MHB Supremum Property (AoC) .... etc .... Yet a further question/Issue ....

[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 ... ...

Theorem 2.1.45 reads as follows:View attachment 7176
View attachment 7177In 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 )

$$\Longrightarrow v - u \gt \frac{1}{ 2^n }$$ ... ... ... ( 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

 

[Opalg]

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)

Yes.

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)

Sohrab is saying that if $\bigcap I_n$ consists of more than one point then it must contain two points, one of which is larger than the other. Call the smaller one $u$ and the larger one $v$.

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 )

$$\Longrightarrow v - u \gt \frac{1}{ 2^n }$$ ... ... ... ( 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)

That is correct. To prove that $2^n>n$, use induction. (I assume that Sohrab will have covered induction by this stage of the book.)

 

[Math Amateur]

Yes.Sohrab is saying that if $\bigcap I_n$ consists of more than one point then it must contain two points, one of which is larger than the other. Call the smaller one $u$ and the larger one $v$.That is correct. To prove that $2^n>n$, use induction. (I assume that Sohrab will have covered induction by this stage of the book.)

Thanks for all your help Opalg ...

Your explanations and help got me to an understanding of the proof!

Thanks again,

Peter