Real Numbers & Sequences of Rationals .... Garling, Corollary 3.2.7 ....

[Math Amateur]

TL;DR

For any given real number x, there exists a strictly increasing sequence of rationals that converges to x as a limit ... same for a strictly decreasing sequence of rationals ...

I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume I: Foundations and Elementary Real Analysis ... ...

I am focused on Chapter 3: Convergent Sequences ... ...

I need some help to fully understand the proof of Corollary 3.2.7 ...Garling's statement and proof of Corollary 3.2.7 (together with Proposition 3.2.6 which is mentioned in Corollary 3.2.7 ... ) reads as follows:

My questions related to the above Corollary are as follows:
Question 1

In the above proof of Corollary 3.2.7 we read the following:

" ... ... Arguing recursively, let $r_n$ be the 'best' rational with$\text{max} ( x - \frac{1}{n} , r_{ n - 1 } ) \lt r_n \lt x$ ... ... "Can someone please explain (preferably in some detail) what is going on here ... how do we arrive at the expression

$\text{max} ( x - \frac{1}{n} , r_{ n - 1 } ) \lt r_n \lt x$ ... ... ?Question 2

In the above proof of Corollary 3.2.7 we read the following:

" ... ... let $s_n$ be the 'best' rational with

$x \lt s_n \lt \text{min} ( x + \frac{1}{n}, s_{ n - 1 } )$ ... ... "Can someone please explain (preferably in some detail) what is going on here ... how do we arrive at the expression

$x \lt s_n \lt \text{min} ( x + \frac{1}{n}, s_{ n - 1 } )$ ... ... ?
Help will be appreciated ...

Peter==========================================================================================The post above mentions Theorem 3.1.1 and alludes to the remarks made after the proof of Theorem 3.1.1 ... so I am providing text of the theorem and the relevant remarks ... as follows:

Hope that helps ...

Peter

 

[member 587159]

This proof uses induction to construct sequences $(s_n)_n, (r_n)_n$. To see what is really happening, just write out what happens when we jump from $1$ to $2$. The following steps $2\to 3, 3 \to 4, \dots$ are just what the induction step makes formal.

So, we have $x-1 < r_1 < x$. Then we have $\max(x-1/1,r_1) < x$ (because both $r_1< x$ and $x-1/1 < x$, thus you can pick a 'best rational' $r_2$ with $\max(x-1/2,r_1) < r_2 < x$.

Now, repeat this construction.

In my opinion, the induction could have been a lot clearer if the author has just added the sentence

Suppose $r_1, \dots, r_{n-1}$ are already constructed. And then proceed with the induction step.

Please let know if this answers your questions.

 

[Math Amateur]

Hmmm ... will spend some time doing what you say ... but basically still struggling ...

Peter

 

[member 587159]

Can you write out how the proof constructs $r_3$ given $r_2$, as constructed in my post?

 

[Math Amateur]

Summary: For any given real number x, there exists a strictly increasing sequence of rationals that converges to x as a limit ... same for a strictly decreasing sequence of rationals ...

I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume I: Foundations and Elementary Real Analysis ... ...

I am focused on Chapter 3: Convergent Sequences ... ...

I need some help to fully understand the proof of Corollary 3.2.7 ...Garling's statement and proof of Corollary 3.2.7 (together with Proposition 3.2.6 which is mentioned in Corollary 3.2.7 ... ) reads as follows:View attachment 244240

My questions related to the above Corollary are as follows:
Question 1

In the above proof of Corollary 3.2.7 we read the following:

" ... ... Arguing recursively, let $r_n$ be the 'best' rational with$\text{max} ( x - \frac{1}{n} , r_{ n - 1 } ) \lt r_n \lt x$ ... ... "Can someone please explain (preferably in some detail) what is going on here ... how do we arrive at the expression

$\text{max} ( x - \frac{1}{n} , r_{ n - 1 } ) \lt r_n \lt x$ ... ... ?Question 2

In the above proof of Corollary 3.2.7 we read the following:

" ... ... let $s_n$ be the 'best' rational with

$x \lt s_n \lt \text{min} ( x + \frac{1}{n}, s_{ n - 1 } )$ ... ... "Can someone please explain (preferably in some detail) what is going on here ... how do we arrive at the expression

$x \lt s_n \lt \text{min} ( x + \frac{1}{n}, s_{ n - 1 } )$ ... ... ?
Help will be appreciated ...

