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Anonymous Sat 21 Aug 11:08:48 2021 No.13543589 View Reply Original Report

Quoted By:

Why is the author talking about the greatest lower bound of the sequence a_n, a_n+1... when trying to prove that there is a point of accumulation between A and B?

Anonymous

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Anonymous Sat 21 Aug 2021 11:24:54 No.13543605 Report

Quoted By:

Theorem 3.1 by the way

Anonymous

Anonymous Sat 21 Aug 2021 11:30:01 No.13543623 Report

Quoted By: >>13543627 >>13543629

Youre still on the same page?
And the answer is that these greater lower bounds converge to an accumulation point.

Anonymous

Anonymous Sat 21 Aug 2021 11:31:36 No.13543627 Report

Quoted By: >>13543632 >>13543636 >>13543637

>>13543623
I don't get why they must be increasing though, what if a_n is decreasing or alternating?

Anonymous

Anonymous Sat 21 Aug 2021 11:32:38 No.13543629 Report

Quoted By:

>>13543623
Also I think it's supposed to be that a_n reaches an accumulation point.

Anonymous

Anonymous Sat 21 Aug 2021 11:33:53 No.13543632 Report

Quoted By:

>>13543627
b_n is increasing, a_n not necessarily. You can prove it be using the definition of greatest lower bound. If you add a number to the set whose greatest lower bound youre calculating, what happens to the result? Think.

Anonymous

Anonymous Sat 21 Aug 2021 11:35:40 No.13543636 Report

Quoted By:

>>13543627
If the a's are decreasing, then the b's are all equal.
You are probably missing that being constant is a special case of being increasing.

Anonymous

Anonymous Sat 21 Aug 2021 11:35:42 No.13543637 Report

Quoted By: >>13543695

>>13543627
It doesn't matter. Since $b_i$ is the greatest lower bound of $\{a_i,a_{i+1},\dots\}$ and of course as you increase $i$ your "options" to choose the lowest $a_j, j\geq i$ are always decreasing. So if you discard a bigger number than the lower bound then the equality holds else the next term in $b$ will be bigger. Thus $b$ is increasing.

Anonymous

Anonymous Sat 21 Aug 2021 11:53:12 No.13543695 Report

Quoted By: >>13543870

>>13543637
But how does that prove a_n reaches an accumulation point, i get intuitively it must if it's bounded below and above, but I can't see how to prove it.

Anonymous

Anonymous Sat 21 Aug 2021 12:39:17 No.13543846 Report

Quoted By:

Bump

Anonymous

Anonymous Sat 21 Aug 2021 12:45:30 No.13543870 Report

Quoted By: >>13543874 >>13543878 >>13543881 >>13543895

>>13543695
Since $b$ is increasing and bounded then it has a limit $L$ .
That means that for all $\epsilon>0$ there exist a number $N$ such that for all $n>N$ it is true that $|b_n-L|<\epsilon$ (1).
Since $b_n$ is the lower bound of $\{a_n,a_{n+1},\dots \}$ there exist a number $j_1\geq n : b_n=a_{j_1}$ so we have $|a_{j_1}-L|<\epsilon$ .
Because of (1) for $n=j_1+1$ we get $|b_{j_1+1}-L|<\epsilon$ so there exist a number $j_2\geq j_1+1$ such that $b_{j_1+1}=a_{j_2}$ etc....
That can be proved rigorously with induction. Thus we found an infinite number of points in $\{a_n\}$ , namely $a_{j_1},a_{j_2}\dots$ so that they are in the $\epsilon$ -neighborhood of L. So L is the accumulation point.

I am not sure about the correctness of this proof because I don't know the exact definition of the accumulation point but I can't think of something else.

Anonymous

Anonymous Sat 21 Aug 2021 12:47:28 No.13543874 Report

Quoted By: >>13543881

>>13543870
If you do not know the definition, you are doing pretty darn well.

Anonymous

Anonymous Sat 21 Aug 2021 12:48:42 No.13543878 Report

Quoted By: >>13543881

>>13543870
Does b_n have to be an element of a_n to be a greatest lower bound?

Anonymous

Anonymous Sat 21 Aug 2021 12:49:11 No.13543881 Report

Quoted By: >>13543895

>>13543870
>>13543874
Sorry I have a mistake. I don't think $b_n$ being the lower bound means that there exist a number in $\{a_n,a_{n+1},\dots\}$ which is equal to $b_n$ but there exist a number such that the difference of it to $b_n$ is small enough so that $|a_{j_i}-L|<\epsilon$ still holds. I haven't proved that but as it its my proof is not correct.

>>13543878
Yeah it doesn't...

Anonymous

Anonymous Sat 21 Aug 2021 12:52:37 No.13543895 Report

Quoted By: >>13544064

>>13543870
>>13543881
Thanks, that makes sense.

Anonymous

Anonymous Sat 21 Aug 2021 14:05:34 No.13544064 Report

Quoted By: >>13544650

>>13543895
Ok wrapped my head around it and came up with something better:
Ok still not sure about it but we can prove that $\forall\epsilon>0$ there exist an infinite number of points in $\{a_n\}$ such that $|a_i-L|<\epsilon$ (Which proves L is an accumulation point).
Proof: Let us choose an arbitrary $\epsilon>0$ .
Because $b$ has limit $L$ ,so for $\epsilon/2$ there exist a number $N$ such that $\forall n>N$ it is true that $|L-b_n|<\frac{\epsilon}{2} \iff -\frac{\epsilon}{2}<L-b_n<\frac{\epsilon}{2}$ (1) . (we will ignore the fact that $|L-b_n|=L-b_n$ )
Because $b_n$ is the greatest lower bound there exist a number $j\geq n$ such that $b_n\leq a_j \leq b_n + \frac{\epsilon}{2}$ , else we have a contradiction. ( $|a_j-b_n| > \frac{\epsilon}{2}$ which can't hold for all $a_j, j\geq n$ since $b_n$ is the greatest lower bound).
So we get $L-a_j\leq L-b_n < \frac{\epsilon}{2} < \epsilon$ and $L-a_j\geq L-b_n-\frac{\epsilon}{2}\geq -\frac{\epsilon}{2} - \frac{\epsilon}{2}=-\epsilon$ so in the end $|L-a_j|<\epsilon$ .
Continuing the algorithm for $n \to j+1$ using induction and we found infinite points in $a$ that are $\epsilon$ -close to $L$ .

Hopefully that is good enough.
Also thank you OP. I got interested in studying more real Analysis now that I have a bunch of free time before uni.

Anonymous

Anonymous Sat 21 Aug 2021 17:13:48 No.13544650 Report

Quoted By:

>>13544064
Thanks.

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