Supremum Property (AoC), Archimedean Property, Nested Intervals Theorem ....

[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:View attachment 7166
https://www.physicsforums.com/attachments/7167My 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 ...https://www.physicsforums.com/attachments/7168

https://www.physicsforums.com/attachments/7169

View attachment 7170

 

[Opalg]

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. What is happening in this proof is that you narrow down the location of $\sup(S)$ by an approximation process. You start from a point $s$ in $S$, and having fixed $n$, you look at the points $s + \frac1{2^n}, s + \frac2{2^n}, s + \frac3{2^n}, \ldots$, until you find the first point $s + \frac{k_n}{2^n}$ that is an upper bound for $S$. That way, you have narrowed down the location of $\sup(S)$ to an interval $I_n$ of length $\frac1{2^n}$. You then increase $n$ to $n+1$, so as to locate $\sup(S)$ within a smaller interval $I_{n+1}$ (which must be either the first half or the second half of the previous interval $I_n$).

Question 2

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 $$ I_n \cap S \ne \emptyset$$?

Since $k_n$ is the smallest integer for which $s + \frac{k_n}{2^n}$ is an upper bound for $S$, it follows that $s + \frac{k_n-1}{2^n}$ is not an upper bound for $S$. Therefore there is an element, call it $t$, of $S$ that is greater than $s + \frac{k_n-1}{2^n}$. But $t\leqslant s + \frac{k_n}{2^n}$ (because $s + \frac{k_n}{2^n}$ is an upper bound for $S$), and so $t$ lies in the interval $I_n$. So $t$ lies in the intersection $I_n\cap S$.

 

[Math Amateur]

No. What is happening in this proof is that you narrow down the location of $\sup(S)$ by an approximation process. You start from a point $s$ in $S$, and having fixed $n$, you look at the points $s + \frac1{2^n}, s + \frac2{2^n}, s + \frac3{2^n}, \ldots$, until you find the first point $s + \frac{k_n}{2^n}$ that is an upper bound for $S$. That way, you have narrowed down the location of $\sup(S)$ to an interval $I_n$ of length $\frac1{2^n}$. You then increase $n$ to $n+1$, so as to locate $\sup(S)$ within a smaller interval $I_{n+1}$ (which must be either the first half or the second half of the previous interval $I_n$).Since $k_n$ is the smallest integer for which $s + \frac{k_n}{2^n}$ is an upper bound for $S$, it follows that $s + \frac{k_n-1}{2^n}$ is not an upper bound for $S$. Therefore there is an element, call it $t$, of $S$ that is greater than $s + \frac{k_n-1}{2^n}$. But $t\leqslant s + \frac{k_n}{2^n}$ (because $s + \frac{k_n}{2^n}$ is an upper bound for $S$), and so $t$ lies in the interval $I_n$. So $t$ lies in the intersection $I_n\cap S$.

Thanks Opalg ... most helpful ...

... indeed, you made it very clear ...

Peter