# I Supremum Property (AoC) ... etc ... Yet a further question

1. Aug 12, 2017

### 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 yet a further issue/problem with Theorem 2.1.45 concerning the Supremum Property (AoC), the Archimedean Property, and the Nested Intervals Theorem ... ...

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

" ... ...The Nested Intervals Theorem now implies that $\bigcap_{ n = 1}^{ \infty } I_n = \{ u \}$ for a unique $u \in \mathbb{R}$. Indeed, if $u \lt v$ and $u,v \in \bigcap_{ n = 1}^{ \infty } I_n$, then $v - u \gt \frac{1}{2n}$ for some $n \in \mathbb{N}$, which contradicts $u, v \in I_n$, since $I_n$ has length $2^{ -n }$. ... ... "

I am unsure of Sohrab's process and assumptions as he is moving through the proof in the above quote ... could someone confirm (or otherwise) my interpretations as follows ... there are essentially 4 questions ( Q1, Q2, Q3 and Q4 respectively ...) ... ...

First issue ... ... I assume that when Sohrab writes: "Indeed, if $u \lt v$ ... etc etc ... " ... he is verifying his statement that $\bigcap_{ n = 1}^{ \infty } I_n = \{ u \}$ for a unique $u \in \mathbb{R}$? Is that right? (Q1)

Second issue ... ... when Sohrab writes: "Indeed, if $u \lt v$ ... etc etc ... " ... ... he could have said $u \gt v$ ... but he is just taking $u \lt v$ as an example ... and we are left to infer that $u \gt v$ works similarly ... in other words there is no reason that $u$ is taken as less than $v$ as against taking $v \lt u$ ... ... Is that right? (Q2)

Third issue ... ... Sohrab then asserts that $v - u \gt \frac{1}{ 2^n }$ ... ... and I am assuming this follows because ...

$u \lt v$

$\Longrightarrow v - u \gt 0$

$\Longrightarrow v - u \gt \frac{1}{n}$ for some $n \in \mathbb{N}$ ... (Corollary 2.1.32 (b) Archimedean Property ... see scanned text insert below)

$\Longrightarrow v - u \gt \frac{1}{ 2^n }$ ... ... ... ( Is this valid? (Q3) ... looks OK ... but justification ?

So indeed ... given we are doing analysis ... how do we justify $\frac{1}{n} \gt \frac{1}{ 2^n }$ or $2^n \gt n$?

and further ... is my interpretation above for the third issue correct (Q4)

Help will be appreciated ...

Peter

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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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• ###### Sohrab - Theorem 2.1.43 ... Nested Intervals Theorem ....png
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2. Aug 12, 2017

### Someone2841

I'm only going to address Q1 and Q2 in this post. Really, since we know the diameter of the nested sequences converges to 0, we also know such a unique $u$ exists. Probably for pedagogical purposes, he shows that there cannot be two elements in the infinite intersection. Because these two elements $u, v$ are arbitrary anyway, there is no need to show both cases $u<v$ AND $v<u$.

3. Aug 12, 2017

### Someone2841

With regard to the fact that $v-u>2^{-n}$, your reasoning is sound. It is easy to show that $2^n>n$ for all natural numbers. First, $2^0=1>0$. Second, $2^n>n \implies 2^{n+1}>n+1$. Adding 1 to the antecedent, $2^n+1>n+1$. Since $2 \cdot 2^n > 2^n+1$ is true whenever $n>0$ (verify this by subtracting $2^n$ and taking $\log_2$), we know that $2^{n+1}>n+1$ as desired.

This reasoning is a bit tedious, though. It's pretty obvious that $(1/2^n)_{n \in \mathbb{N}} \to 0$,and any such sequence must get smaller than any given positive number—this is the spirit of the Archimedean property: there are no arbitrarily small numbers. Therefore, I imagine he took for granted that there existed a $v-u>1/2^n$ for some n.

Let me know if you have follow up questions.

4. Aug 12, 2017