I'm trying to solve the following problem:
Let $\{x_n\}$ denote a bounded sequence. Prove that any number $c \in [a, b]$ is a subsequential limit of $\{x_n\}$ if:
$$
\begin{cases}
\lim_{n\to\infty} (x_n – x_{n+1})=0\\
\lim\inf x_n = a\\
\lim \sup x_n = b\\
a\ne b
\end{cases}
$$
Here are some of my thoughts. We know that $x_n$ is bounded. Then by Bolzano-Weierstrass we may choose some subsequence such that it has a finite limit:
$$
\exists c \in [a, b] : \lim x_{n_k} = c \iff \forall \epsilon_1 > 0 \exists N_1\in\Bbb N: \forall n_k > N_1 \implies |x_{n_k} – c| < \epsilon_1
$$
We are also given that limsup and liminf exist and therefore:
$$
\exists N_2 \in \Bbb N : \forall n_k > N_2 \implies x_{n_k} \ge a \\
\exists N_3 \in \Bbb N : \forall n_k > N_3 \implies x_{n_k} \le b
$$
If we now choose $N$ to be $\max\{N_2, N_3\}$ we obtain:
$$
\exists N = \max\{N_2, N_3\}: \forall n_k > N \implies a \le x_{n_k} \le b \tag1
$$
Also we are given the fact that $\lim (x_n – x_{n+1}) = 0$:
$$
\forall \epsilon_2 > 0, \exists N_4 \in \Bbb N: \forall n_k > N_4\implies |x_n – x_{n+1}| < \epsilon_2
$$
But if $\lim (x_n – x_{n+1}) = 0$, then it is also true for the subsequences:
$$
\forall \epsilon_3 > 0, \exists N_5 \in \Bbb N: \forall n_k > N_5\implies |x_{n_k} – x_{n_k+1}| < \epsilon_3 \tag 2
$$
Now I'm struggling to combine that facts in order to show that any $c \in [a, b]$ is a subsequential limit of $\{x_n\}$, how do I proceed? Feels like i have to consider $(1)$ and $(2)$ in tandem in order to finish the proof.
Best Answer
Pick $c\in (a,b)$ and enumerate $L=\{x_n:x_n\le c\}$ by $\{y_n\}$ and $U=\{x_n:x_n\ge c\}$ by $\{z_n\}$. Notice $\{x_n\}=\{y_n\}\cup\{z_n\}$ and that both $y_n$ and $z_n$ are infinite because $\liminf x_n=a$ and $\liminf x_n=b$. If either $\limsup y_n =c$ or $\liminf z_n=c$ we are done, so suppose neither holds. Then there exists some $N$ s.t. for $m,n\ge N$, $y_m < c-\epsilon<c<c+\epsilon<z_n$ for some \ $\epsilon>0$. If there were only finitely many $k$ s.t. $x_k\le c\le x_{k+1}$, we could not have $\liminf x_n=a$ and $\limsup x_n=b$. Therefore there are infinitely many $k$ s.t. $x_k=y_{n_k}$ and $x_{k+1}=z_{m_k}$. For $k$ large enough, $x_{k+1}-x_k\ge 2\epsilon$, so we can not have $\lim( x_n-x_{n+1})=0$.