Csdrhrt

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:
$$
begin{cases}
lim_{ntoinfty} (x_n - x_{n+1})=0\
liminf x_n = a\
lim sup x_n = b\
ane b
end{cases}
$$

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:
$$
exists c in [a, b] : lim x_{n_k} = c iff forall epsilon_1 > 0 exists N_1inBbb N: forall n_k > N_1 implies |x_{n_k} - c| < epsilon_1
$$

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 $lim (x_n - x_{n+1}) = 0$:
$$
forall epsilon_2 > 0, exists N_4 in Bbb N: forall n_k > N_4implies |x_n - x_{n+1}| < epsilon_2
$$

But if $lim (x_n - x_{n+1}) = 0$, then it is also true for the subsequences:
$$
forall epsilon_3 > 0, exists N_5 in Bbb N: forall n_k > N_5implies |x_{n_k} - x_{n_k+1}| < epsilon_3 tag 2
$$

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 $cin (a,b)$ and enumerate $L={x_n:x_nle c}$ by ${y_n}$ and $U={x_n:x_nge c}$ 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 $m,nge N$, $y_m < c-epsilon<c<c+epsilon<z_n$ for some
$epsilon>0$. If there were only finitely many $k$ s.t. $x_kle cle x_{k+1}$, 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, $x_{k+1}-x_kge 2epsilon$, so we can not have $lim( x_n-x_{n+1})=0$.

Claim: Let $A,B$ be subsets of $Bbb R$ such that $delta:=inf B-sup A>0$. Let $x_nin Acup B$ be a sequence such that $lim_{ntoinfty} (x_n-x_{n+1})=0$, then $exists MinBbb N$ such that either $x_nin A$ for all $nge M$ or $x_nin B$ for all $nge M$.

Proof: Let $n_0inBbb N$ be such that $|x_n-x_{n+1}|<delta/2$ for all $nge n_0$. Suppose for contradiction that such an $M$ cannot be found, then $exists N>n_0$ such that $x_Nin A$ but $x_{N+1}in B$. This, however, implies that
$$
deltale |x_N-x_{N+1}|<delta/2,
$$
a contradiction.

Let's apply the above to your problem:

Assume that $exists cin (a,b)$ such that $c$ is not a subsequential limit of $x_n$, then there is an $varepsilon>0$ and $NinBbb N$ such that for all $nge N$, $x_nnotin (c-varepsilon,c+varepsilon)$. This splits the set in which the sequence $x_n$ (for $nge N$) 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 $x_nin A$ for all $nge M$. Then we have
$$
limsup_{ntoinfty} x_n le sup_{nge M} x_n le c-varepsilon < b,
$$
which contradicts $limsup_{ntoinfty} x_n = b$.

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:
$$
limsup x_n = b = c \
text{or}\
liminf x_n = a = c
$$

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:
$$
lim_{ntoinfty} (x_n - x_{n+1}) = 0
$$

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

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

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:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = c
$$

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:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}(x_{n_k} + x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}x_{n_k} + lim_{ktoinfty}(x_{n_k} - x_{n_k + 1}) = \
= lim_{ktoinfty}x_{n_k} + 0 = \
= lim_{ktoinfty}x_{n_k} = c
$$

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