Supremum Property (AoC) .... etc .... Another 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 another 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 7171
View attachment 7172In the above proof by Sohrab, we read the following:

" ... ... Thus while (by definition) \(\displaystyle s + \frac{k_n}{2^n} = s + \frac{2 k_n}{ 2^{n + 1} }\) is an upper bound of \(\displaystyle S, s + \frac{ 2 k_n - 2 }{ 2^{n + 1} } = s + \frac{k_n - 1 }{2^n}\) is not. Therefore, either \(\displaystyle k_{ n+1 } = 2k_n\) or \(\displaystyle k_{ n+1 } = 2k - 1\) and \(\displaystyle I_{ n + 1 } \subset I_n\) follow. ... ... "I am uncertain and somewhat confused by the logic of the above ...

I am not sure of the argument that either \(\displaystyle k_{ n+1 } = 2k_n\) or \(\displaystyle k_{ n+1 } = 2k - 1\) ... can someone please explain in simple terms why it is valid ... ..

I am also not sure exactly why \(\displaystyle I_{ n + 1 } \subset I_n\) ... if anything it seems that \(\displaystyle I_{ n + 1 } = I_n\) ... again, can someone please explain in simple terms why \(\displaystyle I_{ n + 1 } \subset I_n\) ... that is that \(\displaystyle I_{ n + 1 }\) is a proper subset of \(\displaystyle I_n\) ...[ ***EDIT*** Just checked and found that Sohrab is using \(\displaystyle \subset\) in the sense which includes equality ... so using \(\displaystyle \subset\) as meaning \(\displaystyle \subseteq\) ... ]
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 ...
View attachment 7173

View attachment 7174

View attachment 7175

 

[Opalg]

I am not sure of the argument that either \(\displaystyle k_{ n+1 } = 2k_n\) or \(\displaystyle k_{ n+1 } = 2k - 1\) ... can someone please explain in simple terms why it is valid ... ..

I am also not sure exactly why \(\displaystyle I_{ n + 1 } \subset I_n\) ... if anything it seems that \(\displaystyle I_{ n + 1 } = I_n\) ... again, can someone please explain in simple terms why \(\displaystyle I_{ n + 1 } \subset I_n\) ... that is that \(\displaystyle I_{ n + 1 }\) is a proper subset of \(\displaystyle I_n\) ...

See my comment on your http://mathhelpboards.com/analysis-50/supremum-property-aoc-archimedean-property-nested-intervals-theorem-22069.html#post99642. The way this proof works in that you divide the real line into subintervals of length $1/2^n$ and you locate the subinterval $I_n$ in which $\sup(S)$ (if it exists) must lie. You then increase $n$ to $n+1$, so that you are looking at intervals of half the previous length.

You know that the upper end of $I_n$, namely $s+ \frac{k_n}{2^n} = s + \frac{2k_n}{2^{n+1}}$, is an upper bound for $S$, but that the lower end of $I_n$, namely $s+ \frac{k_n-1}{2^n} = s + \frac{2k_n-2}{2^{n+1}}$, is not an upper bound for $S$.

Now look at the midpoint of $I_n$, namely $s+ \frac{2k_n-1}{2^{n+1}}$. Either that is an upper bound for $S$ or it is not. If it is, then we can take the lower half of $I_n$ to be $I_{n+1}$. If it is not, then we can take the upper half of $I_n$ to be $I_{n+1}$.

 

[Math Amateur]

See my comment on your http://mathhelpboards.com/analysis-50/supremum-property-aoc-archimedean-property-nested-intervals-theorem-22069.html#post99642. The way this proof works in that you divide the real line into subintervals of length $1/2^n$ and you locate the subinterval $I_n$ in which $\sup(S)$ (if it exists) must lie. You then increase $n$ to $n+1$, so that you are looking at intervals of half the previous length.

You know that the upper end of $I_n$, namely $s+ \frac{k_n}{2^n} = s + \frac{2k_n}{2^{n+1}}$, is an upper bound for $S$, but that the lower end of $I_n$, namely $s+ \frac{k_n-1}{2^n} = s + \frac{2k_n-2}{2^{n+1}}$, is not an upper bound for $S$.

Now look at the midpoint of $I_n$, namely $s+ \frac{2k_n-1}{2^{n+1}}$. Either that is an upper bound for $S$ or it is not. If it is, then we can take the lower half of $I_n$ to be $I_{n+1}$. If it is not, then we can take the upper half of $I_n$ to be $I_{n+1}$.

Thanks Opalg ... indeed, thanks to you I am beginning to really understand the proof ...

Most grateful ...

Peter