Prove any number $c \in [a, b]$ is a subsequential limit if $\lim\inf x_n = a$, $\lim \sup x_n = b$, $a\ne b$, $\lim(x_n -x_{n+1})=0$

Question

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.

Answer 1

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 $\lim( x_n-x_{n+1})=0$.

Answer 2

Here's my very informal answer.

The lim sup and lim inf info says that the sequence gets infinitely often arbitrarily close to a and b.

The difference info says that successive terms get eventually arbitrarily close.

So, on the way between a and b, the successive terms are arbitrarily close and so will be, somewhere, arbitrarily close to c.

Since this happens infinitely often, choose these terms close to c as the subsequence.

Answer 3

The answer has already been given, but I will try to add my own thoughts on this, hopefully valid. Please let me know whether the reasoning below is wrong and I will either edit it or delete it completely.

Choose some $c \in [a, b]$. If $c = a$ or $c = b$ then we are done since: $$ \lim\sup x_n = b = c \\ \text{or}\\ \lim\inf x_n = a = c $$

Suppose $ c \notin \{a, b\}$. Then by Bolzano-Weierstrass (and the fact that $x_n$ is bounded and hence $x_{n_k}$ is also) we may choose a convergent subsequence $\{x_{n_k}\}$ from $\{x_n\}$. Suppose its limit is $c$: $$ \lim x_{n_k} = c $$

Using the definition of a limit: $$ \forall \epsilon_1 > 0\ \exists N \in \Bbb N: \forall n_k > N \implies |x_{n_k} - c| < \epsilon_1 \tag1 $$

On the other hand we have the following property: $$ \lim_{n\to\infty} (x_n - x_{n+1}) = 0 $$

This property also applies to subsequences so for $\{x_{n_k}\}$: $$ \lim_{k\to\infty} (x_{n_k} - x_{n_k + 1}) = 0 $$

Rewriting it by definition we obtain: $$ \forall \epsilon_2 > 0\ \exists N \in\Bbb N: \forall n_k > N \implies |x_{n_k} - x_{n_k + 1}| < \epsilon_2 \tag2 $$

Based on $(1)$ and $(2)$ we may choose $N$ such that both statements are satisfied. Since both inequalities in $(1)$ and $(2)$ become valid we may sum them up: $$ \forall \epsilon = \max\{\epsilon_1, \epsilon_2\} > 0\ \exists N = \max\{N_1, N_2\}: n_k > N \implies |x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| < \epsilon $$

By trialnge inequality: $$ |x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| \ge |x_{n_k} - c + x_{n_k} - x_{n_k + 1}| = |2x_{n_k} - x_{n_k + 1} - c| $$

Going from definition to a limit we get: $$ \lim_{k\to\infty}(2x_{n_k} - x_{n_k + 1}) = c $$

We've chosen $\{x_{n_k}\}$ such that $\lim x_{n_k} = c$ exists and we know that $\lim({x_{n_k} - x_{n_k + 1}}) = 0$ also exists, so: $$ \lim_{k\to\infty}(2x_{n_k} - x_{n_k + 1}) = \\ \lim_{k\to\infty}(x_{n_k} + x_{n_k} - x_{n_k + 1}) = \\ \lim_{k\to\infty}x_{n_k} + \lim_{k\to\infty}(x_{n_k} - x_{n_k + 1}) = \\ = \lim_{k\to\infty}x_{n_k} + 0 = \\ = \lim_{k\to\infty}x_{n_k} = c $$

Since $c$ is arbitrary in $[a, b]$ and $x_{n_k}$ is a subsequence of $x_n$ this shows that every $c\in[a, b]$ appears to be a subsequential limit.

Answer 4

Claim: Let $A,B$ be subsets of $\Bbb R$ such that $\delta:=\inf B-\sup A>0$. Let $x_n\in A\cup B$ be a sequence such that $\lim_{n\to\infty} (x_n-x_{n+1})=0$, then $\exists M\in\Bbb N$ such that either $x_n\in A$ for all $n\ge M$ or $x_n\in B$ for all $n\ge M$.

Proof: Let $n_0\in\Bbb N$ be such that $|x_n-x_{n+1}|<\delta/2$ for all $n\ge n_0$. Suppose for contradiction that such an $M$ cannot be found, then $\exists N>n_0$ such that $x_N\in A$ but $x_{N+1}\in B$. This, however, implies that $$ \delta\le |x_N-x_{N+1}|<\delta/2, $$ a contradiction.

Let's apply the above to your problem:

Assume that $\exists c\in (a,b)$ such that $c$ is not a subsequential limit of $x_n$, then there is an $\varepsilon>0$ and $N\in\Bbb N$ such that for all $n\ge N$, $x_n\notin (c-\varepsilon,c+\varepsilon)$. This splits the set in which the sequence $x_n$ (for $n\ge N$) lies into two parts, i.e. $$ A=[a,c-\varepsilon], B=[c+\varepsilon,b]. $$ The sets $A$ and $B$ satisfy the conditions above.

Without loss of generality, let's say $x_n\in A$ for all $n\ge M$. Then we have $$ \limsup_{n\to\infty} x_n \le \sup_{n\ge M} x_n \le c-\varepsilon < b, $$ which contradicts $\limsup_{n\to\infty} x_n = b$.