Supremum Property, Archimedean Property, Nested Intervals

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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:

Printed page from a real analysis textbook showing Theorem 2.1.45, the Supremum Property of R from the Nested Intervals Theorem and the Archimedean Property, followed by the start of its proof.

Printed proof text from a real analysis book showing the proof that the intersection of nested intervals equals a single point u, which is the supremum of S, with a QED box at the end.
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 ...
Textbook page stating the Completeness Axiom (Supremum Property): every nonempty subset of R bounded above has a least upper bound in R, followed by Exercise 2.1.29 asking to show equivalence with the Infimum Property, with a hint about inf(S) = -sup(-S).


Printed text of Theorem 2.1.31, the Archimedean Property of R: for any real x there is an integer n_x in N with x < n_x, with its proof, followed by Corollary 2.1.32 with parts (a), (b) and (c).


Printed textbook page showing Theorem 2.1.43 (Nested Intervals Theorem) with its proof, followed by Exercise 2.1.44 about the intersection of nested closed intervals equalling [alpha, beta], and a closing paragraph on the converse with the Archimedean Property.
 
Attachments
  • Printed textbook page showing Theorem 2.1.45, stating that the Supremum Property of the reals follows from the Nested Intervals Theorem and the Archimedean Property, followed by the beginning of its proof with Greek symbols and fractions.
    Sohrab - 1 - Theorem 2.1.45 ... - PART 1 ... ....png
    37.9 KB · Views: 966
  • Printed proof text from a real analysis book showing the proof of the Intervals Theorem, concluding that the intersection of nested intervals I_n is the single point u, and that u equals sup(S).
    Sohrab - 2 - Theorem 2.1.45 ... - PART 2 ... ....png
    29.9 KB · Views: 1,002
  • Textbook page stating the Completeness Axiom (Supremum Property): every nonempty subset of R bounded above has a least upper bound in R, followed by Exercise 2.1.29 asking to show equivalence with the Infimum Property, with a hint about inf(S) = -sup(-S).
    Sohrab - Axiom of Completeness ... Supremum Property ....png
    31.8 KB · Views: 735
  • Printed text of Theorem 2.1.31 (Archimedean Property of R) with its proof, followed by Corollary 2.1.32 with parts (a), (b) and (c) on positive numbers x and y.
    Sohrab - Theorem 2.1.31 - Archimedean Property ... ....png
    28.3 KB · Views: 2,677
  • Printed textbook page showing Theorem 2.1.43 (Nested Intervals Theorem) with its proof, followed by Exercise 2.1.44 on nested intervals and a closing paragraph about the Supremum Property and the Archimedean Property.
    Sohrab - Theorem 2.1.43 ... Nested Intervals Theorem ....png
    48 KB · Views: 755

Discussion

Math Amateur said:
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)## ... ... ?
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##.

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##?
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##.
 
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andrewkirk said:
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##.
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
 
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##.
 
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Thanks Andrew ... that clarifies the issue ...

Peter
 

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