How do I show that the set of limit points of $\{x_n\}$ is connected? Show that if $\{x_n\}$ is a sequence of real numbers such that $\lim_{n \rightarrow \infty}(x_{n+1}-x_{n})=0$, then the set of limit of points of $\{x_n\}$ is connected, that is either empty, a single point, or an interval.
My attempt:  Since, $\lim_{n \rightarrow \infty}(x_{n+1}-x_{n})=0$ $\implies$ For given $\epsilon>0$, there exist a positive integer $N$ such that for $n>N$,
$|x_{n+1}-x_n|<\epsilon$ $\implies$ $|x_{n+1}|-|x_{n}| \leq |x_{n+1}-x_n|<\epsilon$.
This means for $n>N$, $|x_{n+1}|\leq |x_n|+\epsilon.$
Case 1: If $\{x_{n}\}$ is convergent, say it converges to $x$, and $\lim_{n \rightarrow \infty}(x_{n+1}-x_{n})=x-x=0$. We know that if $\{x_n\}$ converges, then its limit is unique. So, the set of limit points contain a single element.
Case 2: If $\{x_{n+1}\}$ is divergent, then there is a $M$ and $N$ such that $x_{n+1}>M$ for every $n>N$, $\implies$ $x_{n}$ is divergent(by comparison test). In this case, the set of limit points is empty because sequence is divergent.
Case 3: If $x_{n}=k+\frac{1}{n}$ for $k\in \mathbb{R}$, then $\lim_{n \rightarrow \infty}(x_{n+1}-x_{n})=0$, and every point of $\mathbb{R}$ is a limit point of $x_{n}$. So the set of limit points are the interval.
This shows that the set of limit point of a sequence $x_n$ is connected.
Is this solution correct for the given question?
 A: You’ve overlooked the ‘interesting’ cases. For instance, let $x_1=1$, $x_2=1-\frac12=\frac12$, $x_3=x_2-\frac13=\frac16$, and $x_4=x_3-\frac14=-\frac1{12}$. To get the next several terms add $\frac15$, $\frac16$, etc. until the total is at least $1$: $x_5=x_4+\frac15=\frac7{60}$, $x_6=x_5+\frac16=\frac{17}{60}$, and so on. As soon as some $x_n$ is at least $1$, start subtracting again, so that $x_{n+1}=x_n-\frac1{n+1}$, and keep subtracting until you get a term $x_n\le 0$, at which point you start adding again. Continue in this fashion. The resulting sequence will satisfy $\lim_n(x_{n+1}-x_n)=0$, and it should be pretty clear even without an actual proof that every point in $[0,1]$ is a cluster point of it.
To prove your result, you need to show that if $a$ and $b$ are cluster points of the sequence, and $a<c<b$, then $c$ is also a cluster point of the sequence. In this case you’ll have subsequence $\langle x_{n_k}:k\ge 1\rangle$ and $\langle x_{m_k}:k\ge 1\rangle$ converging to $a$ and $b$, respectively, and you can use them to show that the sequence has terms arbitrarily close to $c$.
A: Either the sequence has no limit points.
Or it has one. In this case, the sequence converges to it.
Otherwise, let $a,b$ be any two limit points (we usually use the term 'cluster points') of $(x_n)$, and let $a<c<b$ and $\epsilon>0$. Without loss of generality, ensure $\epsilon<\min(b-c,c-a)$. There exists $N\in\mathbb{N}$ such that $$n\ge N\implies |x_{n+1}-x_n|<\epsilon\tag{1}$$
Since $a,b$ are limit points, there also exist $N_b>N_a\ge N$ such that $$|x_{N_a}-a|<\epsilon,\qquad|x_{N_b}-b|<\epsilon$$
Now consider the sequence $x_n$ for $N_a\le n\le N_b$. It goes from nearby $a$ to nearby $b$ using steps smaller than $\epsilon$. There must be a value of $n$ in this range such that $|x_n-c|<\epsilon$ otherwise one can split the $n$ into those that give $x_n\le c-\epsilon$ and those that give $x_n\ge c+\epsilon$. The largest $n$ in the first group is not $N_b$ and has $n+1$ in the second group, so $$x_{n+1}-x_n\ge(c+\epsilon)-(c-\epsilon)=2\epsilon$$ contradicting (1).
Since the set of limit points $L$ satisfies $a,b\in L\implies [a,b]\subseteq L$, then $L$ must be an interval itself.
