Supremum Property, Archimedean Property, Nested Intervals

[Math Amateur]

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:

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 ...

 

[andrewkirk]

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 $S$.

 

[Math Amateur]

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

 

[andrewkirk]

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 $u\in (Lo,Hi]\subset [Lo,Hi]=I_n$.

 

[Math Amateur]

Thanks Andrew ... that clarifies the issue ...

Peter