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I Supremum Property (AoC) ... etc ... Another question/Issue

  1. Aug 11, 2017 #1
    I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

    I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

    I need help with another issue/problem with Theorem 2.1.45 concerning the Supremum Property (AoC), the Archimedean Property, and the Nested Intervals Theorem ... ...

    Theorem 2.1.45 reads as follows:


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    ?temp_hash=7ce44f56e42685cf4d94406e9e8004ed.png


    In the above proof by Sohrab, we read the following:

    " ... ... Thus while (by definition) ##s + \frac{k_n}{2^n} = s + \frac{2 k_n}{ 2^{n + 1} }## is an upper bound of ##S, s + \frac{ 2 k_n - 2 }{ 2^{n + 1} } = s + \frac{k_n - 1 }{2^n}## is not. Therefore, either ##k_{ n+1 } = 2k_n## or ##k_{ n+1 } = 2k - 1## and ##I_{ n + 1 } \subset I_n## follow. ... ... "


    I am uncertain and somewhat confused by the logic of the above ...

    I am not sure of the argument that either ##k_{ n+1 } = 2k_n## or ##k_{ n+1 } = 2k - 1## ... can someone please explain in simple terms why it is valid ... ..

    I am also not sure exactly why ##I_{ n + 1 } \subset I_n## ... if anything it seems to me that ##I_{ n + 1 } = I_n## ... again, can someone please explain in simple terms why ##I_{ n + 1 } \subset I_n## ... that is that ##I_{ n + 1 }## is a proper subset of ##I_n## ... ...

    [ ***EDIT*** Just checked and found that Sohrab is using \subset in the sense which includes equality ... so using ##\subset## as meaning ##\subseteq## ... ]


    Help will be appreciated ...

    Peter



    ==========================================================================================


    The above theorem concerns the Supremum Property, the Archimedean Property and the Nested Intervals Theorem ... so to give readers the context and notation regarding the above post I am posting the basic information on these properties/theorems ...


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    ?temp_hash=7ce44f56e42685cf4d94406e9e8004ed.png



    ?temp_hash=7ce44f56e42685cf4d94406e9e8004ed.png
     
    Last edited: Aug 11, 2017
  2. jcsd
  3. Aug 12, 2017 #2

    andrewkirk

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    I assume you accept the statement that ##s + \frac{ 2 k_n - 2 }{ 2^{n + 1} }## is not an upper bound of ##S##. If not, this post won't help.

    ##k_{n+1}## is defined as the smallest ##m\in\mathbb N## such that ##s+k_{n+1}/2^{n+1}## is an UB for ##S##. We know that ##s+(2k_n-2)/2^{n+1}## is not an UB, therefore it must be too small to be an UB. So we must have ##k_{n+1}> 2k_n-2##.

    How much bigger does ##k_{n+1}## have to be? We know that ##2k_n## is big enough. So ##k_{n+1}\leq 2k_n##.

    The only integers ##m## that satisfy ##2k_n-2 < m \leq 2k_n## are those two the text names.
     
  4. Aug 12, 2017 #3
    Thanks Andrew ... now understand ...

    Your post was most helpful ...

    Peter
     
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