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

  • #1
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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:

" ... ... is an upper bound of , for some . Let be the smallest such ... ... "Can we argue, based on the above text, that ... ... ?
Question 2

In the above text we read the following:

" ... ... We then have (Why?) ... ... "Is because elements such as belong to ... for example, the element ?

Is that correct ... if not, then why exactly is ?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
 
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  • #2
Peter said:
Question 1

In the above text we read the following:

" ... ... is an upper bound of , for some . Let be the smallest such ... ... "Can we argue, based on the above text, that ... ... ?
No. What is happening in this proof is that you narrow down the location of by an approximation process. You start from a point in , and having fixed , you look at the points , until you find the first point that is an upper bound for . That way, you have narrowed down the location of to an interval of length . You then increase to , so as to locate within a smaller interval (which must be either the first half or the second half of the previous interval ).

Peter said:
Question 2

Is because elements such as belong to ... for example, the element ?

Is that correct ... if not, then why exactly is ?
Since is the smallest integer for which is an upper bound for , it follows that is not an upper bound for . Therefore there is an element, call it , of that is greater than . But (because is an upper bound for ), and so lies in the interval . So lies in the intersection .
 
  • #3
Opalg said:
No. What is happening in this proof is that you narrow down the location of by an approximation process. You start from a point in , and having fixed , you look at the points , until you find the first point that is an upper bound for . That way, you have narrowed down the location of to an interval of length . You then increase to , so as to locate within a smaller interval (which must be either the first half or the second half of the previous interval ).Since is the smallest integer for which is an upper bound for , it follows that is not an upper bound for . Therefore there is an element, call it , of that is greater than . But (because is an upper bound for ), and so lies in the interval . So lies in the intersection .
Thanks Opalg ... most helpful ...

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

Peter
 
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