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I Supremum Property, Archimedean Property, Nested Intervals

  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 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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    My questions regarding the above text from Sohrab are as follows:


    Question 1

    In the above text we read the following:

    " ... ... ##s + \frac{m}{ 2^n}## is an upper bound of ##S##, for some ##m \in \mathbb{N}##. Let ##k_n## be the smallest such ##m## ... ... "


    Can we argue, based on the above text, that ##s + \frac{m}{ 2^n} = \text{Sup}(S)## ... ... ?



    Question 2

    In the above text we read the following:

    " ... ... We then have ##I_n \cap S \ne \emptyset##. (Why?) ... ... "


    Is ## I_n \cap S \ne \emptyset## because elements such as ##s + \frac{ k_n - x }{ 2^n} , \ 0 \lt x \lt 1## belong to ##I_n \cap S## ... for example, the element ##s + \frac{ k_n - 0.5 }{ 2^n} \in I_n \cap S##?

    Is that correct ... if not, then why exactly is ##I_n \cap S \ne \emptyset##?


    Hope someone can help ...

    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=62c764415354f60258c4d51a3600d8e9.png

    ?temp_hash=62c764415354f60258c4d51a3600d8e9.png
     
  2. jcsd
  3. Aug 11, 2017 #2

    andrewkirk

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    No. Consider ##S=\{x\in\mathbb R\ :\ x<\sqrt 2\}## and ##s=1##. The supremum of this ##S## is ##\sqrt 2##, which is irrational, while ##s+m/2^n## is rational for any ##m,n\in\mathbb N##.

    The idea is that the two endpoints of the interval ##I_n## will straddle the supremum of ##S##.

    Because, if the set were empty, then ##s+(k_n-1)/2^n##, the lower bound of ##I_n##, would be an upper bound for ##S##, which would contradict the assumption that ##k_n## is the smallest natural number ##m## such that ##s+m/2^n## is an upper bound for ##S##.
     
  4. Aug 12, 2017 #3

    Hi Andrew,

    Thanks for the help ... but ...

    ... I understand your first statement ... but then you write:

    " ... ... The idea is that the two endpoints of the interval ##I_n## will straddle the supremum of ##S##."

    Surely the supremum will be at or beyond the right endpoint ... and not straddle the left and right endpoints ...

    Can you clarify further ...

    Peter
     
  5. Aug 12, 2017 #4

    andrewkirk

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    The right endpoint is ##Hi=s+k_n/2^n##, which is an upper bound for ##S##, and the left endpoint is ##Lo=s+(k_n-1)/2^n##, which is not.

    If ##u## is the supremum of ##S## then ##u\leq Hi## since a supremum (least upper bound) cannot be greater than any other UB.
    Since ##Lo## is not a UB of ##S## while ##u## is, we must have ##Lo<u##.
    Hence ##Lo<u\leq Hi##, whence ##u\in (Lo,Hi]\subset [Lo,Hi]=I_n##.
     
  6. Aug 12, 2017 #5
    Thanks Andrew ... that clarifies the issue ...

    Peter
     
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