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Upper and Lower Linits (lim sup and lim inf) - Denlinger, Theorem2.9.6 (b) ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Charles G. Denlinger's book: "Elements of Real Analysis".

I am focused on Chapter 2: Sequences ... ...

I need help with the proof of Theorem 2.9.6 (b)


Theorem 2.9.6 reads as follows:


Denlinger - 1 - Theorem 2.9.6 - PART 1 ... .png
Denlinger - 2 - Theorem 2.9.6 - PART 2 ... .png



In the above proof of part (b) we read the following:

" ... ... Then \(\displaystyle B\) is an upper bound for every \(\displaystyle n\)-tail of \(\displaystyle \{ x_n \}\), so \(\displaystyle \overline{ x_n } = \text{sup} \{ x_k \ : \ k \geq n \} \leq B\). Thus \(\displaystyle \lim_{ n \to \infty } \overline{ x_n } \leq B\) ... ... "


My question is as follows:

Can someone please explain exactly how it follows that \(\displaystyle \lim_{ n \to \infty } \overline{ x_n } \leq B\) ... that is, how it follows that \(\displaystyle \overline{ \lim_{ n \to \infty } } x_n \leq B\) ...



(... ... apologies to steep if this is very similar to what has been discussed recently ... )



Hope someone can help ...

Peter



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It may help MHB readers to have access to Denlinger's definitions and notation regarding upper and lower limits ... so I am providing access to the same ... as follows:



Denlinger - 1 - Start of Section 2.9  - PART 1 ... .png
Denlinger - 2 - Start of Section 2.9  - PART 2 .png





Hope that helps ...

Peter
 
Last edited:

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
261
Hi Peter ,

Can someone please explain exactly how it follows that \(\displaystyle \lim_{ n \to \infty } \overline{ x_n } \leq B\) ... that is, how it follows that \(\displaystyle \overline{ \lim_{ n \to \infty } } x_n \leq B\) ...
The newly defined sequence $\left\{\overline{x_{n}}\right\}_{n=1}^{\infty}$ satisfies $\overline{x_{n}}\leq B$ for all $n$. Hence the limit of this sequence cannot exceed the value of $B$. Does this answer your question?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
Hi Peter ,



The newly defined sequence $\left\{\overline{x_{n}}\right\}_{n=1}^{\infty}$ satisfies $\overline{x_{n}}\leq B$ for all $n$. Hence the limit of this sequence cannot exceed the value of $B$. Does this answer your question?



Thanks for the help GJA ...

... your argument gives a plausible account of why \(\displaystyle \overline{ x_n } = \text{sup} \{ x_k \ : \ k \geq n \} \leq B\) implies that \(\displaystyle \lim_{ n \to \infty } \overline{ x_n } \leq B\) ... ...

But does your argument constitute an explicit, formal and rigorous argument/proof that a skeptic would accept ...

What do you think ... am I being too extreme or particular ...

Can you help/comment further ...?

Peter
 
Last edited:

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
261
A more concrete "proof" would go something like this:

Suppose the limit of the sequence is $L$ and that $L>B$. Let $\varepsilon =\dfrac{L-B}{2}$ and choose $N$ such that $|\overline{x_{n}}-L|<\varepsilon$ for all $n\geq N$. Then $\overline{x_{n}}>L-\varepsilon=\dfrac{L+B}{2}>B$ for all $n\geq N$. This, however, contradicts $\overline{x_{n}}\leq B$ for all $n$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
A more concrete "proof" would go something like this:

Suppose the limit of the sequence is $L$ and that $L>B$. Let $\varepsilon =\dfrac{L-B}{2}$ and choose $N$ such that $|\overline{x_{n}}-L|<\varepsilon$ for all $n\geq N$. Then $\overline{x_{n}}>L-\varepsilon=\dfrac{L+B}{2}>B$ for all $n\geq N$. This, however, contradicts $\overline{x_{n}}\leq B$ for all $n$.



Thanks for all your help on this issue GJA ...

Peter