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Basically, I am having trouble understanding this hole page from my complex analysis book. I fail to understand why a point of accumulation of a given set is also its limit supperior and most importantly why it holds the properties mentioned on the page and I am also not understanding the proof provided by the writer.

Complications :

Even assuming that the limit superior of a function is a point of accumulation I see some contradictions. For example it states that this point t of a sequence tn holds

there exist finitely many indices such that $t_n\ge t -e $

=> there exist infinitely indices such that $t_n\le t -e$

=> $t_n -t \le -e$

Now to omit the definition of a point of accumulation I will have to take the modulus of both sides:

$|t_n - t|\le |e|=e$ but I don't think that is a valid step because for example $-3<-2$ but $3<2$ is false. So in order for it to hold this property, I will have to perform an "illegal" operation. I am sorry if this is too confusing if anybody could help me I would greatly appreciate it.

Here is the page: Extract from the book

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  • $\begingroup$ I'm having trouble isolating the source of your difficulties. Do you understand the definition of accumulation point? If you write down the definition of accumulation point, can you prove that $\lambda$ has the required property? If not, where do you get stuck? $\endgroup$
    – saulspatz
    May 15, 2019 at 21:31
  • $\begingroup$ @saulspatz I understand that an accumulation point as a point such that infinetly many indecies satisfy the equation |tn - λ |< epsilon. But the only definition that we are given in the book for limit superrior is least upper bound so I don't know how to prove the property. Furthermore I feel like there is a contradiction and I don't understand where my logic is wrong in the complications part. Could you help me? $\endgroup$
    – Maths Wizzard
    May 15, 2019 at 23:18
  • $\begingroup$ I think you should have a look at this answer : math.stackexchange.com/a/1893725/72031 $\endgroup$
    – Paramanand Singh ♦
    May 16, 2019 at 1:43
  • $\begingroup$ Your statement regarding limit superior is wrong. Probably a typo. The right version is that "there are infinitely many indices $n$ such that $t_n>t-\epsilon$". $\endgroup$
    – Paramanand Singh ♦
    May 16, 2019 at 1:51
  • $\begingroup$ And further there is a specific index $N$ such that $t_n\leq t+\epsilon$ for all $n\geq N$. It should now be obvious that $t$ is an accumulation point. $\endgroup$
    – Paramanand Singh ♦
    May 16, 2019 at 1:53

1 Answer 1

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Suppose $\lambda$ is not itself an accumulation point, but is the least upper bound of the set of accumulation points. Then for every $\varepsilon>0$ there is an accumulation point $a$ within ${\varepsilon\over2}$ of $\lambda.$ But since $a$ is an accumulation point, there are infinitely many indices $n$ such that $t_n$ is within ${\varepsilon\over2}$ of $a$. By the triangle inequality, all these $t_n$ are within $\varepsilon$ of $\lambda$ so that $\lambda is an accumulation point after all.

I don't understand what you are saying in the "Complications" part. Why do you have to take the modulus of both sides? As you correctly point out, this is not a valid operation. The book proves that for any $\varepsilon>0$ there are only finitely many indices $n$ such that $t_n\geq\lambda+\varepsilon.$ Since we know that there are infinitely many $n$ such that $t_n\in(\lambda-\varepsilon, \lambda+\varepsilon)$ All of these are greater than $\lambda-\varepsilon$, so the second part of the staement is clear.

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  • $\begingroup$ Hey, thanks for the answer! I was just wondering why is it epsilon/2? $\endgroup$
    – Maths Wizzard
    May 16, 2019 at 9:54
  • $\begingroup$ Just to make the total come out to less than $\varepsilon$ The requirement is that given any $\varepsilon>0$, thre are infinitely many points of the sequence within $\varepsilon.$ But it's really just a matter of style. It would be enough to show there were infinitely many within $2\varepsilon$, since $\varepsilon$ is arbitrary. $\endgroup$
    – saulspatz
    May 16, 2019 at 11:35
  • $\begingroup$ Hey, I now fully understand what you meant. The book also asks the reader to prove that lambda is the largest such point. Is it enough to say that since lim sup tn = lim n-> infinity (sup (n>=k) tn) if there can not exist greater accumulation point since for greater values of n than for lambda the greatest value you can get is lambda? $\endgroup$
    – Maths Wizzard
    May 16, 2019 at 21:57
  • $\begingroup$ I don't understand what you mean by " for greater values of n than for lambda the greatest value you can get is lambda" $\endgroup$
    – saulspatz
    May 16, 2019 at 22:20
  • $\begingroup$ For example select N such that t(N)=limsup tn . Then consider N+1, but since N is the limsup tn t(N+1)<= t(N)= limsup tn. Does that make sence or is there an error in the logic? $\endgroup$
    – Maths Wizzard
    May 16, 2019 at 22:23

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