Peter==========================================================================================The post above mentions Theorem 3.1.1 and alludes to the remarks made after the proof of Theorem 3.1.1 ... so I am providing text of the theorem and the relevant remarks ... as follows:
View attachment 244242
View attachment 244243Hope that helps ...

Peter

Can you write out how the proof constructs $r_3$ given $r_2$, as constructed in my post?

Thanks Math_QED ... appreciate your help ...

I will try to show Garling's recursion process for n = 1, 2 and 3 ...For $n = 1$ we have ... ...

Let $r_1$ be the 'best' rational with $x - 1 \lt r_1 \lt x$
For $n = 2$ we have ... ...

$x - \frac{1}{n} = x - \frac{1}{2} \lt x$ ... ... and ... ... $r_{ n - 1 } = r_1 \lt x$ ... ...

and so ... $\text{max} ( x - \frac{1}{2} , r_1 ) \lt x$

Therefore we can pick/construct a 'best' rational $r_2$ ...

... such that ... $\text{max} ( x - \frac{1}{2} , r_1 ) \lt r_2 \lt x$ ... ... ... ... ... NOTE: $r_1 \lt r_2$
For $n = 3$ we have ... ...

$x - \frac{1}{n} = x - \frac{1}{3} \lt x$ ... ... and ... ... $r_{ n - 1 } = r_2 \lt x$ ... ...

and so ... $\text{max} ( x - \frac{1}{3} , r_2 ) \lt x$

Therefore we can pick/construct a 'best' rational $r_3$ ...

... such that ... $\text{max} ( x - \frac{1}{3} , r_2 ) \lt r_3 \lt x$ ... ... ... ... ... NOTE: $r_1 \lt r_2 \lt r_3$
The above analysis indicates that $(r_n)_{ n = 1}^\infty$ is strictly increasing and bounded above by $x$ ...

... indeed $\text{sup} \{ r_n \ : \ n \in \mathbb{N} \} = x$ ...

... so $r_n \longrightarrow x$ as $n \longrightarrow \infty$ ...By a similar argument involving the infimum, $s_n \longrightarrow x$ as $n \longrightarrow \infty$ ...Is the above correct?

Peter

 

[member 587159]

Thanks Math_QED ... appreciate your help ...

I will try to show Garling's recursion process for n = 1, 2 and 3 ...For $n = 1$ we have ... ...

Let $r_1$ be the 'best' rational with $x - 1 \lt r_1 \lt x$
For $n = 2$ we have ... ...

$x - \frac{1}{n} = x - \frac{1}{2} \lt x$ ... ... and ... ... $r_{ n - 1 } = r_1 \lt x$ ... ...

and so ... $\text{max} ( x - \frac{1}{2} , r_1 ) \lt x$

Therefore we can pick/construct a 'best' rational $r_2$ ...

... such that ... $\text{max} ( x - \frac{1}{2} , r_1 ) \lt r_2 \lt x$ ... ... ... ... ... NOTE: $r_1 \lt r_2$
For $n = 3$ we have ... ...

$x - \frac{1}{n} = x - \frac{1}{3} \lt x$ ... ... and ... ... $r_{ n - 1 } = r_2 \lt x$ ... ...

and so ... $\text{max} ( x - \frac{1}{3} , r_2 ) \lt x$

Therefore we can pick/construct a 'best' rational $r_3$ ...

... such that ... $\text{max} ( x - \frac{1}{3} , r_2 ) \lt r_3 \lt x$ ... ... ... ... ... NOTE: $r_1 \lt r_2 \lt r_3$
The above analysis indicates that $(r_n)_{ n = 1}^\infty$ is strictly increasing and bounded above by $x$ ...

... indeed $\text{sup} \{ r_n \ : \ n \in \mathbb{N} \} = x$ ...

... so $r_n \longrightarrow x$ as $n \longrightarrow \infty$ ...By a similar argument involving the infimum, $s_n \longrightarrow x$ as $n \longrightarrow \infty$ ...Is the above correct?

