I am having troubles with the following problem. Assume that $u_n$ is a sequence of real numbers such that $$\lim_{n\to \infty} u_{n+1}-u_n=0$$

We define $$A=\{x\in \Bbb R\mid \text{a subsequence $(u_{n_j})_j$ such that }~~ \lim_{j\to \infty }u_{n_j} =x\}$$

A is called set of adherent values of $(u_n)$.

Questions: 1) prove that A is connected.

2) Is the property still true for complex numbers?

I don't know from where to start. thanks


(1) Let's assume $x \le y$ are two adherent values of $(u_n)$ and let $z\in[x,y]$. We're trying to show that $z$ is also an adherent value of $(u_n)$, which is equivalent to say that $\forall \epsilon>0$, there exists infinite $z_n$ such that $|z-u_{z_n}|<\epsilon$.

(2) Since $u_{n+1}-u_n\rightarrow0$, we know that there exists $N$ such that $\forall n>N, |u_{n+1}-u_n|<\epsilon$. This means that if you take $N_1>N$ and $N_2>N$, for every value $v$ in $[u_{N_1}, u_{N_2}]$ there exists an $N_3$ such that $N_1\le N_3\le N_2$ and $|v-u_{N_3}|<\epsilon$, because all the values of $u$ are at most $\epsilon$ apart and they cover the entire range.

(3) $a$ and $b$ are adherent values of $u_n$ so we know there exists infinite $x_n>N$ and $y_n>N$ such that $|u_{x_n}-x|<\epsilon$ and $|u_{y_n}-y|<\epsilon$. Since there's an infinite amount of them, you can pick them such that the intervals $[x_n, y_n]$ don't overlap with each other.

(4) You can now use the result from paragraph (2) and say that for every $n$, there's a $u_{z_n}$ such that $|u_{z_n}-z|<\epsilon$. Indeed, either :

  1. $z\in[u_{x_n}, u_{y_n}]$ and the existence of such $u_{z_n}$ follows directly from (2)
  2. $x \le z < u_{x_n}$ and $|x - u_{x_n}| < \epsilon$ in which case we can take $u_{z_n} = u_{x_n}$

  3. $u_{y_n} < z \le y$ and $|y - u_{y_n}| < \epsilon$ in which case we can take $u_{z_n} = u_{y_n}$.

(5) Because the intervals $[x_n, y_n]$ don't overlap with each other, there's a different $z_n$ for every $n$, so the $(z_n)_n$ set is infinite, which proves that $z$ is an adherent value of $u_n$.

(6) This argument is made possible by the fact that for real values, if you have to go from $x$ to $y$ by doing really small steps, you have to get really close of any value inside $[x,y]$ every time.

This doesn't hold up in the complex plane because there are infinite paths between two points, so you could take a different path every time, thus preventing any accumulation around a third point. You could try constructing a counter-example using this intuition.

  • $\begingroup$ Is $a=x$ and $y=b$? Also the paragraph two is not that clear to me $\endgroup$ – Guy Fsone Oct 26 '17 at 12:14
  • $\begingroup$ @GuyFsone Correct, I edited it for clarity. $\endgroup$ – Rchn Oct 26 '17 at 12:17
  • $\begingroup$ @GuyFsone Oh I'm sorry, do you mean the argument in paragraph (2) is not clear? What part of it? I'm basically saying that $u_{n+1}$ and $u_n$ are at most $\epsilon$ apart so if you're going from a value $u_{N_1}$ to $u_{N_2}$ you're taking steps of size at most $\epsilon$. $\endgroup$ – Rchn Oct 26 '17 at 12:24
  • $\begingroup$ That is ok Now in paragraph 4 why is $z\in [u_{x_n}, u_{y_n}]$? because you this argument from 2. Also what are a and b? $\endgroup$ – Guy Fsone Oct 26 '17 at 13:28
  • $\begingroup$ @GuyFsone I fixed (4), the proof should be correct now. $\endgroup$ – Rchn Oct 26 '17 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.