Show that a sequence is convergent

Question

I found this problem in real analysis and I have no idea how to start, I just need a hint

let $x_n$ be a bounded sequence of real numbers, that satisfies:

1- $\lim_{n\to \infty} x_{n+1} - x_n = 0 $

2- if $ A = \{x_n: \forall n\in \mathbb{N}\} $ then $A'$ is finite

(This defines A as the range of $x_n$ and A' is the set of limit points of A)

prove that $x_n$ is convergent

for $x_n$ to converge it's enough to show that A' has only one element, that if $x \in A' $ and $y \in A'$ then $x = y$ I tried to deduce that from the definition of the set of limit points and no luck there. and since $x_n$ is bounded then it has a convergent subsequence, which converges to a point in A'. I don't know what to do next

Answer 1

If $S\subset \Bbb R$ where $S$ has at least 2 members and $S$ has a member strictly between any 2 of its members, then $S$ cannot be finite.

We show that if $u,v$ are any members of $A'$ with $u<v$ then there exists a member of $A'$ in $(u,v)$ and conclude that if $A'$ has more than 1 member then $A'$ is infinite.

Suppose $u,v\in A'$ with $u<v.$ Let $r=(v-u)/3.$ We show that for any $n\in \Bbb N$ there exists $n'>n$ such that $$(\bullet)\quad x_{n'}\in [u+r,u+2r].$$ So $\{n'\in \Bbb N: x_{n'}\in [u+r,u+2r]\}$ is infinite, so $A'$ has a member in $[u+r,u+2r],$ which is a subset of $(u,v)$.... Here is how:

Given $n\in \Bbb N,$ take $n_1\ge n$ such that $|x_{m+1}-x_m|<r$ whenever $m\ge n_1.$

Now take $n_2\ge n_1$ such that $|u-x_{n_2}|<r,$ which is possible because $u\in A'.$

And $v\in A'$ so take $n_3>n_2$ such that $|v-x_{n_3}|<r. $

We now have $n\le n_1\le n_2<n_3$ and $x_{n_2}<u+r<u+2r<x_{n_3}.$

Finally let $n'$ be the $least$ $j>n_2$ such that $x_j\ge u+r.$

Obviously $n'> n$ (as $n'>n_2\ge n$).

The main point is that $x_{n'-1}<u+r$ and $n'-1\ge n_1$ so $$u+r\le x_{n'}=x_{n'-1} +(x_{n'}-x_{n'-1})<u+r+|x_{n'}-x_{n'-1}|<u+r+r.$$ So $x_{n'}\in [u+r,u+2r] $ as required in $(\bullet)$ above.

Answer 2

It is correct that you have to show that $A'$ has only one element. But for this you have to use that $A'$ is finite – otherwise the statement becomes wrong, see for example A sequence of real numbers such that $\lim_{n\to+\infty}|x_n-x_{n+1}|=0$ but it is not Cauchy.

If $A' = \{ a_1, \ldots, a_k \}$ then you can choose a $\epsilon > 0$ such that all the $k$ intervals $(a_i - 2\epsilon, a_i + 2\epsilon)$ are pairwise disjoint.

Show that all but finitely many $x_n$ are in one of the $k$ intervals $(a_i - \epsilon, a_i + \epsilon)$.

Finally use the condition $\lim_{n\to \infty} x_{n+1} - x_n = 0$ to show that for sufficiently large $n$, all $x_n$ are in the same interval $(a_i - \epsilon, a_i + \epsilon)$. Which means that this $a_i$ is the only limit point of the sequence $(x_n)$.

Answer 3

This is building on Martin R.'s answer, which I think misses a crucial element to show that at some point all the sequence elements must lie "near" the same limit point.

Since $A'$ is finite, let's assume the contrary of what we want to show, so we assume that $|A'| \ge 2$. Then we can choose $a_1,a_2 \in A', a_1 < a_2$ such that no real number in the open interval $(a_1,a_2)$ belongs to $A'$.

Set $\epsilon_0=\frac{a_2-a_1}3$.

Because $a_1,a_2$ are limit points of $A$, there must be two infinite sets of indices $I_1, I_2$ such that $\forall i \in I_1: |a_1-x_i| < \epsilon_0$ and $\forall i \in I_2: |a_2-x_i| < \epsilon_0$.

Let now be $a$ any real number in $[a_1+\epsilon_0, a_2-\epsilon_0]$ and $\epsilon>0$. We can find an $N_\epsilon$ such that $\forall n \ge N_\epsilon:|x_{n+1}-x_n| < \epsilon$ (exists because of condition 1).

Because $I_1$ is infinite, we can find an $i_1 \in I_1$ with $i_1 \ge N_\epsilon$. Because $I_2$ is infinite, we can find $i_2 \in I_2$ with $i_2 > i_1$.

So we have

$$x_{i_1} < a_1+\epsilon_0 \le a \le a_2-\epsilon_0 < x_{i_2}.$$

Now it follows that at least one of $x_{i_1}, x_{i_1+1},\ldots,x_{i_2}$ must lie in the interval $(a-\epsilon,a+\epsilon)$. If either $x_{i_1}$ or $x_{i_2}$ do, we are done. Otherwise $x_{i_1}$ is to the left of that interval and $x_{i_2}$ is to the right of it.

Since $i_1 \ge N_\epsilon$, we know that the "step-length" $|x_{n+1}-x_n|$ is less than $\epsilon$ so the sequence cannot "step over" the interval $(a-\epsilon,a+\epsilon)$ of length $2\epsilon$, even if that interval is open.

So what have we shown? For an arbitrary point $a \in [a_1+\epsilon_0, a_2-\epsilon_0]$ and an arbitrary $\epsilon > 0$ we know that some $x_n \in (a-\epsilon, a+\epsilon)$. But that means that $a$ is a limit point of the sequence!

So contrary to our choice of $a_1,a_2$, the interval $(a_1,a_2)$ contained limit points. That means with finite $A'$, we must have $|A'|=1$.