# Trouble understanding a point of accumulation as a limit superior and understanding its properties

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

• 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? May 15, 2019 at 21:31
• @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? May 15, 2019 at 23:18
• I think you should have a look at this answer : math.stackexchange.com/a/1893725/72031 May 16, 2019 at 1:43
• Your statement regarding limit superior is wrong. Probably a typo. The right version is that "there are infinitely many indices $n$ such that $t_n>t-\epsilon$". May 16, 2019 at 1:51
• 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. May 16, 2019 at 1:53

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. • Hey, thanks for the answer! I was just wondering why is it epsilon/2? May 16, 2019 at 9:54 • 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. May 16, 2019 at 11:35