MHB Limit Superior .... B&S Therem 3.4.11 .... ....

[Math Amateur]

I am reading "Introduction to Real Analysis" (Fourth Edition) b Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 3: Sequences and Series ...

I need help in fully understanding the proof of Theorem 3.4.11 ...

Theorem 3.4.11 and the start of its proof read as follows:https://www.physicsforums.com/attachments/7317In the above proof by Bartle and Sherbert we read the following:

" ... ... If $$\epsilon \gt 0$$, then the fact that $$x^{*}$$ is an infimum implies that there exists a $$\nu$$ such that $$x^* \le \nu \le x^* + \epsilon$$. Therefore $$x^*$$ also belongs to $$V$$, ... ... "Question 1

Could someone please demonstrate rigorously that if $$\epsilon \gt 0$$, then the fact that $$x^*$$ is an infimum implies that there exists a $$\nu$$ such that $$x^* \le \nu \le x^* + \epsilon$$ ... ... ? (... also covering the case where inf $$V$$ = the minimum of $$V$$ ...)Question 2

Can someone please explain why we can then conclude that therefore $$x^*$$ also belongs to $$V$$ ... ... ?Help will be appreciated ...

Peter

===========================================================================It may help readers of the above post to have Bartle and Sherbert's definition of limit superior and limit inferior ... which include the definition of the set V ... as follows ... ...View attachment 7318

 

[GJA]

Hi Peter,

Question 1

I do not have a copy of Bartle and Sherbert with me, so I don't know exactly how they define infimum and supremum. Nevertheless, I will assume they say something to the effect that the infimum for a set is its greatest lower bound.

By Bartle and Shertbert's definition of $\limsup$, $x^{*}$ is the infimum of the set $V$. Exactly how $V$ is constructed is completely irrelevant right now. Since $x^{*}$ is the greatest lower bound for $V$, that means if we add any positive amount to it, then we obtain a number (i.e., $x^{*}+\varepsilon$) that is not a lower bound for $V$. Hence, there is some number $v\in V$ that is strictly smaller than $x^{*}+\varepsilon$.

Question 2

How $V$ is defined is now relevant. Since there are only finitely many elements from the sequence that are greater than $v$ and $x^{*}\leq v$, that means there are only finitely many elements from the sequence that are greater than $x^{*}$.

Edit: I have doubts about the validity of saying $x^{*}$ is in $V$. Perhaps I was too cavalier in my approach before, or am mistaken now in this edit. In either case, I am a little short on time and will revisit this later. For now, I doubt the legitimacy of the argument because we can take as an example $x_{n}=1/n.$ Then $V=(0,\infty)$ and the infimum of this set is $x^{*}=0$, which does not belong to $V$.

 

[Opalg]

Edit: I have doubts about the validity of saying $x^{*}$ is in $V$.

I am sure that you are right, and that $x^*$ need not be in $V$. The explanation seems to be that there is a mistake (or at least a misprint) in the proof of Bartle and Sherbert's Theorem 3.4.11. To be precise, the second sentence in the proof of "(a) implies (b)" should state "Therefore $\color{red}x^* + \varepsilon$ also belongs to $V$, ...". That is what has been established in the first sentence of the proof. It is also what is needed for the rest of the proof, which then makes good sense.

Congratulations to Peter and GJA for spotting the error in the textbook!

 

[GJA]

Well put, Opalg! I agree 100% with your amendment to Bartle and Sherbert's proof.

 

[Math Amateur]

Well put, Opalg! I agree 100% with your amendment to Bartle and Sherbert's proof.

Thanks to Opalg and GJA for the help ...

Just a note to say that I was concerned and a bit perplexed (Q1) by the statement: " ... the fact that $$x^*$$ is an infimum implies that there exists a $$\nu$$ in $$V$$ such that $$x^* \lt \nu \lt x^* + \epsilon$$. ... " in the situation where the infimum is the minimum of the set $$V$$ ... ...

For example ... consider ...

$$( x_n ) $$ where $$x_n = \frac{n}{ n + 1 }$$ for $$ n \in \mathbb{N}$$ ...

Then

$$( x_n ) = ( x_1, x_2, x_3, \ ... \ ... \ ) = ( \frac{1}{ 2 } , \frac{2}{ 3 }, \frac{3}{ 4 } , \ ... \ ... \ ) ... ... $$

Then consider the set ...

$$V = \{ \frac{1}{ 2 } , \frac{2}{ 3 }, \frac{3}{ 4 } , \ ... \ ... \ \}$$

... then $$x^*$$ = inf $$V$$ = $$\frac{1}{ 2 }$$

... and if we take $$\epsilon = \frac{1}{ 30 }$$ ... then consider $$x* + \epsilon$$ ... there is no element $$\nu$$ apart from $$x^* $$itself such that $$x^* \lt \nu \lt x^* + \epsilon$$ ...... but ... maybe ... the answer in this case is to take $$\nu = x^*$$ ... so that $$x^* \lt x^* \lt x^* + \epsilon $$ ... ...Is that a correct interpretation of this situation ... ...

Peter

 

[GJA]

Hi Peter,

For the example $x_{n}=n/(n+1)$, the set $V$ (as defined by Bartle and Sherbert) is $(1,\infty)$, which is why you're having trouble finding a $v$ such that $x^{*}\leq v < x^{*}+1/30$. Nevertheless, it is the case that $1/2$ is the infimum of the set $V$ you gave.

In general, however, it is OK to have $v=x^{*}$, and this can happen when $x^{*}\in V$. The strict inequality only needs to hold for the $x^{*}+\varepsilon$ portion.