Peter

Yes, well done! I see that you understand well what's happening now.

This is key if you don't understand what in induction process does. Write it out for the first couple of steps! It becomes often very obvious what is happening.

 

[Math Amateur]

Summary: For any given real number x, there exists a strictly increasing sequence of rationals that converges to x as a limit ... same for a strictly decreasing sequence of rationals ...

I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume I: Foundations and Elementary Real Analysis ... ...

I am focused on Chapter 3: Convergent Sequences ... ...

I need some help to fully understand the proof of Corollary 3.2.7 ...Garling's statement and proof of Corollary 3.2.7 (together with Proposition 3.2.6 which is mentioned in Corollary 3.2.7 ... ) reads as follows:View attachment 244240

My questions related to the above Corollary are as follows:
Question 1

In the above proof of Corollary 3.2.7 we read the following:

" ... ... Arguing recursively, let $r_n$ be the 'best' rational with$\text{max} ( x - \frac{1}{n} , r_{ n - 1 } ) \lt r_n \lt x$ ... ... "Can someone please explain (preferably in some detail) what is going on here ... how do we arrive at the expression

$\text{max} ( x - \frac{1}{n} , r_{ n - 1 } ) \lt r_n \lt x$ ... ... ?Question 2

In the above proof of Corollary 3.2.7 we read the following:

" ... ... let $s_n$ be the 'best' rational with

$x \lt s_n \lt \text{min} ( x + \frac{1}{n}, s_{ n - 1 } )$ ... ... "Can someone please explain (preferably in some detail) what is going on here ... how do we arrive at the expression

$x \lt s_n \lt \text{min} ( x + \frac{1}{n}, s_{ n - 1 } )$ ... ... ?
Help will be appreciated ...

Peter==========================================================================================The post above mentions Theorem 3.1.1 and alludes to the remarks made after the proof of Theorem 3.1.1 ... so I am providing text of the theorem and the relevant remarks ... as follows:
View attachment 244242
View attachment 244243Hope that helps ...

Peter

Yes, well done! I see that you understand well what's happening now.

This is key if you don't understand what in induction process does. Write it out for the first couple of steps! It becomes often very obvious what is happening.

Thanks again Math_QED

Peter

 

[member 587159]

Thanks Math_QED ... appreciate your help ...

I will try to show Garling's recursion process for n = 1, 2 and 3 ...For $n = 1$ we have ... ...

Let $r_1$ be the 'best' rational with $x - 1 \lt r_1 \lt x$
For $n = 2$ we have ... ...

$x - \frac{1}{n} = x - \frac{1}{2} \lt x$ ... ... and ... ... $r_{ n - 1 } = r_1 \lt x$ ... ...

and so ... $\text{max} ( x - \frac{1}{2} , r_1 ) \lt x$

Therefore we can pick/construct a 'best' rational $r_2$ ...

... such that ... $\text{max} ( x - \frac{1}{2} , r_1 ) \lt r_2 \lt x$ ... ... ... ... ... NOTE: $r_1 \lt r_2$
For $n = 3$ we have ... ...

$x - \frac{1}{n} = x - \frac{1}{3} \lt x$ ... ... and ... ... $r_{ n - 1 } = r_2 \lt x$ ... ...

and so ... $\text{max} ( x - \frac{1}{3} , r_2 ) \lt x$

Therefore we can pick/construct a 'best' rational $r_3$ ...

... such that ... $\text{max} ( x - \frac{1}{3} , r_2 ) \lt r_3 \lt x$ ... ... ... ... ... NOTE: $r_1 \lt r_2 \lt r_3$
The above analysis indicates that $(r_n)_{ n = 1}^\infty$ is strictly increasing and bounded above by $x$ ...

... indeed $\text{sup} \{ r_n \ : \ n \in \mathbb{N} \} = x$ ...

... so $r_n \longrightarrow x$ as $n \longrightarrow \infty$ ...By a similar argument involving the infimum, $s_n \longrightarrow x$ as $n \longrightarrow \infty$ ...Is the above correct?

Peter

Well, maybe one more remark: the convergence of the two sequences follows from the squeeze theorem (as in the proof of author), but what you wrote is correct as well.