Show that the set of adherent values of $u_n$ is connected when $u_{n+1}-u_n\to 0$

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.

• Is $a=x$ and $y=b$? Also the paragraph two is not that clear to me – Guy Fsone Oct 26 '17 at 12:14
• @GuyFsone Correct, I edited it for clarity. – Rchn Oct 26 '17 at 12:17
• @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 $\epsilon$. – Rchn Oct 26 '17 at 12:24
• 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? – Guy Fsone Oct 26 '17 at 13:28
• @GuyFsone I fixed (4), the proof should be correct now. – Rchn Oct 26 '17 at 14:19