Let $a = \liminf x_n$, $b = \limsup x_n$. If $\lim (x_{n+1} - x_n) = 0$ then every element of $(a,b)$ is a subsequential limit of $(x_n)$

Question

I'm fixing an arbitrary $x$ in $a<x<b$ and I must prove that $x$ is the limit of a subsequence of $(x_n)$. I have a theorem that says that $x$ is a subsequential limit of a sequence $(x_n)$ iff every interval $(x-\epsilon, x+\epsilon)$ containts infinitely many terms of $(x_n)$. I also know that $a$ and $b$ are the smallest and largest subsequential limits of $(x_n)$. I don't know where to go from here.

Answer 1

Sketch: Fix $c\in(a,b)$. We will show that for every $\epsilon >0$, small enough, there are infinite many $x_n \in (c-\epsilon, c+\epsilon)$.

Find $n_0$ such that $|x_{n+1} - x_n| <\epsilon$ for all $n\geq n_0$.

Since $\liminf x_n =a $ we can find $n_1 > n_0$ such that $x_{n_1} < c-\epsilon$.

Since $\limsup x_n =b $ we can find $n_2 > n_1$ such that $x_{n_2} > c+\epsilon$.

Therefore, at time $n_1$ $x_{n_1}$ is on the left of $(c-\epsilon, c+\epsilon)$ and at time $n_2$ $x_{n_2}$ is on the right of $(c-\epsilon, c+\epsilon)$. Since the sequence makes jumps of at most length $\epsilon$ there must be a time $\tilde{n}\in (n_1,n_2)$ such that $x_\tilde{n} \in (c-\epsilon, c+\epsilon)$.

We can repeat this procedure to find infinite many $x_n \in (c-\epsilon, c+\epsilon)$, hence $c$ is a limit point of $x_n$.