2. jcsd
  3. Jul 12, 2005 #2

    matt grime

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    i don't think it makes sense until you say what the k_n are. don't bother with the symbols just write it in english.
     
  4. Jul 12, 2005 #3
    Sorry:redface:; the whole mess seems to simplify down to this statement:

    [tex] \forall \left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subset \mathbb{Q}\;{\text{where }}k_1 < k_2 < \ldots < k_n , [/tex]
    [tex] \exists \,\varepsilon > 0\;{\text{such that}}\;\forall n \in \mathbb{N},\;\left\{ {k_1 ,k_2 , \ldots ,k_n } \right\} \subseteq \varepsilon \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{k_n - k_1 }}
    {\varepsilon }} \right\rfloor } \right\} [/tex]

    *|is this True or False?
     
    Last edited: Jul 13, 2005
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