 #1
Math Amateur
Gold Member
MHB
 3,998
 48
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:
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==========================================================================================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 ...
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:
" ... ...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==========================================================================================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 ...
Attachments

Sohrab  1  Theorem 2.1.45 ...  PART 1 ... ....png37.9 KB · Views: 493

Sohrab  2  Theorem 2.1.45 ...  PART 2 ... ....png29.9 KB · Views: 462

Sohrab  Axiom of Completeness ... Supremum Property ....png31.8 KB · Views: 469

Sohrab  Theorem 2.1.31  Archimedean Property ... ....png28.3 KB · Views: 756

Sohrab  Theorem 2.1.43 ... Nested Intervals Theorem ....png48 KB · Views: 466