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.


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

  • $\begingroup$ Could you please elaborate on the last statement? I can't see where the contradiction comes from $\endgroup$ – roman Dec 17 '18 at 15:13
  • $\begingroup$ @roman I edited the answer to be more explicit. $\endgroup$ – Guacho Perez Dec 17 '18 at 15:29

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 $\limsup_{n\to\infty} x_n = b$.


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.


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.

  • $\begingroup$ Remember you are trying to prove the existence of a subsequence that converges to $c$, so when you say "suppose its limit is $c$" after invoking Bolzano-Weierstrass, you are assuming what you are trying to prove. Here, you have only that $x_{n_k}$ converges to some $c$ in $[a,b]$, but you are trying to prove this happens for all $c$. $\endgroup$ – Guacho Perez Dec 17 '18 at 20:58
  • $\begingroup$ @GuachoPerez Well, after going though this answer once again, you are right, thank you for the notices. That problem is driving me mad. I'm going to accept you answer while it seems i'm too dumb to prove it myself $\endgroup$ – roman Dec 18 '18 at 7:50

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